Analisis de Circuitos by Jose LUis Ola

Section 10.7

■

Capacitors in Parallel and Series

EXAMPLE 10–6 A 10-mF, a 15-mF, and a 100-mF capacitor are connected in parallel across a 50-V source. Determine the following: a. Total capacitance. b. Total charge stored. c. Charge on each capacitor. Solution a. CT ⫽ C1 ⫹ C2 ⫹ C3 ⫽ 10 mF ⫹ 15 mF ⫹ 100 mF ⫽ 125 mF b. QT ⫽ CTV ⫽ (125 mF)(50 V) ⫽ 6.25 mC c. Q1 ⫽ C1V ⫽ (10 mF)(50 V) ⫽ 0.5 mC Q2 ⫽ C2V ⫽ (15 mF)(50 V) ⫽ 0.75 mC Q3 ⫽ C3V ⫽ (100 mF)(50 V) ⫽ 5.0 mC Check: QT ⫽ Q1 ⫹ Q2 ⫹ Q3 ⫽ (0.5 ⫹ 0.75 ⫹ 5.0) mC ⫽ 6.25 mC.

1. Three capacitors are connected in parallel. If C1 ⫽ 20 mF, C2 ⫽ 10 mF and CT ⫽ 32.2 mF, what is C3? 2. Three capacitors are paralleled across an 80-V source, with QT ⫽ 0.12 C. If C1 ⫽ 200 mF and C2 ⫽ 300 mF, what is C3? 3. Three capacitors are paralleled. If the value of the second capacitor is twice that of the first and the value of the third is one quarter that of the second and the total capacitance is 70 mF, what are the values of each capacitor? Answers: 1. 2.2 mF

2. 1000 mF

3. 20 mF, 40 mF and 10 mF

Capacitors in Series For capacitors in series (Figure 10–21), the same charge appears on each. Thus, Q ⫽ C1V1, Q ⫽ C2V2, etc. Solving for voltages yields V1 ⫽ Q/C1, V2 ⫽ Q/C2, and so on. Applying KVL, we get V ⫽ V1 ⫹ V2 ⫹ … ⫹ VN. Therefore,

冢

冣

Q Q Q 1 1 1 V ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ … ⫹ ᎏᎏ ⫽ Q ᎏᎏ ⫹ ᎏᎏ ⫹ … ⫹ ᎏᎏ C1 C2 CN C1 C2 CN

V 1

V 2

E CN

V N

(a) Series connection FIGURE 10–21

V CT

(b) Equivalent

1 1 1 1 Capacitors in series: ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ … ⫹ ᎏᎏ. CT C1 C2 CN

PRACTICE PROBLEMS 4

401

402

Chapter 10

■

Capacitors and Capacitance

But V ⫽ Q/CT. Equating this with the right side and cancelling Q yields 1 1 1 1 ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ … ⫹ ᎏᎏ CT C1 C2 CN

(10–15)

For two capacitors in series, this reduces to C C2 CT ⫽ ᎏ1ᎏ C1 ⫹ C2

(10–16)

For N equal capacitors in series, Equation 10–15 yields CT ⫽ C/N.

EXAMPLE 10–7 NOTES... 1. For capacitors in parallel, total capacitance is always larger than the largest capacitance, while for capacitors in series, total capacitance is always smaller than the smallest capacitance. 2. The formula for capacitors in parallel is similar to the formula for resistors in series, while the formula for capacitors in series is similar to the formula for resistors in parallel.

Refer to Figure 10–22(a):

a. Determine CT. b. If 50 V is applied across the capacitors, determine Q. c. Determine the voltage on each capacitor. 30 µF 60 µF 20 µF C1

(a)

10 µF CT (b)

FIGURE 10–22

Solution 1 1 1 1 1 1 1 a. ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ ᎏᎏ CT C1 C2 C3 30 mF 60 mF 20 mF ⫽ 0.0333 ⫻ 106 ⫹ 0.0167 ⫻ 106 ⫹ 0.05 ⫻ 106 ⫽ 0.1 ⫻ 106 Therefore as indicated in (b), 1 CT ⫽ ᎏᎏ6 ⫽ 10 mF 0.1 ⫻ 10 b. Q ⫽ CTV ⫽ (10 ⫻ 10⫺6 F)(50 V) ⫽ 0.5 mC c. V1 ⫽ Q/C1 ⫽ (0.5 ⫻ 10⫺3 C)/(30 ⫻ 10⫺6 F) ⫽ 16.7 V V2 ⫽ Q/C2 ⫽ (0.5 ⫻ 10⫺3 C)/(60 ⫻ 10⫺6 F) ⫽ 8.3 V V3 ⫽ Q/C3 ⫽ (0.5 ⫻ 10⫺3 C)/(20 ⫻ 10⫺6 F) ⫽ 25.0 V Check: V1 ⫹ V2 ⫹ V3 ⫽ 16.7 ⫹ 8.3 ⫹ 25 ⫽ 50 V.

Section 10.7

EXAMPLE 10–8

■

Capacitors in Parallel and Series

403

For the circuit of Figure 10–23(a), determine CT. 45 F

60 F

C2 15 F

12 F

C5 8 F

(a) 60 F

60 F

20 F

(b) 30 F

20 F

12 F

Systematic reduction.

Solution The problem is easily solved through step-by-step reduction. C2 and C3 in parallel yield 45 mF ⫹ 15 mF ⫽ 60 mF. C4 and C5 in parallel total 20 mF. The reduced circuit is shown in (b). The two 60-mF capacitances in series reduce to 30 mF. The series combination of 30 mF and 20 mF can be found from Equation 10–16. Thus, 30 mF ⫻ 20 mF CT ⫽ ᎏ ⫽ 12 mF 30 m F ⫹ 20 m F Alternately, you can reduce (b) directly using Equation 10–15. Try it.

Voltage Divider Rule for Series Capacitors For capacitors in series (Figure 10–24) a simple voltage divider rule can be developed. Recall, for individual capacitors, Q1 ⫽ C1V1, Q2 ⫽ C2V2, etc., and for the complete string, QT ⫽ CTVT. As noted earlier, Q1 ⫽ Q2 ⫽ … ⫽ QT. Thus, C1V1 ⫽ CTVT. Solving for V1 yields

冢冣

C V1 ⫽ ᎏᎏT VT C1

FIGURE 10–24 divider.

V1 V2 V3

Capacitive voltage

404

Chapter 10

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Capacitors and Capacitance

This type of relationship holds for all capacitors. Thus,

冢冣

C Vx ⫽ ᎏᎏT VT Cx

(10–17)

From this, you can see that the voltage across a capacitor is inversely proportional to its capacitance, that is, the smaller the capacitance, the larger the voltage, and vice versa. Other useful variations are

冢冣

C V1 ⫽ ᎏᎏ2 V2, C1 PRACTICE PROBLEMS 5

冢冣

C V1 ⫽ ᎏᎏ3 V3, C1

冢冣

C V2 ⫽ ᎏᎏ3 V3, C2

etc.

1. Verify the voltages of Example 10–7 using the voltage divider rule for capacitors. 2. Determine the voltage across each capacitor of Figure 10–23 if the voltage across C5 is 30 V. Answers: 1. V1 ⫽ 16.7 V V2 ⫽ 8.3 V V3 ⫽ 25.0 V V4 ⫽ V5 ⫽ 30 V

10.8

2. V1 ⫽ 10 V V2 ⫽ V3 ⫽ 10 V

Capacitor Current and Voltage

As noted earlier (Figure 10–2), during charging, electrons are moved from one plate of a capacitor to the other plate. Several points should be noted. 1. This movement of electrons constitutes a current. 2. This current lasts only long enough for the capacitor to charge. When the capacitor is fully charged, current is zero. 3. Current in the circuit during charging is due solely to the movement of electrons from one plate to the other around the external circuit; no current passes through the dielectric between the plates. 4. As charge is deposited on the plates, the capacitor voltage builds. However, this voltage does not jump to full value immediately since it takes time to move electrons from one plate to the other. (Billions of electrons must be moved.) 5. Since voltage builds up as charging progresses, the difference in voltage between the source and the capacitor decreases and hence the rate of movement of electrons (i.e., the current) decreases as the capacitor approaches full charge. Figure 10–25 shows what the voltage and current look like during the charging process. As indicated, the current starts out with an initial surge, then decays to zero while the capacitor voltage gradually climbs from zero to full voltage. The charging time typically ranges from nanoseconds to milliseconds, depending on the resistance and capacitance of the circuit. (We study these relationships in detail in Chapter 11.) A similar surge (but in the opposite direction) occurs during discharge. As Figure 10–25 indicates, current exists only while the capacitor voltage is changing. This observation turns out to be true in general, that is,

Section 10.8

■

Capacitor Current and Voltage

405

iC Current decays as capacitor charges

R iC E

(a)

vC E Time

Before charging

During charging

Voltage builds as capacitor charges

After charging

Time

(b) Current surge during charging. Current is zero when fully charged.

FIGURE 10–25 The capacitor does not charge instantaneously, as a finite amount of time is required to move electrons around the circuit.

current in a capacitor exists only while capacitor voltage is changing. The reason is not hard to understand. As you saw before, a capacitor’s dielectric is an insulator and consequently no current can pass through it (assuming zero leakage). The only charges that can move, therefore, are the free electrons that exist on the capacitor’s plates. When capacitor voltage is constant, these charges are in equilibrium, no net movement of charge occurs, and the current is thus zero. However, if the source voltage is increased, additional electrons are pulled from the positive plate; inversely, if the source voltage is decreased, excess electrons on the negative plate are returned to the positive plate. Thus, in both cases, capacitor current results when capacitor voltage is changed. As we show next, this current is proportional to the rate of change of voltage. Before we do this, however, we need to look at symbols.

Symbols for Time-Varying Voltages and Currents Quantities that vary with time are called instantaneous quantities. Standard industry practice requires that we use lowercase letters for time-varying quantities, rather than capital letters as for dc. Thus, we use vC and iC to represent changing capacitor voltage and current rather than VC and IC. (Often we drop the subscripts and just use v and i.) Since these quantities are functions of time, they may also be shown as vC(t) and iC(t). Capacitor v-i Relationship The relationship between charge and voltage for a capacitor is given by Equation 10–1. For the time-varying case, it is q ⫽ CvC

(10–18)

But current is the rate of movement of charge. In calculus notation, this is iC ⫽ dq/dt. Differentiating Equation 10–18 yields dq d iC ⫽ ᎏᎏ ⫽ ᎏᎏ(CvC) dt dt

(10–19)

NOTES... Calculus is introduced at this point to aid in the development of ideas and to help explain concepts. However, not everyone who uses this book requires calculus. Therefore, the material is presented in such a manner that it never relies entirely on mathematics; thus, where calculus is used, intuitive explanations accompany it. However, to provide the enrichment that calculus offers, optional derivations and problems are included, but they are marked with a ∫ icon so that they may be omitted if desired.

406

Chapter 10

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Capacitors and Capacitance

Since C is constant, we get dvC iC ⫽ Cᎏᎏ dt

FIGURE 10–26 The ⫹ sign for vC goes at the tail of the current arrow.

(A)

(10–20)

Equation 10–20 shows that current through a capacitor is equal to C times the rate of change of voltage across it. This means that the faster the voltage changes, the larger the current, and vice versa. It also means that if the voltage is constant, the current is zero (as we noted earlier). Reference conventions for voltage and current are shown in Figure 10– 26. As usual, the plus sign goes at the tail of the current arrow. If the voltage is increasing, dvC /dt is positive and the current is in the direction of the reference arrow; if the voltage is decreasing, dvC /dt is negative and the current is opposite to the arrow. The derivative dvC /dt of Equation 10–20 is the slope of the capacitor voltage versus time curve. When capacitor voltage varies linearly with time (i.e., the relationship is a straight line as in Figure 10–27), Equation 10–20 reduces to DvC rise iC ⫽ Cᎏᎏ ⫽ Cᎏᎏ ⫽ C ⫻ slope of the line Dt run

(10–21)

EXAMPLE 10–9 A signal generator applies voltage to a 5-mF capacitor with a waveform as in Figure 10–27(a). The voltage rises linearly from 0 to 10 V in 1 ms, falls linearly to ⫺10 V at t ⫽ 3 ms, remains constant until t ⫽ 4 ms, rises to 10 V at t ⫽ 5 ms, and remains constant thereafter. a. Determine the slope of vC in each time interval. b. Determine the current and sketch its graph. vC (V)

Slope = 10 000 V/s

Slope = −10 000 V/s

–10

t(mS)

Slope = 20 000 V/s

(a) iC (mA) 100 mA

50 1

–50

(b) FIGURE 10–27

t (mS)

Section 10.9

■

407

Energy Stored by a Capacitor

Solution a. We need the slope of vC during each time interval where slope ⫽ rise/run ⫽ Dv/Dt. 0 ms to 1 ms: Dv ⫽ 10 V; Dt ⫽ 1 ms; Therefore, slope ⫽ 10 V/1 ms ⫽ 10 000 V/s. 1 ms to 3 ms: Slope ⫽ ⫺20 V/2 ms ⫽ ⫺10 000 V/s. 3 ms to 4 ms: Slope ⫽ 0 V/s. 4 ms to 5 ms: Slope ⫽ 20 V/1 ms ⫽ 20 000 V/s. b. iC ⫽ CdvC/dt ⫽ C times slope. Thus, 0 ms to 1 ms: i ⫽ (5 ⫻ 10⫺6 F)(10 000 V/s) ⫽ 50 mA. 1 ms to 3 ms: i ⫽ ⫺(5 ⫻ 10⫺6 F)(10 000 V/s) ⫽ ⫺50 mA. 3 ms to 4 ms: i ⫽ (5 ⫻ 10⫺6 F)(0 V/s) ⫽ 0 A. 4 ms to 5 ms: i ⫽ (5 ⫻ 10⫺6 F)(20 000 V/s) ⫽ 100 mA. The current is plotted in Figure 10–27(b).

The voltage across a 20-mF capacitor is vC ⫽ 100 t e⫺t V. Determine current iC.

EXAMPLE 10–10

d(uv) dv du Solution Differentiation by parts using ᎏᎏ ⫽ uᎏᎏ ⫹ v ᎏᎏ with u ⫽ 100 d t d t dt t and v ⫽ e⫺t yields

冢

∫

冣

d d d dt iC ⫽ Cᎏᎏ(100 t e⫺t ) ⫽ 100 Cᎏᎏ(t e⫺t ) ⫽ 100 C tᎏᎏ(e⫺t ) ⫹ e⫺t ᎏᎏ dt dt dt dt ⫺6 ⫺t ⫺t ⫺t ⫽ 2000 ⫻ 10 (⫺t e ⫹ e ) A ⫽ 2.0 (1 ⫺ t)e mA

10.9

Energy Stored by a Capacitor

An ideal capacitor does not dissipate power. When power is transferred to a capacitor, all of it is stored as energy in the capacitor’s electric field. When the capacitor is discharged, this stored energy is returned to the circuit. To determine the stored energy, consider Figure 10–28. Power is given by p ⫽ vi watts. Using calculus (see ∫ ), it can be shown that the stored energy is given by 1 W ⫽ ᎏᎏCV 2 2

(J)

Deriving Equation 10–22 Power to the capacitor (Figure 10–28) is given by p ⫽ vi, where i ⫽ Cdv/dt. Therefore, p ⫽ Cvdv/dt. However, p ⫽ dW/dt. Integrating both sides yields

冕 pdt ⫽ C冕 vᎏddᎏvt dt ⫽ C冕 vdv ⫽ ᎏ12ᎏCV t

(10–22)

where V is the voltage across the capacitor. This means that the energy at any time depends on the value of the capacitor’s voltage at that time.

W⫽

FIGURE 10–28 capacitor.

∫

Storing energy in a

408

Chapter 10

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Capacitors and Capacitance

10.10 Capacitor Failures and Troubleshooting Although capacitors are quite reliable, they may fail because of misapplication, excessive voltage, current, temperature, or simply because they age. They can short internally, leads may become open, dielectrics may become excessively leaky, and they may fail catastrophically due to incorrect use. (If an electrolytic capacitor is connected with its polarity reversed, for example, it may explode.) Capacitors should be used well within their rating limits. Excessive voltage can lead to dielectric puncture creating pinholes that short the plates together. High temperatures may cause an increase in leakage and/or a permanent shift in capacitance. High temperatures may be caused by inadequate heat removal, excessive current, lossy dielectrics, or an operating frequency beyond the capacitor’s rated limit.

Basic Testing with an Ohmmeter Some basic (out-of-circuit) tests can be made with an analog ohmmeter. The ohmmeter can detect opens and shorts and, to a certain extent, leaky dielectrics. First, ensure that the capacitor is discharged, then set the ohmmeter to its highest range and connect it to the capacitor. (For electrolytic devices, ensure that the plus (⫹) side of the ohmmeter is connected to the plus (⫹) side of the capacitor.)

Capacitor (a) Measuring C with a DMM. (Not all DMMs can measure capacitance) FIGURE 10–29 Capacitor testing.

(b) Capacitor/inductor analyzer. (Courtesy B + K Precision)

Problems

409

Initially, the ohmmeter reading should be low, then for a good capacitor gradually increase to infinity as the capacitor charges through the ohmmeter circuit. (Or at least a very high value, since most good capacitors, except electrolytics, have a resistance of hundreds of megohms.) For small capacitors, however, the time to charge may be too short to yield useful results. Faulty capacitors respond differently. If a capacitor is shorted, the meter resistance reading will stay low. If it is leaky, the reading will be lower than normal. If it is open circuited, the meter will indicate infinity immediately, without dipping to zero when first connected.

Capacitor Testers Ohmmeter testing of capacitors has its limitations; other tools may be needed. Figure 10–29 shows two of them. The DMM in (a) can measure capacitance and display it directly on its readout. The LCR (inductance, capacitance, resistance) analyzer in (b) can determine capacitance as well as detect opens and shorts. More sophisticated testers are available that determine capacitance value, leakage at rated voltage, dielectric absorption, and so on.

10.1 Capacitance 1. For Figure 10–30, determine the charge on the capacitor, its capacitance, or the voltage across it as applicable for each of the following. a. E ⫽ 40 V, C ⫽ 20 mF b. V ⫽ 500 V, Q ⫽ 1000 mC c. V ⫽ 200 V, C ⫽ 500 nF d. Q ⫽ 3 ⫻ 10⫺4 C, C ⫽ 10 ⫻ 10⫺6 F e. Q ⫽ 6 mC, C ⫽ 40 mF f. V ⫽ 1200 V, Q ⫽ 1.8 mC 2. Repeat Question 1 for the following: a. V ⫽ 2.5 kV, Q ⫽ 375 mC b. V ⫽ 1.5 kV, C ⫽ 0.04 ⫻ 10⫺4 F c. V ⫽ 150 V, Q ⫽ 6 ⫻ 10⫺5 C d. Q ⫽ 10 mC, C ⫽ 400 nF ⫺5 e. V ⫽ 150 V, C ⫽ 40 ⫻ 10 F f. Q ⫽ 6 ⫻ 10⫺9 C, C ⫽ 800 pF 3. The charge on a 50-mF capacitor is 10 ⫻ 10⫺3 C. What is the potential difference between its terminals? 4. When 10 mC of charge is placed on a capacitor, its voltage is 25 V. What is the capacitance? 5. You charge a 5-mF capacitor to 150 V. Your lab partner then momentarily places a resistor across its terminals and bleeds off enough charge that its voltage falls to 84 V. What is the final charge on the capacitor? 10.2 Factors Affecting Capacitance 6. A capacitor with circular plates 0.1 m in diameter and an air dielectric has 0.1 mm spacing between its plates. What is its capacitance? 7. A parallel-plate capacitor with a mica dielectric has dimensions of 1 cm ⫻ 1.5 cm and separation of 0.1 mm. What is its capacitance? 8. For the capacitor of Problem 7, if the mica is removed, what is its new capacitance?

PROBLEMS

FIGURE 10–30

+ V −

410

Chapter 10

■

Capacitors and Capacitance 9. The capacitance of an oil-filled capacitor is 200 pF. If the separation between its plates is 0.1 mm, what is the area of its plates? 10. A 0.01-mF capacitor has ceramic with a dielectric constant of 7500. If the ceramic is removed, the plate separation doubled, and the spacing between plates filled with oil, what is the new value for C? 11. A capacitor with a Teflon dielectric has a capacitance of 33 mF. A second capacitor with identical physical dimensions but with a Mylar dielectric carries a charge of 55 ⫻ 10⫺4 C. What is its voltage? 12. The plate area of a capacitor is 4.5 in2. and the plate separation is 5 mils. If the relative permittivity of the dielectric is 80, what is C?

Plates

Points

(a)

Spheres

Plates

(b)

Points

Spheres

(c) FIGURE 10–31 Source voltage is increased until one of the gaps breaks down. (The source has high internal resistance to limit current following breakdown.)

10.3 Electric Fields 13. a. What is the electric field strength Ᏹ at a distance of 1 cm from a 100-mC charge in transformer oil? b. What is Ᏹ at twice the distance? 14. Suppose that 150 V is applied across a 100-pF parallel-plate capacitor whose plates are separated by 1 mm. What is the electric field intensity Ᏹ between the plates? 10.4 Dielectrics 15. An air-dielectric capacitor has plate spacing of 1.5 mm. How much voltage can be applied before breakdown occurs? 16. Repeat Problem 15 if the dielectric is mica and the spacing is 2 mils. 17. A mica-dielectric capacitor breaks down when E volts is applied. The mica is removed and the spacing between plates doubled. If breakdown now occurs at 500 V, what is E? 18. Determine at what voltage the dielectric of a 200 nF Mylar capacitor with a plate area of 0.625 m2 will break down. 19. Figure 10–31 shows several gaps, including a parallel-plate capacitor, a set of small spherical points, and a pair of sharp points. The spacing is the same for each. As the voltage is increased, which gap breaks down for each case? 20. If you continue to increase the source voltage of Figures 10–31(a), (b), and (c) after a gap breaks down, will the second gap also break down? Justify your answer. 10.5 Nonideal Effects 21. A 25-mF capacitor has a negative temperature coefficient of 175 ppm/°C. By how much and in what direction might it vary if the temperature rises by 50°C? What would be its new value? 22. If a 4.7-mF capacitor changes to 4.8 mF when the temperature rises 40°C, what is its temperature coefficient? 10.7 Capacitors in Parallel and Series 23. What is the equivalent capacitance of 10 mF, 12 mF, 22 mF, and 33 mF connected in parallel? 24. What is the equivalent capacitance of 0.10 mF, 220 nF, and 4.7 ⫻ 10⫺7 F connected in parallel? 25. Repeat Problem 23 if the capacitors are connected in series.

Problems 26. Repeat Problem 24 if the capacitors are connected in series. 27. Determine CT for each circuit of Figure 10–32. 28. Determine total capacitance looking in at the terminals for each circuit of Figure 10–33. 120 µF

C1 C2

8 µF

12 µF

C3 C2 4 µF

80 µF

1 µF

(a)

(b) 500 nF 5 F

6 F

C2 C3

CT C5

1 F 1 F C6

3 F

6 F

4 F b

(c)

2 F

(d)

FIGURE 10–32 0.47 µF

0.0005 µF

0.22 µF

a 1000 pF

500 pF CT

CT 0.15 µF

47 µF

100 pF

100 pF (b)

(a) 10 100 000 pF a 10 µF 2.2 µF

0.1 µF

1 µF

4.7 µF a

47 µF

12 10

16 14

3.2

b (d) All values in F

411

412

Chapter 10

■

Capacitors and Capacitance 29. A 30-mF capacitor is connected in parallel with a 60-mF capacitor, and a 10-mF capacitor is connected in series with the parallel combination. What is CT? 30. For Figure 10–34, determine Cx. 10 F

20 F

40 F

FIGURE 10–34

31. For Figure 10–35, determine C3 and C4. 32. For Figure 10–36, determine CT. 30 F

20 F

FIGURE 10–35

120 V

FIGURE 10–37

60 F

30 F

V 2

20 F

V 3

C3 = 2C4

40 F

50 F

Q = 100 C V = 10 V

75 F

FIGURE 10–36

33. You have capacitors of 22 mF, 47 mF, 2.2 mF and 10 mF. Connecting these any way you want, what is the largest equivalent capacitance you can get? The smallest? 34. A 10-mF and a 4.7-mF capacitor are connnected in parallel. After a third capacitor is added to the circuit, CT ⫽ 2.695 mF. What is the value of the third capacitor? How is it connected? 35. Consider capacitors of 1 mF, 1.5 mF, and 10 mF. If CT ⫽ 10.6 mF, how are the capacitors connected? 36. For the capacitors of Problem 35, if CT ⫽ 2.304 mF, how are the capacitors connected? 37. For Figures 10–32(c) and (d), find the voltage on each capacitor if 100 V is applied to terminals a-b. 38. Use the voltage divider rule to find the voltage across each capacitor of Figure 10–37. 39. Repeat Problem 38 for the circuit of Figure 10–38. 40. For Figure 10–39, Vx ⫽ 50 V. Determine Cx and CT. 41. For Figure 10–40, determine Cx. 42. A dc source is connected to terminals a-b of Figure 10–34. If Cx is 12 mF and the voltage across the 40-mF capacitor is 80 V, a. What is the source voltage? b. What is the total charge on the capacitors?

Problems

60 V

40 F

V 1

16 F

V 2

75 F

100 V V3

25 F

35 F

50 F

FIGURE 10–39

FIGURE 10–38 16 V 500 F 40 F

1200 F

100 V

100 F FIGURE 10–40

c. What is the charge on each individual capacitor? 10.8 Capacitor Current and Voltage 43. The voltage across the capacitor of Figure 10–41(a) is shown in (b). Sketch current iC scaled with numerical values.

iC vC

5 F

(a) vC 30 20 10 0

10 (b) FIGURE 10–41

9 t (ms)

413

414

Chapter 10

■

Capacitors and Capacitance

i (mA)

∫

40 20 0

-20

5 t (mS)

44. The current through a 1-mF capacitor is shown in Figure 10–42. Sketch voltage vC scaled with numerical values. Voltage at t ⫽ 0 s is 0 V. 45. If the voltage across a 4.7-mF capacitor is vC ⫽ 100e⫺0.05t V, what is iC? 10.9 Energy Stored by a Capacitor 46. For the circuit of Figure 10–37, determine the energy stored in each capacitor. 47. For Figure 10–41, determine the capacitor’s energy at each of the following times: t ⫽ 0, 1 ms, 4 ms, 5 ms, 7 ms, and 9 ms.

FIGURE 10–42

10.10 Capacitor Failures and Troubleshooting 48. For each case shown in Figure 10–43, what is the likely fault?

8.92 nF

0.132 µF

DATA HOLD DATA HOLD POWER RANGE

MAJAVG L/C/R

D/0 MHZ/120HZ TOL

POWER

MAJAVG

D/0

MHZ/120HZ

RANGE

L/C/R

TOL

RECALL

UNIVERSAL LCR METER

0.015 µF

0.022 µF

0.068 µF

0.033 µF 0.22 µF C1

C2 C3 0.33 µF

(b)

(a) 0.47 F 0.1 F C1 C2 100 V

0.22 F 68.1 V

Answers to In-Process Learning Checks

In-Process Learning Check 1 1. a. 2.12 nF b. 200 V 2. It becomes 6 times larger. 3. 1.1 nF 4. mica

415

ANSWERS TO IN-PROCESS LEARNING CHECKS

Capacitor Charging, Discharging, and Simple Waveshaping Circuits OBJECTIVES

KEY TERMS

After studying this chapter, you will be able to • explain why transients occur in RC circuits, • explain why an uncharged capacitor looks like a short circuit when first energized, • describe why a capacitor looks like an open circuit to steady state dc, • describe charging and discharging of simple RC circuits with dc excitation, • determine voltages and currents in simple RC circuits during charging and discharging, • plot voltage and current transients, • understand the part that time constants play in determining the duration of transients, • compute time constants, • describe the use of charging and discharging waveforms in simple timing applications, • calculate the pulse response of simple RC circuits, • solve simple RC transient problems using Electronics Workbench and PSpice.

Capacitive Loading Exponential Functions Initial Conditions Pulse Pulse Width (tp) Rise and Fall Times (tr, tf) Step Voltages Time Constant (t RC) Transient Transient Duration (5t)

OUTLINE Introduction Capacitor Charging Equations Capacitor with an Initial Voltage Capacitor Discharging Equations More Complex Circuits An RC Timing Application Pulse Response of RC Circuits Transient Analysis Using Computers

s you saw in Chapter 10, capacitors do not charge or discharge instantaneously. Instead, as illustrated in Figure 10–25, voltages and currents take time to reach their new values. The time taken to reach these new values (i.e., the charge and discharge times) are dependent on the resistance and capacitance of the circuit. During charge, for example, a capacitor charges at a rate determined by its capacitance and the resistance through which it charges, while during discharge, it discharges at a rate determined by its capacitance and the resistance through which it discharges. Since the voltages and currents that exist during these charging and discharging times are transitory in nature, they are called transients. Transients do not last very long, typically only a fraction of a second. However, they are important to us for a number of reasons, some of which you will learn in this chapter. Transients occur in both capacitive and inductive circuits. In capacitive circuits, they occur because capacitor voltage cannot change instantaneously; in inductive circuits, they occur because inductor current cannot change instantaneously. In this chapter, we look at capacitive transients; in Chapter 14, we look at inductive transients. As you will see, many of the basic principles are the same. Note: Optional problems and derivations using calculus are marked by a ∫ icon. They may be omitted without loss of continuity by those who do not require calculus.

Desirable and Undesirable Transients TRANSIENTS OCCUR IN CAPACITIVE and inductive circuits whenever circuit conditions are changed, for example, by the sudden application of a voltage, the switching in or out of a circuit element, or the malfunctioning of a circuit component. Some transients are desirable and useful; others occur under abnormal conditions and are potentially destructive in nature. An example of the latter is the transient that results when lightning strikes a power line. Following a strike, the line voltage, which may have been only a few thousand volts before the strike, momentarily rises to many hundreds of thousands of volts or higher, then rapidly decays, while the current, which may have been only a few hundred amps, suddenly rises to many times its normal value. Although these transients do not last very long, they can cause serious damage. While this is a rather severe example of a transient, it nonetheless illustrates that during transient conditions, many of a circuit or system’s most difficult problems may arise. Some transient effects, on the other hand, are useful. For example, many electronic devices and circuits depend on transient effects; these include timers, oscillators, and waveshaping circuits. As you will see in this chapter and in later electronics courses, the charge/discharge characteristic of RC circuits is fundamental to their operation.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

417

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Chapter 11

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits

11.1

Introduction

A basic switched RC circuit is shown in Figure 11–1. Most of the key ideas concerning charging and discharging and dc transients in RC circuits can be developed from it. Discharge position R

Charge position

1 2

FIGURE 11–1 Circuit for studying capacitor charging and discharging. Transient voltages and currents result when the circuit is switched.

Capacitor Charging First, assume the capacitor is uncharged and that the switch is open. Now move the switch to the charge position, Figure 11–2(a). At the instant the switch is closed the current jumps to E/R amps, then decays to zero, while the voltage, which is zero at the instant the switch is closed, gradually climbs to E volts. This is shown in (b) and (c). The shapes of these curves are easily explained. First, consider voltage. In order to change capacitor voltage, electrons must be moved from one plate to the other. Even for a relatively small capacitor, billions of electrons must be moved. This takes time. Consequently, capacitor voltage cannot change instantaneously, i.e., it cannot jump iC vC

E iC

(a)

E R

E amps R

0A t

0 t

Transient interval Transient interval

(b)

Steady state

(c)

FIGURE 11–2 Capacitor voltage and current during charging. Time t 0 s is defined as the instant the switch is moved to the charge position. The capacitor is initially uncharged.

abruptly from one value to another. Instead, it climbs gradually and smoothly as illustrated in Figure 11–2(b). Now consider current. The movement of electrons noted above is a current. As indicated in Figure 11–2(c), this current jumps abruptly from 0 to E/R amps, i.e., the current is discontinuous. To understand why, consider Figure 11–3(a). Since capacitor voltage cannot change instantaneously, its value just after the switch is closed will be the same as it was just before the switch is closed, namely 0 V. Since the voltage across the capacitor just after the switch is closed is zero (even though there is current through it), the capacitor looks momentarily like a short circuit. This is indicated in (b). This is an important observation and is true in general, that is, an uncharged capacitor looks like a short circuit at the instant of switching. Applying Ohm’s law yields iC E/R amps. This agrees with what we indicated in Figure 11–2(c). Finally, note the trailing end of the current curve Figure 11–2(c). Since the dielectric between the capacitor plates is an insulator, no current can pass through it. This means that the current in the circuit, which is due entirely to

Section 11.1

the movement of electrons from one plate to the other through the battery, must decay to zero as the capacitor charges. E

E Open circuit

(a) vC = E and iC = 0

vC = 0

iC = 0

vC = E

iC = E R

(a) Circuit as it looks just after the switch is moved to the charge position; vC is still zero

iC = 0

419

Introduction R

Steady State Conditions When the capacitor voltage and current reach their final values and stop changing (Figure 11–2(b) and (c)), the circuit is said to be in steady state. Figure 11–4(a) shows the circuit after it has reached steady state. Note that vC E and iC 0. Since the capacitor has voltage across it but no current through it, it looks like an open circuit as indicated in (b). This is also an important observation and one that is true in general, that is, a capacitor looks like an open circuit to steady state dc. R

■

vC = E

iC = E R

vC = 0

Looks like a short (b) Since vC = 0, iC = E/R

(b) Equivalent circuit for the capacitor

FIGURE 11–3 An uncharged capacitor initially looks like a short circuit.

FIGURE 11–4 Charging circuit after it has reached steady state. Since the capacitor has voltage across it but no current, it looks like an open circuit in steady state dc.

Capacitor Discharging Now consider the discharge case, Figures 11–5 and 11–6. First, assume the capacitor is charged to E volts and that the switch is open, Figure 11–5(a). Now close the switch. Since the capacitor has E volts across it just before the switch is closed, and since its voltage cannot change instantaneously, it will still have E volts across it just after as well. This is indicated in (b). The capacitor therefore looks momentarily like a voltage source, (c) and the curReference current direction

R C

(a) Voltage vC equals E just before the switch is closed

vC = E

Actual current direction

(b) Immediately after the switch is closed, vC still equals E

FIGURE 11–5 A charged capacitor looks like a voltage source at the instant of switching. Current is negative since it is opposite in direction to the current reference arrow.

vC = E

Actual current direction

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rent thus jumps immediately to E/R amps. (Note that the current is negative since it is opposite in direction to the reference arrow.) The voltage and current then decay to zero as indicated in Figure 11–6.

vC E t

0 (a) iC

0 E R (b)

FIGURE 11–6 Voltage and current during discharge. Time t 0 s is defined as the instant the switch is moved to the discharge position.

EXAMPLE 11–1 For Figure 11–1, E 40 V, R 10 , and the capacitor is initially uncharged. The switch is moved to the charge position and the capacitor allowed to charge fully. Then the switch is moved to the discharge position and the capacitor allowed to discharge fully. Sketch the voltages and currents and determine the values at switching and in steady state. Solution The current and voltage curves are shown in Figure 11–7. Initially, i 0 A since the switch is open. Immediately after it is moved to the charge position, the current jumps to E/R 40 V/10 4 A; then it decays to zero. At the same time, vC starts at 0 V and climbs to 40 V. When the switch is moved to the discharge position, the capacitor looks momentarily like a 40-V source and the current jumps to 40 V/10 4 A; then it decays to zero. At the same time, vC also decays to zero. vC 40 V 0V

0V t

0 (a)

iC 4A 0A

0A t

0 4 A 10 40 V

10 vC

Charge phase

Discharge phase

(b) Note the circuit that is valid during each time interval FIGURE 11–7

A charge/discharge example.

Section 11.1

■

Introduction

421

PRACTICAL NOTES... Some key points to remember are the following: 1. Capacitor voltage cannot change instantaneously, but current can. 2. To determine currents and voltages in a circuit at the instant of switching, replace uncharged capacitors with short circuits and charged capacitors with dc sources equal to their respective voltages at the instant of switching. 3. To determine currents and voltages in a dc circuit after it has reached steady state, replace capacitors with open circuits. 4. For both the charging and discharging cases, show capacitor current iC such that the plus sign for vC is at the tail of the current reference arrow. For charging, you will find that current is in the same direction as iC and hence is positive, while for discharging, current is opposite in direction to iC and hence is negative.

The Meaning of Time in Transient Analysis The time t used in transient analysis is measured from the instant of switching. Thus, t 0 in Figure 11–2 is defined as the instant the switch is moved to charge, while in Figure 11–6, it is defined as the instant the switch is moved to discharge. Voltages and currents are then represented in terms of this time as vC (t) and iC (t). For example, the voltage across a capacitor at t 0 s is denoted as vC (0), while the voltage at t 10 ms is denoted as vC (10 ms), and so on. A problem arises when a quantity is discontinuous as is the current of Figure 11–2(c). Since its value is changing at t 0 s, iC (0) cannot be defined. To get around this problem, we define two values for 0 s. We define t 0 s as t 0 s just prior to switching and t 0 s as t 0 s just after switching. In Figure 11–2(c), therefore, iC (0 ) 0 A while iC (0 ) E/R amps. For Figure 11–6, iC (0 ) 0 A and iC (0 ) E/R amps. Exponential Functions As we will soon show, the waveforms of Figures 11–2 and 11–6 are exponential and vary according to e x or (1 e x), where e is the base of the natural logarithm. Fortunately, exponential functions are easy to evaluate with modern calculators using their ex function. You will need to be able to evaluate both e x and (1 e x) for any value of x. Table 11–1 shows a tabulation of values for both cases. Note that as x gets larger, e x gets smaller and approaches zero, while (1 e x) gets larger and approaches 1. These observations will be important to you in what follows. 1. Use your calculator and verify the entries in Table 11–1. Be sure to change the sign of x before using the e x function. Note that e 0 e0 1 since any quantity raised to the zeroth power is one. 2. Plot the computed values on graph paper and verify that they yield curves that look like those shown in Figure 11–2(b) and (c).

TABLE 11–1 Table of Exponentials x

eⴚx

1 ⴚ eⴚx

0 1 2 3 4 5

1 0.3679 0.1353 0.0498 0.0183 0.0067

0 0.6321 0.8647 0.9502 0.9817 0.9933

PRACTICE PROBLEMS 1

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits

11.2 E

Capacitor Charging Equations

We will now develop equations for voltages and current during charging. Consider Figure 11–8. KVL yields

vR vC E

(11–1)

But vR RiC and iC CdvC /dt (Equation 10–20). Thus, vR RCdvC /dt. Substituting this into Equation 11–1 yields

FIGURE 11–8 Circuit for the charging case. Capacitor is initially uncharged.

dvC RC vC E dt

(11–2)

Equation 11–2 can be solved for vC using basic calculus (see ∫ ). The result is vC E(1 e t/RC)

(11–3)

where R is in ohms, C is in farads, t is in seconds and e t/RC is the exponential function discussed earlier. The product RC has units of seconds. (This is left as an exercise for the student to show.)

∫

Solving Equation 11–2 (Optional Derivation) First, rearrange Equation 11–2: dvC 1 (E vC) dt RC

Rearrange again: dvC dt E vC RC

Now multiply both sides by 1 and integrate.

冕

dvC 1 vC E RC

冥

ln(vC E)

vC 0

冥

t RC

冕 dt t

t 0

Next, substitute integration limits, t ln(vC E) ln( E) RC vC E t ln E RC

冢

冣

Finally, take the inverse log of both sides. Thus, vC E e t/RC E

When you rearrange this, you get Equation 11–3. That is, vC E(1 e t/RC )

Section 11.2

■

Now consider the resistor voltage. From Equation 11–1, vR E vC. Substituting vC from Equation 11–3 yields vR E E(1 e t/RC ) E E Ee t/RC. After cancellation, you get vR Ee t/RC

vC = E (1 e

t / RC )

vC E

(11–4)

Now divide both sides by R. Since iC iR vR/R, this yields

E iC e t/RC R

(11–5)

The waveforms are shown in Figure 11–9. Values at any time may be determined by substitution.

iC E R

iC = E e t / RC R

EXAMPLE 11–2 Suppose E 100 V, R 10 k , and C 10 mF: a. b. c. d. e.

vR E

Determine the expression for vC. Determine the expression for iC. Compute the capacitor voltage at t 150 ms. Compute the capacitor current at t 150 ms. Locate the computed points on the curves.

vC = 100 (1 e 10t)V

100 V

iC 10 mA

iC = 10 e 10t mA 2.23 mA

77.7 V 0

150 (a)

EWB

FIGURE 11–10

t (ms)

vR = E e

t / RC

FIGURE 11–9 Curves for the circuit of Figure 11–8.

Solution a. RC (10 103 )(10 10 6 F) 0.1 s. From Equation 11–3, vC E(1 e t/RC) 100(1 e t/0.1) 100(1 e 10 t ) V. b. From Equation 11–5, iC (E/R)e t/RC (100 V/10 k )e 10 t 10e 10 t mA. c. At t 0.15 s, vC 100(1 e 10 t) 100(1 e 10(0.15)) 100(1 e 1.5) 100(1 0.223) 77.7 V. d. iC 10e 10 t mA 10e 10(0.15) mA 10e 1.5 mA 2.23 mA. e. The corresponding points are shown in Figure 11–10. vC

423

Capacitor Charging Equations

150

t (ms)

(b) The computed points plotted on the vC and iC curves.

In the above example, we expressed voltage as vC 100(1 e t/0.1) and as 100(1 e 10t ) V. Similarly, current can be expressed as iC 10e t/0.1 or as 10e 10t mA. Although some authors prefer one notation over the other, both are correct and we will use them interchangeably.

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Chapter 11

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PRACTICE PROBLEMS 2

Capacitor Charging, Discharging, and Simple Waveshaping Circuits

1. Determine additional voltage and current points for Figure 11–10 by computing values of vC and iC at values of time from t 0 s to t 500 ms at 100-ms intervals. Plot the results. 2. The switch of Figure 11–11 is closed at t 0 s. If E 80 V, R 4 k , and C 5 mF, determine expressions for vC and iC. Plot the results from t 0 s to t 100 ms at 20-ms intervals. Note that charging takes less time here than for Problem 1. R iC C

FIGURE 11–11 Answers: 1. t(ms) 0 100 200 300 400 500

vC(V)

iC(mA)

0 63.2 86.5 95.0 98.2 99.3

10 3.68 1.35 0.498 0.183 0.067

2. 80(1 e 50t ) V

20e 50t mA

t(ms)

vC(V)

iC(mA)

0 20 40 60 80 100

0 50.6 69.2 76.0 78.6 79.4

20 7.36 2.70 0.996 0.366 0.135

EXAMPLE 11–3 For the circuit of Figure 11–11, E 60 V, R 2 k , and C 25 mF. The switch is closed at t 0 s, opened 40 ms later and left open. Determine equations for capacitor voltage and current and plot. Solution RC (2 k )(25 mF) 50 ms. As long as the switch is closed (i.e., from t 0 s to 40 ms), the following equations hold: vC E(1 e t/RC ) 60(1 e t/50 ms) V iC (E/R)e t/RC 30e t/50 ms mA Voltage starts at 0 V and rises exponentially. At t 40 ms, the switch is opened, interrupting charging. At this instant, vC 60(1 e (40/50)) 60(1 e 0.8) 33.0 V. Since the switch is left open, the voltage remains constant at 33 V thereafter as indicated in Figure 11–12. (The dotted curve shows how the voltage would have kept rising if the switch had remained closed.) Now consider current. The current starts at 30 mA and decays to iC 30e (40/50) mA 13.5 mA at t 40 ms. At this point, the switch is opened, and the current drops instantly to zero. (The dotted line shows how the current would have decayed if the switch had not been opened.)

Section 11.2

■

425

Capacitor Charging Equations

vC iC (mA)

60 V

30 33 V

Capacitor voltage stops changing when the switch opens. t (ms)

13.5 0

(a) EWB

Current drops to zero. t (ms)

40 (b)

FIGURE 11–12 Incomplete charging. The switch of Figure 11–11 was opened at t 40 ms, causing charging to cease.

The Time Constant The rate at which a capacitor charges depends on the product of R and C. This product is known as the time constant of the circuit and is given the symbol t (the Greek letter tau). As noted earlier, RC has units of seconds. Thus, t RC

(seconds, s)

(11–6)

Using t, Equations 11–3 to 11–5 can be written as vC E(1 e t/t )

(11–7)

E iC e t/t R

(11–8)

vR Ee t/t

(11–9)

and

Duration of a Transient The length of time that a transient lasts depends on the exponential function e t/t. As t increases, e t/t decreases, and when it reaches zero, the transient is gone. Theoretically, this takes infinite time. In practice, however, over 99% of the transition takes place during the first five time constants (i.e., transients are within 1% of their final value at t 5 t). This can be verified by direct substitution. At t 5 t, vC E(1 e t/t ) E(1 e 5) E(1 0.0067) 0.993E, meaning that the transient has achieved 99.3% of its final value. Similarly, the current falls to within 1% of its final value in five time constants. Thus, for all practical purposes, transients can be considered to last for only five time constants (Figure 11–13). Figure 11–14 summarizes how transient voltages and currents are affected by the time constant of a circuit—the larger the time constant, the longer the duration of the transient.

vC iC or vR t

0 5 FIGURE 11–13 time constants.

Transients last five

■

Capacitor Charging, Discharging, and Simple Waveshaping Circuits iC

increasing increasing

t (a)

(b)

FIGURE 11–14 Illustrating how voltage and current in an RC circuit are affected by its time constant. The larger the time constant, the longer the capacitor takes to charge.

EXAMPLE 11–4

For the circuit of Figure 11–11, how long will it take for the capacitor to charge if R 2 k and C 10 mF?

Solution t RC (2 k )(10 mF) 20 ms. Therefore, the capacitor charges in 5 t 100 ms.

The transient in a circuit with C 40 mF lasts 0.5 s. What

EXAMPLE 11–5 is R?

Solution 5 t 0.5 s. Thus, t 0.1 s and R t/C 0.1 s/(40 10 6 F) 2.5 k .

Figure 11–15 shows percent capacitor voltage and current plotted versus multiples of time constant. (Points are computed from vC 100(1 e t/t ) and iC 100e t/t. For example, at t t, vC 100(1 e t/t ) 100(1 e t/t ) 100(1 e 1) 63.2 V, i.e., 63.2%, and iC 100e t/t 100e 1 36.8 A, which is 36.8%, and so on.) These curves, referred to as universal time constant curves, provide an easy method to determine voltages and currents with a minimum of computation. vC

iC or vR 99.3%

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

98.2% 95.0%

Percent

Chapter 11

Percent of Full Voltage

426

86.5% 63.2%

2 3 4 5 (a)

FIGURE 11–15

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

100% 36.8% 13.5% 4.98% 1.83%

2 3 4 5 (b)

Universal voltage and current curves for RC circuit.

0.67% t

Section 11.3

■

Capacitor with an Initial Voltage

EXAMPLE 11–6

Using Figure 11–15, compute vC and iC at two time constants into charge for a circuit with E 25 V, R 5 k , and C 4 mF. What is the corresponding value of time?

Solution At t 2 t, vC equals 86.5% of E or 0.865(25 V) 21.6 V. Similarly, iC 0.135I0 0.135(E/R) 0.675 mA. These values occur at t 2 t 2RC 40 ms.

1. If the capacitor of Figure 11–16 is uncharged, what is the current immediately after closing the switch? R = 200

FIGURE 11–16

iC E

40 V

C = 1000 F

2. Given iC 50e 20t mA. a. What is t? b. Compute the current at t 0 s, 25 ms, 50 ms, 75 ms, 100 ms, and 500 ms and sketch it. 3. Given vC 100(1 e 50t ) V, compute vC at the same time intervals as in Problem 2 and sketch. 4. For Figure 11–16, determine expressions for vC and iC. Compute capacitor voltage and current at t 0.6 s. 5. Refer to Figure 11–10: a. What are vC(0 ) and vC(0 )? b. What are iC(0 ) and iC(0 )? c. What are the steady state voltage and current? 6. For the circuit of Figure 11–11, the current just after the switch is closed is 2 mA. The transient lasts 40 ms and the capacitor charges to 80 V. Determine E, R, and C. 7. Find capacitor voltage and current for Figure 11–16 at t 0.6 s using the universal time constant curves of Figure 11–15. (Answers are at the end of the chapter.)

11.3

Capacitor with an Initial Voltage

Suppose a previously charged capacitor has not been discharged and thus still has voltage on it. Let this voltage be denoted as V0. If the capacitor is now placed in a circuit like that in Figure 11–16, the voltage and current

IN-PROCESS

LEARNING CHECK 1

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits

during charging will be affected by the initial voltage. In this case, Equations 11–7 and 11–8 become vC E (V0 E)e t/t

(11–10)

E V iC 0 e t/t R

(11–11)

Note that these revert to their original forms when you set V0 0 V.

EXAMPLE 11–7

Suppose the capacitor of Figure 11–16 has 25 volts on it with polarity shown at the time the switch is closed.

a. Determine the expression for vC. b. Determine the expression for iC. c. Compute vC and iC at t 0.1 s. d. Sketch vC and iC. Solution t RC (200 )(1000 mF) 0.2 s a. From Equation 11–10, vC E (V0 E)e t/t 40 (25 40)e t/0.2 40 15e 5t V b. From Equation 11–11, E V 40 25 iC 0 e t/t e 5t 75e 5t mA R 200 c. At t 0.1 s, vC 40 15e 5t 40 15e 0.5 30.9 V iC 75e 5t mA 75e 0.5 mA 45.5 mA d. The waveforms are shown in Figure 11–17 with the above points plotted. vC

vC = 40 15 e 5t V

75 mA

25 V

0.1 s (a) V0 = 25V.

EWB

PRACTICE PROBLEMS 3

FIGURE 11–17

iC = 75 e 5t mA 45.5 mA

30.9 V 0

40 V

t (s)

0.1 s

t (s)

(b) Capacitor with an initial voltage.

Repeat Example 11–7 for the circuit of Figure 11–16 if V0 150 V. Answers: a. 40 190e 5t V b. 0.95e 5t A c. 75.2 V; 0.576 A d. Curves are similar to Figure 11–17 except that vC starts at 150 V and rises to 40 V while iC starts at 0.95 A and decays to zero.

Section 11.4

11.4

■

429

Capacitor Discharging Equations

Capacitor Discharging Equations vR

To determine the discharge equations, move the switch to the discharge position (Figure 11–18). KVL yields vR vC 0. Substituting vR RCdvC /dt from Section 11.2 yields d vC RC vC 0 dt

iR 2

(11–12)

This can be solved for vC using basic calculus. The result is vC V0 e t/RC

(11–13)

where V0 is the voltage on the capacitor at the instant the switch is moved to discharge. Now consider the resistor voltage. Since vR vC 0, vR vC and vR V0 e t/RC

(11–14)

Now divide both sides by R. Since iC iR vR/R, V iC 0 e t/RC R

FIGURE 11–18 Discharge case. Initial capacitor voltage is V0. Note the reference direction for iC. (To conform to the standard voltage/current reference convention, iC must be drawn in this direction so that the sign for vC is at the tail of the current arrow.) Since the actual current direction is opposite to the reference direction, iC will be negative. This is indicated in Figure 11–19(b).

(11–15)

Note that this is negative, since, during discharge, the current is opposite in direction to the reference arrow of Figure 11–18. (If you need to refresh your memory, see again Figure 11–5.) Voltage vC and current iC are shown in Figure 11–19. As in the charging case, discharge transients last five time constants. In Equations 11–13 to 11–15, V0 represents the voltage on the capacitor at the instant the switch is moved to the discharge position. If the switch has been in the charge position long enough for the capacitor to fully charge, V0 E and Equations 11–13 and 11–15 become vC Ee t/RC and iC (E/R)e t/RC respectively.

vC vC = V 0 e

t / RC

V0 t 0

5 (a) iC t

EXAMPLE 11–8

For the circuit of Figure 11–18, assume the capacitor is charged to 100 V before the switch is moved to the discharge position. Suppose R 5 k and C 25 mF. After the switch is moved to discharge, a. Determine the expression for vC. b. Determine the expression for iC. c. Compute the voltage and current at 0.375 s. Solution RC (5 k )(25 mF) 0.125 s and V0 100 V. Therefore, a. vC V0e t/RC 100e t/0.125 100e 8t V. b. iC (V0 /R)e t/RC 20e 8t mA. c. At t 0.375 s, vC 100e 8t 100e 3 4.98 V iC 20e 8t mA 20e 3 mA 0.996 mA

V0 R

iC =

V0 e R

t / RC

(b) During discharge, iC is negative as determined in Figure 11–18 FIGURE 11–19 Capacitor voltage and current for the discharge case.

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits

The universal time constant curve of Figure 11–15(b) may also be used to solve discharge problems. For example, for the circuit of Example 11–8, at t 3 t, capacitor voltage has fallen to 4.98% of E, which is (0.0498)(100 V) 4.98 V and current has decayed to 4.98% of 20 mA which is (0.0498)( 20 mA) 0.996 mA. (These agree with Example 11–8 since 3 t was also the value of time used there.)

11.5

More Complex Circuits

The charge and discharge equations described previously apply only to circuits of the forms shown in Figures 11–2 and 11–5 respectively. Fortunately, many circuits can be reduced to these forms using standard circuit reduction techniques such as series and parallel combinations, source conversions, Thévenin’s theorem, and so on. Once a circuit has been reduced to its series equivalent, you can use any of the equations that we have developed so far.

EXAMPLE 11–9 For the circuit of Figure 11–20(a), determine expressions for vC and iC. Capacitors are initially uncharged. R1 = 3 k iC R2 = 6 k E

C1 8 F

100 V

C2 2 F

Req = 2 k

100 V Ceq = 10 F

(a)

(b)

FIGURE 11–20

Solution

Reduce circuit (a) to circuit (b). Req R1㥋R2 2.0 k ; 6

ReqCeq (2 k )(10 10

Ceq C1 C2 10 mF. F) 0.020 s

Thus, vC E(1 e t/ReqCeq) 100(1 e t/0.02) 100(1 e 50t) V E 100 iC e t/ReqCeq e t/0.02 50 e 50t mA Req 2000

iC vC

Section 11.5

EXAMPLE 11–10 The capacitor of Figure 11–21 is initially uncharged. Close the switch at t 0 s. a. Determine the expression for vC. b. Determine the expression for iC. c. Determine capacitor current and voltage at t 5 ms.

R1 = 240

R2 800

100 V

C = 50 F b

R3 200

R4 104

R'2 FIGURE 11–21

Solution

Reduce the circuit to its series equivalent using Thévenin’s theorem: R⬘2 R2㥋R3 160

From Figure 11–22(a), RTh R1㥋R⬘2 R4 240㥋160 104 96 104 200 From Figure 11–22(b), R⬘2 160 V⬘2 E 100 V 40 V R1 R⬘2 240 160

冢

冣

R1 = 240

R1 = 240 b

R4 104

(a) Finding RTh FIGURE 11–22 switch closure.

冣

RTh a

R'2 160

冢

R'2 100 V160

ETh a b R4 V'2 104

(b) Finding ETh

Determining the Thévenin equivalent of Figure 11–21 following

From KVL, ETh V⬘2 40 V. The resultant equivalent circuit is shown in Figure 11–23. t RThC (200 )(50 mF) 10 ms

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a. vC ETh(1 e t/t ) 40(1 e 100t ) V E h t/t 40 b. iC T e e t/0.01 200e 100t mA RTh 200 c. iC 200e 100(5 ms) 121 mA. Similarly, vC 15.7 V RTh = 200

ETh

FIGURE 11–23

PRACTICE PROBLEMS 4

40 V 50 F

iC vC

The Thévenin equivalent of Figure 11–21.

1. For Figure 11–21, if R1 400 , R2 1200 , R3 300 , R4 50 , C 20 mF, and E 200 V, determine vC and iC. 2. Using the values shown in Figure 11–21, determine vC and iC if the capacitor has an initial voltage of 60 V. 3. Using the values of Problem 1, determine vC and iC if the capacitor has an initial voltage of 50 V. Answers: 1. 75(1 e 250t ) V;

0.375e 250t A

2. 40 20e 100t V;

0.1e 100t A

250t

3. 75 125e

0.625e 250t A

PRACTICAL NOTES... Notes About Time References 1. So far, we have dealt with charging and discharging problems separately. For these, we define t 0 s as the instant the switch is moved to the charge position for charging problems and to the discharge position for discharging problems. 2. When you have both charge and discharge cases in the same example, you need to establish clearly what you mean by “time.” We use the following procedure: a. Define t 0 s as the instant the switch is moved to the first position, then determine corresponding expressions for vC and iC. These expressions and the corresponding time scale are valid until the switch is moved to its new position. b. When the switch is moved to its new position, shift the time reference and make t 0 s the time at which the switch is moved to its new

Section 11.5

■

More Complex Circuits

433

position, then determine corresponding expressions for vC and iC. These new expressions are only valid from the new t 0 s reference point. The old expressions are valid only on the old time scale. c. We now have two time scales for the same graph. However, we generally only show the first scale explicitly; the second scale is implied rather than shown. d. Use tC to represent the time constant for charging and td to represent the time constant for discharging. Since the equivalent resistance and capacitance for discharging may be different than that for charging, the time constants may be different for the two cases.

EXAMPLE 11–11

The capacitor of Figure 11–24(a) is uncharged. The switch is moved to position 1 for 10 ms, then to position 2, where it remains.

a. b. c. d. e.

Determine vC during charge. Determine iC during charge. Determine vC during discharge. Determine iC during discharge. Sketch the charge and discharge waveforms. R1

200

800 2 100 V R3

300

2 F

(a) Full circuit RTc = 1000 iC 100 V

2 F

(b) Charging circuit

iC 500

2 F

FIGURE 11–24

Solution Figure 11–24(b) shows the equivalent charging circuit. Here, tc (R1 R2)C (1 k )(2 mF) 2.0 ms.

NOTES… When solving a transient problem, always draw the circuit as it looks during each time interval of interest. (It doesn’t take long to do this, and it helps clarify just what it is you need to look at for each part of the solution.) This is illustrated in Example 11–11. Here, we have drawn the circuit in Figure 11–24(b) as it looks during charging and in (c) as it looks during discharging. It is now clear which components are relevant to the charging phase and which are relevant to the discharging phase.

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a. vC E(1 e t/tc) 100(1 e 500t ) V E 100 b. iC e t/tc e 500t 100e 500t mA RTc 1000 Since 5tc 10 ms, charging is complete by the time the switch is moved to discharge. Thus, V0 100 V when discharging begins. c. Figure 11–24(c) shows the equivalent discharge circuit. Note V0 100 V. td (500 )(2 mF) 1.0 ms vC V0e t/td 100e 1000t V where t 0 s has been redefined for discharge as noted above. V0 100 t/td d. iC e e 1000t 200e 1000t mA 500 R2 R3 e. See Figure 11–25. Note that discharge is more rapid than charge since td tc. vC (V) 100

iC (mA) 100 0 100

t (ms)

0 2 4 6 8 10 12 14 5 c

2 4 6 8 10

t (ms)

200 (b)

(a) FIGURE 11–25 than tc.

Waveforms for the circuit of Figure 11–24. Note that td is shorter

EXAMPLE 11–12 The capacitor of Figure 11–26 is uncharged. The switch is moved to position 1 for 5 ms, then to position 2 and left there. EWB

FIGURE 11–26

1000

1 2 E1

10 V

iC 30 V

a. Determine vC while the switch is in position 1. b. Determine iC while the switch is in position 1. c. Compute vC and iC at t 5 ms.

4 F

Section 11.5

d. e. f. g.

Determine vC while the switch is in position 2. Determine iC while the switch is in position 2. Sketch the voltage and current waveforms. Determine vC and iC at t 10 ms.

Solution tc td RC (1 k )(4 mF) 4 ms a. vC E1(1 e t/tc) 10(1 e 250t ) V E1 t/tc 10 e 250t 10e 250t mA b. iC e R 1000 c. At t 5 ms, vC 10(1 e 250 0.005) 7.14 V iC 10e 250 0.005 mA 2.87 mA d. In position 2, E2 30 V, and V0 7.14 V. Use Equation 11–10: vC E2 (V0 E2)e t/td 30 (7.14 30)e 250t 30 22.86e 250t V where t 0 s has been redefined for position 2. E2 V0 t/td 30 7.14 250t e. iC e e 22.86e 250t mA R 1000 f. See Figure 11–27. g. t 10 ms is 5 ms into the new time scale. Thus, vC 30 22.86e 250(5 ms) 23.5 V and iC 22.86e 250(5 ms) 6.55 mA. Values are plotted on the graph. iC (mA)

vC (V)

23.5 V

20 10 0

7.14 V 5

FIGURE 11–27

22.86 mA

25 20 15 10 5

t (ms)

2.87 mA 6.55 mA 5

t (ms)

Capacitor voltage and current for the circuit of Figure 11–26.

EXAMPLE 11–13

In Figure 11–28(a), the capacitor is initially uncharged. The switch is moved to the charge position, then to the discharge position, yielding the current shown in (b). The capacitor discharges in 1.75 ms. Determine the following: a. E. b. R1. c. C.

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R3 = 10

iC (A)

Fully charged here

E R2

t (s)

0 3 (b)

(a) FIGURE 11–28

Solution a. Since the capacitor charges fully, it has a value of E volts when switched to discharge. The discharge current spike is therefore E 3 A 10 25 Thus, E 105 V. b. The charging current spike has a value of E 7 A 10 R1 Since E 105 V, this yields R1 5 . c. 5 td 1.75 ms. Therefore td 350 ms. But td (R2 R3)C. Thus, C 350 ms/35 10 mF.

RC Circuits in Steady State DC When an RC circuit reaches steady state dc, its capacitors look like open circuits. Thus, a transient analysis is not needed.

EXAMPLE 11–14 The circuit of Figure 11–29(a) has reached steady state. Determine the capacitor voltages. 40

90 V 40

100 F 200 V 60

(a) EWB

FIGURE 11–29

Continues

20 F

Section 11.5

■

More Complex Circuits

Solution Replace all capacitors with open circuits. Thus, V1

90 V

120 V 12

18 V

I1 200 V 60

12 V V2

I2 I1 = 2 A

I2 = 1.5 A (b)

EWB

FIGURE 11–29

Continued

90 V I2 1.5 A 40 8 12

200 V I1 2 A, 40 60

KVL: V1 120 18 0. Therefore, V1 138 V. Further, V2 (8 )(1.5 A) 12 V

1. The capacitor of Figure 11–30(a) is initially unchanged. At t 0 s, the switch is moved to position 1 and 100 ms later, to position 2. Determine vC and iC for position 2. 2. Repeat for Figure 11–30(b). Hint: Use Thévenin’s theorem. 40

iC 20

2 20 V 80

(a) C = 500 F

1 1 k iC 12 V 2 mA

4 k

(b) C = 20 F FIGURE 11–30

2 k

2 vC

PRACTICE PROBLEMS 5

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits 3. The circuit of Figure 11–31 has reached steady state. Determine source currents I1 and I2. 50 F 4

I1 E1

I2 11 F

10 V

6 1 F

30 V

FIGURE 11–31 Answers: 1. 20e 20t V; 25t

0.2e 20t A 6.3e 25t mA

2. 12.6e

3. 0 A;

1.67 A

11.6

An RC Timing Application

RC circuits are used to create delays for alarm, motor control, and timing applications. Figure 11–32 shows an alarm application. The alarm unit contains a threshold detector, and when the input to this detector exceeds a preset value, the alarm is turned on. Input from sensor

Alarm unit

R C

Audio horn

(a) Delay circuit

Input from sensor

E Threshold

vC Alarm on Delay (b) FIGURE 11–32

Creating a time delay with an RC circuit.

Section 11.6

■

An RC Timing Application

EXAMPLE 11–15 The circuit of Figure 11–32 is part of a building security system. When an armed door is opened, you have a specified number of seconds to disarm the system before the alarm goes off. If E 20 V, C 40 mF, the alarm is activated when vC reaches 16 V, and you want a delay of at least 25 s, what value of R is needed? Solution vC E(1 e t/RC). After a bit of manipulation, you get E vC e t/RC E Taking the natural log of both sides yields

EWB

t E vC ln RC E

冢

冣

At t 25 s, vC 16 V. Thus, t 20 16 ln ln 0.2 1.6094 RC 20

冢

冣

Substituting t 25 s and C 40 mF yields 25 s t R 388 k 1.6094 40 10 6 1.6094C Choose the next higher standard value, namely 390 k .

1. Suppose you want to increase the disarm time of Example 11–15 to at least 35 s. Compute the new value of R. 2. If, in Example 11–15, the threshold is 15 V and R 1 M , what is the disarm time?

PRACTICE PROBLEMS 6

Answers: 1. 544 k . Use 560 k . 2. 55.5 s

1. Refer to Figure 11–16: a. Determine the expression for vC when V0 80 V. Sketch vC. b. Repeat (a) if V0 40 V. Why is there no transient? c. Repeat (a) if V0 60 V. 2. For Part (c) of Question 1, vC starts at 60 V and climbs to 40 V. Determine at what time vC passes through 0 V, using the technique of Example 11–15. 3. For the circuit of Figure 11–18, suppose R 10 k and C 10 mF: a. Determine the expressions for vC and iC when V0 100 V. Sketch vC and iC. b. Repeat (a) if V0 100 V. 4. Repeat Example 11–12 if voltage source 2 is reversed, i.e., E2 30 V.

IN-PROCESS

LEARNING CHECK 2

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits 5. The switch of Figure 11–33(a) is closed at t 0 s. The Norton equivalent of the circuit in the box is shown in (b). Determine expressions for vC and iC. The capacitor is initially uncharged. 30

Circuit

0.6 A

C = 1000 F (a) FIGURE 11–33

(b) Norton equivalent

Hint: Use a source transformation. (Answers are at the end of the chapter.)

11.7

Pulse Response of RC Circuits

In previous sections, we looked at the response of RC circuits to switched dc inputs. In this section, we consider the effect that RC circuits have on pulse waveforms. Since many electronic devices and systems utilize pulse or rectangular waveforms, including computers, communications systems, and motor control circuits, these are important considerations.

Pulse Basics A pulse is a voltage or current that changes from one level to the other and back again as in Figure 11–34(a) and (b). A pulse train is a repetitive stream of pulses, as in (c). If a waveform’s high time equals its low time, as in (d), it is called a square wave. The length of each cycle of a pulse train is termed its period, T, and the number of pulses per second is defined as its pulse repetition rate (PRR) or pulse repetition frequency (PRF). For example, in (e), there are two complete cycles in one second; therefore, the PRR 2 pulses/s. With two cycles every second, the time for one cycle is T 1⁄ 2 s. Note that this is 1/PRR. This is true in general. That is, 1 T s PRR

(11–16)

The width, tp, of a pulse relative to its period, [Figure 11–34(c)] is its duty cycle. Thus, tp duty cycle 100% T

(11–17)

A square wave [Figure 11–34(d)] therefore has a 50% duty cycle, while a waveform with tp 1.5 ms and a period of 10 ms has a duty cycle of 15%.

Section 11.7

■

441

Pulse Response of RC Circuits

In practice, waveforms are not ideal, that is, they do not change from low to high or high to low instantaneously. Instead, they have finite rise and fall times. Rise and fall times are denoted as tr and tf and are measured between the 10% and 90% points as indicated in Figure 11–35(a). Pulse width is measured at the 50% point. The difference between a real waveform and an ideal waveform is often slight. For example, rise and fall times of real pulses may be only a few nanoseconds and when viewed on an oscilloscope, as in Figure 11–35(b), appear to be ideal. In what follows, we will assume ideal waveforms.

V 0 Time (a) Positive pulse V 0 Time

90%

90% 10% tr Rise time

tp Pulse width

(b) Negative pulse 10% tf Fall time

(a) Pulse definitions

V 0

Time

T 2

V 0

Time

(d) Square wave 1s

(e) PRR = 2 pulses/s FIGURE 11–34 Ideal pulses and pulse waveforms. (b) Pulse waveform viewed on an oscilloscope. FIGURE 11–35

Practical pulse waveforms.

The Effect of Pulse Width The width of a pulse relative to a circuit’s time constant determines how it is affected by an RC circuit. Consider Figure 11–36. In (a), the circuit has been drawn to focus on the voltage across C; in (b), it has been drawn to focus on the voltage across R. (Otherwise, the circuits are identical.) An easy way to visualize the operation of these circuits is to assume that the pulse is generated by a switch that is moved rapidly back and forth between V and common as in (c). This alternately creates a charge and discharge circuit, and thus all of the ideas developed in this chapter apply directly.

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits C R vin

(a) Output across C

vin

(b) Output across R

RC circuits with pulse input.

Pulse Width tp ⬎⬎ 5 ␶

First, consider the ouput of circuit (a). When the pulse width and time between pulses are very long compared with the circuit time constant, the capacitor charges and discharges fully, Figure 11–37(b). (This case is similar to what we have already seen in this chapter.) Note, that charging and discharging occur at the transitions of the pulse. The transients therefore increase the rise and fall times of the output. In high-speed circuits, this may be a problem. (You will learn more about this in your digital electronics courses.)

vin V 0

T 2

T 2 (a)

vC V

Circuit (a)

0 5

(b) vR V

Circuit (b)

0 V

EXAMPLE 11–16 A square wave is applied to the input of Figure 11–36(a). If R 1 k and C 100 pF, estimate the rise and fall time of the output signal using the universal time constant curve of Figure 11–15(a). Solution Here, t RC (1 103)(100 10 12) 100 ns. From Figure 11–15(a), note that vC reaches the 10% point at about 0.1 t, which is (0.1)(100 ns) 10 ns. The 90% point is reached at about 2.3 t, which is (2.3)(100 ns) 230 ns. The rise time is therefore approximately 230 ns 10 ns 220 ns. The fall time will be the same.

(c) FIGURE 11–37 Pulse width much greater than 5 t. Note that the shaded areas indicate where the capacitor is charging and discharging. Spikes occur on the input voltage transitions.

Now consider the circuit in Figure 11–36(b). Here, current iC will be similar to that of Figure 11–27(b), except that the pulse widths will be narrower. Since voltage vR R iC, the output will be a series of short, sharp spikes that occur at input transitions as in Figure 11–37(c). Under the conditions here (i.e., pulse width much greater than the circuit time constant), vR is

Section 11.7

■

an approximation to the derivative of vin and the circuit is called a differentiator circuit. Such circuits have important practical uses.

Pulse Response of RC Circuits V

vin T 2

Pulse Width tp ⴝ 5 ␶

These waveforms are shown in Figure 11–38. Since the pulse width is 5t, the capacitor fully charges and discharges during each pulse. Thus waveforms here will be similar to what we have seen previously. Pulse Width tp ⬍⬍ 5 ␶

This case differs from what we have seen so far in this chapter only in that the capacitor does not have time to charge and discharge significantly between pulses. The result is that switching occurs on the early (nearly straight line) part of the charging and discharging curves and thus, vC is roughly triangular in shape, Figure 11–39(a). As shown below, it has an average value of V/2. Under the conditions here, vC is the approximate integral of vin and the circuit is called an integrator circuit. It should be noted that vC does not reach the steady state shown in Figure 11–39 immediately. Instead, it works its way up over a period of five time constants (Figure 11–40). To illustrate, assume an input square wave of 5 V with a pulse width of 0.1 s and t 0.1 s.

0.1

vR V

3.59

3.16 1.16

5τ

V FIGURE 11–38 Pulse width equal to 5 t. These are the same as Figure 11–37 except that the transients last relatively longer. V vin

3.65 1.34

steady state

(b) Output voltage vC EWB

FIGURE 11–40

T 2

V V 2 0 Circuit (a)

3.65 1.32

Circuit (b)

(a) Input waveform 5V

V Circuit (a)

T 2

vC V

0.1

443

Circuit takes five time constants to reach a steady state.

vC E(1 e t/t ). At the end of the first pulse (t 0.1 s), vC has climbed to vC 5(1 e 0.1/0.1) 5(1 e 1) 3.16 V. From the end of pulse 1 to the beginning of pulse 2 (i.e., over an interval of 0.1 s), vC decays from 3.16 V to 3.16e 0.1/0.1 3.16e 1 1.16 V. Pulse 1:

Pulse 2: vC starts at 1.16 V and 0.1 s later has a value of vC E (V0 E)e t/t 5 (1.16 5)e 0.1/0.1 5 3.84e 1 3.59 V. It then decays to 3.59e 1 1.32 V over the next 0.1 s. Continuing in this manner, the remaining values for Figure 11–40(b) are determined. After 5 t, vC cycles between 1.34 and 3.65 V, with an average of (1.34 3.65)/2 2.5 V, or half the input pulse amplitude.

V vR 0 V Circuit (b) FIGURE 11–39 Pulse width much less than 5 t. The circuit does not have time to charge or discharge substantially.

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PRACTICE PROBLEMS 7

Verify the remaining points of Figure 11–40(b).

(a) Unloaded driver

Load

(b) Distorted signal FIGURE 11–41 Distortion caused by capacitive loading.

Capacitive Loading Capacitance occurs whenever conductors are separated by insulating material. This means that capacitance exists between wires in cables, between traces on printed circuit boards, and so on. In general, this capacitance is undesirable but it cannot be avoided. It is called stray capacitance. Fortunately, stray capacitance is often so small that it can be neglected. However, in high-speed circuits, it may cause problems. To illustrate, consider Figure 11–41. The electronic driver of (a) produces square pulses. However, when it drives a long line as in (b), stray capacitance loads it and increases the signal’s rise and fall times (since capacitance takes time to charge and discharge). If the rise and fall times become excessively long, the signal reaching the load may be so degraded that the system malfunctions. (Capacitive loading is a serious issue but we will leave it for future courses to deal with.)

11.8

ELECTRONICS WORKBENCH

PSpice

Transient Analysis Using Computers

Electronics Workbench and PSpice are well suited for studying transients as they both incorporate easy to use graphing facilities that you can use to plot results directly on the screen. When plotting transients, you must specify the time scale for your plot—i.e., the length of time that you expect the transient to last. A good value to start with is 5 t where t is the time constant of the circuit. (For complex circuits, if you do not know t, make an estimate, run a simulation, adjust the time scale, and repeat until you get an acceptable plot.)

Electronics Workbench Workbench provides two ways to view waveforms—via its oscilloscope or via its analysis graphing facility. Since the analysis graphing facility is easier to use and yields better output, we will begin with it. (In Chapter 14, we introduce the oscilloscope.) As a first example, consider the RC charging circuit of Figure 11–42. Determine capacitor voltage at t 50 ms and t 150 ms. (You don’t need a switch; you simply tell Workbench to perform a transient analysis.) Proceed as follows: • Create the circuit of Figure 11–42 on the screen. (To display node numbering, select Circuit/Schematic Options, enable Show Nodes, then click OK). • Select Analysis/Transient and in the dialog box, click Initial Conditions Set to Zero. Set End Time (TSTOP) to 0.25, highlight Node 2 (to display the voltage across the capacitor), then click on Add. • Click Simulate. When the analysis is finished, the graph of Figure 11– 43(b) should appear. Click the Toggle Grid icon on the Analysis Graphs menu bar. Expand to full screen.

Section 11.8

■

Transient Analysis Using Computers

445

NOTES…

FIGURE 11–42 Electronics Workbench example. The switch is not required since the transient solution is initiated by software.

• Click the Toggle Cursor icon and drag the cursors to the specific time at which you want to read voltages.

1. Be careful when connecting multiple lines to a node or Workbench may cause a line to be dropped. For example, in Figure 11–42 be sure that the ground is actually connected, not simply butted up against the bottom wire. If you have trouble, go to the Basic Parts bin, get a connector dot, (•) and use it to make the connection. 2. A shorthand notation is used here. For example, “select Circuit/Schematic Options” means to click Circuit on the toolbar and select “Schematic Options” from the drop-down menu that appears.

Analysis of Results

As indicated in Figure 11–43(a), vC 6.32 V at t 50 ms and 9.50 V at t 150 ms. (Check by substitution into vC 10(1 e 20t). You will find that results agree exactly.)

(a) Cursor readings

(b) Capacitor voltage waveform

FIGURE 11–43 Solution for the circuit of Figure 11–42. Since t 50 ms, run the simulation to at least 250 ms (i.e., 5 t).

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits Initial Conditions in Electronics Workbench

Let us change the above problem to include an initial voltage of 20 V on the capacitor. Get the circuit of Fig. 11–42 back on the screen, click the Basic Parts bin icon, drag the connector dot (●) and insert it into the circuit wiring slightly above the capacitor. Double click the dot and in the Connector Properties box that opens, click Use Initial Conditions, type 20 into the Transient Analysis (IC) box, then click OK. Click Analysis, Transient and under Initial Conditions, click User-Defined, select the capacitor node voltage for display, then click Simulate. Note that the transient starts at 20 V and decays to its steady state value of 10 V in five time constants as expected. Another Example

Using the clock source from the Sources bin, build the circuit of Figure 11– 44. (The clock, with its default settings, produces a square wave that cycles

FIGURE 11–44

Applying a square wave source.

FIGURE 11–45

Output waveform for Figure 11–44. Compare to Figure 11–37(c).

Section 11.8

■

Transient Analysis Using Computers

between 0 V and 5 V with a cycle length of T 1 ms.) This means that its on time tp is T/2 500 ms. Since the time constant of Figure 11–44 is t RC 50 ms, tp is greater than 5t and a waveform similar to that of Figure 11–37(c) should result. To verify, follow the procedure of the previous example, except set End Time (TSTOP) to 0.0025 in the Analysis/Transient dialog box. The waveform of Figure 11–45 appears. Note that output spikes occur on the transitions of the input waveform as predicted.

OrCAD PSpice As a first example, consider Figure 11–2 with R 200 , C 50 mF and E 40 V. Let the capacitor be initially uncharged (i.e., Vo 0 V). First, read the PSpice Operational Notes, then proceed as follows: • Create the circuit on the screen as in Figure 11–46. (The switch can be found in the EVAL library as part Sw_tClose.) Remember to rotate the capacitor three times as discussed in Appendix A, then set its initial condition (IC) to zero. To do this, double click the capacitor symbol, type 0V into the Property editor cell labeled IC, click Apply, then close the editor. Click the New Simulation Profile icon, enter a name (e.g., Figure 11–47) then click Create. In the Simulation Settings box, click the Analysis tab, select Time Domain (Transient) and in Options, select General Settings. Set the duration of the transient (TSTOP) to 50ms (i.e., five time constants). Find the voltage marker on the toolbar and place as shown.

FIGURE 11–46 PSpice example. The voltage marker displays voltage with respect to ground, which, in this case, is the voltage across C1.

• Click the Run icon. When simulation is complete, a trace of capacitor voltage versus time (the green trace of Figure 11–47) appears. Click Plot (on the toolbar) then add Y Axis to create the second axis. Activate the additional toolbar icons described in Operational Note 5, then click the Add Trace icon on the new toolbar. In the dialog box, click I(C1) (assuming your capacitor is designated C1), then OK. This adds the current trace.

447

NOTES... PSpice Operational Notes 1. Do not use a space between a value and its unit. Thus, use 50ms, not 50 ms, etc. 2. When instructed to enter data via a Properties editor, first click the Parts tab at the bottom of the screen; scroll right until you find the cell that you want, and then type in its value. 3. Sometimes you get choppy waveforms. If this happens, enter a suitable value for Maximum step size in the Simulation Profile box. If you make the value too large, the waveform will be choppy, while if you make it too small, simulation time will be too lengthy. Values usually aren’t critical, but you may need to experiment a bit. (To illustrate, you can smooth the waveforms of Figure 11–47 using a Maximum step size of 10us, and the curves of Figure 11–50 with a step size of 20ms.) 4. For transient problems you need to specify an initial condition (IC) for each capacitor and inductor. The procedure is described in the examples. 5. To activate the icons used for adding and viewing waveforms in these examples, you may need to set up additional toolbar icons. Proceed as follows: When the display of Figure 11–47 appears, click Tools, Customize, select the Toolbars tab, select all toolbars shown, i.e., File, Edit, Simulate, Probe and Cursor, then click OK. Position the pointer over various icons to note their function.

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FIGURE 11–47

Waveforms for the circuit of Figure 11–46.

Analysis of Results

Click the Toggle cursor icon, then use the cursor to determine values from the screen. For example, at t 5 ms, you should find vC 15.7 V and iC 121 mA. (An analytic solution for this circuit (which is Figure 11–23) may be found in Example 11–10, part (c). It agrees exactly with the PSpice solution.) As a second example, consider the circuit of Figure 11–21 (shown as Figure 11–48). Create the circuit using the same general procedure as in the previous example, except do not rotate the capacitor. Again, be sure to set V0 (the initial capacitor voltage) to zero. In the Simulation Profile box, set TSTOP to 50ms. Place differential voltage markers (found on the toolbar at the top of the screen) across C to graph the capacitor voltage. Run the analysis, create a second axis, then add the current plot. You should get the same graph (i.e., Figure 11–47) as you got for the previous example, since its circuit is the Thévenin equivalent of this one.

FIGURE 11–48

Differential markers are used to display the voltage across C1.

As a final example, consider Figure 11–49(a), which shows double switching action.

Section 11.8

■

Transient Analysis Using Computers

EXAMPLE 11–17 The capacitor of Figure 11–49(a) has an initial voltage of 10 V. The switch is moved to the charge position for 1 s, then to the discharge position where it remains. Determine curves for vC and iC. R = 5 k iC 20 V C = 40 F

(a) Circuit to be modelled 20 V 1s

(b) The applied pulse

Creating a charge/discharge waveform using PSpice.

Solution PSpice has no switch that implements the above switching sequence. However, moving the switch first to charge then to discharge is equivalent to placing 20 V across the RC combination for the charge time, then 0 V thereafter as indicated in (b). You can do this with a pulse source (VPULSE) as indicated in (c). (VPULSE is found in the SOURCE library.) To set pulse parameters, double click the VPULSE symbol, then scroll its Property editor as described in the Operational Notes until you find a group of cells labeled PER, PW, etc. Enter 5s for PER, 1s for PW, 1us for TF, 1us

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for TR, 0 for TD, 20V for V2, and 0V for V1. (This defines a pulse with a period of 5 s, a width of 1 s, rise and fall times of 1 ms, amplitude of 20 V, and an initial value of 0 V.) Click Apply, then close the Property editor. Double click the capacitor symbol and set IC to 10V in the Properties Editor. Set TSTOP to 2s. Place a Voltage Marker as shown, then click Run. You should get the voltage trace of Figure 11–50 on the screen. Add the second axis and the current trace as described in the previous examples. The red current curve should appear.

FIGURE 11–50

Waveforms for the circuit of Figure 11–49 with V0 10 V.

Note that voltage starts at 10 V and climbs to 20 V while current starts at (E V0)/R 30 V/5 k 6 mA and decays to zero. When the switch is turned to the discharge position, the current drops from 0 A to 20 V/5 k 4 mA and then decays to zero while the voltage decays from 20 V to zero. Thus the solution checks.

PUTTING IT INTO PRACTICE

n electronic device employs a timer circuit of the kind shown in Figure 11–32(a), i.e., an RC charging circuit and a threshold detector. (Its timing waveforms are thus identical to those of Figure 11–32(b)). The input to the RC circuit is a 0 to 5 V 4% step, R 680 k 10%, C 0.22 mF 10%, the threshold detector activates at vC 1.8 V 0.05 V and the required delay is 67 ms 18 ms. You test a number of units as they come off the production line and find that some do not meet the timing spec. Perform a design review and determine the cause. Redesign the timing portion of the circuit in the most economical way possible.

Problems

11.1 Introduction 1. The capacitor of Figure 11–51 is uncharged. a. What are the capacitor voltage and current just after the switch is closed? b. What are the capacitor voltage and current after the capacitor is fully charged? 2. Repeat Problem 1 if the 20-V source is replaced by a 60-V source. 3. a. What does an uncharged capacitor look like at the instant of switching? b. What does a charged capacitor look like at the instant of switching? c. What does a capacitor look like to steady state dc? d. What do we mean by i(0 )? By i(0 )? 4. For a charging circuit, E 25 V, R 2.2 k , and the capacitor is initially uncharged. The switch is closed at t 0. What is i(0 )?

PROBLEMS

R 4 iC E

FIGURE 11–51

11.2 Capacitor Charging Equations 6. The switch of Figure 11–51 is closed at t 0 s. The capacitor is initially uncharged. a. Determine the equation for charging voltage vC. b. Determine the equation for charging current iC. c. By direct substitution, compute vC and iC at t 0 s, 40 ms, 80 ms, 120 ms, 160 ms, and 200 ms. d. Plot vC and iC on graph paper using the results of (c). Hint: See Example 11–2. 7. Repeat Problem 6 if R 500 , C 25 mF, and E 45 V, except compute and plot values at t 0 s, 20 ms, 40 ms, 60 ms, 80 ms, and 100 ms. 8. The switch of Figure 11–52 is closed at t 0 s. Determine the equations for capacitor voltage and current. Compute vC and iC at t 50 ms. 9. Repeat Problem 8 for the circuit of Figure 11–53. vC 10 k iC

FIGURE 11–52

V0 0 V, C 10 mF.

3.9 k

FIGURE 11–53

C 10 mF, V0 0 V.

10. The capacitor of Figure 11–2 is uncharged at the instant the switch is closed. If E 80 V, C 10 mF, and iC(0 ) 20 mA, determine the equations for vC and iC.

20 V

C = 10 F

5. For a charging circuit, R 5.6 k and vC(0 ) 0 V. If i(0 ) 2.7 mA, what is E?

20 V

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40 V

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits 11. Determine the time constant for the circuit of Figure 11–51. How long (in seconds) will it take for the capacitor to charge? 12. A capacitor takes 200 ms to charge. If R 5 k , what is C? 13. For Figure 11–51, the capacitor voltage with the switch open is 0 V. Close the switch at t 0 and determine capacitor voltage and current at t 0 , 40 ms, 80 ms, 120 ms, 160 ms, and 200 ms using the universal time constant curves. 14. If iC 25e 40t A, what is the time constant t and how long will the transient last? 15. For Figure 11–2, the current jumps to 3 mA when the switch is closed. The capacitor takes 1 s to charge. If E 75 V, determine R and C. 16. For Figure 11–2, if vC 100(1 e 50t) V and iC 25e 50t mA, what are E, R, and C? 17. For Figure 11–2, determine E, R, and C if the capacitor takes 5 ms to charge, the current at 1 time constant after the switch is closed is 3.679 mA, and the capacitor charges to 45 volts. 18. For Figure 11–2, vC (t) 41.08 V and iC (2t) 219.4 mA. Determine E and R. 11.3 Capacitor with an Initial Voltage 19. The capacitor of Figure 11–51 has an initial voltage. If V0 10 V, what is the current just after the switch is closed? 20. Repeat Problem 19 if V0 10 V. 21. For the capacitor of Figure 11–51, V0 30 V. a. Determine the expression for charging voltage vC. b. Determine the expression for current iC. c. Sketch vC and iC. 22. Repeat Problem 21 if V0 5 V.

iC R

25 k

C = 20 F FIGURE 11–54

11.4 Capacitor Discharging Equations 23. For the circuit of Figure 11–54, assume the capacitor is charged to 50 V before the switch is closed. a. Determine the equation for discharge voltage vC. b. Determine the equation for discharge current iC. c. Determine the time constant of the circuit. d. Compute vC and iC at t 0 s, t t, 2t, 3t, 4t, and 5t. e. Plot the results of (d) with the time axis scaled in seconds and time constants. 24. The initial voltage on the capacitor of Figure 11–54 is 55 V. The switch is closed at t 0. Determine capacitor voltage and current at t 0 , 0.5 s, 1 s, 1.5 s, 2 s, and 2.5 s using the universal time constant curves. 25. A 4.7-mF capacitor is charged to 43 volts. If a 39-kV resistor is then connected across the capacitor, what is its voltage 200 ms after the resistor is connected?

Problems 26. The initial voltage on the capacitor of Figure 11–54 is 55 V. The switch is closed at t 0 s and opened 1 s later. Sketch vC. What is the capacitor’s voltage at t 3.25 s? 27. For Figure 11–55, let E 200 V, R2 1 k , and C 0.5 mF. After the capacitor has fully charged in position 1, the switch is moved to position 2. a. What is the capacitor voltage immediately after the switch is moved to position 2? What is its current? b. What is the discharge time constant? c. Determine discharge equations for vC and iC. R2 2 10 k R1

15 k R3

iC C

FIGURE 11–55

28. For Figure 11–55, C is fully charged before the switch is moved to discharge. Current just after it is moved is iC 4 mA and C takes 20 ms to discharge. If E 80 V, what are R2 and C? 11.5 More Complex Circuits 29. The capacitors of Figure 11–56 are uncharged. The switch is closed at t 0. Determine the equation for vC. Compute vC at one time constant using the equation and the universal time constant curve. Compare answers.

47 k

5.6 k 4.7 µF vC

39 k

45 V

225 V

C = 100 F

2.2 µF FIGURE 11–56

FIGURE 11–57

30. For Figure 11–57, the switch is closed at t 0. Given V0 0 V. a. Determine the equations for vC and iC. b. Compute the capacitor voltage at t 0 , 2, 4, 6, 8, 10, and 12 ms.

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits c. Repeat (b) for the capacitor current. d. Why does 225 V/30 also yield i(0 )? 31. Repeat Problem 30, parts (a) to (c) for the circuit of Figure 11–58.

80 20

90 V

10 F

vC 40 F

4 FIGURE 11–58

iC (mA) 3.6 t (s)

0 3 12.5 FIGURE 11–59

τd

32. Consider again Figure 11–55. Suppose E 80 V, R2 25 k , and C 0.5 mF: a. What is the charge time constant? b. What is the discharge time constant? c. With the capacitor initially discharged, move the switch to position 1 and determine equations for vC and iC during charge. d. Move the switch to the discharge position. How long does it take for the capacitor to discharge? e. Sketch vC and iC from the time the switch is placed in charge to the time that the capacitor is fully discharged. Assume the switch is in the charge position for 80 ms. 33. For the circuit of Figure 11–55, the capacitor is initially uncharged. The switch is first moved to charge, then to discharge, yielding the current shown in Figure 11–59. The capacitor fully charges in 12.5 s. Determine E, R2, and C. 34. Refer to the circuit of Figure 11–60: a. What is the charge time constant? b. What is the discharge time constant? c. The switch is in position 2 and the capacitor is uncharged. Move the switch to position 1 and determine equations for vC and iC. d. After the capacitor has charged for two time constants, move the switch to position 2 and determine equations for vC and iC during discharge. e. Sketch vC and iC. 35. Determine the capacitor voltages and the source current for the circuit of Figure 11–61 after it has reached steady state.

Problems

2 mA

5V 1

10 k iC

0.22 F

15 k

FIGURE 11–60 20 IT

C2 80 F

80 V

C1 10 F

C3 20 F

110

FIGURE 11–61

36. A black box containing dc sources and resistors has open-circuit voltage of 45 volts as in Figure 11–62(a). When the output is shorted as in (b), the short-circuit current is 1.5 mA. A switch and an uncharged 500-mF capacitor are connected as in (c). Determine the capacitor voltage and current 25 s after the switch is closed.

DC sources & resistors

1.5 mA

45 V

(a)

b (b)

(c)

FIGURE 11–62

11.6 An RC Timing Application 37. For the alarm circuit of Figure 11–32, if the input from the sensor is 5 V, R 750 k , and the alarm is activated at 15 s when vC 3.8 V, what is C? 38. For the alarm circuit of Figure 11–32, the input from the sensor is 5 V, C 47 mF, and the alarm is activated when vC 4.2 V. Choose the nearest standard resistor value to achieve a delay of at least 37 s.

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0 1 2 3 4 5 6 7 8 910 12 14 16 t ( s)

FIGURE 11–63

11.7 Pulse Response of RC Circuits 39. Consider the waveform of Figure 11–63. a. What is the period? b. What is the duty cycle? c. What is the PRR? 40. Repeat Problem 39 for the waveform of Figure 11–64. v

0 1 2

5 6 3 4

7 8

30 20 10

9 10 t (ms) 11 12

0 12345678

FIGURE 11–64

t ( s)

FIGURE 11–65

41. Determine the rise time, fall time, and pulse width for the pulse in Figure 11–65. 42. A single pulse is input to the circuit of Figure 11–66. Assuming that the capacitor is initially uncharged, sketch the output for each set of values below: a. R 2 k , C 1 mF. b. R 2 k , C 0.1 mF. C R 10 ms 10 V 0

10 ms

vout

10 V 0

vout

20 ms FIGURE 11–66

FIGURE 11–67

43. A step is applied to the circuit of Figure 11–67. If R 150 and C 20 pF, estimate the rise time of the output voltage. 44. A pulse train is input to the circuit of Figure 11–67. Assuming that the capacitor is initially uncharged, sketch the output for each set of values below after the circuit has reached steady state: a. R 2 k , C 0.1 mF. b. R 20 k , C 1.0 mF. 11.8

Transient Analysis Using Computers 45. EWB Graph capacitor voltage for the circuit of Figure 11–2 with E 25 V, R 40 , V0 0 V, and C 400 mF. Scale values from the plot at t 20 ms using the cursor. Compare to the results you get using Equation 11–3 or the curve of Figure 11–15(a).

Problems 46. EWB Obtain a plot of voltage versus time across R for the circuit of Figure 11–68. Assume an initially uncharged capacitor. Use the cursor to read voltage t 50 ms and use Ohm’s law to compute current. Compare to the values determined analytically. Repeat if V0 100 V. 50 µF

10 k 10 V

40 V

50 µF

1000

FIGURE 11–68

FIGURE 11–69

47. EWB The switch of Figure 11–69 is initially in the discharge position and the capacitor is uncharged. Move the switch to charge for 1 s, then to discharge where it remains. Solve for vC. (Hint: Use Workbench’s time delay (TD) switch. Double click it and set TON to 1 s and TOFF to 0. This will cause the switch to move to the charge position for 1 s, then return to discharge where it stays.) With the cursor, determine the peak voltage. Using Equation 11–3, compute vC at t 1 s and compare it to the value obtained above. 48. EWB Use Workbench to graph capacitor voltage for the circuit of Problem 34. With the cursor, determine vC at t 10 ms and 12 ms. Compare to the theoretical answers of 29.3 V and 15.3 V. 49. PSpice Graph capacitor voltage and current for a charging circuit with E 25 V, R 40 , V0 0 V, and C 400 mF. Scale values from the plot using the cursor. Compare to the results you get using Equations 11–3 and 11–5 or the curves of Figure 11–15. 50. PSpice Repeat the problem of Question 46 using PSpice. Plot both voltage and current. 51. PSpice The switch of Figure 11–70 is closed at t 0 s. Plot voltage and current waveforms. Use the cursor to determine vC and iC at t 10 ms. 5V

10 k

0.22 µF 2mA

15 k V0 = 0 V

FIGURE 11–70

52. PSpice Redo Example 11–17 with the switch in the charge position for 0.5 s and everything else the same. With your calculator, compute vC and iC

457

NOTE... Using its default setting, Electronics Workbench generates plotting time steps automatically. Sometimes, however, it does not generate enough and you get a jagged curve. To specify more points, under Analysis/Transient, click the “Generate time steps automatically” box, then click the “Minimum number of time points” button and type in a suitable value (for example, 1000). Experiment until you get a suitably smooth curve.

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Capacitor Charging, Discharging, and Simple Waveshaping Circuits at 0.5 s and compare them to PSpice’s plot. Repeat for iC just after moving the switch to the discharge position. 53. PSpice Use PSpice to solve for voltages and currents in the circuit of Figure 11–61. From this, determine the final (steady state) voltages and currents and compare them to the answers of Problem 35.

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 1. 0.2 A 2. a. 50 ms b. t(ms)

0 25 50 75 100 500

iC (mA) 50 30.3 18.4 11.2 6.8 0.0

3. a. 20 ms b. t(ms)

vC (V)

0 25 50 75 100 500

0 71.3 91.8 97.7 99.3 100

4. 40(1 e 5t ) V;

200e 5t mA; 38.0 V; 9.96 mA 5. a. vC(0 ) vC(0 ) 0 b. iC(0 ) 0; iC(0 ) 10 mA 6. 80 V 40 k 0.2 mF 7. 38.0 V 9.96 mA

c. 100 V 0 A

In-Process Learning Check 2 1. a. 40 40e 5t V. vC starts at 80 V and decays exponentially to 40 V. b. There is no transient since initial value final value. c. 40 100e 5t V. vC starts at 60 V and climbs exponentially to 40 V. 2. 0.1833 s 3. a. 100e 10t V; 10e 10t mA; vC starts at 100 V and decays to 0 in 0.5 s (i.e., 5 time constants); iC starts at 10 mA and decays to 0 in 0.5 s b. 100e 10t V; 10e 10t mA; vC starts at 100 V and decays to 0 in 0.5 s (i.e., 5 time constants); iC starts at 10 mA and decays to 0 in 0.5 s

Answers to In-Process Learning Checks 4. a., b., and c. Same as Example 11â&#x20AC;&#x201C;12 d. 30 37.14e 250t V e. 37.14e 250t mA f. iC (mA)

vC (V) 5 0 5 10 15 20 25 30

7.14 V 5

5. 6(1 e 25t ) V;

t (ms) 20

150e 25t mA

10 5 0 5 10 15 20 25 30 35 40

2.87 mA t (ms) 5

37.14 mA

459

Magnetism and Magnetic Circuits OBJECTIVES After studying this chapter, you will be able to • represent magnetic fields using Faraday’s flux concept, • describe magnetic fields quantitatively in terms of flux and flux density, • explain what magnetic circuits are and why they are used, • determine magnetic field intensity or magnetic flux density from a B-H curve, • solve series magnetic circuits, • solve series-parallel magnetic circuits, • compute the attractive force of an electromagnet, • explain the domain theory of magnetism, • describe the demagnetization process.

KEY TERMS Ampere’s Law Ampere-Turns Domain Theory Ferromagnetic Flux Flux Density Fringing Hall Effect Hysteresis Magnetic Circuit

Magnetic Field Magnetic Field Intensity Magnetomotive Force Permeability Reluctance Residual Magnetism Right-Hand Rule Saturation Tesla Weber

OUTLINE The Nature of a Magnetic Field Electromagnetism Flux and Flux Density Magnetic Circuits Air Gaps, Fringing, and Laminated Cores Series Elements and Parallel Elements Magnetic Circuits with DC Excitation Magnetic Field Intensity and Magnetization Curves Ampere’s Circuital Law Series Magnetic Circuits: Given , Find NI Series-Parallel Magnetic Circuits Series Magnetic Circuits: Given NI, Find Force Due to an Electromagnet Properties of Magnetic Materials Measuring Magnetic Fields

any common devices rely on magnetism. Familiar examples include computer disk drives, tape recorders, VCRs, transformers, motors, generators, and so on. To understand their operation, you need a knowledge of magnetism and magnetic circuit principles. In this chapter, we look at fundamentals of magnetism, relationships between electrical and magnetic quantities, magnetic circuit concepts, and methods of analysis. In Chapter 13, we look at electromagnetic induction and inductance, and in Chapter 24, we apply magnetic principles to the study of transformers.

Magnetism and Electromagnetism WHILE THE BASIC FACTS about magnetism have been known since ancient times, it was not until the early 1800s that the connection between electricity and magnetism was made and the foundations of modern electromagnetic theory laid down. In 1819, Hans Christian Oersted, a Danish scientist, demonstrated that electricity and magnetism were related when he showed that a compass needle was deflected by a current-carrying conductor. The following year, Andre Ampere (1775â&#x20AC;&#x201C;1836) showed that current-carrying conductors attract or repel each other just like magnets. However, it was Michael Faraday (recall Chapter 10) who developed our present concept of the magnetic field as a collection of flux lines in space that conceptually represent both the intensity and the direction of the field. It was this concept that led to an understanding of magnetism and the development of important practical devices such as the transformer and the electric generator. In 1873, James Clerk Maxwell (see photo), a Scottish scientist, tied the then known theoretical and experimental concepts together and developed a unified theory of electromagnetism that predicted the existence of radio waves. Some 30 years later, Heinrich Hertz, a German physcist, showed experimentally that such waves existed, thus verifying Maxwellâ&#x20AC;&#x2122;s theories and paving the way for modern radio and television.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

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12.1

The Nature of a Magnetic Field

Magnetism refers to the force that acts between magnets and magnetic materials. We know, for example, that magnets attract pieces of iron, deflect compass needles, attract or repel other magnets, and so on. This force acts at a distance and without the need for direct physical contact. The region where the force is felt is called the “field of the magnet” or simply, its magnetic field. Thus, a magnetic field is a force field.

NOTES... Flux is perhaps an unfortunate name to apply to a magnetic field. Flux suggests a flow, but in a magnetic field, nothing actually flows; a magnetic field is simply a condition of space, i.e., a region in which magnetic force exists. Nonetheless, the concept of flux is enormously helpful as an aid to visualizing magnetic phenomenon, and we will continue to use it for that purpose.

Magnetic Flux Faraday’s flux concept (recall Putting It Into Perspective, Chapter 10) helps us visualize this field. Using Faraday’s representation, magnetic fields are shown as lines in space. These lines, called flux lines or lines of force, show the direction and intensity of the field at all points. This is illustrated in Figure 12–1 for the field of a bar magnet. As indicated, the field is strongest at the poles of the magnet (where flux lines are most dense), its direction is from north (N) to south (S) external to the magnet, and flux lines never cross. The symbol for magnetic flux (Figure 12–1) is the Greek letter (phi).

Φ N

FIGURE 12–1

Field of a bar magnet. Flux is denoted by the Greek letter .

Figure 12–2 shows what happens when two magnets are brought close together. In (a), unlike poles attract, and flux lines pass from one magnet to the other. In (b), like poles repel, and the flux lines are pushed back as indicated by the flattening of the field between the two magnets. Φ N

(a) Attraction FIGURE 12–2

Φ N

(b) Repulsion

Field patterns due to attraction and repulsion.

Ferromagnetic Materials Magnetic materials (materials that are attracted by magnets such as iron, nickel, cobalt, and their alloys) are called ferromagnetic materials. Ferromagnetic materials provide an easy path for magnetic flux. This is illustrated

Section 12.1

■

463

The Nature of a Magnetic Field

in Figure 12–3 where the flux lines take the longer (but easier) path through the soft iron, rather than the shorter path that they would normally take (recall Figure 12–1). Note, however, that nonmagnetic materials (plastic, wood, glass, and so on) have no effect on the field. Figure 12–4 shows an application of these principles. Part (a) shows a simplified representation of a loudspeaker, and part (b) shows expanded

Soft iron Φ

Speaker cone

Voice coils (see Figure 12–11) N

Magnet

Plastic (no effect) FIGURE 12–3 Magnetic field follows the longer (but easier) path through the iron. The plastic has no effect on the field.

Magnetic flux (see expanded detail below) (a) Simplified representation of the magnetic field. Here, the complex field of (b) is represented symbolically by a single line

Magnetic gap: Note the intense field.

Magnet

(b) Magnetic field pattern for the loud speaker. (Courtesy JBL Professional) FIGURE 12–4 Magnetic circuit of a loudspeaker. The magnetic structure and voice coil are called a “speaker motor”. The field is created by the permanent magnet.

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details of its magnetic field. (Since the speaker is symmetrical, only half its structure is shown in (b).) The field is created by the permanent magnet, and the iron pole pieces guide the field and concentrate it in the gap where the speaker coil is placed. (For a description of how the speaker works, see Section 12.4.) Within the iron structure, the flux crowds together at sharp interior corners, spreads apart at exterior corners, and is essentially uniform elsewhere. This is characteristic of magnetic fields in iron.

12.2 Φ

(a) Magnetic field produced by current. Field is proportional to I

Electromagnetism

Most applications of magnetism involve magnetic effects due to electric currents. We look first at some basic principles. Consider Figure 12–5. The current, I, creates a magnetic field that is concentric about the conductor, uniform along its length, and whose strength is directly proportional to I. Note the direction of the field. It may be remembered with the aid of the right-hand rule. As indicated in (b), imagine placing your right hand around the conductor with your thumb pointing in the direction of current. Your fingers then point in the direction of the field. If you reverse the direction of the current, the direction of the field reverses. If the conductor is wound into a coil, the fields of its individual turns combine, producing a resultant field as in Figure 12–6. The direction of the coil flux can also be remembered by means of a simple rule: curl the fingers of your right hand around the coil in the direction of the current and your thumb will point in the direction of the field. If the direction of the current is reversed, the field also reverses. Provided no ferromagnetic material is present, the strength of the coil’s field is directly proportional to its current.

(b) Right-hand rule

Φ N

FIGURE 12–5 Field about a currentcarrying conductor. If the current is reversed, the field reverses direction.

S FIGURE 12–6

Field produced by a coil.

If the coil is wound on a ferromagnetic core as in Figure 12–7 (transformers are built this way), almost all flux is confined to the core, although a small amount (called stray or leakage flux) passes through the surrounding air. However, now that ferromagnetic material is present, the core flux is no longer proportional to current. The reason for this is discussed in Section 12.14.

Section 12.3

■

Flux and Flux Density

465

Core flux (simplified representation)

Leakage flux I

Iron core FIGURE 12–7

12.3

For ferromagnetic materials, most flux is confined to the core.

Flux and Flux Density

As noted in Figure 12–1, magnetic flux is represented by the symbol . In the SI system, the unit of flux is the weber (Wb), in honor of pioneer researcher Wilhelm Eduard Weber, 1804–1891. However, we are often more interested in flux density B (i.e., flux per unit area) than in total flux . Since flux is measured in Wb and area A in m2, flux density is measured as Wb/m2. However, to honor Nikola Tesla (another early researcher, 1856– 1943) the unit of flux density is called the tesla (T) where 1 T ⫽ 1 Wb/m2. Flux density is found by dividing the total flux passing perpendicularly through an area by the size of the area, Figure 12–8. That is, B ⫽ ᎏᎏ (tesla, T) A

(12–1)

Thus, if ⫽ 600 mWb of flux pass perpendicularly through an area A ⫽ 20 ⫻ 10⫺4 m2, the flux density is B ⫽ (600 ⫻ 10⫺6 Wb)/(20 ⫻ 10⫺4 m2) ⫽ 0.3 T. The greater the flux density, the stronger the field.

EXAMPLE 12–1

For the magnetic core of Figure 12–9, the flux density at cross section 1 is B1 ⫽ 0.4 T. Determine B2. FIGURE 12–9

A1 = 2 10 2 m2

A2 = 1 10 2 m2

teslas

A Iron FIGURE 12–8 Concept of flux density. 1 T ⫽ 1 Wb/m2.

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Solution ⫽ B1 ⫻ A1 ⫽ (0.4 T)(2 ⫻ 10⫺2 m2) ⫽ 0.8 ⫻ 10⫺2 Wb. Since all flux is confined to the core, the flux at cross section 2 is the same as at cross section 1. Therefore, B2 ⫽ /A2 ⫽ (0.8 ⫻ 10⫺2 Wb)/(1 ⫻ 10⫺2 m2) ⫽ 0.8 T

PRACTICE PROBLEMS 1

1. Refer to the core of Figure 12–8: a. If A is 2 cm ⫻ 2.5 cm and B ⫽ 0.4 T, compute in webers. b. If A is 0.5 inch by 0.8 inch and B ⫽ 0.35 T, compute in webers. 2. In Figure 12–9, if ⫽ 100 ⫻ 10⫺4 Wb, compute B1 and B2. Answers: 1. a. 2 ⫻ 10⫺4 Wb

b. 90.3 mWb

2. 0.5 T; 1.0 T

To gain a feeling for the size of magnetic units, note that the strength of the earth’s field is approximately 50 mT near the earth’s surface, the field of a large generator or motor is on the order of 1 or 2 T, and the largest fields yet produced (using superconducting magnets) are on the order of 25 T. Other systems of units (now largely superseded) are the CGS system and the English systems. In the CGS system, flux is measured in maxwells and flux density in gauss. In the English system, flux is measured in lines and flux density in lines per square inch. Conversion factors are given in Table 12–1. We use only the SI system in this book. TABLE 12–1 Magnetic Units Conversion Table System

IN-PROCESS

LEARNING CHECK 1

Flux (⌽)

Flux Density (B)

webers (Wb)

teslas (T) 1 T ⫽ 1 Wb/m2

English

lines 1 Wb ⫽ 108 lines

lines/in2 1 T ⫽ 6.452 ⫻ 104 lines/in2

CGS

maxwells 1 Wb ⫽ 108 maxwells

gauss 1 gauss ⫽ 1 maxwell/cm2 1 T ⫽ 104 gauss

1. A magnetic field is a 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 field. 2. With Faraday’s flux concept, the density of lines represents the 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 of the field and their direction represents the 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 of the field. 3. Three ferromagnetic materials are 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮, 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮, and 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮. 4. The direction of a magnetic field is from 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 to 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 outside a magnet. 5. For Figures 12–5 and 12–6, if the direction of current is reversed, sketch what the fields look like. 6. If the core shown in Figure 12–7 is plastic, sketch what the field will look like.

Section 12.4 7. Flux density B is defined as the ratio /A, where A is the area (parallel, perpendicular) to . 8. For Figure 12–9, if A1 is 2 cm ⫻ 2.5 cm, B1 is 0.5 T, and B2 ⫽ 0.25 T, what is A2? (Answers are at the end of the chapter.)

12.4

Magnetic Circuits

Most practical applications of magnetism use magnetic structures to guide and shape magnetic fields by providing a well-defined path for flux. Such structures are called magnetic circuits. Magnetic circuits are found in motors, generators, computer disk drives, tape recorders, and so on. The speaker of Figure 12–4 illustrates the concept. It uses a powerful magnet to create flux and an iron circuit to guide the flux to the air gap to provide the intense field required by the voice coil. Note how effectively it does its job; almost the entire flux produced by the magnet is confined to the iron path with little leakage into the air. A second example is shown in Figures 12–10 and 12–11. Tape recorders, VCRs, and computer disk drives all store information magnetically on iron oxide coated surfaces for later retrieval and use. The basic tape recorder scheme is shown symbolically in Figure 12–10. Sound picked up by a microphone is converted to an electrical signal, amplified, and the output applied to the record head. The record head is a small magnetic circuit. Current from the amplifier passes through its coil, creating a magnetic field that magnetizes the moving tape. The magnetized patterns on the tape correspond to the original sound input. Sound waves

I Amplifier

Tape I

Microphone Tape

(b) Record head Direction of tape movement (a) Recording System FIGURE 12–10

The recording head of a tape recorder is a magnetic circuit.

During playback the magnetized tape is passed by a playback head, as shown in Figure 12–11(a). Voltages induced in the playback coil are amplified and applied to a speaker. The speaker (b) utilizes a flexible cone to reproduce sound. A coil of fine wire attached at the apex of this cone is placed in the field of the speaker air gap. Current from the amplifier passes through this coil, creating a varying field that interacts with the fixed field of

■

Magnetic Circuits

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Magnetism and Magnetic Circuits

Amplifier

Speaker Tape

Direction of tape movement (a) Playback System

Coil vibrates

Voice coil

Vibrating cone

Sound waves

Permanent magnet

Φ Frame mount (b) Speaker FIGURE 12–11 (a) Fringing at gap A

the speaker magnet, causing the cone to vibrate. Since these vibrations correspond to the magnetized patterns on the tape, the original sound is reproduced. Computer disk drives use a similar record/playback scheme; in this case, binary logic patterns are stored and retrieved rather than music and voice.

12.5 (b) Laminated section. Effective magnetic area is less than the physical area FIGURE 12–12 nations.

Fringing and lami-

Both the playback system and the speaker use magnetic circuits.

Air Gaps, Fringing, and Laminated Cores

For magnetic circuits with air gaps, fringing occurs, causing a decrease in flux density in the gap as in Figure 12–12(a). For short gaps, fringing can usually be neglected. Alternatively, correction can be made by increasing each cross-sectional dimension of the gap by the size of the gap to approximate the decrease in flux density.

Section 12.6

■

Series Elements and Parallel Elements

EXAMPLE 12–2 A core with cross-sectional dimensions of 2.5 cm by 3 cm has a 0.1-mm gap. If flux density B ⫽ 0.86 T in the iron, what is the approximate (corrected) flux density in the gap? Solution ⫽ BA ⫽ (0.86 T)(2.5 ⫻ 10⫺2 m)(3 ⫻ 10⫺2 m) ⫽ 0.645 mWb Ag ⯝ (2.51 ⫻ 10⫺2 m)(3.01 ⫻ 10⫺2 m) ⫽ 7.555 ⫻ 10⫺4 m2 Thus, in the gap Bg ⯝ 0.645 mWb/7.555 ⫻ 10⫺4 m2 ⫽ 0.854 T

Now consider laminations. Many practical magnetic circuits (such as transformers) use thin sheets of stacked iron or steel as in Figure 12–12(b). Since the core is not a solid block, its effective cross-sectional area (i.e., the actual area of iron) is less than its physical area. A stacking factor, defined as the ratio of the actual area of ferrous material to the physical area of the core, permits you to determine the core’s effective area. A laminated section of core has cross-sectional dimensions of 0.03 m by 0.05 m and a stacking factor of 0.9. a. What is the effective area of the core? b. Given ⫽ 1.4 ⫻ 10⫺3 Wb, what is the flux density, B? Answers: a. 1.35 ⫻ 10⫺3 m2

12.6

b. 1.04 T

Series Elements and Parallel Elements

Magnetic circuits may have sections of different materials. For example, the circuit of Figure 12–13 has sections of cast iron, sheet steel, and an air gap. For this circuit, flux is the same in all sections. Such a circuit is called a series magnetic circuit. Although the flux is the same in all sections, the flux density in each section may vary, depending on its effective cross-sectional area as you saw earlier.

Laminated sheet steel

I N turns 1

Cast iron Air gap FIGURE 12–13 Series magnetic circuit. Flux is the same throughout.

FIGURE 12–14 The sum of the flux entering a junction equals the sum leaving. Here, 1 ⫽ 2 ⫹ 3.

PRACTICE PROBLEMS 2

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A circuit may also have elements in parallel (Figure 12–14). At each junction, the sum of fluxes entering is equal to the sum leaving. This is the counterpart of Kirchhoff’s current law. Thus, for Figure 12–14, if 1 ⫽ 25 mWb and 2 ⫽ 15 mWb, then 3 ⫽ 10 mWb. For cores that are symmetrical about the center leg, 2 ⫽ 3. IN-PROCESS

LEARNING CHECK 2

1. Why is the flux density in each section of Figure 12–13 different? 2. For Figure 12–13, ⫽ 1.32 mWb, the cross section of the core is 3 cm by 4 cm, the laminated section has a stacking factor of 0.8, and the gap is 1 mm. Determine the flux density in each section, taking fringing into account. 3. If the core of Figure 12–14 is symmetrical about its center leg, B1 ⫽ 0.4 T, and the cross-sectional area of the center leg is 25 cm2, what are 2 and 3? (Answers are at the end of the chapter.)

12.7

Magnetic Circuits with DC Excitation

We now look at the analysis of magnetic circuits with dc excitation. There are two basic problems to consider: (1) given the flux, to determine the current required to produce it and (2) given the current, to compute the flux produced. To help visualize how to solve such problems, we first establish an analogy between magnetic circuits and electric circuits.

MMF: The Source of Magnetic Flux Current through a coil creates magnetic flux. The greater the current or the greater the number of turns, the greater will be the flux. This flux-producing ability of a coil is called its magnetomotive force (mmf). Magnetomotive force is given the symbol Ᏺ and is defined as Ᏺ ⫽ NI (ampere-turns, At)

(12–2)

Thus, a coil with 100 turns and 2.5 amps will have an mmf of 250 ampereturns, while a coil with 500 turns and 4 amps will have an mmf of 2000 ampere-turns.

Reluctance, ᑬ: Opposition to Magnetic Flux Flux in a magnetic circuit also depends on the opposition that the circuit presents to it. Termed reluctance, this opposition depends on the dimensions of the core and the material of which it is made. Like the resistance of a wire, reluctance is directly proportional to length and inversely proportional to cross-sectional area. In equation form, ᐉ ᑬ ⫽ ᎏᎏ (At/Wb) mA

(12–3)

where m is a property of the core material called its permeability (discussed in Section 12.8). Permeability is a measure of how easy it is to establish flux in a material. Ferromagnetic materials have high permeability and hence low ᑬ, while nonmagnetic materials have low permeability and high ᑬ.

Section 12.8

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471

Magnetic Field Intensity and Magnetization Curves

Ohm’s Law for Magnetic Circuits The relationship between flux, mmf, and reluctance is ⫽ Ᏺ/ᑬ (Wb)

(12–4)

This relationship is similar to Ohm’s law and is depicted symbolically in Figure 12–15. (Remember however that flux, unlike electric current, does not flow—see note in Section 12.1.) Ᏺ

ᑬ

EXAMPLE 12–3 For Figure 12–16, if the reluctance of the magnetic circuit is ᑬ ⫽ 12 ⫻ 104 At/Wb, what is the flux in the circuit? FIGURE 12–16

0.5 A

FIGURE 12–15 Electric circuit analogy of a magnetic circuit. ⫽ Ᏺ/ᑬ.

N = 300 turns

Solution Ᏺ ⫽ NI ⫽ (300)(0.5 A) ⫽ 150 At ⫽ Ᏺ/ᑬ ⫽ (150 At)/(12 ⫻ 104 At/Wb) ⫽ 12.5 ⫻ 10⫺4 Wb

In Example 12–3, we assumed that the reluctance of the core was constant. This is only approximately true under certain conditions. In general, it is not true, since ᑬ is a function of flux density. Thus Equation 12–4 is not really very useful, since for ferromagnetic material, ᑬ depends on flux, the very quantity that you are trying to find. The main use of Equations 12–3 and 12–4 is to provide an analogy between electric and magnetic circuit analysis.

12.8

mmf = NI I

Coil

Magnetic Field Intensity and Magnetization Curves

We now look at a more practical approach to analyzing magnetic circuits. First, we require a quantity called magnetic field intensity, H (also known as magnetizing force). It is a measure of the mmf per unit length of a circuit. To get at the idea, suppose you apply the same mmf (say 600 At) to two circuits with different path lengths (Figure 12–17). In (a), you have 600 ampere-turns of mmf to “drive” flux through 0.6 m of core; in (b), you have the same mmf but it is spread across only 0.15 m of path length. Thus the mmf per unit length in the second case is more intense. Based on this idea, one can define magnetic field intensity as the ratio of applied mmf to the length of path that it acts over. Thus, H ⫽ Ᏺ/ᐉ ⫽ NI/ᐉ

(At/m)

(12–5)

For the circuit of Figure 12–17(a), H ⫽ 600 At/0.6 m ⫽ 1000 At/m, while for the circuit of (b), H ⫽ 600 At/0.15 ⫽ 4000 At/m. Thus, in (a) you have

l = 0.6 m (a) A long path

mmf = NI I l = 0.15 m (b) A short path FIGURE 12–17 By definition, H ⫽ mmf/length ⫽ NI/ᐉ.

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1000 ampere-turns of “driving force” per meter of length to establish flux in the core, whereas in (b) you have four times as much. (However, you won’t get four times as much flux, since the opposition to flux varies with the density of the flux.) Rearranging Equation 12–5 yields an important result: NI ⫽ Hᐉ

(At)

(12–6)

In an analogy with electric circuits (Figure 12–18), the NI product is an mmf source, while the Hᐉ product is an mmf drop. NI

NI = Hl FIGURE 12–18 model.

Circuit analogy, Hᐉ

The Relationship between B and H From Equation 12–5, you can see that magnetizing force, H, is a measure of the flux-producing ability of the coil (since it depends on NI). You also know that B is a measure of the resulting flux (since B ⫽ /A). Thus, B and H are related. The relationship is B ⫽ mH

(12–7)

where m is the permeability of the core (recall Equation 12–3). It was stated earlier that permeability is a measure of how easy it is to establish flux in a material. To see why, note from Equation 12–7 that the larger the value of m, the larger the flux density for a given H. However, H is proportional to current; therefore, the larger the value of m, the larger the flux density for a given magnetizing current. From this, it follows that the larger the permeability, the more flux you get for a given magnetizing current. In the SI system, m has units of webers per ampere-turn-meter. The permeability of free space is m0 ⫽ 4p ⫻ 10⫺7. For all practical purposes, the permeability of air and other nonmagnetic materials is the same as for a vacuum. Thus, in air gaps, Bg ⫽ m0Hg ⫽ 4p ⫻ 10⫺7 ⫻ Hg

(12–8)

Rearranging Equation 12–8 yields Bg Hg ⫽ ᎏᎏ ⫽ 7.96 ⫻ 105Bg 4p ⫻ 10⫺7 PRACTICE PROBLEMS 3

(At/m)

(12–9)

For Figure 12–16, the core cross section is 0.05 m ⫻ 0.08 m. If a gap is cut in the core and H in the gap is 3.6 ⫻ 105 At/m, what is the flux in the core? Neglect fringing. Answer: 1.81 mWb

B-H Curves For ferromagnetic materials, m is not constant but varies with flux density and there is no easy way to compute it. In reality, however, it isn’t m that you are interested in: What you really want to know is, given B, what is H, and vice versa. A set of curves, called B-H or magnetization curves, provides this information. (These curves are obtained experimentally and are available in

Section 12.8

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Magnetic Field Intensity and Magnetization Curves

473

handbooks. A separate curve is required for each material.) Figure 12–19 shows typical curves for cast iron, cast steel, and sheet steel. B (T) 1.6 Sheet steel Cast steel

1.4

1.2

1.0

0.8 Cast iron

0.6

0.4

0.2

H (At/m) 0

500

FIGURE 12–19

1000

1500

2000

2500

3000

3500

4000

B-H curves for selected materials.

EXAMPLE 12–4

If B ⫽ 1.4 T for sheet steel, what is H?

Solution Enter Figure 12–19 on the axis at B ⫽ 1.4 T, continue across until you encounter the curve for sheet steel, then read the corresponding value for H as indicated in Figure 12–20: H ⫽ 1000 At/m. B (T) Sheet steel

1.4

1000 FIGURE 12–20

H (At/m)

For sheet steel, H ⫽ 1000 At/m when B ⫽ 1.4 T.

4500

5000

PRACTICE PROBLEMS 4

■

Magnetism and Magnetic Circuits The cross section of a sheet steel core is 0.1 m ⫻ 0.1 m and its stacking factor is 0.93. If H ⫽ 1500 At/m, compute flux density B and magnetic flux . Answer: 1.45 T

12.9

13.5 mWb

Ampere’s Circuital Law

One of the key relationships in magnetic circuit theory is Ampere’s circuital law. Ampere’s law was determined experimentally and is a generalization of the relationship Ᏺ ⫽ NI ⫽ Hᐉ that we developed earlier. Ampere showed that the algebraic sum of mmfs around a closed loop in a magnetic circuit is zero, regardless of the number of sections or coils. That is,

Ᏺ⫽0

(12–10)

This can be rewrittten as

NI ⫽

Hᐉ

(12–11)

which states that the sum of applied mmfs around a closed loop equals the sum of the mmf drops. The summation is algebraic and terms are additive or subtractive, depending on the direction of flux and how the coils are wound. To illustrate, consider again Figure 12–13. Here, NI ⫺ Hironᐉiron ⫺ Hsteelᐉsteel ⫺ Hgᐉg ⫽ 0

Thus, NI ⫽ Hironᐉiron ⫹ Hsteelᐉsteel ⫹ Hgᐉg

        

Chapter 12

{

474

Impressed mmf

sum of mmf drops

which states that the applied mmf NI is equal to the sum of the Hᐉ drops around the loop. The path to use for the Hᐉ terms is the mean (average) path. You now have two magnetic circuit models (Figure 12–21). While the reluctance model (a) is not very useful for solving problems, it helps relate magnetic circuit problems to familiar electrical circuit concepts. The Ampere’s law model, on the other hand, permits us to solve practical problems. We look at how to do this in the next section. Hiron liron

ᑬiron

Ᏺ

ᑬsteel

ᑬg (a) Reluctance model FIGURE 12–21

Ᏺ

Hsteel lsteel

Hg lg (b) Ampere’s circuital law model

Two models for the magnetic circuit of Figure 12–13.

Section 12.10

■

Series Magnetic Circuits: Given , Find NI

1. If the mmf of a 200-turn coil is 700 At, the current in the coil is 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 amps. 2. For Figure 12–17, if H ⫽ 3500 At/m and N ⫽ 1000 turns, then for (a), I is 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 A, while for (b), I is 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 A. 3. For cast iron, if B ⫽ 0.5 T, then H ⫽ 㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮㛮 At/m. 4. A circuit consists of one coil, a section of iron, a section of steel, and two air gaps (of different sizes). Draw the Ampere’s law model. 5. Which is the correct answer for the circuit of Figure 12–22? a. Ampere’s law around loop 1 yields (NI ⫽ H1ᐉ1⫹ H2ᐉ2, or NI ⫽ H1ᐉ1 ⫺ H2ᐉ2). b. Ampere’s law around loop 2 yields (0 ⫽ H2ᐉ2 ⫹ H3ᐉ3, or 0 ⫽ H2ᐉ2 ⫺ H3ᐉ3). Path l2 = lca H = H2 1 cm

6 cm

c Path l1 = labc H = H1 FIGURE 12–22

1 cm Path l3 = lcda H = H3

1 cm

8 cm

1 cm

FIGURE 12–23

6. For the circuit of Figure 12–23, the length ᐉ to use in Ampere’s law is (0.36 m, 0.32 m, 0.28 m). Why? (Answers are at the end of the chapter.)

12.10 Series Magnetic Circuits: Given ⌽, Find NI You now have the tools needed to solve basic magnetic circuit problems. We will begin with series circuits where is known and we want to find the excitation to produce it. Problems of this type can be solved using four basic steps: 1. Compute B for each section using B ⫽ /A. 2. Determine H for each magnetic section from the B-H curves. Use Hg ⫽ 7.96 ⫻ 105Bg for air gaps. 3. Compute NI using Ampere’s circuital law. 4. Use the computed NI to determine coil current or turns as required. (Circuits with more than one coil are handled as in Example 12–6.) Be sure to use the mean path through the circuit when applying Ampere’s law. Unless directed otherwise, neglect fringing.

IN-PROCESS

LEARNING CHECK 3

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PRACTICAL NOTES... Magnetic circuit analysis is not as precise as electric circuit analysis because (1) the assumption of uniform flux density breaks down at sharp corners as you saw in Figure 12–4, and (2) the B-H curve is a mean curve and has considerable uncertainty as discussed later (Section 12–14). Although the answers are approximate, they are adequate for most purposes.

If the core of Figure 12–24 is cast iron and ⫽ 0.1 ⫻ 10⫺3 Wb, what is the coil current?

EXAMPLE 12–5

Area A

B= A

Mean length abcda = 0.25 m N = 500 turns A = 0.2 10 3 m2 FIGURE 12–24

Solution

Following the four steps outlined above:

1. The flux density is 0.1 ⫻ 10⫺3 ⫽ 0.5 T B ⫽ ᎏᎏ ⫽ ᎏᎏ A 0.2 ⫻ 10⫺3 2. From the B-H curve (cast iron), Figure 12–19, H ⫽ 1550 At/m. 3. Apply Ampere’s law. There is only one coil and one core section. Length ⫽ 0.25 m. Thus, NI ⫽ Hᐉ ⫽ 1550 ⫻ 0.25 ⫽ 388 At 4. Divide by N: I ⫽ 388/500 ⫽ 0.78 amps

Section 12.10

■

Series Magnetic Circuits: Given , Find NI

EXAMPLE 12–6 A second coil is added as shown in Figure 12–25. If ⫽ 0.1 ⫻ 10⫺3 Wb as before, but I1 ⫽ 1.5 amps, what is I2? FIGURE 12–25 I1

N1 = 500

N2 = 200

Solution From the previous example, you know that a current of 0.78 amps in coil 1 produces ⫽ 0.1 ⫻ 10⫺3 Wb. But you already have 1.5 amps in coil 1. Thus, coil 2 must be wound in opposition so that its mmf is subtractive. Applying Ampere’s law yields N1I1 ⫺ N2I2 ⫽ Hᐉ. Hence, (500)(1.5 A) ⫺ 200I2 ⫽ 388 At and so I2 ⫽ 1.8 amps.

Note. Since magnetic circuits are nonlinear, you cannot use superposition, that is, you cannot consider each coil of Figure 12–25 by itself, then sum the results. You must consider them simultaneously as we did in Example 12–6.

More Examples If a magnetic circuit contains an air gap, add another element to the conceptual models (recall Figure 12–21). Since air represents a poor magnetic path, its reluctance will be high compared with that of iron. Recalling our analogy to electric circuits, this suggests that the mmf drop across the gap will be large compared with that of the iron. You can see this in the following example.

EXAMPLE 12–7

The core of Figure 12–24 has a 0.008-m gap cut as shown in Figure 12–26. Determine how much the current must increase to maintain the original core flux. Neglect fringing. Hiron liron I

Hg lg

lg = 0.008 m (a) FIGURE 12–26

(b)

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Magnetism and Magnetic Circuits

Solution Iron ᐉiron ⫽ 0.25 ⫺ 0.008 ⫽ 0.242 m. Since does not change, B and H will be the same as before. Thus, Biron ⫽ 0.5 T and Hiron ⫽ 1550 At/m. Air Gap Bg is the same as Biron. Thus, Bg ⫽ 0.5 T and Hg ⫽ 7.96 ⫻ 105Bg ⫽ 3.98 ⫻ 105 At/m. Ampere’s Law NI ⫽ Hironᐉiron ⫹ Hgᐉg ⫽ (1550)(0.242) ⫹ (3.98 ⫻ 105)(0.008) ⫽ 375 ⫹ 3184 ⫽ 3559 At. Thus, I ⫽ 3559/500 ⫽ 7.1 amps. Note that the current had to increase from 0.78 amp to 7.1 amps in order to maintain the same flux, over a ninefold increase.

EXAMPLE 12–8

The laminated sheet steel section of Figure 12–27 has a stacking factor of 0.9. Compute the current required to establish a flux of ⫽ 1.4 ⫻ 10⫺4 Wb. Neglect fringing. Cast iron I

e f

N = 150 d

c b

lg = 0.2"

Laminated sheet steel (SF = 0.9) lef = 2.5" lde = 2"

a h 0.5"

0.8"

Cross section = 0.5" 0.8" (all members) = 1.4 10 4 Wb FIGURE 12–27

Solution Convert all dimensions to metric. Cast Iron ᐉiron ⫽ ᐉab ⫹ ᐉcdef ⫽ 2.5 ⫹ 2 ⫹ 2.5 ⫺ 0.2 ⫽ 6.8 in ⫽ 0.173 m Airon ⫽ (0.5 in)(0.8 in) ⫽ 0.4 in2 ⫽ 0.258 ⫻ 10⫺3 m2 Biron ⫽ /Airon ⫽ (1.4 ⫻ 10⫺4)/(0.258 ⫻ 10⫺3) ⫽ 0.54 T Hiron ⫽ 1850 At/m (from Figure 12–19) Sheet Steel ᐉsteel ⫽ ᐉfg ⫹ ᐉgh ⫹ ᐉha ⫽ 0.25 ⫹ 2 ⫹ 0.25 ⫽ 2.5 in ⫽ 6.35 ⫻ 10⫺2 m Asteel ⫽ (0.9)(0.258 ⫻ 10⫺3) ⫽ 0.232 ⫻ 10⫺3 m2 Bsteel ⫽ /Asteel ⫽ (1.4 ⫻ 10⫺4)/(0.232 ⫻ 10⫺3) ⫽ 0.60 T Hsteel ⫽ 125 At/m (from Figure 12–19)

Section 12.10

■

Series Magnetic Circuits: Given , Find NI

Air Gap ᐉg ⫽ 0.2 in ⫽ 5.08 ⫻ 10⫺3 m Bg ⫽ Biron ⫽ 0.54 T Hg ⫽ (7.96 ⫻ 105)(0.54) ⫽ 4.3 ⫻ 105 At/m Ampere’s Law NI ⫽ Hironᐉiron ⫹ Hsteelᐉsteel ⫹ Hgᐉg ⫽ (1850)(0.173) ⫹ (125)(6.35 ⫻ 10⫺2) ⫹ (4.3 ⫻ 105)(5.08 ⫻ 10⫺3) ⫽ 320 ⫹ 7.9 ⫹ 2184 ⫽ 2512 At I ⫽ 2512/N ⫽ 2512/150 ⫽ 16.7 amps

Figure 12–28 shows a portion of a solenoid. Flux ⫽ 4 ⫻ 10⫺4 Wb when I ⫽ 2.5 amps. Find the number of turns on the coil.

EXAMPLE 12–9

l plunger = 0.1 m Spring Plunger

2.5 cm

Gap

2 cm

lg Coil

0.4 cm Yoke

Cross section 2.5 cm 2.5 cm

lyoke = 0.2 m I FIGURE 12–28

Solenoid. All parts are cast steel.

Solution Yoke Ayoke ⫽ 2.5 cm ⫻ 2.5 cm ⫽ 6.25 cm2 ⫽ 6.25 ⫻ 10⫺4 m2 4 ⫻ 10⫺4 Byoke ⫽ ᎏᎏ ⫽ ᎏᎏ ⫽ 0.64 T Ayoke 6.25 ⫻ 10⫺4 Hyoke ⫽ 410 At/m (from Figure 12–19) Plunger Aplunger ⫽ 2.0 cm ⫻ 2.5 cm ⫽ 5.0 cm2 ⫽ 5.0 ⫻ 10⫺4 m2 4 ⫻ 10⫺4 Bplunger ⫽ ᎏᎏ ⫽ ᎏᎏ ⫽ 0.8 T Aplunger 5.0 ⫻ 10⫺4 Hplunger ⫽ 500 At/m (from Figure 12–19)

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Air Gap There are two identical gaps. For each, Bg ⫽ Byoke ⫽ 0.64 T Thus, Hg ⫽ (7.96 ⫻ 105)(0.64) ⫽ 5.09 ⫻ 105 At/m The results are summarized in Table 12–2. Ampere’s Law NI ⫽ Hyokeᐉyoke ⫹ Hplungerᐉplunger ⫹ 2Hgᐉg ⫽ 82 ⫹ 50 ⫹ 2(2036) ⫽ 4204 At N ⫽ 4204/2.5 ⫽ 1682 turns TABLE 12–2 Material Cast steel

Section yoke

A (m2)

Length (m)

6.25 ⫻ 10

0.2

B (T) ⫺4

⫺4

H (At/m)

Hᐉ (At)

0.64

410

Cast steel

plunger

0.1

5 ⫻ 10

0.8

500

Air

gap

0.4 ⫻ 10⫺2

6.25 ⫻ 10⫺4

0.64

5.09 ⫻ 105

2036

12.11 Series-Parallel Magnetic Circuits Series-parallel magnetic circuits are handled using the sum of fluxes principle (Figure 12–14) and Ampere’s law.

EXAMPLE 12–10

The core of Figure 12–29 is cast steel. Determine the current to establish an air-gap flux g ⫽ 6 ⫻ 10⫺3 Wb. Neglect fringing. Cast steel A = 2 10 2 m2 lg = lbc = 0.25 10 3 m g = 3 1

I N = 200

3 b

d lab = lcd = 0.25 m lda = 0.2 m ldea = 0.35 m FIGURE 12–29

Section 12.11

■

Series-Parallel Magnetic Circuits

Solution Consider each section in turn. Air Gap Bg ⫽ g /Ag ⫽ (6 ⫻ 10⫺3)/(2 ⫻ 10⫺2) ⫽ 0.3 T Hg ⫽ (7.96 ⫻ 105)(0.3) ⫽ 2.388 ⫻ 105 At/m Sections ab and cd Bab ⫽ Bcd ⫽ Bg ⫽ 0.3 T Hab ⫽ Hcd ⫽ 250 At/m (from Figure 12–19) Ampere’s Law (Loop 2) NI ⫽ Hᐉ. Since you are going opposite to flux in leg da, the corresponding term (i.e., Hdaᐉda) will be subtractive. Also, NI ⫽ 0 for loop 2. Thus, 0⫽

loop2

Hᐉ

0 ⫽ Habᐉab ⫹ Hgᐉg ⫹ Hcdᐉcd ⫺ Hdaᐉda ⫽ (250)(0.25) ⫹ (2.388 ⫻ 105)(0.25 ⫻ 10⫺3) ⫹ (250)(0.25) ⫺ 0.2Hda ⫽ 62.5 ⫹ 59.7 ⫹ 62.5 ⫺ 0.2Hda ⫽ 184.7 ⫺ 0.2Hda Thus, 0.2Hda ⫽ 184.7 and Hda ⫽ 925 At/m. From Figure 12–19, Bda ⫽ 1.12 T. 2 ⫽ Bda A ⫽ 1.12 ⫻ 0.02 ⫽ 2.24 ⫻ 10⫺2 Wb 1 ⫽ 2 ⫹ 3 ⫽ 2.84 ⫻ 10⫺2 Wb. Bdea ⫽ 1/A ⫽ (2.84 ⫻ 10⫺2)/0.02 ⫽ 1.42 T Hdea ⫽ 2125 At/m (from Figure 12–19) Ampere’s Law (Loop 1) NI ⫽ Hdeaᐉdea ⫹ Hadᐉad ⫽ (2125)(0.35) ⫹ 184.7 ⫽ 929 At I ⫽ 929/200 ⫽ 4.65 A The cast-iron core of Figure 12–30 is symmetrical. Determine current I. Hint: To find NI, you can write Ampere’s law around either loop. Be sure to make use of symmetry. 2 = 30 Wb

FIGURE 12–30 a

m 2

k I

N = 400

1 f

lab = lbc = lcd = 4 cm Gap: lg = 0.5 cm lek = 3 cm Core dimensions: 1 cm 1 cm Answer: 6.5 A

PRACTICE PROBLEMS 5

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12.12 Series Magnetic Circuits: Given NI, Find ⌽ In previous problems, you were given the flux and asked to find the current. We now look at the converse problem: given NI, find the resultant flux. For the special case of a core of one material and constant cross section (Example 12–11) this is straightforward. For all other cases, trial and error must be used.

EXAMPLE 12–11

mine .

For the circuit of Figure 12–31, NI ⫽ 250 At. Deter-

FIGURE 12–31 l = 0.2 m I

N turns

Cross-sectional area A = 0.01 m2

Cast steel

Solution Hᐉ ⫽ NI. Thus, H ⫽ NI/ᐉ ⫽ 250/0.2 ⫽ 1250 At/m. From the B-H curve of Figure 12–19, B ⫽ 1.24 T. Therefore, ⫽ BA ⫽ 1.24 ⫻ 0.01 ⫽ 1.24 ⫻ 10⫺2 Wb.

For circuits with two or more sections, the process is not so simple. Before you can find H in any section, for example, you need to know the flux density. However, in order to determine flux density, you need to know H. Thus, neither nor H can be found without knowing the other first. To get around this problem, use trial and error. First, take a guess at the value for flux, compute NI using the 4-step procedure of Section 12.10, then compare the computed NI against the given NI. If they agree, the problem is solved. If they don’t, adjust your guess and try again. Repeat the procedure until you are within 5% of the given NI. The problem is how to come up with a good first guess. For circuits of the type of Figure 12–32, note that NI ⫽ Hsteelᐉsteel ⫹ Hgᐉg. As a first guess, assume that the reluctance of the air gap is so high that the full mmf drop appears across the gap. Thus, NI ⯝ Hgᐉg, and Hg ⯝ NI/ᐉg

Ᏺ = Hsteel lsteel Hg lg ⯝ Hg lg if Hg lg >> Hsteel lsteel lg = 0.002 m lsteel = 0.2 m FIGURE 12–32

(12–12)

You can now apply Ampere’s law to see how close to the given NI your trial guess is. Since we know that some of the mmf drop appears across the steel, we will start at less than 100% for the gap. Common sense and a bit of experience helps. The relative size of the mmf drops also depends on the core material. For cast iron, the percentage drop across the iron is larger than the

Section 12.12

■

Series Magnetic Circuits: Given NI, Find

percentage across a similar piece of sheet steel or cast steel. This is illustrated in the next two examples.

EXAMPLE 12–12 The core of Figure 12–32 is cast steel, NI ⫽ 1100 At, and the cross-sectional area everywhere is 0.0025 m2. Determine the flux in the core. Solution Initial Guess Assume that 90% of the mmf appears across the gap. The applied mmf is 1100 At. Ninety percent of this is 990 At. Thus, Hg ⯝ 0.9NI/ᐉ ⫽ 990/0.002 ⫽ 4.95 ⫻ 105 At/m and Bg ⫽ m0Hg ⫽ (4p ⫻ 10⫺7)(4.95 ⫻ 105) ⫽ 0.62 T. Trial 1 Since the area of the steel is the same as that of the gap, the flux density is the same, neglecting fringing. Thus, Bsteel ⫽ Bg ⫽ 0.62 T. From the B-H curve, Hsteel ⫽ 400 At/m. Now apply Ampere’s law: NI ⫽ Hsteelᐉsteel ⫹ Hgᐉg ⫽ (400)(0.2) ⫹ (4.95 ⫻ 105)(0.002) ⫽ 80 ⫹ 990 ⫽ 1070 At This answer is 2.7% lower than the given NI of 1100 At and is therefore acceptable. Thus, ⫽ BA ⫽ 0.62 ⫻ 0.0025 ⫽ 1.55 ⫻ 10⫺3 Wb.

The initial guess in Example 12–12 yielded an acceptable answer on the first trial. (You are seldom this lucky.)

EXAMPLE 12–13

If the core of Figure 12–32 is cast iron instead of steel,

compute .

Solution Because cast iron has a larger H for a given flux density (Figure 12–19), it will have a larger Hᐉ drop and less will appear across the gap. Assume 75% across the gap. Initial Guess Hg ⯝ 0.75 NI/ᐉ ⫽ (0.75)(1100)/0.002 ⫽ 4.125 ⫻ 105 At/m. Bg ⫽ m0Hg ⫽ (4m ⫻ 10⫺7)(4.125 ⫻ 105) ⫽ 0.52 T. Trial 1 Biron ⫽ Bg. Thus, Biron ⫽ 0.52 T. From the B-H curve, Hiron ⫽ 1700 At/m. Ampere’s Law NI ⫽ Hironᐉiron ⫹ Hgᐉg ⫽ (1700)(0.2) ⫹ (4.125 ⫻ 105)(0.002) ⫽ 340 ⫹ 825 ⫽ 1165 At (high by 5.9%) Trial 2 Reduce the guess by 5.9% to Biron ⫽ 0.49 T. Thus, Hiron ⫽ 1500 At/m (from the B-H curve) and Hg ⫽ 7.96 ⫻ 105 Bg ⫽ 3.90 ⫻ 105 At/m.

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Ampere’s Law NI ⫽ Hironᐉiron ⫹ Hgᐉg ⫽ (1500)(0.2) ⫹ (3.90 ⫻ 105)(0.002) ⫽ 300 ⫹ 780 ⫽ 1080 At The error is now 1.82%, which is excellent. Thus, ⫽ BA ⫽ (0.49)(2.5 ⫻ 10⫺3 ) ⫽ 1.23 ⫻ 10⫺3 Wb. If the error had been larger than 5%, a new trial would have been needed.

12.13 Force Due to an Electromagnet Electromagnets are used in relays, door bells, lifting magnets, and so on. For an electromagnetic relay as in Figure 12–33, it can be shown that the force created by the magnetic field is B2g Ag F ⫽ ᎏᎏ 2m0

(12–13)

where Bg is flux density in the gap in teslas, Ag is gap area in square meters, and F is force in newtons.

EXAMPLE 12–14

Figure 12–33 shows a typical relay. The force due to the current-carrying coil pulls the pivoted arm against spring tension to close the contacts and energize the load. If the pole face is 1⁄ 4 inch square and ⫽ 0.5 ⫻ 10⫺4 Wb, what is the pull on the armature in pounds?

Contacts

Coil

FIGURE 12–33 A typical relay.

Solution

Convert to metric units. Ag ⫽ (0.25 in)(0.25 in) ⫽ 0.0625 in2 ⫽ 0.403 ⫻ 10⫺4 m2 Bg ⫽ /Ag ⫽ (0.5 ⫻ 10⫺4)/(0.403 ⫻ 10⫺4) ⫽ 1.24 T

Section 12.14

Thus,

■

Properties of Magnetic Materials

485

(1.24)2(0.403 ⫻ 10⫺4) B2g A F ⫽ ᎏᎏ ⫽ ᎏᎏᎏ ⫽ 24.66 N ⫽ 5.54 lb 2(4p ⫻ 10⫺7) 2m0

Figure 12–34 shows how a relay is used in practice. When the switch is closed, the energized coil pulls the armature down. This closes the contacts and energizes the load. When the switch is opened, the spring pulls the contacts open again. Schemes like this use relatively small currents to control large loads. In addition, they permit remote control, as the relay and load may be a considerable distance from the actuating switch. Armature

Insulation

Pivot point

Spring

Contacts

Coil

Load

FIGURE 12–34 Controlling a load with a relay.

12.14 Properties of Magnetic Materials Magnetic properties are related to atomic structure. Each atom of a substance, for example, produces a tiny atomic-level magnetic field because its moving (i.e., orbiting) electrons constitute an atomic-level current and currents create magnetic fields. For nonmagnetic materials, these fields are randomly oriented and cancel. However, for ferromagnetic materials, the fields in small regions, called domains (Figure 12–35), do not cancel. (Domains are of microscopic size, but are large enough to hold from 1017 to 1021 atoms.) If the domain fields in a ferromagnetic material line up, the material is magnetized; if they are randomly oriented, the material is not magnetized.

Magnetizing a Specimen A nonmagnetized specimen can be magnetized by making its domain fields line up. Figure 12–36 shows how this can be done. As current through the coil is increased, the field strength increases and more and more domains

Ferromagnetic material

FIGURE 12–35 Random orientation of microscopic fields in a nonmagnetized ferromagnetic material. The small regions are called domains.

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Magnetism and Magnetic Circuits

align themselves in the direction of the field. If the field is made strong enough, almost all domain fields line up and the material is said to be in saturation (the almost flat portion of the B-H curve). In saturation, the flux density increases slowly as magnetization intensity increases. This means that once the material is in saturation, you cannot magnetize it much further no matter how hard you try. Path 0-a traced from the nonmagnetized state to the saturated state is termed the dc curve or normal magnetization curve. (This is the B-H curve that you used earlier when you solved magnetic circuit problems.)

Saturation a

Ferromagnetic material

0 (a) The magnetizing circuit B

(b) Progressive change in the domain orientations as the field is increased. H is proportional to current I

FIGURE 12–36 The magnetization process.

Residual magnetism

b H

Residual magnetism FIGURE 12–37 Hysteresis loop. B

FIGURE 12–38 Demagnetization by successively shrinking the hysteresis loop.

Hysteresis If you now reduce the current to zero, you will find that the material still retains some magnetism, called residual magnetism (Figure 12–37, point b). If now you reverse the current, the flux reverses and the bottom part of the curve can be traced. By reversing the current again at d, the curve can be traced back to point a. The result is called a hysteresis loop. A major source of uncertainty in magnetic circuit behavior should now be apparent: As you can see, flux density depends not just on current, it also depends on which arm of the curve the sample is magnetized on, i.e., it depends on the circuit’s past history. For this reason, B-H curves are the average of the two arms of the hysteresis loop, i.e., the dc curve of Figure 12–36. The Demagnetization Process As indicated above, simply turning the current off does not demagnetize ferromagnetic material. To demagnetize it, you must successively decrease its hysteresis loop to zero as in Figure 12–38. You can place the specimen inside a coil that is driven by a variable ac source and gradually decrease the coil current to zero, or you can use a fixed ac supply and gradually withdraw the specimen from the field. Such procedures are used by service personnel to “degauss” TV picture tubes.

Problems

12.15 Measuring Magnetic Fields One way to measure magnetic field strength is to use the Hall effect (after E. H. Hall). The basic idea is illustrated in Figure 12–39. When a strip of semiconductor material such as indium arsenide is placed in a magnetic field, a small voltage, called the Hall voltage, VH, appears across opposite edges. For a fixed current I, VH is proportional to magnetic field strength B. Instruments using this principle are known as Hall-effect gaussmeters. To measure a magnetic field with such a meter, insert its probe into the field perpendicular to the field (Figure 12–40). The meter indicates flux density directly.

Flux

VH I

Current source FIGURE 12–39 The Hall effect.

Hall effect probe

473

Gaussmeter

FIGURE 12–40 Magnetic field measurement.

12.3 Flux and Flux Density 1. Refer to Figure 12–41: a. Which area, A1 or A2, do you use to calculate flux density? b. If ⫽ 28 mWb, what is flux density in teslas? 2. For Figure 12–41, if ⫽ 250 mWb, A1 ⫽ 1.25 in2, and A2 ⫽ 2.0 in2, what is the flux density in the English system of units? 3. The toroid of Figure 12–42 has a circular cross section and ⫽ 628 mWb. If r1 ⫽ 8 cm and r2 ⫽ 12 cm, what is the flux density in teslas? 4. If r1 of Figure 12–42 is 3.5 inches and r2 is 4.5 inches, what is the flux density in the English system of units if ⫽ 628 mWb?

PROBLEMS

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Chapter 12

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Magnetism and Magnetic Circuits

Portion of magnetic circuit (cast steel)

0.02 m2

0.024 m2

FIGURE 12–41

FIGURE 12–42

12.5 Air Gaps, Fringing, and Laminated Cores 5. If the section of core in Figure 12–43 is 0.025 m by 0.04 m, has a stacking factor of 0.85, and B ⫽ 1.45 T, what is in webers? 6. If the core of Figure 12–43 is 3 cm by 2 cm, has a stacking factor of 0.9, and B ⫽ 8 ⫻ 103 gauss, what is in maxwells?

FIGURE 12–43

12.6 Series Elements and Parallel Elements 7. For the iron core of Figure 12–44, flux density B2 ⫽ 0.6 T. Compute B1 and B3. A1 = 0.02 m2

A1 = 0.02 m2

A3 = 0.01 m2 FIGURE 12–44

Ᏺ

ᑬ

(a) ᑬ =

l A

Ᏺ

(b) B = H FIGURE 12–46 Ᏺ ⫽ NI.

A2 = 0.015 m2 B2 = 0.6 T

A3 = 0.016 m2

A2 = 0.01 m2 FIGURE 12–45

8. For the section of iron core of Figure 12–45, if 1 ⫽ 12 mWb and 3 ⫽ 2 mWb, what is B2? 9. For the section of iron core of Figure 12–45, if B1 ⫽ 0.8 T and B2 ⫽ 0.6 T, what is B3? 12.8 Magnetic Field Intensity and Magnetization Curves 10. Consider again Figure 12–42. If I ⫽ 10 A, N ⫽ 40 turns, r1 ⫽ 5 cm, and r2 ⫽ 7 cm, what is H in ampere-turns per meter? 11. Figure 12–46 shows the two electric circuit equivalents for magnetic circuits. Show that m in ᑬ ⫽ ᐉ/mA is the same as m in B ⫽ mH. 12.9 Ampere’s Circuital Law 12. Let H1 and ᐉ1 be the magnetizing force and path length respectively, where flux 1 exists in Figure 12–47 and similarly for 2 and 3. Write Ampere’s law around each of the windows.

489

Problems 13. Assume that a coil N2 carrying current I2 is added on leg 3 of the core shown in Figure 12–47 and that it produces flux directed upward. Assume, however, that the net flux in leg 3 is still downward. Write the Ampere’s law equations for this case. 14. Repeat Problem 13 if the net flux in leg 3 is upward but the directions of 1 and 2 remain as in Figure 12–47. 12.10 Series Magnetic Circuits: Given , Find NI 15. Find the current I in Figure 12–48 if ⫽ 0.16 mWb. 16. Let everything be the same as in Problem 15 except that the cast steel portion is replaced with laminated sheet steel with a stacking factor of 0.85. 17. A gap of 0.5 mm is cut in the cast steel portion of the core in Figure 12–48. Find the current for ⫽ 0.128 mWb. Neglect fringing. 18. Two gaps, each 1 mm, are cut in the circuit of Figure 12–48, one in the cast steel portion and the other in the cast iron portion. Determine current for ⫽ 0.128 mWb. Neglect fringing. 19. The cast iron core of Figure 12–49 measures 1 cm ⫻ 1.5 cm, ᐉg ⫽ 0.3 mm, the air gap flux density is 0.426 T and N ⫽ 600 turns. The end pieces are half circles. Taking into account fringing, find current I.

H2 and l2 this path 1

I1 N1

H3 and l3 this path

H1 and l1 this path FIGURE 12–47

Cast iron

Cast steel I

l steel = 0.14 m

1.5 cm

l iron = 0.06 m

A = 3.2 10 4 m2 N = 300 turns N

3.2 cm

1 cm

FIGURE 12–49

20. For the circuit of Figure 12–50, ⫽ 141 mWb and N ⫽ 400 turns. The bottom member is sheet steel with a stacking factor of 0.94, while the remainder is cast steel. All pieces are 1 cm ⫻ 1 cm. The length of the cast steel path is 16 cm. Find current I. 21. For the circuit of Figure 12–51, ⫽ 30 mWb and N ⫽ 2000 turns. Neglecting fringing, find current I. 22. For the circuit of Figure 12–52, ⫽ 25,000 lines. The stacking factor for the sheet steel portion is 0.95. Find current I. 23. A second coil of 450 turns with I2 ⫽ 4 amps is wound on the cast steel portion of Figure 12–52. Its flux is in opposition to the flux produced by the

FIGURE 12–48

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Magnetism and Magnetic Circuits

N Sheet steel

10 cm FIGURE 12–50

l1 = 2 in A1 = 1 in2 l iron = 3 cm l steel = 8 cm

l1 I

Cast iron

Cast iron l2

N = 600

l2 = 3.5 in lg = 0.2 in l3 = 5.8 in

l4 Cast steel lg = 2 mm A (everywhere) = 0.5 cm2 FIGURE 12–51

Sheet steel

l4 = 7.5 in

Cast steel Area of all sections (except A1) = 2 in2 FIGURE 12–52

Cast steel 1

2 c

100 turns

lg = lxy = 0.001 m labc = 0.14 m lcda = 0.16 m lax = lcy = 0.039 m A = 4 cm2 everywhere FIGURE 12–53

original coil. The resulting flux is 35 000 lines in the counterclockwise direction. Find the current I1. 12.11 Series-Parallel Magnetic Circuits 24. For Figure 12–53, if g ⫽ 80 mWb, find I. 25. If the circuit of Figure 12–53 has no gap and 3 ⫽ 0.2 mWb, find I. 12.12 Series Magnetic Circuits: Given NI, Find 26. A cast steel magnetic circuit with N ⫽ 2500 turns, I ⫽ 200 mA, and a crosssectional area of 0.02 m2 has an air gap of 0.00254 m. Assuming 90% of the mmf appears across the gap, estimate the flux in the core. 27. If NI ⫽ 644 At for the cast steel core of Figure 12–54, find the flux, . 28. A gap ᐉ ⫽ 0.004 m is cut in the core of Figure 12–54. Everything else remains the same. Find the flux, .

Answers to In-Process Learning Checks

491

Diameter = 2 cm I

Cast steel

Radius = 6 cm FIGURE 12–54

12.13 Force Due to an Electromagnet 29. For the relay of Figure 12–34, if the pole face is 2 cm by 2.5 cm and a force of 2 pounds is required to close the gap, what flux (in webers) is needed? 30. For the solenoid of Figure 12–28, ⫽ 4 ⫻ 10⫺4 Wb. Find the force of attraction on the plunger in newtons and in pounds.

In-Process Learning Check 1 1. Force 2. Strength, direction 3. Iron, nickel, cobalt 4. North, south 5. Same except direction of flux reversed 6. Same as Figure 12–6 7. Perpendicular 8. 10 cm2 In-Process Learning Check 2 1. While flux is the same throughout, the effective area of each section differs. 2. Biron ⫽ 1.1 T; Bsteel ⫽ 1.38 T; Bg ⫽ 1.04 T 3. 2 ⫽ 3 ⫽ 0.5 mWb In-Process Learning Check 3 1. 3.5 A 2. a. 2.1 A b. 0.525 A 3. 1550 At/m 4. Same as Figure 12–21(b) except add Hg2ᐉg2. 5. a. NI ⫽ H1ᐉ1 ⫹ H2ᐉ2 b. 0 ⫽ H2ᐉ2 ⫺ H3ᐉ3 6. 0.32 m; use the mean path length.

ANSWERS TO IN-PROCESS LEARNING CHECKS

Inductance and Inductors OBJECTIVES After studying this chapter, you will be able to • describe what an inductor is and what its effect on circuit operation is, • explain Faraday’s law and Lenz’s law, • compute induced voltage using Faraday’s law, • define inductance, • compute voltage across an inductance, • compute inductance for series and parallel configurations, • compute inductor voltages and currents for steady state dc excitation, • compute energy stored in an inductance, • describe common inductor problems and how to test for them.

KEY TERMS Back Voltage Choke Counter EMF

Faraday’s Law Flux Linkage Henry Induced Voltage Inductance Inductor L Lenz’s Law Nf Stray Inductance

OUTLINE Electromagnetic Induction Induced Voltage and Induction Self-Inductance Computing Induced Voltage Inductances in Series and Parallel Practical Considerations Inductance and Steady State DC Energy Stored by an Inductance Inductor Troubleshooting Hints

n this chapter, we look at self-inductance and inductors. Self-inductance (usually just called inductance) is a circuit property that is due entirely to the magnetic field created by current in a circuit. The effect that inductance has on circuit operation is to oppose any change in current—thus, in a sense, inductance can be likened to inertia in a mechanical system. A circuit element built to possess inductance is called an inductor. In its simplest form an inductor is simply a coil of wire, Figure 13–1(a). Ideally, inductors have only inductance. However, since they are made of wire, practical inductors also have some resistance. Initially, however, we assume that this resistance is negligible and treat inductors as ideal (i.e., we assume that they have no property other than inductance). (Coil resistance is considered in Sections 13.6 and 13.7.) In practice, inductors are also referred to as chokes (because they try to limit or “choke” current change), or as reactors (for reasons to be discussed in Chapter 16). In this chapter, we refer to them mainly as inductors.

CHAPTER PREVIEW

I L

(a) A basic inductor FIGURE 13–1

(b) Ideal inductor symbol

Inductance is due to the magnetic field created by an electric current.

On circuit diagrams and in equations, inductance is represented by the letter L. Its circuit symbol is a coil as shown in Figure 13–1(b). The unit of inductance is the henry. Inductors are used in many places. In radios, they are part of the tuning circuit that you adjust when you select a station. In fluorescent lamps, they are part of the ballast circuit that limits current when the lamp is turned on; in power systems, they are part of the protection circuitry used to control short-circuit currents during fault conditions.

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Inductance and Inductors

PUTTING IT IN PERSPECTIVE

The Discovery of Electromagnetic Induction MOST OF OUR IDEAS CONCERNING INDUCTANCE and induced voltages are due to Michael Faraday (recall Chapter 12) and Joseph Henry (1797–1878). Working independently (Faraday in England and Henry—shown at left—in the USA), they discovered, almost simultaneously, the fundamental laws governing electromagnetic induction. While experimenting with magnetic fields, Faraday developed the transformer. He wound two coils on an iron ring and energized one of them from a battery. As he closed the switch energizing the first coil, Faraday noticed that a momentary voltage was induced in the second coil, and when he opened the switch, he found that a momentary voltage was again induced but with opposite polarity. When the current was steady, no voltage was produced at all. Faraday explained this effect in terms of his magnetic lines of flux concept. When current was first turned on, he visualized the lines as springing outward into space; when it was turned off, he visualized the lines as collapsing inward. He then visualized that voltage was produced by these lines as they cut across circuit conductors. Companion experiments showed that voltage was also produced when a magnet was passed through a coil or when a conductor was moved through a magnetic field. Again, he visualized these voltages in terms of flux cutting a conductor. Working independently in the United States, Henry discovered essentially the same results. In fact, Henry’s work preceded Faraday’s by a few months, but because he did not publish them first, credit was given to Faraday. However, Henry is credited with the discovery of self-induction, and in honor of his work the unit of inductance was named the henry.

13.1 NOTES… Since we work with time-varying flux linkages in this chapter, we use f rather than ⌽ for flux (as we did in Chapter 12). This is in keeping with the standard practice of using lowercase symbols for time-varying quantities and uppercase symbols for dc quantities.

Electromagnetic Induction

Since inductance depends on induced voltage, we begin with a review of electromagnetic induction. First, we look at Faraday’s and Henry’s results. Consider Figure 13–2. In (a), a magnet is moved through a coil of wire, and this action induces a voltage in the coil. When the magnet is thrust into the coil, the meter deflects upscale; when it is withdrawn, the meter deflects downscale, indicating that polarity has changed. The voltage magnitude is proportional to how fast the magnet is moved. In (b), when the conductor is moved through the field, voltage is induced. If the conductor is moved to the right, its far end is positive; if it is moved to the left, the polarity reverses and its far end becomes negative. Again, the voltage magnitude is proportional to how fast the wire is moved. In (c), voltage is induced in coil 2 due to the magnetic field created by the current in coil 1. At the instant the switch is closed, the meter kicks upscale; at the instant it is opened, the meter kicks downscale. In (d) voltage is induced in a coil by its own current. At the instant the switch is closed, the top end of the coil becomes positive, while at the instant it is opened, the polarity reverses and the top end becomes negative.

Section 13.1

⫹

■

495

Electromagnetic Induction

⫹

⫺

⫺ 0

⫺

⫹

⫺

⫹

⫺

⫹ N N

Motion

S ⫺

(b) Motional emf

(a) Motional emf

⫹

⫺

φ i

⫹ ⫹

⫺ Coil 1

⫺

Coil 2 (c) Mutually induced voltage

FIGURE 13–2 Principle of electromagnetic induction. Voltage is induced as long as the flux linkage of the circuit is changing.

Faraday’s Law Based on these observations, Faraday concluded that voltage is induced in a circuit whenever the flux linking (i.e., passing through) the circuit is changing and that the magnitude of the voltage is proportional to the rate of change of the flux linkages. This result, known as Faraday’s law, is also sometimes stated in terms of the rate of cutting flux lines. We look at this viewpoint in Chapter 15.

(d) Self-induced voltage

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Inductance and Inductors

Lenz’s Law Heinrich Lenz (a Russian physicist, 1804–1865) determined a companion result. He showed that the polarity of the induced voltage is such as to oppose the cause producing it. This result is known as Lenz’s law.

13.2

Induced Voltage and Induction

We now turn our attention to inductors. As noted earlier, inductance is due entirely to the magnetic field created by current-carrying conductors. Consider Figure 13–3 (which shows an inductor at three instants of time). In (a) the current is constant, and since the magnetic field is due to this current, the magnetic field is also constant. Applying Faraday’s law, we note that, because the flux linking the coil is not changing, the induced voltage is zero. Now consider (b). Here, the current (and hence the field) is increasing. According to Faraday’s law, a voltage is induced that is proportional to how fast the field is changing and according to Lenz’s law, the polarity of this voltage must be such as to oppose the increase in current. Thus, the polarity of the voltage is as shown. Note that the faster the current increases, the larger the opposing voltage. Now consider (c). Since the current is decreasing, Lenz’s law shows that the polarity of the induced voltage reverses, that is, the collapsing field produces a voltage that tries to keep the current going. Again, the faster the rate of change of current, the larger is this voltage.

Constant current

Increasing current

⫹

i ⫺

Induced voltage

OV ⫺ (a) Steady current: Induced voltage is zero

Decreasing current

(b) Increasing current: The induced voltage opposes the current build-up

Induced voltage ⫹ (c) Decreasing current: The induced voltage opposes the current decay

FIGURE 13–3 Self-induced voltage due to a coil’s own current. The induced voltage opposes the current change. Note carefully the polarities in (b) and (c).

Counter EMF Because the induced voltage in Figure 13–3 tries to counter (i.e., opposes) changes in current, it is called a counter emf or back voltage. Note carefully, however, that this voltage does not oppose current, it opposes only changes in current. It also does not prevent the current from changing; it only prevents it from changing abruptly. The result is that current in an inductor changes gradually and smoothly from one value to another as indicated in Figure 13–4(b). The effect of inductance is thus similar to the effect

Section 13.2

■

Induced Voltage and Induction

497

Current

of inertia in a mechanical system. The flywheel used on an engine, for example, prevents abrupt changes in engine speed but does not prevent the engine from gradually changing from one speed to another.

Time

(a) Current cannot jump from one value to another like this FIGURE 13–4

Time

(b) Current must change smoothly with no abrupt jumps

Current in inductance.

Iron-Core and Air-Core Inductors As Faraday discovered, the voltage induced in a coil depends on flux linkages, and flux linkages depend on core materials. Coils with ferromagnetic cores (called iron-core coils) have their flux almost entirely confined to their cores, while coils wound on nonferromagnetic materials do not. (The latter are sometimes called air-core coils because all nonmagnetic core materials have the same permeability as air and thus behave magnetically the same as air.) First, consider the iron-core case, Figure 13–5. Ideally, all flux lines are confined to the core and hence pass through (link) all turns of the winding. The product of flux times the number of turns that it passes through is defined as the flux linkage of the coil. For Figure 13–5, f lines pass through N turns yielding a flux linkage of Nf. By Faraday’s law, the induced voltage is proportional to the rate of change of Nf. In the SI system, the constant of proportionality is one and Faraday’s law for this case may therefore be stated as e ⫽ N ⫻ the rate of change of f

(13–1)

In calculus notation,

i ⫹ e ⫺

φ N turns

Iron core FIGURE 13–5 When flux f passes through all N turns, the flux linking the coil is Nf.

NOTES... df e ⫽ N ᎏᎏ dt

(volts, V)

(13–2)

where f is in webers, t in seconds, and e in volts. Thus if the flux changes at the rate of 1 Wb/s in a 1 turn coil, the voltage induced is 1 volt.

EXAMPLE 13–1

If the flux through a 200-turn coil changes steadily from 1 Wb to 4 Wb in one second, what is the voltage induced? Solution The flux changes by 3 Wb in one second. Thus, its rate of change is 3 Wb/s. e ⫽ N ⫻ rate of change of flux ⫽ (200 turns)(3 Wb/s) ⫽ 600 volts

Equation 13–2 is sometimes shown with a minus sign. However, the minus sign is unnecessary. In circuit theory, we use Equation 13–2 to determine the magnitude of the induced voltage and Lenz’s law to determine its polarity.

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Inductance and Inductors

Now consider an air-core inductor (Figure 13–6). Since not all flux lines pass through all windings, it is difficult to determine flux linkages as above. However, (since no ferromagnetic material is present) flux is directly proportional to current. In this case, then, since induced voltage is proportional to the rate of change of flux, and since flux is proportional to current, induced voltage will be proportional to the rate of change of current. Let the constant of proportionality be L. Thus, e ⫽ L ⫻ rate of change of current

(13–3)

In calculus notation, this can be written as di e ⫽ Lᎏᎏ (volts, V) dt

(13–4)

L is called the self-inductance of the coil, and in the SI system its unit is the henry. (This is discussed in more detail in Section 13.3.)

i ⫹ e ⫺

FIGURE 13–6

The flux linking the coil is proportional to current. Flux linkage is LI.

We now have two equations for coil voltage. Equation 13–4 is the more useful form for this chapter, while Equation 13–2 is the more useful form for the circuits of Chapter 24. We look at Equation 13–4 in the next section. IN-PROCESS

LEARNING CHECK 1

1. Which of the current graphs shown in Figure 13–7 cannot be the current in an inductor? Why?

(a)

(b)

(c)

FIGURE 13–7

2. Compute the flux linkage for the coil of Figure 13–5, given f ⫽ 500 mWb and N ⫽ 1200 turns. 3. If the flux f of Question 2 changes steadily from 500 mWb to 525 mWb in 1 s, what is the voltage induced in the coil? 4. If the flux f of Question 2 changes steadily from 500 mWb to 475 mWb in 100 ms, what is the voltage induced? (Answers are at the end of the chapter.)

Section 13.3

13.3

■

Self-Inductance

499

Self-Inductance

In the preceding section, we showed that the voltage induced in a coil is e ⫽ Ldi/dt; where L is the self-inductance of the coil (usually referred to simply as inductance) and di/dt is the rate of change of its current. In the SI system, L is measured in henries. As can be seen from Equation 13–4, it is the ratio of voltage induced in a coil to the rate of change of current producing it. From this, we get the definition of the henry. By definition, the inductance of a coil is one henry if the voltage created by its changing current is one volt when its current changes at the rate of one ampere per second. In practice, the voltage across an inductance is denoted by vL rather than by e. Thus, di vL ⫽ Lᎏᎏ (V) dt

(13–5)

Voltage and current references are as shown in Figure 13–8. i L

EXAMPLE 13–2 If the current through a 5-mH inductance changes at the rate of 1000 A/s, what is the voltage induced? Solution

⫹ vL = L di dt ⫺

FIGURE 13–8 Voltage-current reference convention. As usual, the plus sign for voltage goes at the tail of the current arrow.

vL ⫽ L ⫻ rate of change of current ⫽ (5 ⫻ 10⫺3 H)(1000 A/s) ⫽ 5 volts

1. The voltage across an inductance is 250 V when its current changes at the rate of 10 mA/ms. What is L? 2. If the voltage across a 2-mH inductance is 50 volts, how fast is the current changing? Answers: 1. 25 mH

PRACTICE PROBLEMS 1

2. 25 ⫻ 103 A/s

Inductance Formulas Inductance for some simple shapes can be determined using the principles of Chapter 12. For example, the approximate inductance of the coil of Figure 13–9 can be shown to be mN 2A L ⫽ ᎏᎏ ᐉ

(H)

l A

(13–6)

where ᐉ is in meters, A is in square meters, N is the number of turns, and m is the permeability of the core. (Details can be found in many physics books.)

␮N2A H l

FIGURE 13–9 Inductance formula for a single-layer coil.

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Chapter 13

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Inductance and Inductors

EXAMPLE 13–3 A 0.15-m-long air-core coil has a radius of 0.006 m and 120 turns. Compute its inductance. Solution A ⫽ pr 2 ⫽ 1.131 ⫻ 10⫺4 m2 m ⫽ m0 ⫽ 4p ⫻ 10⫺7 Thus, L ⫽ 4p ⫻ 10⫺7 (120)2 (1.131 ⫻ 10⫺4)/0.15 ⫽ 13.6 mH

Laminated core FIGURE 13–10 Iron-core coil with an air gap to control saturation.

The accuracy of Equation 13–6 breaks down for small ᐉ/d ratios. (If ᐉ/d is greater than 10, the error is less than 4%.) Improved formulas may be found in design handbooks, such as the Radio Amateur’s Handbook published by the American Radio Relay League (ARRL). To provide greater inductance in smaller spaces, iron cores are sometimes used. Unless the core flux is kept below saturation, however, permeability varies and inductance is not constant. To get constant inductance an air gap may be used (Figure 13–10). If the gap is wide enough to dominate, coil inductance is approximately m0 N2Ag L ⫽ ᎏᎏ ᐉg

(H)

(13–7)

where m0 is the permeability of air, Ag is the area of the air gap, and ᐉg is its length. (See end-of-chapter Problem 11.) Another way to increase inductance is to use a ferrite core (Section 13.6).

EXAMPLE 13–4

The inductor of Figure 13–10 has 1000 turns, a 5-mm gap, and a cross-sectional area at the gap of 5 ⫻ 10⫺4 m2. What is its inductance? Solution L ⫽ (4p ⫻ 10⫺7)(1000)2(5 ⫻ 10⫺4)/(5 ⫻ 10⫺3) ⫽ 0.126 H

PRACTICAL NOTES... 1. Since inductance is due to a conductor’s magnetic field, it depends on the same factors that the magnetic field depends on. The stronger the field for a given current, the greater the inductance. Thus, a coil of many turns will have more inductance than a coil of a few turns (L is proportional to N 2) and a coil wound on a magnetic core will have greater inductance than a coil wound on a nonmagnetic form. 2. However, if a coil is wound on a magnetic core, the core’s permeability m may change with flux density. Since flux density depends on current, L becomes a function of current. For example, the inductor of Figure 13–11

Section 13.4

■

Computing Induced Voltage

has a nonlinear inductance due to core saturation. All inductors encountered in this book are assumed to be linear, i.e., of constant value. B

H Iron core FIGURE 13–11 This coil does not have a fixed inductance because its flux is not proportional to its current.

1. The voltage across an inductance whose current changes uniformly by 10 mA in 4 ms is 70 volts. What is its inductance? 2. If you triple the number of turns in the inductor of Figure 13–10, but everything else remains the same, by what factor does the inductance increase? (Answers are at the end of the chapter.)

13.4

Computing Induced Voltage

Earlier, we determined that the voltage across an inductance is given by vL ⫽ Ldi/dt, where the voltage and current references are shown in Figure 13–8. Note that the polarity of vL depends on whether the current is increasing or decreasing. For example, if the current is increasing, di/dt is positive and so vL is positive, while if the current is decreasing, di/dt is negative and vL is negative. To compute voltage, we need to determine di/dt. In general, this requires calculus. However, since di/dt is slope, you can determine voltage without calculus for currents that can be described by straight lines, as in Figure 13–12. For any ⌬t segment, slope ⫽ ⌬i/⌬t, where ⌬i is the amount that the current changes during time interval ⌬t.

EXAMPLE 13–5 Figure 13–12 is the current through a 10-mH inductance. Determine voltage vL and sketch it. i (A) 4 3 2 1 0 ⫺1 ⫺2 ⫺3 ⫺4 FIGURE 13–12

⌬i 1 2 3 4 5 6 ⌬t

t (ms)

IN-PROCESS

LEARNING CHECK 2

501

502

Chapter 13

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Inductance and Inductors

Solution Break the problem into intervals over which the slope is constant, determine the slope for each segment, then compute voltage using vL ⫽ L ⫻ slope for that interval: Slope ⫽ 0. Thus, vL ⫽ 0 V. Slope ⫽ ⌬i/⌬t ⫽ 4 A/(1 ⫻ 10 ⫺3 s) ⫽ 4 ⫻ 10 3 A/s. Thus, vL ⫽ L⌬i/⌬t ⫽ (0.010 H)(4 ⫻ 103 A/s) ⫽ 40 V. Slope ⫽ ⌬i/⌬t ⫽ ⫺8 A/(2 ⫻ 10⫺3 s) ⫽ ⫺4 ⫻ 103 A/s. Thus, vL ⫽ L⌬i/⌬t ⫽ (0.010 H)(⫺4 ⫻ 103 A/s) ⫽ ⫺40 V. Slope ⫽ 0. Thus, vL ⫽ 0 V. Same slope as from 1 ms to 2 ms. Thus, vL ⫽ 40 V.

0 to 1 ms: 1 ms to 2 ms: 2 ms to 4 ms: 4 ms to 5 ms: 5 ms to 6 ms:

The voltage waveform is shown in Figure 13–13. i (A) 4 0

1 2 3 4 5 6

t (ms)

⫺4 vL (V) 40 0

1 2 3 4 5 6

t (ms)

⫺40 FIGURE 13–13

For currents that are not linear functions of time, you need to use calculus as illustrated in the following example.

∫

EXAMPLE 13–6

What is the equation for the voltage across a 12.5 H inductance whose current is i ⫽ te⫺t amps?

Solution

Differentiate by parts using d(uv) dv du ᎏᎏ ⫽ uᎏᎏ ⫹ v ᎏᎏ with u ⫽ t and v ⫽ e⫺t dt dt dt

Thus, di d vL ⫽ Lᎏᎏ ⫽ Lᎏᎏ(te⫺t) ⫽ L[t(⫺e⫺t) ⫹ e⫺t] ⫽ 12.5e⫺t (1 ⫺ t) volts dt dt

Section 13.5

■

Inductances in Series and Parallel

1. Figure 13–14 shows the current through a 5-H inductance. Determine voltage vL and sketch it.

PRACTICE PROBLEMS 2

i (mA) 6 0

t (ms)

FIGURE 13–14

2. If the current of Figure 13–12 is applied to an unknown inductance and the voltage from 1 ms to 2 ms is 28 volts, what is L? 3. ∫ The current in a 4-H inductance is i ⫽ t2e⫺5t A. What is voltage vL? Answers: 1. vL is a square wave. Between 0 and 2 ms, its value is 15 V; between 2 ms and 4 ms, its value is ⫺15 V, etc. 2. 7 mH

3. 4e⫺5t(2t ⫺ 5t2) V

1. An inductance L1 of 50 mH is in series with an inductance L2 of 35 mH. If the voltage across L1 at some instant is 125 volts, what is the voltage across L2 at that instant? Hint: Since the same current passes through both inductances, the rate of change of current is the same for both. 2. Current through a 5-H inductance changes linearly from 10 A to 12 A in 0.5 s. Suppose now the current changes linearly from 2 mA to 6 mA in 1 ms. Although the currents are significantly different, the induced voltage is the same in both cases. Why? Compute the voltage. (Answers are at the end of the chapter.)

13.5

Inductances in Series and Parallel

For inductances in series or parallel, the equivalent inductance is found by using the same rules that you used for resistance. For the series case (Figure 13–15) the total inductance is the sum of the individual inductances: LT ⫽ L1 ⫹ L2 ⫹ L3 ⫹ … ⫹ LN

FIGURE 13–15

L T ⫽ L1 ⫹ L2 ⫹ … ⫹ LN

L3 LN

(13–8)

IN-PROCESS

LEARNING CHECK 3

503

504

Chapter 13

■

Inductance and Inductors

For the parallel case (Figure 13–16), LT

1 1 1 1 1 ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ ᎏᎏ ⫹ … ⫹ ᎏᎏ LT L1 L2 L3 LN

(13–9)

For two inductances, Equation 13–9 reduces to FIGURE 13–16 … ⫹ ᎏ1ᎏ LN

L L2 LT ⫽ ᎏ1ᎏ L1 ⫹ L2

1 1 1 ᎏᎏ ⫽ ᎏᎏ ⫹ ᎏᎏ ⫹ LT L1 L2

EXAMPLE 13–7

(13–10)

Find LT for the circuit of Figure 13–17. L2 L1

2.5 H

11 H

Leq = 1.5 H FIGURE 13–17

Solution

The parallel combination of L2 and L3 is L L3 6⫻2 Leq ⫽ ᎏ2ᎏ ⫽ ᎏᎏ ⫽ 1.5 H 6⫹2 L2 ⫹ L3

This is in series with L1 and L4. Thus, LT ⫽ 2.5 ⫹ 1.5 ⫹ 11 ⫽ 15 H.

PRACTICE PROBLEMS 3

1. For Figure 13–18, LT ⫽ 2.25 H. Determine Lx. 3H

FIGURE 13–18 7H

3H 3H

2.25 H

2. For Figure 13–15, the current is the same in each inductance and v1 ⫽ L1di/dt, v2 ⫽ L2di/dt, and so on. Apply KVL and show that LT ⫽ L1 ⫹ L2 ⫹ L3 ⫹ … ⫹ LN. Answer: 1. 3 H

13.6

Practical Considerations

Core Types The type of core used in an inductor depends to a great extent on its intended use and frequency range. (Although you have not studied frequency yet, you can get a feel for frequency by noting that the electrical

Section 13.6

■

Practical Considerations

505

power system operates at low frequency [60 cycles per second, called 60 hertz], while radio and TV systems operate at high frequency [hundreds of megahertz].) Inductors used in audio or power supply applications generally have iron cores (because they need large inductance values), while inductors for radio-frequency circuits generally use air or ferrite cores. (Ferrite is a mixture of iron oxide in a ceramic binder. It has characteristics that make it suitable for high-frequency work.) Iron cannot be used, however, since it has large power losses at high frequencies (for reasons discussed in Chapter 17).

Variable Inductors Inductors can be made so that their inductance is variable. In one approach, inductance is varied by changing coil spacing with a screwdriver adjustment. In another approach (Figure 13–19), a threaded ferrite slug is screwed in or out of the coil to vary its inductance.

(a) Iron-core

(b) Ferrite-core

FIGURE 13–19

A variable inductor with its ferrite core removed for viewing.

Circuit Symbols Figure 13–20 shows inductor symbols. Iron-cores are identified by double solid lines, while dashed lines denote a ferrite core. (Air-core inductors have no core symbol.) An arrow indicates a variable inductor.

Coil Resistance Ideally, inductors have only inductance. However, since inductors are made of imperfect conductors (e.g., copper wire), they also have resistance. (We can view this resistance as being in series with the coil’s inductance as indicated in Figure 13–21(a). Also shown is stray capacitance, considered next.) Although coil resistance is generally small, it cannot always be ignored and thus, must sometimes be included in the analysis of a circuit.

(d) Air-core FIGURE 13–20 inductors.

Circuit symbols for

506

Chapter 13

■

Inductance and Inductors

In Section 13.7, we show how this resistance is taken into account in dc analysis; in later chapters, you will learn how to take it into account in ac analysis.

Cw Rl

(a) Equivalent circuit FIGURE 13–21

(b) Separating coil into sections helps reduce stray capacitance

A ferrite-core choke.

Stray Capacitance Because the turns of an inductor are separated from each other by insulation, a small amount of capacitance exists from winding to winding. This capacitance is called stray or parasitic capacitance. Although this capacitance is distributed from turn to turn, its effect can be approximated by lumping it as in Figure 13–21(a). The effect of stray capacitance depends on frequency. At low frequencies, it can usually be neglected; at high frequencies, it may have to be taken into account. Some coils are wound in multiple sections as in Figure 13–21(b) to reduce stray capacitance. Stray Inductance Because inductance is due entirely to the magnetic effects of electric current, all current-carrying conductors have inductance. This means that leads on circuit components such as resistors, capacitors, transistors, and so on, all have inductance, as do traces on printed circuit boards and wires in cables. We call this inductance “stray inductance.” Fortunately, in many cases, the stray inductance is so small that it can be neglected.

PRACTICAL NOTES... Although stray inductance is small, it is not always negligible. In general, stray inductance will not be a problem for short wires at low to moderate frequencies. However, even a short piece of wire can be a problem at high frequencies, or a long piece of wire at low frequencies. For example, the inductance of just a few centimeters of conductor in a high-speed logic system may be nonnegligible because the current through it changes at such a high rate.

Section 13.7

13.7

■

507

Inductance and Steady State DC

We now look at inductive circuits with constant dc current. Consider Figure 13–22. The voltage across an ideal inductance with constant dc current is zero because the rate of change of current is zero. This is indicated in (a). Since the inductor has current through it but no voltage across it, it looks like a short circuit, (b). This is true in general, that is, an ideal inductor looks like a short circuit in steady state dc. (This should not be surprising since it is just a piece of wire to dc.) For a nonideal inductor, its dc equivalent is its coil resistance (Figure 13–23). For steady state dc, problems can be solved using simple dc analysis techniques.

vL = 0 ⫹ ⫺ Constant dc current (a) Since the field is constant, induced voltage is zero vL = 0 ⫹ ⫺ (b) Equivalent of inductor to dc is a short circuit FIGURE 13–22 Inductance looks like a short circuit to steady state dc.

EXAMPLE 13–8 In Figure 13–24(a), the coil resistance is 14.4 ⍀. What is the steady state current I? 6⍀

Coil

I L Rl 9⍀

120 V

(a) Coil

(b) DC equivalent

L Coil Rl = 14.4 ⍀ (a)

Thévenin equivalent I R Th

3.6 ⍀ 14.4 ⍀ ETh

72 V

Coil equivalent (b) Coil replaced by its dc equivalent FIGURE 13–24

FIGURE 13–23 Steady state dc equivalent of a coil with winding resistance.

508

Chapter 13

■

Inductance and Inductors

Solution

Reduce the circuit as in (b). ETh ⫽ (9/15)(120) ⫽ 72 V RTh ⫽ 6⍀㥋9⍀ ⫽ 3.6 ⍀

Now replace the coil by its dc equivalent circuit as in (b). Thus, I ⫽ ETh/RT ⫽ 72/(3.6 ⫹ 14.4) ⫽ 4 A

EXAMPLE 13–9 The resistance of coil 1 in Figure 13–25(a) is 30 ⍀ and that of coil 2 is 15 ⍀. Find the voltage across the capacitor assuming steady state dc. Coil 2 30 ⍀

L1 15 ⍀

Coil 1

⫹ VC ⫺

60 V

(a)

R1 30 ⍀ ⫹ VC ⫺

E 60 V

R2 15 ⍀

(b) dc equivalent FIGURE 13–25

Solution Replace each coil inductance with a short circuit and the capacitor with an open circuit. As you can see from (b), the voltage across C is the same as the voltage across R2. Thus, R2 15 ⍀ VC ⫽ ᎏᎏE ⫽ ᎏᎏ (60 V) ⫽ 20 V 45 ⍀ R1 ⫹ R2

冢

冣

Section 13.8

■

Energy Stored by an Inductance

PRACTICE PROBLEMS 4

For Figure 13–26, find I, VC1, and VC2 in the steady state. I

20 V 15 ⍀

10 ⍀

⫹ V ⫺ C1

150 V Coil 1

L2 Coil 2

5⍀

EWB

509

15 ⍀

⫹ VC ⫺ 2

FIGURE 13–26

Answers: 10.7 A; 63 V; 31.5 V

13.8

Energy Stored by an Inductance

When power flows into an inductor, energy is stored in its magnetic field. When the field collapses, this energy is returned to the circuit. For an ideal inductor, Rᐉ ⫽ 0 ohm and hence no power is dissipated; thus, an ideal inductor has zero power loss. To determine the energy stored by an ideal inductor, consider Figure 13– 27. Power to the inductor is given by p ⫽ vLi watts, where vL ⫽ Ldi/dt. By summing this power (see next ∫ ), the energy is found to be 1 W ⫽ ᎏᎏLi2 2

(J)

(13–11)

where i is the instantaneous value of current. When current reaches its steady state value I, W ⫽ 1⁄2LI 2 J. This energy remains stored in the field as long as the current continues. When the current goes to zero, the field collapses and the energy is returned to the circuit.

The coil of Figure 13–28(a) has a resistance of 15 ⍀. When the current reaches its steady state value, the energy stored is 12 J. What is the inductance of the coil?

EXAMPLE 13–10

i ⫹ vL ⫺ p

1 W = 2 Li2

FIGURE 13–27 Energy is stored in the magnetic field of an inductor.

510

Chapter 13

■

Inductance and Inductors

I 10 ⍀ Coil

15 ⍀

100 V

10 ⍀

15 ⍀

100 V W = 12 J (a)

(b)

FIGURE 13–28

Solution

From (b), I ⫽ 100 V/25 ⍀ ⫽ 4 A 1 W ⫽ ᎏᎏLI 2 J 2 1 12 J ⫽ ᎏᎏL(4 A)2 2

Thus, L ⫽ 2(12)/42 ⫽ 1.5 H

∫

Deriving Equation 13–11 The power to the inductor in Figure 13–27 is given by p ⫽ vLi, where vL ⫽ Ldi/dt. Therefore, p ⫽ Lidi/dt. However, p ⫽ dW/dt. Integrating yields W⫽

13.9

i 1 冕 pdt ⫽ 冕 Liᎏddᎏdt ⫽ L冕 idi ⫽ ᎏᎏLi t 2 t

Inductor Troubleshooting Hints

Inductors may fail by either opening or shorting. Failures may be caused by misuse, defects in manufacturing, or faulty installation.

Open Coil Opens can be the result of poor solder joints or broken connections. First, make a visual inspection. If nothing wrong is found, disconnect the inductor and check it with an ohmmeter. An open-circuited coil has infinite resistance. Shorts Shorts can occur between windings or between the coil and its core (for an iron-core unit). A short may result in excessive current and overheating. Again, check visually. Look for burned insulation, discolored components,

Problems

511

an acrid odor, and other evidence of overheating. An ohmmeter can be used to check for shorts between windings and the core. However, checking coil resistance for shorted turns is often of little value, especially if only a few turns are shorted. This is because the shorting of a few windings will not change the overall resistance enough to be measurable. Sometimes the only conclusive test is to substitute a known good inductor for the suspected one.

Unless otherwise indicated, assume ideal inductors and coils.

PROBLEMS

13.2 Induced Voltage and Inductance 1. If the flux linking a 75-turn coil changes at the rate of 3 Wb/s, what is the voltage across the coil? 2. If 80 volts is induced when the flux linking a coil changes at a uniform rate from 3.5 mWb to 4.5 mWb in 0.5 ms, how many turns does the coil have? 3. Flux changing at a uniform rate for 1 ms induces 60 V in a coil. What is the induced voltage if the same flux change takes place in 0.01 s? 13.3 Self-Inductance 4. The current in a 0.4-H inductor is changing at the rate of 200 A/s. What is the voltage across it? 5. The current in a 75-mH inductor changes uniformly by 200 mA in 0.1 ms. What is the voltage across it? 6. The voltage across an inductance is 25 volts when the current changes at 5 A/s. What is L? 7. The voltage induced when current changes uniformly from 3 amps to 5 amps in a 10-H inductor is 180 volts. How long did it take for the current to change from 3 to 5 amps? 8. Current changing at a uniform rate for 1 ms induces 45 V in a coil. What is the induced voltage if the same current change takes place in 100 ms? 9. Compute the inductance of the air-core coil of Figure 13–29, given ᐉ ⫽ 20 cm, N ⫽ 200 turns, and d ⫽ 2 cm. 10. The iron-core inductor of Figure 13–30 has 2000 turns, a cross-section of 1.5 ⫻ 1.2 inches, and an air gap of 0.2 inch. Compute its inductance. 11. The iron-core inductor of Figure 13–30 has a high-permeability core. Therefore, by Ampere’s law, NI ⯝ Hgᐉg. Because the air gap dominates, saturation does not occur and the core flux is proportional to the current, i.e., the flux linkage equals LI. In addition, since all flux passes through the coil, the flux linkage equals N⌽. By equating the two values of flux linkage and using ideas from Chapter 12, show that the inductance of the coil is m0 N2Ag L ⫽ ᎏᎏ ᐉg 13.4 Computing Induced Voltage 12. Figure 13–31 shows the current in a 0.75-H inductor. Determine vL and plot its waveform.

l d

FIGURE 13–29

⌽ lg

FIGURE 13–30

512

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■

Inductance and Inductors i (mA) 50

i (mA) 30 20 10 0 ⫺10 ⫺20

0 1 2 3 4 5 6 7 8 9 10

⫺20

1 2 3 4 5 6 7 8 9 10 t (ms)

t (ms)

FIGURE 13–31

FIGURE 13–32

13. Figure 13–32 shows the current in a coil. If the voltage from 0 to 2 ms is 100 volts, what is L? 14. Why is Figure 13–33 not a valid inductor current? Sketch the voltage across L to show why. Pay particular attention to t ⫽ 10 ms. vL (V) i (A)

FIGURE 13–33

∫

t (ms)

⫺4

5 t (s)

⫺8 FIGURE 13–34

15. Figure 13–34 shows the graph of the voltage across an inductance. The current changes from 4 A to 5 A during the time interval from 4 s to 5 s. a. What is L? b. Determine the current waveform and plot it. c. What is the current at t ⫽ 10 s? 16. If the current in a 25-H inductance is iL ⫽ 20e⫺12t mA, what is vL? 13.5 Inductances in Series and Parallel 17. What is the equivalent inductance of 12 mH, 14 mH, 22 mH, and 36 mH connected in series? 18. What is the equivalent inductance of 0.010 H, 22 mH, 86 ⫻ 10⫺3 H, and 12000 mH connected in series? 19. Repeat Problem 17 if the inductances are connected in parallel. 20. Repeat Problem 18 if the inductances are connected in parallel. 21. Determine LT for the circuits of Figure 13–35. 22. Determine LT for the circuits of Figure 13–36. 23. A 30-mH inductance is connected in series with a 60-mH inductance, and a 10-mH inductance is connected in parallel with the series combination. What is LT?

Problems L1 = 10 H

L2 5 H

L1 3 H

L2 6 H

L3 6H (a)

(b)

18 H

14 H

(c)

2H 1.6 H LT

(d)

6 mH

28.5 mH

600 ␮H LT

19 mH

(e) FIGURE 13–35

24. For Figure 13–37, determine Lx. 25. For the circuits of Figure 13–38, determine L3 and L4. 26. You have inductances of 24 mH, 36 mH, 22 mH and 10 mH. Connecting these any way you want, what is the largest equivalent inductance you can get? The smallest?

513

514

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Inductance and Inductors 6H 27 H 2H

1H LT

6H 2H

4H 18 H (a) 12 H

1.8 H 18 H

14 H

1H 1.2 H

1.4 H

4H 18.7 H 11 H

(b) FIGURE 13–36 2H

27 H L3 = 4L4

1.5 H LT = 6 H

18 H

Lx 6H

11.25 H

12.4 H

L4 = 4L3

FIGURE 13–37 (a)

(b)

FIGURE 13–38

i i2

i1 ⫹ v ⫺

⫹ v ⫺

L1 FIGURE 13–39

i3 ⫹ v ⫺ L3

⫹ v ⫺ LN

27. A 6-H and a 4-H inductance are connected in parallel. After a third inductance is added, LT ⫽ 4 H. What is the value of the third inductance and how was it connected? 28. Inductances of 2 H, 4 H, and 9 H are connected in a circuit. If LT ⫽ 3.6 H, how are the inductors connected? 29. Inductances of 8 H, 12 H, and 1.2 H are connected in a circuit. If LT ⫽ 6 H, how are the inductors connected? 30. For inductors in parallel (Figure 13–39), the same voltage appears across each. Thus, v ⫽ L1di1/dt, v ⫽ L2di2/dt, etc. Apply KCL and show that 1/LT ⫽ 1/L1 ⫹ 1/L2 ⫹ … ⫹ 1/LN.

Problems 31. By combining elements, reduce each of the circuits of Figure 13–40 to their simplest form. 7 µF 4.5 µF

2 µF

6H 1H 2 µF 3H (a)

8 µF (b)

10 µF 10 ⍀

10 ⍀

10 µF

10 H

60 µF

40 ⍀

40 H

5 µF 30 µF (c)

(d)

FIGURE 13–40

13.7 Inductance and Steady State DC 32. For each of the circuits of Figure 13–41, the voltages and currents have reached their final (steady state) values. Solve for the quantities indicated. 4.5 ⍀

4⍀

3H 1H

9⍀

6⍀ 18 V

13.5 V

11 ⍀

−

4A (a) Find E FIGURE 13–41

13.8 Energy Stored by an Inductance 33. Find the energy stored in the inductor of Figure 13–42.

(b) Find Rx

6⍀ Rx 3H

515

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Inductance and Inductors 11 ⍀

5⍀

6⍀

20 ⍀

Rl 10 ⍀

4⍀

5⍀

40 V

100 V L 4H

Coil FIGURE 13–42

FIGURE 13–43

34. In Figure 13–43, L1 ⫽ 2L2. The total energy stored is WT ⫽ 75 J. Find L1 and L2. 13.9 Inductor Troubleshooting Hints 35. In Figure 13–44, an inductance meter measures 7 H. What is the likely fault?

H

⫹

⫺

FIGURE 13–44

36. Referring to Figure 13–45, an inductance meter measures LT ⫽ 8 mH. What is the likely fault? L1 6 mH L2 5 mH

LT L4

2 mH FIGURE 13–45

L3 11 mH

Answers to In-Process Learning Checks

In-Process Learning Check 1 1. Both a and b. Current cannot change instantaneously. 2. 600 Wb-turns 3. 30 V 4. âŤş300 V In-Process Learning Check 2 1. 28 mH 2. 9 times In-Process Learning Check 3 1. 87.5 V 2. Rate of change of current is the same; 20 V

517

ANSWERS TO IN-PROCESS LEARNING CHECKS

Inductive Transients OBJECTIVES

KEY TERMS

After studying this chapter, you will be able to • explain why transients occur in RL circuits, • explain why an inductor looks like an open circuit when first energized, • compute time constants for RL circuits, • compute voltage and current transients in RL circuits during the current buildup phase, • compute voltage and current transients in RL circuits during the current decay phase, • solve moderately complex RL transient problems using circuit simplification techniques, • solve RL transient problems using Electronics Workbench and PSpice.

Continuity of Current De-energizing Transient Energizing Transient Initial Condition Circuit RL Transients Time Constant

OUTLINE Introduction Current Buildup Transients Interrupting Current in an Inductive Circuit De-energizing Transients More Complex Circuits RL Transients Using Computers

n Chapter 11, you learned that transients occur in capacitive circuits because capacitor voltage cannot change instantaneously. In this chapter, you will learn that transients occur in inductive circuits because inductor current cannot change instantaneously. Although the details differ, you will find that many of the basic ideas are the same. Inductive transients result when circuits containing inductance are disturbed. More so than capacitive transients, inductive transients are potentially destructive and dangerous. For example, when you break the current in an inductive circuit, an extremely large and damaging voltage may result. In this chapter, we study basic RL transients. We look at transients during current buildup and decay and learn how to calculate the voltages and currents that result.

Inductance, the Dual of Capacitance INDUCTANCE IS THE DUAL of capacitance. This means that the effect that inductance has on circuit operation is identical with that of capacitance if you interchange the term current for voltage, open circuit for short circuit, and so on. For example, for simple dc transients, current in an RL circuit has the same form as voltage in an RC circuit: they both rise to their final value exponentially according to 1 e t/t. Similarly, voltage across inductance decays in the same manner as current through capacitance, i.e., according to e t/t. Duality applies to steady state and initial condition representations as well. To steady state dc, for example, a capacitor looks like an open circuit, while an inductor looks like a short circuit. Similarly, the dual of a capacitor that looks like a short circuit at the instant of switching is an inductor that looks like an open circuit. Finally, the dual of a capacitor that has an initial condition of V0 volts is an inductance with an initial condition of I0 amps. The principle of duality is helpful in circuit analysis as it lets you transfer the principles and concepts learned in one area directly into another. You will find, for example, that many of the ideas learned in Chapter 11 reappear here in their dual form.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

519

520

Chapter 14

■

Inductive Transients

14.1

Introduction

As you saw in Chapter 11, when a circuit containing capacitance is disturbed, voltages and currents do not change to their new values immediately, but instead pass through a transitional phase as the circuit capacitance charges or discharges. The voltages and currents during this transitional interval are called transients. In a dual fashion, transients occur when circuits containing inductances are disturbed. In this case, however, transients occur because current in inductance cannot change instantaneously. To get at the idea, consider Figure 14–1. In (a), we see a purely resistive circuit. At the instant the switch is closed, current jumps from 0 to E/R as required by Ohm’s law. Thus, no transient (i.e., transitional phase) occurs because current reaches its final value immediately. Now consider (b). Here, we have added inductance. At the instant the switch is closed, a counter emf appears across the inductance. This voltage attempts to stop the current from changing and consequently slows its rise. Current thus does not jump to E/R immediately as in (a), but instead climbs gradually and smoothly as in (b). The larger the inductance, the longer the transition takes. R

NOTES... The continuity statement for inductor current has a sound mathematical basis. Recall, induced voltage is proportional to the rate of change of current. In calculus notation, di vL L dt This means that the faster that current changes, the larger the induced voltage. If inductor current could change from one value to another instantaneously as in Figure 14–1(a), its rate of change (i.e., di/dt) would be infinite and hence the induced voltage would be infinite. But infinite voltage is not possible. Thus, inductor current cannot change instantaneously.

Final (steady state) value

i E R

⫹ vL = L di dt ⫺

Increasing L 0

(a) No transient occurs in a purely resistive circuit

(b ) Adding inductance causes a transient to appear. R is held constant here

FIGURE 14–1 Transient due to inductance. Adding inductance to a resistive circuit slows the current rise and fall, thus creating a transient.

Continuity of Current As Figure 14–1(b) illustates, current through an inductance cannot change instantaneously, i.e., it cannot jump abruptly from one value to another, but must be continuous at all values of time. This observation is known as the statement of continuity of current for inductance. You will find this statement of great value when analyzing circuits containing inductance. We will use it many times in what follows.

Section 14.1

■

Introduction

521

Inductor Voltage Now consider inductor voltage. When the switch is open as in Figure 14– 2(a), the current in the circuit and voltage across L are both zero. Now close the switch. Immediately after the switch is closed, the current is still zero, (since it cannot change instantaneously). Since vR Ri, the voltage across R is also zero and thus the full source voltage appears across L as shown in (b). The inductor voltage therefore jumps from 0 V just before the switch is closed to E volts just after. It then decays to zero, since, as we saw in Chapter 13, the voltage across inductance is zero for steady state dc. This is indicated in (c). vR = Ri = 0 ⫹ ⫺

R E

⫹ vL = 0 ⫺

i=0

⫹ vL = E ⫺

t = 0⫹ (a) Circuit with switch open. Current i = 0

FIGURE 14–2

Voltage transient

0V 0

(b) Circuit just after the switch has been closed. Current is still equal to zero. Thus, vL = E

Voltage across L.

Open-Circuit Equivalent of an Inductance Consider again Figure 14–2(b). Note that just after the switch is closed, the inductor has voltage across it but no current through it. It therefore momentarily looks like an open circuit. This is indicated in Figure 14–3. This observation is true in general, that is, an inductor with zero initial current looks like an open circuit at the instant of switching. (Later, we extend this statement to include inductors with nonzero initial currents.) Initial Condition Circuits Voltages and currents in circuits immediately after switching must sometimes be calculated. These can be determined with the aid of the open-circuit equivalent. By replacing inductances with open circuits, you can see what a circuit looks like just after switching. Such a circuit is called an initial condition circuit.

EXAMPLE 14–1

A coil and two resistors are connected to a 20-V source as in Figure 14–4(a). Determine source current i and inductor voltage vL at the instant the switch is closed.

vR = 0 ⫹ ⫺ i=0 E

⫹ vL = E ⫺

FIGURE 14–3 Inductor with zero initial current looks like an open circuit at the instant the switch is closed.

522

Chapter 14

■

Inductive Transients

6⍀

R1 6⍀ E 20 V

i R2

⫹ vL ⫺

4⍀

⫹ v2 ⫺8 V

4⍀

20 V

⫹ vL ⫺8V

(b) Initial condition network

(a) Original circuit FIGURE 14–4

Solution Replace the inductance with an open circuit. This yields the network shown in (b). Thus i E/RT 20 V/10 ⍀ 2 A and the voltage across R2 is v2 (2 A)(4 ⍀) 8 V. Since vL v2, vL 8 volts as well.

vL This is the value determined by the initial condition network

FIGURE 14–5 The initial condition network yields only the value at t 0 s.

PRACTICE PROBLEMS 1

Initial condition networks yield voltages and currents only at the instant of switching, i.e., at t 0 s. Thus, the value of 8 V calculated in Example 14–1 is only a momentary value as illustrated in Figure 14–5. Sometimes such an initial value is all that you need. In other cases, you need the complete solution. This is considered next, in Section 14.2.

Determine all voltages and currents in the circuit of Figure 14–6 immediately after the switch is closed and in steady state. iT

30 ⍀ 160 V L1

EWB

+ vL2 −

+ vR2 −

i1 20 ⍀ + vR1 − + vL1 −

L2 i3 60 ⍀

+ vR3 −

i4 5⍀

+ vR4 −

FIGURE 14–6

Answers: Initial: vR1 0 V; vR2 40 V; vR3 120 V; vR4 0 V; vL1 160 V; vL2 120 V; iT 2 A; i1 0 A; i2 2 A; i3 2 A; i4 0 A. Steady State: vR1 160 V; vR2 130 V; vR3 vR4 30 V; vL1 vL2 0 V; iT 11.83 A; i1 5.33 A; i2 6.5 A; i3 0.5 A; i4 6.0 A

Section 14.2

14.2

■

Current Buildup Transients

Current We will now develop equations to describe voltages and current during energization. Consider Figure 14–7. KVL yields vL vR E

di L Ri E dt

⫺

R E

⫹ vL ⫺

(14–2)

Equation 14–2 can be solved using basic calculus in a manner similar to what we did for RC circuits in Chapter 11. The result is E i (1 e Rt/L) (A) R

⫹

(14–1)

Substituting vL Ldi/dt and vR Ri into Equation 14–1 yields

(14–3)

where R is in ohms, L is in henries, and t is in seconds. Equation 14–3 describes current buildup. Values of current at any point in time can be found by direct substitution as we illustrate next. Note that E/R is the final (steady state) current.

EXAMPLE 14–2 For the circuit of Figure 14–7, suppose E 50 V, R 10 ⍀, and L 2 H: a. b. c. d.

523

Current Buildup Transients

Determine the expression for i. Compute and tabulate values of i at t 0 , 0.2, 0.4, 0.6, 0.8, and 1.0 s. Using these values, plot the current. What is the steady state current?

Solution a. Substituting the values into Equation 14–3 yields E 50 V i (1 e Rt/L) (1 e 10t/2) 5 (1 e 5t ) amps R 10 ⍀ b. At t 0 s, i 5(1 e 5t ) 5(1 e0) 5(1 1) 0 A. At t 0.2 s, i 5(1 e 5(0.2)) 5(1 e 1) 3.16 A. At t 0.4 s, i 5(1 e 5(0.4)) 5(1 e 2) 4.32 A. Continuing in this manner, you get Table 14–1. c. Values are plotted in Figure 14–8. Note that this curve looks exactly like the curves we determined intuitively in Figure 14–1(b). d. Steady state current is E/R 50 V/10 ⍀ 5 A. This agrees with the curve of Figure 14–8.

FIGURE 14–7 vR E.

KVL yields vL

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Chapter 14

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Inductive Transients

TABLE 14–1 Time

Current

0 0.2 0.4 0.6 0.8 1.0

0 3.16 4.32 4.75 4.91 4.97

i (A) 5.0 4.32 4.0 3.16

3.0

4.75 4.91

4.97

Final value = E = 5 A R

2.0 i = 5(1 − e−5t)A

1.0 0 0 EWB

FIGURE 14–8

0.2 0.4 0.6 0.8 1.0

t (s)

Current buildup transient.

Circuit Voltages With i known, circuit voltages can be determined. Consider voltage vR. Since vR Ri, when you multiply R times Equation 14–3, you get vR E(1 e Rt/L) (V)

(14–4)

Note that vR has exactly the same shape as the current. Now consider vL. Voltage vL can be found by subtracting vR from E as per Equation 14–1: vL E vR E E(1 e Rt/L) E E Ee Rt/L

Thus, vL Ee Rt/L

(14–5)

An examination of Equation 14–5 shows that vL has an initial value of E at t 0 s and then decays exponentially to zero. This agrees with our earlier observation in Figure 14–2(c).

EXAMPLE 14–3

Repeat Example 14–2 for voltage vL.

Solution a. From equation 14–5, vL Ee Rt/L 50e 5t volts

Section 14.2

■

Current Buildup Transients

b. At t 0 s, vL 50e 5t 50e0 50(1) 50 V. At t 0.2 s, vL 50e 5(0.2) 50e 1 18.4 V. At t 0.4 s, vL 50e 5(0.4) 50e 2 6.77 V. Continuing in this manner, you get Table 14–2. c. The waveform is shown in Figure 14–9. d. Steady state voltage is 0 V, as you can see in Figure 14–9. vL (V) 50

TABLE 14–2 Time (s)

Voltage (V)

0 0.2 0.4 0.6 0.8 1.0

50.0 18.4 6.77 2.49 0.916 0.337

Initial value = E vL = 50e−5t V

18.4 6.77 2.49 0.916 0.337

10 0 EWB

0.2 0.4 0.6 0.8 1.0

t (s)

FIGURE 14–9 Inductor voltage transient.

For the circuit of Figure 14–7, with E 80 V, R 5 k⍀, and L 2.5 mH: a. Determine expressions for i, vL, and vR. b. Compute and tabulate values at t 0 , 0.5, 1.0, 1.5, 2.0, and 2.5 ms. c. At each point in time, does vL vR E? d. Plot i, vL, and vR using the values computed in (b). 6

Answers: a. i 16(1 e 2 10 t ) mA; vL 80e 2 10 t V; vR 80(1 e 2 10 t ) b.

t (␮s)

vL (V)

iL (mA)

vR (V)

0 0.5 1.0 1.5 2.0 2.5

80 29.4 10.8 3.98 1.47 0.539

0 10.1 13.8 15.2 15.7 15.9

0 50.6 69.2 76.0 78.5 79.5

c. Yes d. i and vR have the shape shown in Figure 14–8, while vL has the shape shown in Figure 14–9, with values according to the table shown in b.

Time Constant In Equations 14–3 to 14–5 L/R is the time constant of the circuit. L t R

(s)

(14–6)

PRACTICE PROBLEMS 2

525

526

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Inductive Transients

Note that t has units of seconds. (This is left as an exercise for the student.) Equations 14–3, 14–4, and 14–5 may now be written as E i (1 e t/t) (A) R

i or vR 100 Percent

99.3 95.0 98.2 Final value 86.5 63.2 36.8 13.5 vL 0

␶

2␶

3␶

4␶

5␶

vL Ee t/t (V)

(14–8)

vR E(1 e t/t) (V)

(14–9)

Curves are plotted in Figure 14–10 versus time constant. As expected, transitions take approximately 5t; thus, for all practical purposes, inductive transients last five time constants.

EXAMPLE 14–4 Final value 4.98 1.83 0.674

(14–7)

In a circuit where L 2 mH, transients last 50 ms. What

is R? Solution Transients last five time constants. Thus, t 50 ms/5 10 ms. Now t L/R. Therefore, R L/t 2 mH/10 ms 200 ⍀.

FIGURE 14–10 Universal time constant curves for the RL circuit.

EXAMPLE 14–5

For an RL circuit, i 40(1 e 5t) A and vL 100e 5t V.

a. What are E and t? b. What is R? c. Determine L. Solution a. From Equation 14–8, vL Ee t/t 100e 5t. Therefore, E 100 V and 1 t 0.2s. 5 b. From Equation 14–7, E i (1 e t/t) 40(1 e 5t). R Therefore, E/R 40 A and R E/40 A 100 V/40 A 2.5 ⍀. c. t L/R. Therefore, L Rt (2.5)(0.2) 0.5 H.

vL Increasing L (fixed R) Increasing R (fixed L) t FIGURE 14–11 Effect of R and L on transient duration.

It is sometimes easier to solve problems using the time constant curves than it is to solve the equation. (Be sure to convert curve percentages to a decimal value first, e.g., 63.2% to 0.632.) To illustrate, consider the problem of Examples 14–2 and 14–3. From Figure 14–10 at t t 0.2 s, i 0.632E/R and v L 0.368E. Thus, i 0.632(5 A) 3.16 A and v L 0.368(50 V) 18.4 V as we found earlier. The effect of inductance and resistance on transient duration is shown in Figure 14–11. The larger the inductance, the longer the transient for a given resistance. Resistance has the opposite effect: for a fixed inductance, the

Section 14.3

■

Interrupting Current in an Inductive Circuit

larger the resistance, the shorter the transient. [This is not hard to understand. As R increases, the circuit looks more and more resistive. If you get to a point where inductance is negligible compared with resistance, the circuit looks purely resistive, as in Figure 14–1(a), and no transient occurs.] 1. For the circuit of Figure 14–12, the switch is closed at t 0 s. a. Determine expressions for vL and i. b. Compute vL and i at t 0 , 10 ms, 20 ms, 30 ms, 40 ms, and 50 ms. c. Plot curves for vL and i. FIGURE 14–12

8 k⍀ i ⫹ vL ⫺

20 V 2 k⍀

L = 100 mH

2. For the circuit of Figure 14–7, E 85 V, R 50 ⍀, and L 0.5 H. Use the universal time constant curves to determine vL and i at t 20 ms. 3. For a certain RL circuit, transients last 25 s. If L 10 H and steady state current is 2 A, what is E? 4. An RL circuit has E 50 V and R 10 ⍀. The switch is closed at t 0 s. What is the current at the end of 1.5 time constants? (Answers are at the end of the chapter.)

14.3

Interrupting Current in an Inductive Circuit

We now look at what happens when inductor current is interrupted. Consider Figure 14–13. At the instant the switch is opened, the field begins to collapse, which induces a voltage in the coil. If inductance is large and current

Collapsing field

Spark

i E

FIGURE 14–13 The sudden collapse of the magnetic field when the switch is opened causes a large induced voltage across the coil. (Several thousand volts may result.) The switch arcs over due to this voltage.

IN-PROCESS

LEARNING CHECK 1

527

528

Chapter 14

■

Inductive Transients

NOTES... The intuitive explanation here has a sound mathematical basis. Recall, back emf (induced voltage) across a coil is given by di ⌬i vL L 艐 L dt ⌬t where ⌬i is the change in current and ⌬t is the time interval over which the change takes place. When you open the switch, current begins to drop immediately toward zero. Since ⌬i is finite, and ⌬t → 0, the voltage across L rises to a very large value, causing a flashover to occur. After the flashover, current has a path through which to decay and thus ⌬t, although small, no longer approaches zero. The result is a large but finite voltage spike across L.

Discharge resistor

1. Flashovers, as in Figure 14–13, are generally undesirable. However, they can be controlled through proper engineering design. (One way is to use a discharge resistor, as in the next example; another way is to use a diode, as you will see in your electronics course.) 2. On the other hand, the large voltages created by breaking inductive currents have their uses. One is in the ignition system of automobiles, where current in the primary winding of a transformer coil is interrupted at the appropriate time by a control circuit to create the spark needed to fire the engine. 3. It is not possible to rigorously analyze the circuit of Figure 14–13 because the resistance of the arc changes as the switch opens. However, the main ideas can be established by studying circuits using fixed resistors as we see next.

The Basic Ideas We begin with the circuit of Figure 14–14. Assume the switch is closed and the circuit is in steady state. Since the inductance looks like a short circuit [Figure 14–15(a)], its current is iL 120 V/30 ⍀ 4 A.

is high, a great deal of energy is released in a very short time, creating a huge voltage that may damage equipment and create a shock hazard. (This induced voltage is referred to as an inductive kick.) For example, abruptly breaking the current through a large inductor (such as a motor or generator field coil) can create voltage spikes up to several thousand volts, a value large enough to draw long arcs as indicated in Figure 14–13. Even moderate sized inductances in electronic systems can create enough voltage to cause damage if protective circuitry is not used. The dynamics of the switch flashover are not hard to understand. When the field collapses, the voltage across the coil rises rapidly. Part of this voltage appears across the switch. As the switch voltage rises, it quickly exceeds the breakdown strength of air, causing a flashover between its contacts. Once struck, the arc is easily maintained, as it creates ionized gases that provide a relatively low resistance path for conduction. As the contacts spread apart, the arc elongates and eventually disappears as the coil energy is dissipated and coil voltage drops below that required to sustain the arc. There are several important points to note here:

⫹ vL ⫺

R1 = 30 ⍀

SW i2 = 0.2 A 120 V

iL = 4 A 600 ⍀

vL = 0

FIGURE 14–14 Discharge resistor R2 helps limit the size of the induced voltage. (a) Circuit just before the switch is opened

EWB

vR1 = 120 V ⫹ ⫺

⫹ vSW ⫺ ⫹ ⫺

v R2 ⫺ 2400 V ⫹

R1 R2

⫹ v L ⫺ ⫺2520 V

(b) Circuit just after SW is opened. Since coil voltage polarity is opposite to that shown, vL is negative

FIGURE 14–15 Circuit of Figure 14–14 immediately before and after the switch is opened. Coil voltage changes abruptly from 0 V to 2520 V for this example.

Section 14.4

Now open the switch. Just prior to opening the switch, iL 4 A; therefore, just after opening the switch, it must still be 4 A. As indicated in (b), this 4 A passes through resistances R1 and R2, creating voltages vR1 4 A 30 ⍀ 120 V and vR2 4 A 600 ⍀ 2400 V with the polarity shown. From KVL, vL vR1 vR2 0. Therefore at the instant the switch is opened,

■

vL t

vL (vR1 vR2) 2520 volts

appears across the coil, yielding a negative voltage spike as in Figure 14–16. Note that this spike is more than 20 times larger than the source voltage. As we see in the next section, the size of this spike depends on the ratio of R2 to R1; the larger the ratio, the larger the voltage. Consider again Figure 14–15. Note that current i2 changes abruptly from 0.2 A just prior to switching to 4 A just after. This is permissible, however, since i2 does not pass through the inductor and only currents through inductance cannot change abruptly.

Figure 14–16 shows the voltage across the coil of Figure 14–14. Make a similar sketch for the voltage across the switch and across resistor R2. Hint: Use KVL to find vSW and vR2.

529

De-energizing Transients

⫹ vL ⫺

Voltage at the instant the switch is opened

⫺2520 V

FIGURE 14–16 Voltage spike for the circuit of Figure 14–14. This voltage is more than 20 times larger than the source voltage.

PRACTICE PROBLEMS 3

Answer: vSW: With the switch closed, vSW 0 V; When the switch is opened, vSW jumps to 2520 V, then decays to 120 V. vR2: Identical to Figure 14–16 except that vR2 begins at 2400 V instead of 2520 V.

Inductor Equivalent at Switching Figure 14–17 shows the current through L of Figure 14–15. Because the current is the same immediately after switching as it is immediately before, it is constant over the interval from t 0 s to t 0 s. Since this is true in general, we see that an inductance with an initial current looks like a current source at the instant of switching. Its value is the value of the current at switching. This is shown in Figure 14–18.

iL = I0

(a) Current at switching FIGURE 14–18 of switching.

14.4

(b) Current source equivalent

An inductor carrying current looks like a current source at the instant

De-energizing Transients

We now look at equations for the voltages and currents described in the previous section. As we go through this material, you should focus on the basic principles involved, rather than just the resulting equations. You will probably

4 amps 0

t Switch opened here

FIGURE 14–17 Inductor current for the circuit of Figure 14–15.

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■

i = I0

forget formulas, but if you understand principles, you should be able to reason your way through many problems using just the basic current and voltage relationships. Consider Figure 14–19(a). Let the initial current in the inductor be denoted as I0 amps. Now open the switch as in (b). KVL yields vL vR1 vR2 0. Substituting vL Ldi/dt, vR1 R1i, and vR2 R2i yields Ldi/dt (R1 R2)i 0. Now using calculus, it can be shown that

i I0 e t/t⬘ (A)

(14–10)

where L L t⬘ RT R1 R2

(a) Before the switch is opened ⫹

vR1

vR2 ⫹

⫺

(14–11)

is the time constant of the discharge circuit. If the circuit is in steady state before the switch is opened, initial current I0 E/R1 and Equation 14–10 becomes

R1 ⫺

(s)

E i e t/t⬘ (A) R1

⫹ vL ⫺

(b) Decay circuit FIGURE 14–19 Circuit for studying decay transients.

(14–12)

EXAMPLE 14–6 For Figure 14–19(a), assume the current has reached steady state with the switch closed. Suppose that E 120 V, R1 30 ⍀, R2 600 ⍀, and L 126 mH: a. b. c. d.

Determine I0. Determine the decay time constant. Determine the equation for the current decay. Compute the current i at t 0 s and t 0.5 ms.

Solution a. Consider Figure 14–19(a). Since the circuit is in a steady state, the inductor looks like a short circuit to dc. Thus, I0 E/R1 4 A. b. Consider Figure 14–19(b). t⬘ L/(R1 R2) 126 mH/630 ⍀ 0.2 ms. c. i I0e t/t⬘ 4e t/0.2 ms A. d. At t 0 s, i 4e 0 4 A. At t 0.5 ms, i 4e 0.5 ms/0.2 ms 4e 2.5 0.328 A.

Now consider voltage vL. It can be shown to be vL V0 e t/t⬘

(14–13)

where V0 is the voltage across L just after the switch is opened. Letting i I0 in Figure 14–19(b), you can see that V0 I0(R1 R2) I0RT. Thus Equation 14–13 can be written as vL I0RT e t/t⬘

(14–14)

Section 14.5

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531

More Complex Circuits

Finally, if the current has reached steady state before the switch is opened, I0 E/R1, and Equation 14–14 becomes

冢

冣

R2 vL E 1 e t/t⬘ R1

(14–15)

Note that vL starts at V0 volts (which is negative) and decays to zero as shown in Figure 14–20. Now consider the resistor voltages. Each is the product of resistance times current (Equation 14–10). Thus, vR1 R1I0 e t/t⬘

t 0

(14–16)

and

vL = V0e

⫺t/␶'

V0 t/t⬘

vR2 R2I0 e

(14–17) FIGURE 14–20 Inductor voltage during decay phase. V0 is negative.

If current has reached steady state before switching, these become vR1 Ee t/t⬘

(14–18)

R2 t/t⬘ vR2 Ee R1

(14–19)

and

Substituting the values of Example 14–6 into these equations, we get for the circuit of Figure 14–19 vL 2520e t/0.2 ms V, vR1 120e t/0.2 ms V and vR2 2400e t/0.2 ms V. These can also be written as vL 2520e 5000t V and so on if desired. Decay problems can also be solved using the decay portion of the universal time constant curves shown in Figure 14–10.

EXAMPLE 14–7 For Figure 14–19, let E 120 V, R1 40 ⍀ and R2 20 ⍀. The circuit is in steady state with the switch closed. Use Figure 14–10 to find i and vL at t 2 t after the switch is opened. Solution I0 E/R1 3 A. At t 2 t, current will have decayed to 13.5%. Therefore, i 0.135I0 0.405 A and vL (R1 R1)i (60 ⍀)(0.405 A) 24.3 V. (Alternately, V0 (3 A)(60 ⍀) 180 V. At t 2 t, this has decayed to 13.5%. Therefore, vL 0.135( 180 V) 24.3 V as above.)

14.5

More Complex Circuits

The equations developed so far apply only to circuits of the forms of Figures 14–7 or 14–19. Fortunately, many circuits can be reduced to these forms using circuit reduction techniques such as series and parallel combinations, source conversions, Thévenin’s theorem, and so on.

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EXAMPLE 14–8

Determine iL for the circuit of Figure 14–21(a) if L 5 H.

240 ⍀

100 V

R2 800 ⍀

R3 200 ⍀

R4 104 ⍀

(a) Circuit

200 ⍀ RTh ETh

L iL

40 V

(b) Thévenin equivalent FIGURE 14–21

Solution The circuit can be reduced to its Thévenin equivalent (b) as you saw in Chapter 11 (Section 11.5). For this circuit, t L/RTh 5 H/200 ⍀ 25 ms. Now apply Equation 14–7. Thus, E 40 iL T h (1 e t/t) (1 e t/25 ms) 0.2 (1 e 40t) (A) RTh 200

EXAMPLE 14–9

For the circuit of Example 14–8, at what time does current reach 0.12 amps? Solution iL 0.2(1 e 40t ) (A) Thus, 0.12 0.2(1 e 40t ) (Figure 14–22) 0.6 1 e 40t e 40t 0.4

Section 14.5

■

More Complex Circuits

Taking the natural log of both sides, ln e 40t ln 0.4 40t 0.916 t 22.9 ms iL (A) 0.2 0.12

FIGURE 14–22

Time

1. For the circuit of Figure 14–21, let E 120 V, R1 600 ⍀, R2 3 k⍀, R3 2 k⍀, R4 100 ⍀, and L 0.25 H: a. Determine iL and sketch it. b. Determine vL and sketch it. 2. Let everything be as in Problem 1 except L. If iL 0.12 A at t 20 ms, what is L? Answers: 1. a. 160(1 e 2000t ) mA b. 80e 2000t V. iL climbs from 0 to 160 mA with the waveshape of Figure 14–1(b), reaching steady state in 2.5 ms. vL looks like Figure 14–2(c). It starts at 80 V and decays to 0 V in 2.5 ms. 2. 7.21 H

A Note About Time Scales Until now, we have considered energization and de-energization phases separately. When both occur in the same problem, we must clearly define what we mean by time. One way to handle this problem (as we did with RC circuits) is to define t 0 s as the beginning of the first phase and solve for voltages and currents in the usual manner, then shift the time axis to the beginning of the second phase, redefine t 0 s and then solve the second part. This is illustrated in Example 14–10. Note that only the first time scale is shown explicitly on the graph.

EXAMPLE 14–10 Refer to the circuit of Figure 14–23: a. Close the switch at t 0 and determine equations for iL and vL. b. At t 300 ms, open the switch and determine equations for iL and vL during the decay phase. c. Determine voltage and current at t 100 ms and at t 350 ms. d. Sketch iL and vL. Mark the points from (c) on the sketch.

PRACTICE PROBLEMS 4

533

534

Chapter 14

■

Inductive Transients

R3 80 ⍀ R1

10 A

30 ⍀

60 ⍀

⫹ vL ⫺

L=5H EWB

FIGURE 14–23

Solution a. Convert the circuit to the left of L to its Thévenin equivalent. As indicated in Figure 14–24(a), RTh 60㥋30 80 100 ⍀. From (b), ETh V2, where V2 (10 A)(20 ⍀) 200 V R3 80 ⍀ R1

30 ⍀

RTh = 100 ⍀

60 ⍀

20 ⍀ (a) 80 ⍀

10 A

20 ⍀

⫹

⫹ V2 ⫺

ETh = V2 = 200 V ⫺

R1 || R2 (b) FIGURE 14–24

The Thévenin equivalent circuit is shown in Figure 14–25(a). t L/RTh 50 ms. Thus during current buildup, Eh 200 iL T (1 e t/t) (1 e t/50 ms) 2 (1 e 20t ) A RTh 100 vL EThe t/t 200 e 20t V b. Current build-up is sketched in Figure 14–25(b). Since 5t 250 ms, current is in steady state when the switch is opened at 300 ms. Thus I0

Section 14.5 RTh = 100 ⍀ iL ETh

200 V

⫹ vL ⫺

t (ms)

L=5H

300 (b)

(a) Thévenin equivalent of Figure 14–23 FIGURE 14–25

Open switch here (see Figure 14–23)

Circuit and current during the buildup phase.

2 A. When the switch is opened, current decays to zero through a resistance of 60 80 140 ⍀ as shown in Figure 14–26. Thus, t⬘ 5H/140 ⍀ 35.7 ms. If t 0 s is redefined as the instant the switch is opened, the equation for the decay is iL I0 e t/t⬘ 2e t/35.7 ms 2e 28t A R3 = 80 ⍀

160 V ⫹ ⫺ ⫹ vL ⫺

60 ⍀

L=5H (a) Decay circuit

FIGURE 14–26

R1 ⫺ 120 V ⫹

⫹ vL = ⫺280 V ⫺

I0 = 2 A (b) As it looks immediately after the switch is opened. KVL yields vL = ⫺280 V

The circuit of Figure 14–23 as it looks during the decay phase.

Now consider voltage. As indicated in Figure 14–26(b), the voltage across L just after the switch is open is V0 280 V. Thus vL V0 e t/t⬘ 280e 28t V c. You can use the universal time constant curves at t 100 ms since 100 ms represents 2t. At 2t, current has reached 86.5% of its final value. Thus, iL 0.865(2 A) 1.73 A. Voltage has fallen to 13.5 %. Thus vL 0.135(200 V) 27.0 V. Now consider t 350 ms: Note that this is 50 ms into the decay portion of the curve. However, since 50 ms is not a multiple of t⬘, it is difficult to use the curves. Therefore, use the equations. Thus, iL 2 A e 28(50 ms) 2 A e 1.4 0.493 A vL ( 280 V)e 28(50 ms) ( 280 V)e 1.4 69.0 V d. The above points are plotted on the waveforms of Figure 14–27.

■

More Complex Circuits

535

536

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■

Inductive Transients

iL (A) 2.0 0.493 A

1.73 A

1.0 0 0

t (ms)

50 100 150 200 250 300 350 400

vL (V) Switch opened here

200 100

27.0 V

t (ms)

50 100 150 200 250 300 350 400

−100

−69.0 V

−200

−280 V

−300

FIGURE 14–27

The basic principles that we have developed in this chapter permit us to solve problems that do not correspond exactly to the circuits of Figure 14–7 and 14–19. This is illustrated in the following example.

EXAMPLE 14–11

The circuit of Figure 14–28(a) is in steady state with the switch open. At t 0 s, the switch is closed. R1

10 ⍀

40 ⍀

E 100 V

t=0

⫹

⫺

R2 = 40 ⍀

L = 100 mH (a) Steady state current with the 100 V switch open is =2A 50 ⍀ EWB

vR 2

⫹ vL ⫺

L = 100 mH (b) Decay circuit L ␶' = = 2.5 ms R2

FIGURE 14–28

a. Sketch the circuit as it looks after the switch is closed and determine t⬘. b. Determine current iL at t 0 s. c. Determine the expression for iL.

Section 14.6

d. e. f. g.

■

RL Transients Using Computers

537

Determine vL at t 0 s. Determine the expression for vL. How long does the transient last? Sketch iL and vL.

Solution a. When you close the switch, you short out E and R1, leaving the decay circuit of (b). Thus t⬘ L/R2 100 mH/40 ⍀ 2.5 ms. b. In steady state with the switch open, iL I0 100 V/50 ⍀ 2 A. This is the current just before the switch is closed. Therefore, just after the switch is closed, iL will still be 2 A. c. iL decays from 2 A to 0. From Equation 14–10, iL I0e t/t⬘ 2e t/2.5 ms 2e 400t A. d. KVL yields vL vR2 R2I0 (40 ⍀) (2A) 80 V. Thus, V0 80 V. e. vL decays from 80 V to 0. Thus, vL V0e t/t⬘ 80e 400t V. f. Transients last 5t⬘ 5(2.5 ms) 12.5 ms. iL

FIGURE 14–29

iL = 2Ae−400t 2A t 12.5 ms vL (V) t vL = −80Ve−400t ⫺80 V

14.6

RL Transients Using Computers

Electronics Workbench Workbench can easily plot voltage, but it has no simple way to plot current. If you want to determine current, use Ohm’s law and the applicable voltage waveform. To illustrate, consider Figure 14–21(a). Since the inductor current passes through R4, we can use the voltage across R4 to monitor current. Suppose we want to know at what time iL reaches 0.12 A. (This corresponds to 0.12 A 104 ⍀ 12.48 V across R4.) Create the circuit as in Figure 14–30.

NOTES... 1. Only basic steps are given for the following computer examples, as procedures here are similar to those of Chapter 11. If you need help, refer back to Chapter 11 or to Appendix A. 2. When scaling values from a computer plot, it is not always possible to place the cursor exactly where you want it (because of the nature of simulation programs). Consequently, you may have to set it as closely as you can get, then estimate the value you are trying to measure.

538

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Inductive Transients

Select Analysis/Transient, click Initial Conditions Set to Zero, set End Time (TSTOP) to 0.1, select Node 3, click Add, then Simulate. You should get the waveform shown on the screen of Figure 14–30. Expand to full size, then use the cursor to determine the time at which voltage equals 12.48 V. Figure 14–31 shows the result. Rounded to 3 figures, the answer is 22.9 ms (which agrees with the answer we obtained earlier in Example 14–9).

FIGURE 14–30 Electronics Workbench representation of Figure 14–21. No switch is required as the transient solution is initiated by software.

FIGURE 14–31 Values scaled from the waveform of Figure 14–30.

Using Electronics Workbench’s Oscilloscope

Waveforms may also be observed using Electronics Workbench’s oscilloscope. Close the Analysis Graphs window, click the Instruments Parts bin, position the scope and connect as in Figure 14–32(a). Double click the scope icon, set Time base to 5 ms/div, Channel A to 5 V/div and Y position to 1 (to better fit the trace to the screen.) Select Analysis/Analysis Options, click on the Instruments tab, select Pause after each screen, select Initial Conditions Set to Zero, click OK, then activate the circuit by clicking the ON/OFF power switch in the upper right-hand corner of the screen. The trace shown in Figure 14–32(b) should appear. Click the Expand button on the oscilloscope and drag

(a) Connecting the oscilloscope FIGURE 14–32

Using the Electronics Workbench oscilloscope.

(b) Expanded oscilloscope detail

Section 14.6

■

RL Transients Using Computers

the cursor until VA reads 12.48 volts (or as close as you can get). The corresponding value of time should be 22.9 ms as determined previously.

OrCAD PSpice RL transients are handled much like RC transients. As a first example, consider Figure 14–33. The circuit is in steady state with the switch closed. At t 0, the switch is opened. Use PSpice to plot inductor voltage and current, then use the cursor to determine values at t 100 ms. Verify manually. (Since the process is similar to that of Chapter 11, abbreviated instructions only are given.)

FIGURE 14–33 PSpice can easily determine both voltage and current transients. The voltage marker displays voltage across the inductance. The current display is added after the simulation is run.

Preliminary: First, determine the initial current in the inductor, i.e., the current I0 that exists at the instant the switch is opened. This is done by noting that the inductor looks like a short circuit to steady state dc. Thus, I0 12 V/ 4⍀ 3 A. Next, note that after the switch is opened, current builds up through R1, R2, and R3 in series. Thus, the time constant of the circuit is t L1/RT 3 H/30 ⍀ 0.1 s. Now proceed as follows:

• Build the circuit on the screen. Double click the inductor symbol and using the procedure from Chapter 11, set its initial condition IC to 3A. Click the New Profile icon and name the file fig14-33. In the Simulations Settings box, select transient analysis and set TSTOP to 0.5 (five time constants). Click OK. • Click the Run icon. When simulation is complete, a trace of capacitor voltage versus time appears. Create a second Y Axis, then add the current trace I(L1). You should now have the curves of Figure 14–34 on the screen. (The Y-axes can be labeled if desired as described in Appendix A.) Consider Figure 14–33. With the switch open, steady state current is 180 V/30 ⍀ 6 A and the initial current is 3 A. Thus, the current should start at 3 A and rise to 6 A in 5 time constants. (It does.) Inductor voltage

Results:

539

540

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Inductive Transients

FIGURE 14–34

Inductor voltage and current for the circuit of Figure 14–33.

should start at 180 V (3 A)(30 ⍀) 90 V and decay to 0 V in 5 time constants. (It does). Thus, the solution checks. Now, with the cursor, scale voltage and current values at t 100 ms. You should get 33.1 V for vL and 4.9 A for iL. (To check, note that the equations for inductor voltage and current are vL 90 e 10 t V and iL 6 3 e 10 t A respectively. Substitute t 100 ms into these and verify results.)

EXAMPLE 14–12

Consider the circuit of Figure 14–23, Example 14–10. The switch is closed at t 0 and opened 300 ms later. Prepare a PSpice analysis of this problem and determine vL and iL at t 100 ms and at t 350 ms.

Solution PSpice doesn’t have a switch that both opens and closes. However, you can simulate such a switch by using two switches as in Figure 14– 35. Begin by creating the circuit on the screen using IDC for the current source. Now double click TOPEN of switch U2 and set it to 300ms, then double click the inductor symbol and set IC to 0A. Click the New Profile icon

FIGURE 14–35 Simulating the circuit of Example 14–10. Two switches are used to model the closing and opening of the switch of Figure 14–23.

Section 14.6

■

and name the file fig14-35. In the Simulation Settings box, select transient analysis then type in a value of 0.5 for TSTOP. Run the simulation, create a second Y-axis, then add the current trace I(L1). You should now have the curves of Figure 14–36 on the screen. (Compare to Figure 14–27.) Using the cursor, read values at t 100 ms and 350 ms. You should get approximately 27 V and 1.73 mA at t 100 ms and 69 V and 490 mA at t 350 ms. Note how well these agree with the results of Example 14–10.

FIGURE 14–36

Inductor voltage and current for the circuit of Figure 14–35.

PUTTING IT INTO PRACTICE

he first sample of a new product that your company has designed has an indicator light that fails. (Symptom: When you turn a new unit on, the indicator light comes on as it should. However, when you turn the power off and back on, the lamp does not come on again.) You have been asked to investigate the problem and design a fix. You acquire a copy of the schematic and study the portion of the circuit where the indicator lamp is located. As shown in the accompanying figure, the lamp is used to indicate the status of the coil; the light is to be on when the coil is energized and off when it is not. Immediately, you see the problem, solder in one component and the problem is fixed. Write a short note to your supervisor outlining the nature of the problem, explaining why the lamp burned out and why your design modification fixed the problem. Note also that your modification did not result in any substantial increase in power consumption (i.e., you did not use a resistor). Note: This problem requires a diode. If you have not had an introduction to electronics, you will need to obtain a basic electronics book and read about it.

30 ⍀ 18 V 4H

RL Transients Using Computers

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PROBLEMS

14.1 Introduction 1. a. What does an inductor carrying no current look like at the instant of switching? b. For each circuit of Figure 14–37, determine iS and vL immediately after the switch is closed. iS

40 V

25 ⍀

(a) Purely resistive circuit

(b)

30 ⍀

iS ⫹ L vL ⫺

15 ⍀

90 V

⫹ L vL ⫺

10 ⍀

60 V

⫹ L vL ⫺

(d)

The value of L does not affect the solution.

2. Determine all voltages and currents in Figure 14–38 immediately after the switch is closed. + vR2 −

iT i1 10 ⍀

i2 40 ⍀ + v R1 −

E 180 V L1

+ v L1

60 ⍀

+ i4

v R4 −

i3 + v R3 −

16 ⍀ i5

14 ⍀

18 ⍀ + v R5 − L2

−

FIGURE 14–38

3. Repeat Problem 2 if L1 is replaced with an uncharged capacitor. 14.2 Current Buildup Transients 4. a. If iL 8(1 e 500t ) A, what is the current at t 6 ms? b. If vL 125e 500t V, what is the voltage vL at t 5 ms?

+ v R6 − + vL2 −

543

Problems 5. The switch of Figure 14–39 is closed at t 0 s. a. What is the time constant of the circuit? b. How long is it until current reaches its steady value? c. Determine the equations for iL and vL. d. Compute values for iL and vL at intervals of one time constant from t 0 to 5 t. e. Sketch iL and vL. Label the axis in t and in seconds. 6. Close the switch at t 0 s and determine equations for iL and vL for the circuit of Figure 14–40. Compute iL and vL at t 1.8 ms. 7. Repeat Problem 5 for the circuit of Figure 14–41 with L 4 H. R1 4⍀

20 V

16 ⍀

t = 0s

⫹ vL ⫺

R = 60 ⍀

180 V

⫹ vL ⫺

L=3H FIGURE 14–39 390 ⍀ iL ⫹ vL 600 mH ⫺

18 V

t=0s FIGURE 14–40

FIGURE 14–41

8. For the circuit of Figure 14–39, determine inductor voltage and current at t 50 ms using the universal time constant curve of Figure 14–10. 9. Close the switch at t 0 s and determine equations for iL and vL for the circuit of Figure 14–42. Compute iL and vL at t 3.4 ms. 10. Using Figure 14–10, find vL at one time constant for the circuit of Figure 14–42. 11. For the circuit of Figure 14–1(b), the voltage across the inductance at the instant the switch is closed is 80 V, the final steady state current is 4 A, and the transient lasts 0.5 s. Determine E, R, and L. 12. For an RL circuit, iL 20(1 e t/t) mA and vL 40e t/t V. If the transient lasts 0.625 ms, what are E, R, and L? 13. For Figure 14–1(b), if vL 40e 2000t V and the steady state current is 10 mA, what are E, R, and L? 14.4 De-energizing Transients 14. For Figure 14–43, E 80 V, R1 200 ⍀, R2 300 ⍀, and L 0.5 H. a. When the switch is closed, how long does it take for iL to reach steady state? b. When the switch is opened, how long does it take for iL to reach steady state? c. After the circuit has reached steady state with the switch closed, it is opened. Determine equations for iL and vL. 15. For Figure 14–43, R1 20 ⍀, R2 230 ⍀, and L 0.5 H, and the inductor current has reached a steady value of 5 A with the switch closed. At t 0 s, the switch is opened. a. What is the decay time constant?

220 ⍀ t=0s 40 V ⫹ vL ⫺ iL

560 mH FIGURE 14–42

⫹ vL ⫺

L FIGURE 14–43

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■

16. 17. 18.

iL (A)

19. 20. t (s) 5s

b. Determine equations for iL and vL. c. Compute values for iL and vL at intervals of one time constant from t 0 to 5 t. d. Sketch iL and vL. Label the axis in t and in seconds. Using the values from Problem 15, determine inductor voltage and current at t 3t using the universal time constant curves shown in Figure 14–10. Given vL 2700 Ve 100t. Using the universal time constant curve, find vL at t 20 ms. For Figure 14–43, the inductor voltage at the instant the switch is closed is 150 V and iL 0 A. After the circuit has reached steady state, the switch is opened. At the instant the switch is opened, iL 3 A and vL jumps to 750 V. The decay transient lasts 5 ms. Determine E, R1, R2, and L. For Figure 14–43, L 20 H. The current during buildup and decay is shown in Figure 14–44. Determine R1 and R2. For Figure 14–43, when the switch is moved to energization, iL 2 A (1 e 10t). Now open the switch after the circuit has reached steady state and redefine t 0 s as the instant the switch is opened. For this case, vL 400 Ve 25t. Determine E, R1, R2, and L.

14.5 More Complex Circuits 21. For the coil of Figure 14–45 R ᐉ 1.7 ⍀ and L 150 mH. Determine coil current at t 18.4 ms.

FIGURE 14–44

5.6 ⍀ 3.9 ⍀

iL Rl

4.7 ⍀ 67 V

⫹ vL L ⫺ Coil

t=0s

FIGURE 14–45 10.5 k⍀

10 k⍀ ⫹ vL ⫺

30 k⍀

120 V

L = 0.36 H FIGURE 14–46

22. Refer to Figure 14–46: a. What is the energizing circuit time constant? b. Close the switch and determine the equation for iL and vL during current buildup. c. What is the voltage across the inductor and the current through it at t 20 ms? 23. For Figure 14–46, the circuit has reached steady state with the switch closed. Now open the switch. a. Determine the de-energizing circuit time constant. b. Determine the equations for iL and vL. c. Find the voltage across the inductor and current through it at t 17.8 ms using the equations determined above.

545

Problems 24. Repeat Part (c) of Problem 23 using the universal time constant curves shown in Figure 14–10. 25. a. Repeat Problem 22, Parts (a) and (b) for the circuit of Figure 14–47. b. What are iL and vL at t 25 ms? 0.5 A

280 ⍀

40 V

200 ⍀

⫹ ⫺

300 ⍀

4H FIGURE 14–47

25 V 2

15 ⍀

50 ⍀ iL

3 30 ⍀

⫹ 5 H vL ⫺

FIGURE 14–49

14.6 RL Transients Using Computers 29. EWB or PSpice The switch of Figure 14–46 is closed at t 0 and remains closed. Graph the voltage across L and find vL at 20 ms using the cursor. 30. EWB or PSpice For the circuit of Figure 14–47, close the switch at t 0 and find vL at t 10 ms. (For PSpice, use current source IDC.)

100 ⍀ Unknown circuit

26. Repeat Problem 23 for the circuit of Figure 14–47, except find vL and iL at t 13.8 ms. 27. An unknown circuit containing dc sources and resistors has an open-circuit voltage of 45 volts. When its output terminals are shorted, the short-circuit current is 0.15 A. A switch, resistor, and inductance are connected (Figure 14–48). Determine the inductor current and voltage 2.5 ms after the switch is closed. 28. The circuit of Figure 14–49 is in steady state with the switch in position 1. At t 0, it is moved to position 2, where it remains for 1.0 s. It is then moved to position 3, where it remains. Sketch curves for iL and vL from t 0 until the circuit reaches steady state in position 3. Compute the inductor voltage and current at t 0.1 s and at t 1.1 s.

L = 0.4 H FIGURE 14–48

⫹ vL ⫺

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Inductive Transients 31. EWB or PSpice For Figure 14–6, let L1 30 mH and L2 90 mH. Close the switch at t 0 and find the current in the 30 ⍀ resistor at t 2 ms. (Answer: 4.61 A) [Hint for Workbench users: Redraw the circuit with L1 and the 30 ⍀ resistor interchanged. Finally, use Ohm’s law.] 32. EWB or PSpice For Figure 14–41, let L 4 H. Solve for vL and, using the cursor, measure values at t 200 ms and 500 ms. For PSpice users, also find current at these times. 33. PSpice We solved the circuit of Figure 14–21(a) by reducing it to its Thévenin equivalent. Using PSpice, analyze the circuit in its original form and plot the inductor current. Check a few points on the curve by computing values according to the solution of Example 14–8 and compare to values obtained from screen. 34. PSpice The circuit of Figure 14–46 is in steady state with the switch open. At t 0, the switch is closed. It remains closed for 150 ms and is then opened and left open. Compute and plot iL and vL. With the cursor, determine values at t 60 ms and at t 165 ms.

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 1. a. 20e 100 000t V; 2(1 e 100 000t ) mA b. t(␮s)

vL (V)

iL (mA)

0 10 20 30 40 50

20 7.36 2.71 0.996 0.366 0.135

0 1.26 1.73 1.90 1.96 1.99

c. 20 V

iss = 2 mA i vL

5␶ = 50 µs

2. 11.5 V; 3. 4 V 4. 3.88 A

1.47 A

vss = 0 V t

AC Fundamentals OBJECTIVES After studying this chapter, you will be able to • explain how ac voltages and currents differ from dc, • draw waveforms for ac voltage and currents and explain what they mean, • explain the voltage polarity and current direction conventions used for ac, • describe the basic ac generator and explain how ac voltage is generated, • define and compute frequency, period, amplitude, and peak-to-peak values, • compute instantaneous sinusoidal voltage or current at any instant in time, • define the relationships between q, T, and f for a sine wave, • define and compute phase differences between waveforms, • use phasors to represent sinusoidal voltages and currents, • determine phase relationships between waveforms using phasors, • define and compute average values for time-varying waveforms, • define and compute effective values for time-varying waveforms, • use Electronics Workbench and PSpice to study ac waveforms.

KEY TERMS ac Alternating Voltage Alternating Current

Amplitude Angular Velocity Average Value Cycle Effective Value Frequency Hertz Instantaneous Value Oscilloscope Peak Value Period Phase Shifts Phasor RMS Sine Wave

OUTLINE Introduction Generating AC Voltages Voltage and Current Conventions for AC Frequency, Period, Amplitude, and Peak Value Angular and Graphic Relationships for Sine Waves Voltage and Currents as Functions of Time Introduction to Phasors AC Waveforms and Average Value Effective Values Rate of Change of a Sine Wave AC Voltage and Current Measurement Circuit Analysis Using Computers

lternating currents (ac) are currents that alternate in direction (usually many times per second), passing first in one direction, then in the other through a circuit. Such currents are produced by voltage sources whose polarities alternate between positive and negative (rather than being fixed as with dc sources). By convention, alternating currents are called ac currents and alternating voltages are called ac voltages. The variation of an ac voltage or current versus time is called its waveform. Since waveforms vary with time, they are designated by lowercase letters v(t), i(t), e(t), and so on, rather than by uppercase letters V, I, and E as for dc. Often we drop the functional notation and simply use v, i, and e. While many waveforms are important to us, the most fundamental is the sine wave (also called sinusoidal ac). In fact, the sine wave is of such importance that many people associate the term ac with sinusoidal, even though ac refers to any quantity that alternates with time. In this chapter, we look at basic ac principles, including the generation of ac voltages and ways to represent and manipulate ac quantities. These ideas are then used throughout the remainder of the book to develop methods of analysis for ac circuits.

Thomas Alva Edison NOWADAYS WE TAKE IT FOR GRANTED that our electrical power systems are ac. (This is driven home every time you see a piece of equipment rated “60 hertz ac”, for example.) However, this was not always the case. In the late 1800s, a fierce battle—the so-called “war of the currents”—raged in the emerging electrical power industry. The forces favoring the use of dc were led by Thomas Alva Edison, and those favoring the use of ac were led by George Westinghouse (Chapter 24) and Nikola Tesla (Chapter 23). Edison, a prolific inventor who gave us the electric light, the phonograph, and many other great inventions as well, fought vigorously for dc. He had spent a considerable amount of time and money on the development of dc power and had a lot at stake, in terms of both money and prestige. So unscrupulous was Edison in this battle that he first persuaded the state of New York to adopt ac for its newly devised electric chair, and then pointed at it with horror as an example of how deadly ac was. Ultimately, however, the combination of ac’s advantages over dc and the stout opposition of Tesla and Westinghouse won the day for ac. Edison was born in 1847 in Milan, Ohio. Most of his work was done at two sites in New Jersey—first at a laboratory in Menlo Park, and later at a much larger laboratory in West Orange, where his staff at one time numbered around 5,000. He received patents as inventor or co-inventor on an astonishing 1,093 inventions, making him probably the greatest inventor of all time. Thomas Edison died at the age of 84 on October 18, 1831.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

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15.1

Introduction

Previously you learned that dc sources have fixed polarities and constant magnitudes and thus produce currents with constant value and unchanging direction, as illustrated in Figure 15–1. In contrast, the voltages of ac sources alternate in polarity and vary in magnitude and thus produce currents that vary in magnitude and alternate in direction.

12 V 6

12 V

12 V 2A

(b) Voltage and current versus time for dc

(a) FIGURE 15–1

Voltage or Current

I=2A

In a dc circuit, voltage polarities and current directions do not change.

Sinusoidal AC Voltage To illustrate, consider the voltage at the wall outlet in your home. Called a sine wave or sinusoidal ac waveform (for reasons discussed in Section 15.5), this voltage has the shape shown in Figure 15–2. Starting at zero, the voltage increases to a positive maximum, decreases to zero, changes polarity, increases to a negative maximum, then returns again to zero. One complete variation is referred to as a cycle. Since the waveform repeats itself at regular intervals as in (b), it is called a periodic waveform. Voltage is positive Voltage is negative

t Polarity change

1 cycle (a) Variation of voltage versus time

e(t)

FIGURE 15–3 Symbol for a sinusoidal voltage source. Lowercase letter e is used to indicate that the voltage varies with time.

Voltage

t (b) A continuous stream of cycles

FIGURE 15–2 Sinusoidal ac waveforms. Values above the axis are positive while values below are negative.

Symbol for an AC Voltage Source The symbol for a sinusoidal voltage source is shown in Figure 15–3. Note that a lowercase e is used to represent voltage rather than E, since it is a function of time. Polarity marks are also shown although, since the polarity of the source varies, their meaning has yet to be established.

Section 15.2

■

551

Generating AC Voltages

Sinusoidal AC Current Figure 15–4 shows a resistor connected to an ac source. During the first halfcycle, the source voltage is positive; therefore, the current is in the clockwise direction. During the second half-cycle, the voltage polarity reverses; therefore, the current is in the counterclockwise direction. Since current is proportional to voltage, its shape is also sinusoidal (Figure 15–5).

Actual current direction during first half-cycle Actual source polarity during first half-cycle

(a) FIGURE 15–4

15.2

Actual current direction during second half-cycle Actual source polarity during second half-cycle

Voltage Current t

(b)

Current direction reverses when the source polarity reverses.

Generating AC Voltages

One way to generate an ac voltage is to rotate a coil of wire at constant angular velocity in a fixed magnetic field, Figure 15–6. (Slip rings and brushes connect the coil to the load.) The magnitude of the resulting voltage is proportional to the rate at which flux lines are cut (Faraday’s law, Chapter 13), and its polarity is dependent on the direction the coil sides move through the field. Since the rate of cutting flux varies with time, the resulting voltage will also vary with time. For example in (a), since the coil sides are moving parallel to the field, no flux lines are being cut and the induced voltage at this instant (and hence the current) is zero. (This is defined as the 0° position of the coil.) As the coil rotates from the 0° position, coil sides AA⬘ and BB⬘ cut across flux lines; hence, voltage builds, reaching a peak when flux is cut at the maximum rate in the 90° position as in (b). Note the polarity of the voltage and the direction of current. As the coil rotates further, voltage decreases, reaching zero at the 180° position when the coil sides again move parallel to the field as in (c). At this point, the coil has gone through a half-revolution. During the second half-revolution, coil sides cut flux in directions opposite to that which they did in the first half revolution; hence, the polarity of the induced voltage reverses. As indicated in (d), voltage reaches a peak at the 270° point, and, since the polarity of the voltage has changed, so has the direction of current. When the coil reaches the 360° position, voltage is again zero and the cycle starts over. Figure 15–7 shows one cycle of the resulting waveform. Since the coil rotates continuously, the voltage produced will be a repetitive, periodic waveform as you saw in Figure 15–2(b).

FIGURE 15–5 Current has the same wave shape as voltage.

Chapter 15

■

AC Fundamentals Rotation A'

A A

B' B

Brushes

B 0V

Load

(a) 0° Position: Coil sides move parallel to flux lines. Since no flux is being cut, induced voltage is zero.

(b) 90° Position: Coil end A is positive with respect to B. Current direction is out of slip ring A.

B B

S B

0V (c) 180° Position: Coil again cutting no flux. Induced voltage is zero.

(d) 270° Position: Voltage polarity has reversed, therefore, current direction reverses.

FIGURE 15–6 Generating an ac voltage. The 0° position of the coil is defined as in (a) where the coil sides move parallel to the flux lines.

Generator Voltage

552

+ 270° 0

90°

180°

360° Coil Position

FIGURE 15–7 Coil voltage versus angular position.

PRACTICAL NOTES... In practice, the coil of Figure 15–6 consists of many turns wound on an iron core. The coil, core, and slip rings rotate as a unit. In Figure 15–6, the magnetic field is fixed and the coil rotates. While small generators are built this way, large ac generators usually have the oppo-

Section 15.2

■

553

Generating AC Voltages

site construction, that is, their coils are fixed and the magnetic field is rotated instead. In addition, large ac generators are usually made as three-phase machines with three sets of coils instead of one. This is covered in Chapter 23. However, although its details are oversimplified, the generator of Figure 15–6 gives a true picture of the voltage produced by a real ac generator.

Instantaneous Value As Figure 15–8 shows, the coil voltage changes from instant to instant. The value of voltage at any point on the waveform is referred to as its instantaneous value. This is illustrated in Figure 15–9. Figure 15–9(a) shows a photograph of an actual waveform, and (b) shows it redrawn, with values scaled from the photo. For this example, the voltage has a peak value of 40 volts and a cycle time of 6 ms. From the graph, we see that at t 0 ms, the voltage is zero. At t 0.5 ms, it is 20 V. At t 2 ms, it is 35 V. At t 3.5 ms, it is 20 V, and so on.

e(t) Generator Voltage

Time Scales The horizontal axis of Figure 15–7 is scaled in degrees. Often we need it scaled in time. The length of time required to generate one cycle depends on the velocity of rotation. To illustrate, assume that the coil rotates at 600 rpm (revolutions per minute). Six hundred revolutions in one minute equals 600 rev/60 s 10 revolutions in one second. At ten revolutions per second, the time for one revolution is one tenth of a second, i.e., 100 ms. Since one cycle is 100 ms, a half-cycle is 50 ms, a quarter-cycle is 25 ms, and so on. Figure 15–8 shows the waveform rescaled in time.

Half-cycle point 0

t (ms) 25 50

100

Cycle

FIGURE 15–8 Cycle scaled in time. At 600 rpm, the cycle length is 100 ms.

e (V) 40 30 20 10 0 10 20 30 40 (a) Sinusoidal voltage FIGURE 15–9 Instantaneous values.

35 V 20 V 1

2 3

20 V

t (ms)

28 V

(b) Values scaled from the photograph

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AC Fundamentals

Electronic Signal Generators AC waveforms may also be created electronically using signal generators. In fact, with signal generators, you are not limited to sinusoidal ac. The general-purpose lab signal generator of Figure 15–10, for example, can produce a variety of variable-frequency waveforms, including sinusoidal, square wave, triangular, and so on. Waveforms such as these are commonly used to test electronic gear.

(a) A typical signal generator Sine wave t Square wave t Triangle wave t (b) Sample waveforms FIGURE 15–10 Electronic signal generators produce waveforms of different shapes.

15.3

Voltage and Current Conventions for AC

In Section 15.1, we looked briefly at voltage polarities and current directions. At that time, we used separate diagrams for each half-cycle (Figure 15–4). However, this is unnecessary; one diagram and one set of references is all that is required. This is illustrated in Figure 15–11. First, we assign reference polarities for the source and a reference direction for the current. We then use the convention that, when e has a positive value, its actual polarity is the same as the reference polarity, and when e has a negative value, its actual polarity is opposite to that of the reference. For current, we use the convention that when i has a positive value, its actual direction is the same as the reference arrow, and when i has a negative value, its actual direction is opposite to that of the reference.

Section 15.3

■

Voltage and Current Conventions for AC

i e

e i t

(a) References for voltage and current.

(b) During the first half-cycle, voltage polarity and current direction are as shown in (a). Therefore, e and i are positive. During the second half-cycle, voltage polarity and current direction are opposite to that shown in (a). Therefore, e and i are negative.

FIGURE 15–11 AC voltage and current reference conventions.

To illustrate, consider Figure 15–12. At time t1, e has a value of 10 volts. This means that at this instant, the voltage of the source is 10 V and its top end is positive with respect to its bottom end. This is indicated in (b). With a voltage of 10 V and a resistance of 5 , the instantaneous value of current is i e/R 10 V/5 2 A. Since i is positive, the current is in the direction of the reference arrow. ( ) 10 V e i t1 2 A

i = 2 A

i=2A 2A t2

10 V

( ) (a)

(b) Time t1: e = 10 V and i = 2 A. Thus voltage and current have the polarity and direction indicated

(c) Time t2: e = 10 V and i = 2 A. Thus, voltage polarity is opposite to that indicated and current direction is opposite to the arrow direction.

FIGURE 15–12 Illustrating the ac voltage and current convention.

Now consider time t2. Here, e 10 V. This means that source voltage is again 10 V, but now its top end is negative with respect to its bottom end. Again applying Ohm’s law, you get i e/R 10 V/5 2 A. Since i is negative, current is actually opposite in direction to the reference arrow. This is indicated in (c). The above concept is valid for any ac signal, regardless of waveshape.

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EXAMPLE 15–1 Figure 15–13(b) shows one cycle of a triangular voltage wave. Determine the current and its direction at t 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 ms and sketch. e (V) i e

90 60 30 0 30 60 90

20 k

1 2 3 4 5 6 7 8 9 10 11 12 t ( s)

(b) Voltage

(a) i (ma) 4.5

7 4.5

9 10 11 12

t (µs)

Direction (c) Current

FIGURE 15–13

Solution Apply Ohm’s law at each point in time. At t 0 ms, e 0 V, so i e/R 0 V/20 k 0 mA. At t 1 ms, e 30 V. Thus, i e/R 30 V/20 k 1.5 mA. At t 2 ms, e 60 V. Thus, i e/R 60 V/20 k 3 mA. Continuing in this manner, you get the values shown in Table 15–1. The waveform is plotted as Figure 15–13(c). TABLE 15–1

Values for Example 15–1

t (␮s)

e (V)

i (mA)

0 1 2 3 4 5 6 7 8 9 10 11 12

0 30 60 90 60 30 0 30 60 90 60 30 0

0 1.5 3.0 4.5 3.0 1.5 0 1.5 3.0 4.5 3.0 1.5 0

Section 15.4

■

Frequency, Period, Amplitude, and Peak Value

PRACTICE PROBLEMS 1

1. Let the source voltage of Figure 15–11 be the waveform of Figure 15–9. If R 2.5 k , determine the current at t 0, 0.5, 1, 1.5, 3, 4.5, and 5.25 ms. 2. For Figure 15–13, if R 180 , determine the current at t 1.5, 3, 7.5, and 9 ms. Answers: 1. 0, 8, 14, 16, 0, 16, 11.2 (all mA) 2. 0.25, 0.5, 0.25, 0.5 (all A)

15.4

Frequency, Period, Amplitude, and Peak Value

Periodic waveforms (i.e., waveforms that repeat at regular intervals), regardless of their waveshape, may be described by a group of attributes such as frequency, period, amplitude, peak value, and so on.

Frequency The number of cycles per second of a waveform is defined as its frequency. In Figure 15–14(a), one cycle occurs in one second; thus its frequency is one cycle per second. Similarly, the frequency of (b) is two cycles per second and that of (c) is 60 cycles per second. Frequency is denoted by the lowercase letter f. In the SI system, its unit is the hertz (Hz, named in honor of pioneer researcher Heinrich Hertz, 1857–1894). By definition, 1 Hz 1 cycle per second

(15–1)

Thus, the examples depicted in Figure 15–14 represent 1 Hz, 2 Hz, and 60 Hz respectively. Cycle

Cycle

Cycle •• •• ••

t 1 second (a) 1 cycle per second = 1 Hz

60 Cycles

1 second

1 second (b) 2 cycles per second = 2 Hz

FIGURE 15–14 Frequency is measured in hertz (Hz).

The range of frequencies is immense. Power line frequencies, for example, are 60 Hz in North America and 50 Hz in many other parts of the world. Audible sound frequencies range from about 20 Hz to about 20 kHz. The standard AM radio band occupies from 550 kHz to 1.6 MHz, while the FM band extends from 88 MHz to 108 MHz. TV transmissions occupy several bands in the 54-MHz to 890-MHz range. Above 300 GHz are optical and X-ray frequencies.

Period The period, T, of a waveform, (Figure 15–15) is the duration of one cycle. It is the inverse of frequency. To illustrate, consider again Figure 15–14. In (a), the frequency is 1 cycle per second; thus, the duration of each cycle is T 1 s.

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In (b), the frequency is two cycles per second; thus, the duration of each cycle is T 1⁄ 2 s, and so on. In general, 1 T f

Period, T FIGURE 15–15 Period T is the duration of one cycle, measured in seconds.

(s)

(15–2)

1 f (Hz) T

(15–3)

and

Note that these definitions are independent of wave shape.

EXAMPLE 15–2 a. What is the period of a 50-Hz voltage? b. What is the period of a 1-MHz current? Solution 1 1 (a) T 20 ms f 50 Hz 1 1 (b) T 1 ms f 1 106 Hz

EXAMPLE 15–3

Figure 15–16 shows an oscilloscope trace of a square wave. Each horizontal division represents 50 ms. Determine the frequency.

FIGURE 15–16 The concepts of frequency and period apply to nonsinusoidal waveforms.

Section 15.4

■

559

Frequency, Period, Amplitude, and Peak Value

Solution Since the wave repeats itself every 200 ms, its period is 200 ms and 1 f 5 kHz 200 10 6 s

The period of a waveform can be measured between any two corresponding points (Figure 15–17). Often it is measured between zero points because they are easy to establish on an oscilloscope trace.

EXAMPLE 15–4

T (Between peaks) t T T (Between zero (Any two points) identical points)

Determine the period and frequency of the waveform of

Figure 15–18. FIGURE 15–18

FIGURE 15–17 Period may be measured between any two corresponding points.

T2 = 10 ms t T1 = 8 ms

Solution Time interval T1 does not represent a period as it is not measured between corresponding points. Interval T2, however, is. Thus, T 10 ms and 1 1 f 100 Hz T 10 10 3 s

v Em

Amplitude and Peak-to-Peak Value The amplitude of a sine wave is the distance from its average to its peak. Thus, the amplitude of the voltage in Figures 15–19(a) and (b) is Em. Peak-to-peak voltage is also indicated in Figure 15–19(a). It is measured between minimum and maximum peaks. Peak-to-peak voltages are denoted Ep-p or Vp-p in this book. (Some authors use Vpk-pk or the like.) Similarly, peak-to-peak currents are denoted as Ip-p. To illustrate, consider again Figure 15–9. The amplitude of this voltage is Em 40 V, and its peak-topeak voltage is Ep-p 80 V.

(a) v E

Peak Value The peak value of a voltage or current is its maximum value with respect to zero. Consider Figure 15–19(b). Here, a sine wave rides on top of a dc value, yielding a peak that is the sum of the dc voltage and the ac waveform amplitude. For the case indicated, the peak voltage is E Em.

t Ep–p

Em Em

E Em E Em

(b) FIGURE 15–19 Definitions.

1. What is the period of the commercial ac power system voltage in North America? 2. If you double the rotational speed of an ac generator, what happens to the frequency and period of the waveform?

IN-PROCESS

LEARNING CHECK 1

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AC Fundamentals 3. If the generator of Figure 15–6 rotates at 3000 rpm, what is the period and frequency of the resulting voltage? Sketch four cycles and scale the horizontal axis in units of time. 4. For the waveform of Figure 15–9, list all values of time at which e 20 V and e 35 V. Hint: Sine waves are symmetrical. 5. Which of the waveform pairs of Figure 15–20 are valid combinations? Why? i e

(a) Circuit

(d)

(c)

(b)

FIGURE 15–20 Which waveform pairs are valid?

6. For the waveform in Figure 15–21, determine the frequency. v 3 0 2

1 2 3 4 5 6 7 8 9 10 1112 13

t (ms)

FIGURE 15–21

7. Two waveforms have periods of T1 10 ms and T2 30 ms respectively. Which has the higher frequency? Compute the frequencies of both waveforms. 8. Two sources have frequencies f1 and f2 respectively. If f2 20f1, and T2 is 1 ms, what is f1? What is f2? 9. Consider Figure 15–22. What is the frequency of the waveform? FIGURE 15–22

i t 100 ms

10. For Figure 15–11, if f 20 Hz, what is the current direction at t 12 ms, 37 ms, and 60 ms? Hint: Sketch the waveform and scale the horizontal axis in ms. The answers should be apparent.

Section 15.5

■

561

Angular and Graphic Relationships for Sine Waves

11. A 10-Hz sinusoidal current has a value of 5 amps at t 25 ms. What is its value at t 75 ms? See hint in Problem 10. (Answers are at the end of the chapter.)

15.5

Angular and Graphic Relationships for Sine Waves

Rotation

The Basic Sine Wave Equation Consider again the generator of Figure 15–6, reoriented and redrawn in end view as Figure 15–23. The voltage produced by this generator is e Emsin a (V)

(15–4)

where Em is the maximum coil voltage and a is the instantaneous angular position of the coil. (For a given generator and rotational velocity, Em is constant.) Note that a 0° represents the horizontal position of the coil and that one complete cycle corresponds to 360°. Equation 15–4 states that the voltage at any point on the sine wave may be found by multiplying Em times the sine of the angle at that point.

0°

(a) End view showing coil position e

EXAMPLE 15–5

If the amplitude of the waveform of Figure 15–23(b) is Em 100 V, determine the coil voltage at 30° and 330°.

Em 0

Solution At a 30°, e Em sin a 100 sin 30° 50 V. At 330°, e 100 sin 330° 50 V. These are shown on the graph of Figure 15–24.

e = Emsin α Rotation α

90° 180°

270°

360°

(b) Voltage waveform e (V) FIGURE 15–23 Coil voltage versus angular position.

100 50 V 0

330° 30°

50 V 100 FIGURE 15–24

Table 15–2 is a tabulation of voltage versus angle computed from e 100 sin a. Use your calculator to verify each value, then plot the result on graph paper. The resulting waveshape should look like Figure 15–24.

PRACTICE PROBLEMS 2

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AC Fundamentals TABLE 15–2 Data for Plotting e 100 sin a Angle ␣

Voltage e

0 30 60 90 120 150 180 210 240 270 300 330 360

0 50 86.6 100 86.6 50 0 50 86.6 100 86.6 50 0

Angular Velocity, ␻ The rate at which the generator coil rotates is called its angular velocity. If the coil rotates through an angle of 30° in one second, for example, its angular velocity is 30° per second. Angular velocity is denoted by the Greek letter q (omega). For the case cited, q 30°/s. (Normally angular velocity is expressed in radians per second instead of degrees per second. We will make this change shortly.) When you know the angular velocity of a coil and the length of time that it has rotated, you can compute the angle through which it has turned. For example, a coil rotating at 30°/s rotates through an angle of 30° in one second, 60° in two seconds, 90° in three seconds, and so on. In general, a qt

(15–5)

Expressions for t and q can now be found. They are a t q

(s)

a q t

(15–6) (15–7)

EXAMPLE 15–6 If the coil of Figure 15–23 rotates at q 300°/s, how long does it take to complete one revolution? Solution

One revolution is 360°. Thus, 360 degrees a t 1.2 s degrees q 300 s

Since this is one period, we should use the symbol T. Thus, T 1.2 s, as in Figure 15–25.

Section 15.5

■

Angular and Graphic Relationships for Sine Waves

563

0 FIGURE 15–25

T = 1.2 s

PRACTICE PROBLEMS 3

If the coil of Figure 15–23 rotates at 3600 rpm, determine its angular velocity, q, in degrees per second. Answer: 21 600 deg/s

Radian Measure In practice, q is usually expressed in radians per second, where radians and degrees are related by the identity 2p radians 360°

(15–8)

One radian therefore equals 360°/2p 57.296°. A full circle, as shown in Figure 15–26(a), can be designated as either 360° or 2p radians. Likewise, the cycle length of a sinusoid, shown in Figure 15–26(b), can be stated as either 360° or 2p radians; a half-cycle as 180° or p radians, and so on. To convert from degrees to radians, multiply by p/180, while to convert from radians to degrees, multiply by 180/p. (15–9)

180° adegrees aradians p

(15–10)

or 2 radi 0°

°36

p aradians adegrees 180°

s an

(a) 360° = 2 radians

Table 15–3 shows selected angles in both measures. TABLE 15–3 Selected Angles in Degrees and Radians Degrees

Radians

30 45 60 90 180 270 360

p/6 p/4 p/3 p/2 p 3p/2 2p

EXAMPLE 15–7 a. Convert 315° to radians. b. Convert 5p/4 radians to degrees.

90° 180° 270° 360° (degrees)

3 /2

2 (radians)

(b) Cycle length scaled in degrees and radians FIGURE 15–26 Radian measure.

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360° 0

90 180

270

360

α (°)

Solution a. aradians (p/180°)(315°) 5.5 rad b. adegrees (180°/p)(5p/4) 225°

(a) Degrees

Scientific calculators can perform these conversions directly. You will find this more convenient than using the above formulas.

2 rad 0

3 2

α (rad)

(b) Radians

T (s)

T 4

T 2

3T 4

t T

(c) Period FIGURE 15–27 Comparison of various horizontal scales. Cycle length may be scaled in degrees, radians or period. Each of these is independent of frequency.

Graphing Sine Waves A sinusoidal waveform can be graphed with its horizontal axis scaled in degrees, radians, or time. When scaled in degrees or radians, one cycle is always 360° or 2p radians (Figure 15–27); when scaled in time, it is frequency dependent, since the length of a cycle depends on the coil’s velocity of rotation. However, if scaled in terms of period T instead of in seconds, the waveform is also frequency independent, since one cycle is always T, as shown in Figure 15–27(c). When graphing a sine wave, you don’t actually need many points to get a good sketch: Values every 45° (one eighth of a cycle) are generally adequate. Table 15–4 shows corresponding values for sin a at this spacing. TABLE 15–4

Values for Rapid Sketching

␣ (deg)

␣ (rad)

t (T)

Value of sin ␣

0 45 90 135 180 225 270 315 360

0 p/4 p/2 3p/4 p 5p/4 3p/2 7p/4 2p

0 T/8 T/4 3T/8 T/2 5T/8 3T/4 7T/8 T

0.0 0.707 1.0 0.707 0.0 0.707 1.0 0.707 0.0

EXAMPLE 15–8 Sketch the waveform for a 25-kHz sinusoidal current that has an amplitude of 4 mA. Scale the axis in seconds. Solution The easiest approach is to use T 1/f, then scale the graph accordingly. For this waveform, T 1/25 kHz 40 ms. Thus, 1. Mark the end of the cycle as 40 ms, the half-cycle point as 20 ms, the quartercycle point as 10 ms, and so on (Figure 15–28). 2. The peak value (i.e., 4 mA) occurs at the quarter-cycle point, which is 10 ms on the waveform. Likewise, 4 mA occurs at 30 ms. Now sketch. 3. Values at other time points can be determined easily. For example, the value at 5 ms can be calculated by noting that 5 ms is one eighth of a cycle, or 45°. Thus, i 4 sin 45° mA 2.83 mA. Alternately, from Table 15–4,

Section 15.6

■

Voltages and Currents as Functions of Time

at T/8, i (4 mA)(0.707) 2.83 mA. As many points as you need can be computed and plotted in this manner. 4. Values at particular angles can also be located easily. For instance, if you want a value at 30°, the required value is i 4 sin 30° mA 2.0 mA. To locate this point, note that 30° is one twelfth of a cycle or T/12 (40 ms)/12 3.33 ms. The point is shown on Figure 15–28. FIGURE 15–28

i (mA) 4 2.83 0 2.83 4

15.6

i = 2 mA at t = 3.33 s — see text 5 10 15 20 25 30 35 40 3.33 s

t ( s)

T = 40 s

Voltages and Currents as Functions of Time

Relationship between ␻, T, and f Earlier you learned that one cycle of sine wave may be represented as either a 2p rads or t T s, Figure 15–27. Substituting these into a qt (Equation 15–5), you get 2p qT. Transposing yields qT 2p (rad)

(15–11)

2p q T

(15–12)

Thus, (rad/s)

Recall, f 1/T Hz. Substituting this into Equation 15–12 you get q 2pf (rad/s)

(15–13)

EXAMPLE 15–9 In some parts of the world, the power system frequency is 60 Hz; in other parts, it is 50 Hz. Determine q for each. Solution For 60 Hz, q 2pf 2p(60) 377 rad/s. For 50 Hz, q 2pf 2p(50) 314.2 rad/s.

1. If q 240 rad/s, what are T and f? How many cycles occur in 27 s? 2. If 56 000 cycles occur in 3.5 s, what is q? Answers: 1. 26.18 ms, 38.2 Hz, 1031 cycles 2. 100.5 103 rad/s

PRACTICE PROBLEMS 4

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Sinusoidal Voltages and Currents as Functions of Time Recall from Equation 15–4, e Em sin a, and from Equation 15–5, a qt. Combining these equations yields e Emsin qt

(15–14a)

v Vmsin qt

(15–14b)

i Imsin qt

(15–14c)

Similarly,

EXAMPLE 15–10 A 100-Hz sinusoidal voltage source has an amplitude of 150 volts. Write the equation for e as a function of time. Solution q 2pf 2p(100) 628 rad/s and Em 150 V. Thus, e Emsin qt 150 sin 628t V.

Equations 15–14 may be used to compute voltages or currents at any instant in time. Usually, q is in radians per second, and thus qt is in radians. You can work directly in radians or you can convert to degrees. For example, suppose you want to know the voltage at t 1.25 ms for e 150 sin 628t V. With your calculator in the RAD mode, e 150 sin(628)(1.25 10 3) 150 sin 0.785 rad 106 V.

Working in Rads.

0.785 rad 45°. Thus, e 150 sin 45° 106 V as

Working in Degree.

before.

EXAMPLE 15–11 For v 170 sin 2450t, determine v at t 3.65 ms and show the point on the v waveform. Solution q 2450 rad/s. Therefore qt (2450)(3.65 10 3) 8.943 rad 512.4°. Thus, v 170 sin 512.4° 78.8 V. Alternatively, v 170 sin 8.943 rad 78.8 V. The point is plotted on the waveform in Figure 15–29. FIGURE 15–29

v (V) 170 78.8 V 3.65

PRACTICE PROBLEMS 5

t (ms)

A sinusoidal current has a peak amplitude of 10 amps and a period of 120 ms. a. Determine its equation as a function of time using Equation 15–14c. b. Using this equation, compute a table of values at 10-ms intervals and plot one cycle of the waveform scaled in seconds.

Section 15.6

■

Voltages and Currents as Functions of Time

c. Sketch one cycle of the waveform using the procedure of Example 15–8. (Note how much less work this is.) Answers: a. i 10 sin 52.36t A c. Mark the end of the cycle as 120 ms, 1⁄ 2 cycle as 60 ms, 1⁄ 4 cycle as 30 ms, etc. Draw the sine wave so that it is zero at t 0, 10 A at 30 ms, 0 A at 60 ms, 10 A at 90 ms and ends at t 120 ms. (See Figure 15–30.)

Determining when a Particular Value Occurs Sometimes you need to know when a particular value of voltage or current occurs. Given v Vmsin a. Rewrite this as sin a v/Vm. Then, v a sin 1 Vm

(15–15)

Compute the angle a at which the desired value occurs using the inverse sine function of your calculator, then determine the time from t a/q

EXAMPLE 15–12

A sinusoidal current has an amplitude of 10 A and a period of 0.120 s. Determine the times at which a. i 5.0 A, b. i 5 A.

Solution a. Consider Figure 15–30. As you can see, there are two points on the waveform where i 5 A. Let these be denoted t1 and t2 respectively. First, determine q: 2p 2p q 52.36 rad/s T 0.120 s Let i 10 sin a A. Now, find the angle a1 at which i 5 A: i 5A a1 sin 1 sin 1 sin 10.5 30° 0.5236 rad Im 10 A Thus, t1 a1/q (0.5236 rad)/(52.36 rad/s) 0.01 s 10 ms. This is indicated in Figure 15–30. Now consider t2. Note that t2 is the same distance back from the half-cycle point as t1 is in from the beginning of the cycle. Thus, t2 60 ms 10 ms 50 ms. b. Similarly, t3 (the first point at which i 5 A occurs) is 10 ms past midpoint, while t4 is 10 ms back from the end of the cycle. Thus, t3 70 ms and t4 110 ms.

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i (A) 10 8 i=5A 6 i=5A 4 t t t4 1 2 2 0 t (ms) 2 10 30 50 70 90 110 4 t3 6 8 i = 5 A i = 5 A 10 FIGURE 15–30

PRACTICE PROBLEMS 6

Given v 10 sin 52.36t, determine both occurrences of v 8.66 V. Answer: 80 ms

100 ms

Voltages and Currents with Phase Shifts If a sine wave does not pass through zero at t 0 s as in Figure 15–30, it has a phase shift. Waveforms may be shifted to the left or to the right (see Figure 15–31). For a waveform shifted left as in (a), NOTES... When applying equations 15– 16(a) and (b), it is customary to express qt in radians and v in degrees, yielding mixed angular units (as indicated in the following examples). Although this is acceptable when the equations are written in symbolic form, you must convert both angles to the same unit before you make numerical computations.

v Vmsin(qt v)

(15–16a)

while, for a waveform shifted right as in (b), v Vmsin(qt v)

(15–16b)

ωt (rad)

ωt (rad) θ

(a) v = Vm sin(ωt θ)

(b) v = Vm sin(ωt θ)

FIGURE 15–31 Waveforms with phase shifts. Angle v is normally measured in degrees, yielding mixed angular units. (See note.)

Demonstrate that v 20 sin(qt 60°), where q p/6 rad/s (i.e, 30°/s), yields the shifted waveform shown in Figure 15– 32.

EXAMPLE 15–13

Solution 1. Since qt and 60° are both angles, (qt 60°) is also an angle. Let us define it as x. Then v 20 sin x, which means that the shifted wave is also sinusoidal. 2. Consider v sin(qt 60°). At t 0 s, v 20 sin(0 60°) 20 sin ( 60°) 17.3 V as indicated in Figure 15–32.

Section 15.6

■

Voltages and Currents as Functions of Time

3. Since q 30°/s, it takes 2 s for qt to reach 60°. Thus, at t 2 s, v 20 sin(60° 60°) 0 V, and the waveform passes through zero at t 2 s as indicated. v 20

v 20 sin (ωt 60°) 60°

0 t 2s 20

EWB

α ωt

20 sin ( 60°) 17.3V

FIGURE 15–32

Summary: Since v 20 sin(qt 60°) is a sine wave and since it passes through zero at t 2 s, where qt 60°, it represents the shifted wave shown in Figure 15–32.

EXAMPLE 15–14 a. Determine the equation for the waveform of Figure 15–33(a), given f 60 Hz. Compute current at t 4 ms. b. Repeat (a) for Figure 15–33(b). Solution a. Im 2 A and q 2p(60) 377 rad/s. This waveform corresponds to Figure 15–31(b). Therefore, i Im sin(qt v) 2 sin(377t 120°) A At t 4 ms, current is i 2 sin(377 4 ms 120°) 2 sin(1.508 rad 120°) 2 sin(86.4° 120°) 2 sin( 33.64°) 1.11 A. b. This waveform matches Figure 15–31(a) if you extend the waveform back 90° from its peak as in (c). Thus, i 2 sin(377t 40°) A At t 4 ms, current is i 2 sin(377 4 ms 40°) 2 sin(126.4°) 1.61 A.

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i (A)

i (A) 2 1 0 1 2

2 1 ωt 0 1 2

120°

ωt

90°

50°

(b)

(a)

50°

FIGURE 15–33

PRACTICE PROBLEMS 7

1. Given i 2 sin(377t 60°), compute the current at t 3 ms. 2. Sketch each of the following: a. v 10 sin(qt 20°) V. b. i 80 sin(qt 50°) A. c. i 50 sin(qt 90°) A. d. v 5 sin(qt 180°) V. 3. Given i 2 sin(377t 60°), determine at what time i 1.8 A. Answers: 1. 1.64 A 2. a. Same as Figure 15–31(a) with Vm 10 V, v 20°. b. Same as Figure 15–31(b) with Im 80 A, v 50°. c. Same as Figure 15–39(b) except use Im 50 A instead of Vm. d. A negative sine wave with magnitude of 5 V. 3. 0.193 ms

Probably the easiest way to deal with shifted waveforms is to use phasors. We introduce the idea next.

15.7

Introduction to Phasors

A phasor is a rotating line whose projection on a vertical axis can be used to represent sinusoidally varying quantities. To get at the idea, consider the red line of length Vm shown in Figure 15–34(a). (It is the phasor.) The vertical v

Amplitude is the same as the length of the phasor 2

Vm Vm

Vm sin α (a) Phasor

α = ωt

Vm (b) Resulting sine wave

FIGURE 15–34 As the phasor rotates about the origin, its vertical projection creates a sine wave. (Figure 15–35 illustrates the process.)

Section 15.7

■

Introduction to Phasors

571

projection of this line (indicated in dotted red) is Vm sin a. Now, assume that the phasor rotates at angular velocity of q rad/s in the counterclockwise direction. Then, a qt, and its vertical projection is Vm sin qt. If we designate this projection (height) as v, we get v Vmsin qt, which is the familiar sinusoidal voltage equation. If you plot a graph of v versus a, you get the sine wave of Figure 15– 34(b). Figure 15–35 illustrates the process. It shows snapshots of the phasor and the evolving waveform at various instants of time for a phasor of magnitude Vm 100 V rotating at q 30°/s. For example, consider t 0, 1, 2, and 3s: 1. At t 0 s, a 0, the phasor is at its 0° position, and its vertical projection is v Vmsin qt 100 sin 0° 0 V. The point is at the origin. 2. At t 1 s, the phasor has rotated 30° and its vertical projection is v 100 sin 30° 50 V. This point is plotted at a 30° on the horizontal axis. ω

t=2s t=1s

100 V

FIGURE 15–35 Evolution of the sine wave of Figure 15–34.

100 V

87 V

t=3s

50 V 0V

t=0s

100 V

60 90

α (°)

NOTES… ω

t=6s

87 V 50 V

100 V

t=4s t=5s

0V 0

100 V

90 120 150 180

210 240 270 100 V

100 V t=7s

180 50 V

t=8s t=9s ω

100 V

α (°)

100 V

87 V

0V 270 300 330 360

t = 12 s

α (°)

t = 11 s t = 10 s 0

50 V 87 V

1. Although we have indicated phasor rotation in Figure 15– 35 by a series of “snapshots,” this is too cumbersome; in practice, we show only the phasor at its t 0 s (reference) position and imply rotation rather than show it explicitly. 2. Although we are using maximum values (Em and Im ) here, phasors are normally drawn in terms of effective values (considered in Section 15.9). For the moment, we will continue to use maximum values. We make the change in Chapter 16.

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3. At t 2 s, a 60° and v 100 sin 60° 87 V, which is plotted at a 60° on the horizontal axis. Similarly, at t 3 s, a 90°, and v 100 V. Continuing in this manner, the complete waveform is evolved. From the foregoing, we conclude that a sinusoidal waveform can be created by plotting the vertical projection of a phasor that rotates in the counterclockwise direction at constant angular velocity q. If the phasor has a length of Vm , the waveform represents voltage; if the phasor has a length of Im , it represents current. Note carefully: Phasors apply only to sinusoidal waveforms.

EXAMPLE 15–15

for f 100 Hz.

Draw the phasor and waveform for current i 25 sin qt mA

Solution The phasor has a length of 25 mA and is drawn at its t 0 position, which is zero degrees as indicated in Figure 15–36. Since f 100 Hz, the period is T 1/f 10 ms. ω

i (mA)

25 Im = 25 mA

t (ms)

25 FIGURE 15–36 The reference position of the phasor is its t 0 position.

Shifted Sine Waves Phasors may be used to represent shifted waveforms, v Vm sin(qt v) or i Imsin(qt v) as indicated in Figure 15–37. Angle v is the position of the phasor at t 0 s. ω

Im sin θ Im

j θ

Im sin θ

2 θ

θ ωt

ωt

Im (a) i = Im sin(ωt θ)

Im sin( θ) (b) i = Im sin(ωt θ)

FIGURE 15–37 Phasors for shifted waveforms. Angle v is the position of the phasor at t 0 s.

Section 15.7

EXAMPLE 15–16 Consider v 20 sin(qt 60°), where q p/6 rad/s (i.e., 30°/s). Show that the phasor of Figure 15–38(a) represents this waveform.

v (V)

20 V

60°

t=2s ωt

0 20 V

60°

20 sin ( 60°) (b) v = 20 sin (ωt 60°), ω = 30°/s

(a) Phasor FIGURE 15–38

Solution The phasor has length 20 V and at time t 0 is at 60° as indicated in (a). Now, as the phasor rotates, it generates a sinusoidal waveform, oscillating between 20 V as indicated in (b). Note that the zero crossover point occurs at t 2 s, since it takes 2 seconds for the phasor to rotate from 60° to 0° at 30 degrees per second. Now compare the waveform of (b) to the waveform of Figure 15–32, Example 15–13. They are identical. Thus, the phasor of (a) represents the shifted waveform v 20 sin(qt 60°).

With the aid of a phasor, sketch the waveform for v

EXAMPLE 15–17

Vmsin(qt 90°).

Solution Place the phasor at 90° as in Figure 15–39(a). Note that the resultant waveform (b) is a cosine waveform, i.e., v Vmcos qt. From this, we conclude that sin(qt 90°) cos qt ω

v j

v = Vm sin (ωt 90°) Vm

ωt

90° (a) Phasor at 90° position

(b) Waveform can also be described as a cosine wave

FIGURE 15–39 Demonstrating that sin(qt 90°) cos qt.

■

Introduction to Phasors

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PRACTICE PROBLEMS 8

With the aid of phasors, show that a. sin(qt 90°) cos qt, b. sin(qt 180°) sin qt,

Phase Difference Phase difference refers to the angular displacement between different waveforms of the same frequency. Consider Figure 15–40. If the angular displacement is 0° as in (a), the waveforms are said to be in phase; otherwise, they are out of phase. When describing a phase difference, select one waveform as reference. Other waveforms then lead, lag, or are in phase with this reference. For example, in (b), for reasons to be discussed in the next paragraph, the current waveform is said to lead the voltage waveform, while in (c) the current waveform is said to lag.

v i

(a) In phase

(b) Current leads

FIGURE 15–40 Illustrating phase difference. In these examples, voltage is taken as reference.

The terms lead and lag can be understood in terms of phasors. If you observe phasors rotating as in Figure 15–41(a), the one that you see passing first is leading and the other is lagging. By definition, the waveform generated by the leading phasor leads the waveform generated by the lagging phasor and vice versa. In Figure 15–41, phasor Im leads phasor Vm; thus current i(t) leads voltage v(t).

NOTES… If you have trouble determining which waveform leads and which lags when you are solving a problem, make a quick sketch of their phasors, and the answer will be apparent. Note also that the terms lead and lag are relative. In Figure 15–41, we said that current leads voltage; you can just as correctly say that voltage lags current.

j v(t) i(t)

Im θ

(a) Im leads Vm FIGURE 15–41 Defining lead and lag.

Time θ

(b) Therefore, i(t) leads v(t)

Section 15.7

EXAMPLE 15–18

Voltage and current are out of phase by 40°, and voltage lags. Using current as the reference, sketch the phasor diagram and the corresponding waveforms. Solution Since current is the reference, place its phasor in the 0° position and the voltage phasor at 40°. Figure 15–42 shows the phasors and corresponding waveforms. ω

v(t) = Vm sin (ωt 40°)

i(t) = Im sin ωt Im 40°

ωt

40° Vm sin( 40°)

Vm FIGURE 15–42

EXAMPLE 15–19 Given v 20 sin(qt 30°) and i 18 sin(qt 40°), draw the phasor diagram, determine phase relationships, and sketch the waveforms. Solution The phasors are shown in Figure 15–43(a). From these, you can see that v leads i by 70°. The waveforms are shown in (b). j Vm = 20 V 30° 30° 40° 70° Im = 18 A (a)

v = 20 sin (ωt 30°) i = 18 sin (ωt 40°)

ωt

40° 70° (b)

FIGURE 15–43

EXAMPLE 15–20 Figure 15–44 shows a pair of waveforms v1 and v2 on an oscilloscope. Each major vertical division represents 20 V and each major division on the horizontal (time) scale represents 20 ms. Voltage v1 leads. Prepare a phasor diagram using v1 as reference. Determine equations for both voltages.

■

Introduction to Phasors

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60°

Vm1 = 60 V

Vm2 = 40 V (a) EWB

(b)

FIGURE 15–44

Solution From the photograph, the magnitude of v1 is Vm1 3 div 20 V/div 60 V. Similarly, Vm2 40 V. Cycle length is T 6 20 ms 120 ms, and the displacement between waveforms is 20 ms which is 1⁄ 6 of a cycle (i.e., 60°). Selecting v1 as reference and noting that v2 lags yields the phasors shown in (b). Angular frequency q 2p/T 2p/(120 10 6 s) 52.36 103 rad/s. Thus, v1 Vm1 sin qt 60 sin(52.36 103t) V and v2 40 sin(52.36 103t 60°) V.

Sometimes voltages and currents are expressed in terms of cos qt rather than sin qt. As Example 15–17 shows, a cosine wave is a sine wave shifted by 90°, or alternatively, a sine wave is a cosine wave shifted by 90°. For sines or cosines with an angle, the following formulas apply. cos(qt v) sin(qt v 90°)

(15–17a)

sin(qt v) cos(qt v 90°)

(15–17b)

To illustrate, consider cos(qt 30°). From Equation 15–17a, cos(qt 30°) sin(qt 30° 90°) sin(qt 120°). Figure 15–45 illustrates this relationship graphically. The red phasor in (a) generates cos qt as was shown

120°

30°

0 Generates cos (ωt 30°)

60°

ωt

Generates sin (ωt 120°) cos (ωt 30°) = sin (ωt 120°)

(a)

(b)

(c)

FIGURE 15–45 Using phasors to show that cos(qt 30°) sin(qt 120°).

Section 15.7

in Example 15–17. Therefore, the green phasor generates a waveform that leads it by 30°, namely cos(qt 30°). For (b), the red phasor generates sin qt, and the green phasor generates a waveform that leads it by 120°, i.e., sin(qt 120°). Since the green phasor is the same in both cases, you can see that cos(qt 30°) sin(qt 120°). Note that this process is easier than trying to remember equations 15–17(a) and (b).

Determine the phase angle between v 30 cos(qt 20°) and i 25 sin(qt 70°).

EXAMPLE 15–21

Solution i 25 sin(qt 70°) may be represented by a phasor at 70°, and v 30 cos(qt 20°) by a phasor at (90° 20°) 110°, Figure 15–46(a). Thus, v leads i by 40°. Waveforms are shown in (b). 20° 40° Phase displacement Im = 25 A

Vm = 30 V

ωt

70° 110°

40° 20°

110° 70°

i = 25 sin (ωt + 70°)

v = 30 cos (ωt + 20°) = 30 sin (ωt + 110°)

(a)

(b)

FIGURE 15–46

Sometimes you encounter negative waveforms such as i Im sin qt. To see how to handle these, refer back to Figure 15–36, which shows the waveform and phasor for i Im sin qt. If you multiply this waveform by 1, you get the inverted waveform Im sin qt of Figure 15–47(a) with corresponding phasor (b). Note that the phasor is the same as the original phasor except that it is rotated by 180°. This is always true—thus, if you multiply a waveform by 1, the phasor for the new waveform is 180° rotated from the original phasor, regardless of the angle of the original phasor. i

Im Im

t Im (a) i −Im sin ωt

(b) Phasor

FIGURE 15–47 The phasor for a negative sine wave is at 180°.

■

Introduction to Phasors

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AC Fundamentals Find the phase relationship between i 4 sin(qt 50°) and v 120 sin(qt 60°).

EXAMPLE 15–22

Solution i 4 sin(qt 50°) is represented by a phasor at (50° 180°) 130° and v 120 sin (qt 60°) by a phasor at 60°, Figure 15–48. The phase difference is 70° and voltage leads. Note also that i can be written as i 4 sin(qt 130°). j

50° 60°

130°

Im 4 A

70° Vm 120 V

FIGURE 15–48

The importance of phasors to ac circuit analysis cannot be overstated— you will find that they are one of your main tools for representing ideas and for solving problems in later chapters. We will leave them for the moment, but pick them up again in Chapter 16.

IN-PROCESS

LEARNING CHECK 2

1. If i 15 sin a mA, compute the current at a 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°, and 360°. 2. Convert the following angles to radians: a. 20° b. 50° c. 120° d. 250° 3. If a coil rotates at q p/60 radians per millisecond, how many degrees does it rotate through in 10 ms? In 40 ms? In 150 ms? 4. A current has an amplitude of 50 mA and q 0.2p rad/s. Sketch the waveform with the horizontal axis scaled in a. degrees b. radians c. seconds 5. If 2400 cycles of a waveform occur in 10 ms, what is q in radians per second? 6. A sinusoidal current has a period of 40 ms and an amplitude of 8 A. Write its equation in the form of i Imsin qt, with numerical values for Im and q. 7. A current i Imsin qt has a period of 90 ms. If i 3 A at t 7.5 ms, what is its equation? 8. Write equations for each of the waveforms in Figure 15–49 with the phase angle v expressed in degrees.

Section 15.8

■

AC Waveforms and Average Value

579

i (A) i = Im sin (ωt 30°)

i (A)

v (V)

ωt

125 A (a) f = 40 Hz

v (V)

ωt

4 (b) T = 100 ms

47 V ωt

6 (c) f = 100 Hz

FIGURE 15–49

9. Given i 10 sin qt, where f 50 Hz, find all occurrences of a. i 8 A between t 0 and t 40 ms b. i 5 A between t 0 and t 40 ms 10. Sketch the following waveforms with the horizontal axis scaled in degrees: a. v1 80 sin(qt 45°) V b. v2 40 sin(qt 80°) V c. i1 10 cos qt mA d. i2 5 cos(qt 20°) mA 11. Given q p/3 rad/s, determine when voltage first crosses through 0 for a. v1 80 sin(qt 45°) V b. v2 40 sin(qt 80°) V 12. Consider the voltages of Question 10: a. Sketch phasors for v1 and v2 b. What is the phase difference between v1 and v2? c. Determine which voltage leads and which lags. 13. Repeat Question 12 for the currents of Question 10. (Answers are at the end of the chapter.)

15.8

AC Waveforms and Average Value

While we can describe ac quantities in terms of frequency, period, instantaneous value, etc., we do not yet have any way to give a meaningful value to an ac current or voltage in the same sense that we can say of a car battery that it has a voltage of 12 volts. This is because ac quantities constantly change and thus there is no one single numerical value that truly represents a waveform over its complete cycle. For this reason, ac quantities are generally described by a group of characteristics, including instantaneous, peak, average, and effective values. The first two of these we have already seen. In this section, we look at average values; in Section 15.9, we consider effective values.

Average Values Many quantities are measured by their average, for instance, test and examination scores. To find the average of a set of marks for example, you add them, then divide by the number of items summed. For waveforms, the process is conceptually the same. For example, to find the average of a waveform, you can sum the instantaneous values over a full cycle, then

t 2 µs (d) f = 50 kHz

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divide by the number of points used. The trouble with this approach is that waveforms do not consist of discrete values.

Average in Terms of the Area Under a Curve An approach more suitable for use with waveforms is to find the area under the curve, then divide by the baseline of the curve. To get at the idea, we can use an analogy. Consider again the technique of computing the average for a set of numbers. Assume that you earn marks of 80, 60, 60, 95, and 75 on a group of tests. Your average mark is therefore average (80 60 60 95 75)/5 74

Test Marks

An alternate way to view these marks is graphically as in Figure 15–50. The area under this curve can be computed as area (80 1) (60 2) (95 1) (75 1)

Grades

100 80 60 40 20

Now divide this by the length of the base, namely 5. Thus, 1

2 3 4 Base = 5

FIGURE 15–50 Determining average by area.

(80 1) (60 2) (95 1) (75 1) 74 5

which is exactly the answer obtained above. That is, area under curve average length of base

(15–18)

This result is true in general. Thus, to find the average value of a waveform, divide the area under the waveform by the length of its base. Areas above the axis are counted as positive, while areas below the axis are counted as negative. This approach is valid regardless of waveshape. Average values are also called dc values, because dc meters indicate average values rather than instantaneous values. Thus, if you measure a nondc quantity with a dc meter, the meter will read the average of the waveform, i.e., the value calculated according to Equation 15–18.

EXAMPLE 15–23 a. Compute the average for the current waveform of Figure 15–51. b. If the negative portion of Figure 15–51 is 3 A instead of 1.5 A, what is the average? c. If the current is measured by a dc ammeter, what will the ammeter indicate? i (A) 2 1 0 1 2 FIGURE 15–51

1 2 3 4 5 6 7 8 1 cycle

t (ms)

Section 15.8

■

AC Waveforms and Average Value

Solution a. The waveform repeats itself after 7 ms. Thus, T 7 ms and the average is (2 A 3 ms) (1.5 A 4 ms) 6 6 Iavg 0 A 7 ms 7 (2 A 3 ms) (3 A 4 ms) 6 A b. Iavg 0.857 A 7 ms 7 c. A dc ammeter measuring (a) will indicate zero, while for (b) it will indicate 0.857 A.

EXAMPLE 15–24

Compute the average value for the waveforms of Figures 15–52(a) and (c). Sketch the averages for each. v (V) 30 20 10 0

1 2 3 4 5 6 7 8 1 cycle

t (s)

20 10 0

(a)

Vavg = 16.7 V

(b)

i (mA) 40 20 0 20

1 2 3 4 5 6 7 8 9

t (s)

0 10

40 (c)

Iavg = 7.5 mA (d)

FIGURE 15–52

Solution For the waveform of (a), T 6 s. Thus, (10V 2s) (20V 1s) (30V 2s) (0V 1s) 100 V-s Vavg 16.7V 6s 6s The average is shown as (b). A dc voltmeter would indicate 16.7 V. For the waveform of (c), T 8 s and 1 (40mA 3s) (20mA 2s) (40mA 2s) 2 60 Iavg mA 7.5mA 8s 8 In this case, a dc ammeter would indicate 7.5 mA.

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Determine the averages for Figures 15–53(a) and (b).

PRACTICE PROBLEMS 9

i (A)

v (V)

20 10 0

0 −20

t ( s)

1 2 3 45 6 7 8

Half circles

1 23 4 5 67 8 9

t (s)

(b)

(a) FIGURE 15–53 Answers: a. 1.43 A

i Im

Area = 2Im

FIGURE 15–54 Area under a halfcycle.

i Iavg = 0.637Im

Im 0

Base

FIGURE 15–55 Full-wave average.

Sine Wave Averages Because a sine wave is symmetrical, its area below the horizontal axis is the same as its area above the axis; thus, over a full cycle its net area is zero, independent of frequency and phase angle. Thus, the average of sin qt, sin(qt v), sin 2qt, cos qt, cos(qt v), cos 2qt, and so on are each zero. The average of half a sine wave, however, is not zero. Consider Figure 15–54. The area under the half-cycle may be found using calculus as area

冕 I sina da I cos a 冤 p

2Im

(15–19)

Similarly, the area under a half-cycle of voltage is 2Vm. (If you haven’t studied calculus, you can approximate this area using numerical methods as described later in this section.) Two cases are important; full-wave average and half-wave average. The full-wave case is illustrated in Figure 15–55. The area from 0 to 2p is 2(2Im) and the base is 2p. Thus, the average is 2(2Im ) 2Im Iavg 0.637Im 2p p

For the half-wave case (Figure 15–56),

i Im

b. 6.67 V

I 2Im Iavg m 0.318Im p 2p

Iavg = 0.318Im

Base

The corresponding expressions for voltage are Vavg 0.637Vm (full-wave) Vavg 0.318Vm (half-wave)

FIGURE 15–56 Half-wave average.

Numerical Methods If the area under a curve cannot be computed exactly, it can be approximated. One method is to approximate the curve by straight line segments as in Figure 15–57. (If the straight lines closely fit the curve, the accuracy is very good.) Each element of area is a trapezoid (b) whose area is its average

Section 15.8

■

AC Waveforms and Average Value

height times its base. Thus, A1 1⁄ 2(y0 y1) x, A2 1⁄ 2(y1 y2) x, etc. Summing areas and combining terms yields

冢

冣

y y area 0 y1 y2 … yk 1 k x 2 2

(15–20)

This result is known as the trapezoidal rule. Example 15–25 illustrates its use. f(x)

Actual curve

A1 A 2 y0 y1 y2 0

(a) Approximating the curve

(b) Element of area

FIGURE 15–57 Calculating area using the trapezoidal rule.

Approximate the area under y sin(qt 30°), Figure 15–58. Use an increment size of p/6 rad, i.e., 30°.

EXAMPLE 15–25

y y3 = 0.866 y2 = 0.5 y1 = 0 0

6 y0 = 0.5

y4 = 1.0

y5 = 0.866 y6 = 0.5 y7 = 0

FIGURE 15–58

Solution Points on the curve sin(qt 30°) have been computed by calculator and plotted as Figure 15–58. Substituting these values into Equation 15–20 yields

冢

冣冢冣

1 1 p area ( 0.5) 0 0.5 0.866 1.0 0.866 0.5 (0) 1.823 2 2 6

The exact area (found using calculus) is 1.866; thus, the above value is in error by 2.3%.

1. Repeat Example 15–25 using an increment size of p/12 rad. What is the percent error?

PRACTICE PROBLEMS 10

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AC Fundamentals 2. Approximate the area under v 50 sin(qt 30°) from qt 0° to qt 210°. Use an increment size of p/12 rad. Answers: 1. 1.855; 0.59% 2. 67.9 (exact 68.3; error 0.6%)

Superimposed AC and DC Sometimes ac and dc are used in the same circuit. For example, amplifiers are powered by dc but the signals they amplify are ac. Figure 15–59 shows a simple circuit with combined ac and dc. Figure 15–60(c) shows superimposed ac and dc. Since we know that the average of a sine wave is zero, the average value of the combined waveform will be its dc component, E. However, peak voltages depend on both components as illustrated in (c). Note for the case illustrated that although the waveform varies sinusoidally, it does not alternate in polarity since it never changes polarity to become negative.

v=E+e

FIGURE 15–59

v Em

v=e T

0 Em

v=E e

v=E

Time

T 2

E Em E Em

Time

(a) AC alone. E = 0 V. Vavg = 0 V

(b) DC alone. e = 0 V. Vavg = E

Time

FIGURE 15–60 Superimposed dc and ac.

EXAMPLE 15–26 Draw the voltage waveform for the circuit of Figure 15–61(a). Determine average, peak, and minimum voltages. e = 15 sin ωt

v=E+e

10 V

(a)

v (V) 15 V 30 20 10 0 10

25 V Vavg = 10 V ωt

15 V 5 V (b)

FIGURE 15–61 v 10 15 sin qt.

Solution The waveform consists of a 10-V dc value with 15 V ac riding on top of it. The average is the dc value, Vavg 10 V. The peak voltage is 10 15 25 V, while the minimum voltage is 10 15 5 V. This waveform alternates in polarity, although not symmetrically (as is the case when there is no dc component).

Section 15.9

Repeat Example 15–26 if the dc source of Figure 15–61 is E 5 V.

■

Effective Values

585

PRACTICE PROBLEMS 11

Answers: Vavg 5 V; positive peak 10 V; negative peak 20 V

15.9

Effective Values

While instantaneous, peak, and average values provide useful information about a waveform, none of them truly represents the ability of the waveform to do useful work. In this section, we look at a representation that does. It is called the waveform’s effective value. The concept of effective value is an important one; in practice, most ac voltages and currents are expressed as effective values. Effective values are also called rms values for reasons discussed shortly.

What Is an Effective Value? An effective value is an equivalent dc value: it tells you how many volts or amps of dc that a time-varying waveform is equal to in terms of its ability to produce average power. Effective values depend on the waveform. A familiar example of such a value is the value of the voltage at the wall outlet in your home. In North America its value is 120 Vac. This means that the sinusoidal voltage at the wall outlets of your home is capable of producing the same average power as 120 volts of steady dc. Effective Values for Sine Waves The effective value of a waveform can be determined using the circuits of Figure 15–62. Consider a sinusoidally varying current, i(t). By definition, the effective value of i is that value of dc current that produces the same average power. Consider (b). Let the dc source be adjusted until its average power is the same as the average power in (a). The resulting dc current is then the effective value of the current of (a). To determine this value, determine the average power for both cases, then equate them. p(t) p(t) e

Im2R Pavg =

i R

Im2R 2

I E

P R t

i(t) p(t) = i2R. Therefore, p(t) varies cyclically.

P = I 2R. Therefore, P is constant.

(a) AC circuit

(b) DC circuit

FIGURE 15–62 Determining the effective value of sinusoidal ac.

First, consider the dc case. Since current is constant, power is constant, and average power is Pavg P I 2R

(15–21)

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Now consider the ac case. Power to the resistor at any value of time is p(t) i 2R, where i is the instantaneous value of current. A sketch of p(t) is shown in Figure 15–62(a), obtained by squaring values of current at various points along the axis, then multiplying by R. Average power is the average of p(t). Since i Im sin qt, p(t) i2R (Im sin qt)2R I m2R sin2qt 1 I m2R (1 cos 2qt) 2

冤

冥

where we have used the trigonometric identity sin2qt 1⁄2(1 cos 2qt), from the mathematics tables inside the front cover to expand sin2qt. Thus,

NOTES... Because ac currents alternate in direction, you might expect average power to be zero, with power during the negative half-cycle being equal and opposite to power during the positive halfcycle and hence cancelling. However, as Equation 15–22 shows, current is squared, and hence power is never negative. This is consistent with the idea that insofar as power dissipation is concerned, the direction of current through a resistor does not matter (Figure 15–63).

(a) P =

(15–22)

(4)2(80)

I=4A

= 1280 W

I=4A

(b) P = (4)2(80) = 1280 W FIGURE 15–63 Since power depends only on current magnitude, it is the same for both directions.

Im2R Im2R p(t) cos 2qt 2 2

(15–23)

To get the average of p(t), note that the average of cos 2qt is zero and thus the last term of Equation 15–23 drops off leaving Im2R Pavg average of p(t) 2

(15–24)

Now equate Equations 15–21 and 15–24, then cancel R. I 2 I 2 m 2

Now take the square root of both sides. Thus, I

冪莦莦

I 2 I m m 0.707Im 2 兹2苶

Current I is the value that we are looking for; it is the effective value of current i. To emphasize that it is an effective value, we will initially use subscripted notation Ieff. Thus, I Ieff m 0.707Im 兹2苶

(15–25)

Effective values for voltage are found in the same way: Em Eeff 0.707Em 兹2苶

(15–26a)

Vm Veff 0.707Vm 兹2苶

(15–26b)

As you can see, effective values for sinusoidal waveforms depend only on magnitude.

EXAMPLE 15–27 Determine the effective values of a. i 10 sin qt A, b. i 50 sin(qt 20°) mA, c. v 100 cos 2qt V

Section 15.9

■

Effective Values

587

Solution Since effective values depend only on magnitude, a. Ieff (0.707)(10 A) 7.07 A, b. Ieff (0.707)(50 mA) 35.35 mA, c. Veff (0.707)(100 V) 70.7 V.

To obtain peak values from effective values, rewrite Equations 15–25 and 15–26. Thus, Im 兹2 苶Ieff 1.414Ieff

(15–27)

Em 兹2 苶Eeff 1.414Veff

(15–28a)

Vm 兹2 苶Veff 1.414Veff

(15–28b)

It is important to note that these relationships hold only for sinusoidal waveforms. However, the concept of effective value applies to all waveforms, as we soon see. Consider again the ac voltage at the wall outlet in your home. Since Eeff 120 V, Em (兹2苶)(120 V) 170 V. This means that a sinusoidal voltage alternating between 170 V produces the same average power as 120 V of steady dc (Figure 15–64). e (V)

FIGURE 15–64 120 V of steady dc is capable of producing the same average power as sinusoidal ac with Em 170 V.

170 0

Time

120 V

170

Time (b)

(a)

General Equation for Effective Values The 兹2 苶 relationship holds only for sinusoidal waveforms. For other waveforms, you need a more general formula. Using calculus, it can be shown that for any waveform Ieff

冪莦莦冕莦i莦dt莦 1 T

(15–29)

with a similar equation for voltage. This equation can be used to compute effective values for any waveform, including sinusoidal. In addition, it leads to a graphic approach to finding effective values. In Equation 15–29, the integral of i 2 represents the area under the i 2 waveform. Thus, Ieff

冪莦莦 area under the i2 curve base

(15–30)

To compute effective values using this equation, do the following: Step 1: Square the current (or voltage) curve. Step 2: Find the area under the squared curve.

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Step 3: Divide the area by the length of the curve. Step 4: Find the square root of the value from Step 3. This process is easily carried out for rectangular-shaped waveforms since the area under their squared curves is easy to compute. For other waveforms, you have to use calculus or approximate the area using numerical methods. For the special case of superimposed ac and dc (Figure 15–60), Equation 15–29 leads to the following formula: 2 Ieff 兹I苶d苶苶Iac 苶2苶 c 苶

(15–31)

where Idc is the dc current value, Iac is the effective value of the ac component, and Ieff is the effective value of the combined ac and dc currents. Equations 15–30 and 15–31 also hold for voltage when V is substituted for I.

RMS Values Consider again Equation 15–30. To use this equation, we compute the root of the mean square to obtain the effective value. For this reason, effective values are called root mean square or rms values and the terms effective and rms are synonymous. Since, in practice, ac quantities are almost always expressed as rms values, we shall assume from here on that, unless otherwise noted, all ac voltages and currents are rms values.

EXAMPLE 15–28

One cycle of a voltage waveform is shown in Figure 15–65(a). Determine its effective value. v2 (V)

v (V) 30 20 10 0 10

900 100

400 1 2 3 4 5 6 7 8 9 10 T

(a) Voltage waveform

0 10

T = 10 (b) Squared waveform

FIGURE 15–65

Solution Square the voltage waveform and plot it as in (b). Apply Equation 15–30: Veff

(400 4) (900 2) (100 2) (0 2) 冪莦莦莦 10

冪莦36莦1 0莦00莦 19.0 V

The waveform of Figure 15–65(a) has the same effective value as 19.0 V of steady dc.

Section 15.9

EXAMPLE 15–29

■

Effective Values

Determine the effective value of the waveform of Figure

15–66(a). i2 9 i (A) 3 2 1 0 1 2

4 1 2 3 4 5 6 7 8

t (s)

1 0

1 cycle (a)

1 2 3 4 5 6 7 8

t (s)

(b)

FIGURE 15–66

Solution Square the curve, then apply Equation 15–30. Thus, Ieff

EXAMPLE 15–30

(9 3) (1 2) (4 3) 冪莦莦 8

冪莦 481 莦 2.26 A

Compute the effective value of the waveform of Figure

15–61(b). Solution Use Equation 15–31 (with I replaced by V). First, compute the rms value of the ac component. Vac 0.707 15 10.61 V. Now substitute this into Equation 15–31. Thus, Vrms 兹V 苶2dc 苶苶苶 V 2ac 苶兹(1 苶0苶苶 )2苶苶(1苶0苶.6 苶1苶苶 )2 14.6 V

1. Determine the effective value of the current of Figure 15–51. 2. Repeat for the voltage graphed in Figure 15–52(a). Answers: 1. 1.73 A 2. 20 V

One Final Note The subscripts eff and rms are not used in practice. Once the concept is familiar, we drop them.

PRACTICE PROBLEMS 12

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15.10 Rate of Change of a Sine Wave (Derivative)

NOTES... ∫ The Derivative of a Sine Wave The result developed intuitively here can be proven easily using calculus. To illustrate, consider the waveform sin qt shown in Figure 15–67. The slope of this function is its derivative. Thus, d Slope sin qt qcos qt dt Therefore, the slope of a sine wave is a cosine wave as depicted in Figure 15–68.

Several important circuit effects depend on the rate of change of sinusoidal quantities. The rate of change of a quantity is the slope (i.e., derivative) of its waveform versus time. Consider the waveform of Figure 15–67. As indicated, the slope is maximum positive at the beginning of the cycle, zero at both its peaks, maximum negative at the half-cycle crossover point, and maximum positive at the end of the cycle. This slope is plotted in Figure 15–68. Note that the original waveform and its slope are 90° out of phase. Thus, if sinusoidal waveform A is taken as reference, its slope B leads it by 90°, whereas, if the slope B is taken as reference, A lags it by 90°. Thus, if A is a sine wave, B is a cosine wave, and so on. (This result is important to us in Chapter 16.) Maximum positive slope

Zero slope Maximum positive slope

Waveform A t

Maximum negative slope

Slope (Waveform B) Zero slope

FIGURE 15–67 Slope at various places for a sine wave.

FIGURE 15–68 Showing the 90° phase shift.

15.11 AC Voltage and Current Measurement Two of the most important instruments for measuring ac quantities are the multimeter and the oscilloscope. Multimeters read voltage, current, and sometimes frequency. Oscilloscopes show waveshape and period and permit determination of frequency, phase difference, and so on.

Meters for Voltage and Current Measurement There are two basic classes of ac meters: one measures rms correctly for sinusoidal waveforms only (called “average responding” instruments); the other measures rms correctly regardless of waveform (called “true rms” meters). Most common meters are average responding meters. Average Responding Meters

Average responding meters use a rectifier circuit to convert incoming ac to dc. They then respond to the average value of the rectified input, which, as shown in Figure 15–55, is 0.637Vm for a “full-wave” rectified sine wave. However, the rms value of a sine wave is 0.707Vm. Thus, the scale of such a meter is modified by the factor 0.707Vm/0.637Vm 1.11 so that it indicates rms values directly. Other meters use a “half-wave” circuit, which yields the waveform of Figure 15–56 for a sine wave input. In this case, its average is 0.318Vm , yielding a scale factor of 0.707Vm/0.318Vm 2.22. Figure 15–69 shows a typical average responding DMM.

Section 15.11

■

AC Voltage and Current Measurement

FIGURE 15–69 A DMM. While all DMMs measure voltage, current, and resistance, this one also measures frequency.

True RMS Measurement

To measure the rms value of a nonsinusoidal waveform, you need a true rms meter. A true rms meter indicates true rms voltages and currents regardless of waveform. For example, for the waveform of Figure 15–64(a), any ac meter will correctly read 120 V (since it is a sine wave). For the waveform of 15–61(b), a true rms meter will correctly read 14.6 V (the rms value that we calculated earlier, in Example 15–30) but an average responding meter will yield only a meaningless value. True rms instruments are more expensive than standard meters.

Oscilloscopes Oscilloscopes (frequently referred to as scopes, Figure 15–70) are used for time domain measurement, i.e., waveshape, frequency, period, phase difference, and so on. Usually, you scale values from the screen, although some higher-priced models can compute and display them for you on a digital readout. Oscilloscopes measure voltage. To measure current, you need a currentto-voltage converter. One type of converter is a clip-on device, known as a current gun that clamps over the current-carrying conductor and monitors its magnetic field. (It works only with ac.) The varying magnetic field induces a voltage which is then displayed on the screen. With such a device, you can monitor current waveshapes and make current-related measurements. Alternately, you can place a small resistor in the current path, mea-

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FIGURE 15–70 An oscilloscope may be used for waveform analysis.

sure voltage across it with the oscilloscope, then use Ohm’s law to determine the current.

A Final Note AC meters measure voltage and current only over a limited frequency range, typically from 50 Hz to a few kHz, although others are available that work up to the 100-kHz range. Note, however, that accuracy may be affected by frequency. (Check the manual.) Oscilloscopes, on the other hand, can measure very high frequencies; even moderately priced oscilloscopes work at frequencies up to hundreds of MHz.

15.12 Circuit Analysis Using Computers

ELECTRONICS WORKBENCH

PSpice

Electronics Workbench and PSpice both provide a convenient way to study the phase relationships of this chapter, as they both incorporate easy-to-use graphing facilities. You simply set up sources with the desired magnitude and phase values and instruct the software to compute and plot the results. To illustrate, let us graph e1 100 sin qt V and e2 80 sin(qt 60°) V. Use a frequency of 500 Hz.

Electronics Workbench For Electronics Workbench, you must specify rms rather than peak values. Thus, to plot e1 100 sin qt V, key in 70.71 V and 0 degrees. Similarly, for e2, use 56.57 V and 60°. Here are the steps: Create the circuit of Figure 15–71 on the screen; Double click Source 1 and enter 70.71 V, 0 deg, and 500 Hz in the dialog box; Similarly, set Source 2 to 56.57 V, 60 deg, and

Section 15.12

■

593

Circuit Analysis Using Computers

FIGURE 15–71 Studying phase relationships using Electronics Workbench.

500 Hz; Click Analysis, Transient, set TSTOP to 0.002 (to run the solution out to 2 ms so that you display a full cycle) and TMAX to 2e-06 (to avoid getting a choppy waveform); Highlight Node 1 (to display e1) and click Add; Repeat for Node 2 (to display e2); Click the Simulate icon; Following simulation, graphs e1 and e2 (Figure 15–72) appear.

FIGURE 15–73

FIGURE 15–72

You can verify the angle between the waveforms using cursors. First, note that the period T 2 ms 2000 ms. (This corresponds to 360°.) Expand the graph to full screen, click the Grid icon, then the Cursors icon. Using the cursors, measure the time between crossover points, as indicated in Figure 15–73. You should get 333 ms. This yields an angular displacement of 333 ms v 360° 60° 2000 ms

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OrCAD PSpice For this problem, you need a sinusoidal time-varying ac voltage source. Use VSIN (it is found in the SOURCE library.) For VSIN, you must specify the magnitude, phase, and frequency of the source, as well as its offset (since we do not want an offset in this problem, we will set it to zero.) Proceed as follows. Build the circuit of Figure 15–74 on the screen. Double click source 1 and in the Properties editor, select the Parts tab. Scroll right until you find a list of source properties: then in cell VAMPL, enter 100V; in cell PHASE, enter 0deg; in cell FREQ, enter 500Hz; and in cell VOFF, enter 0V (this sets source ei 100 sin qt V with q 2p(500 Hz)). Click Apply, then close the Properties editor window. Similarly, set up source 2 (making sure you use a phase angle of 60°). Click the New Simulation icon and enter a name (e.g., fig15-74). In the Simulation Settings box, select Time Domain and General Settings. Set TSTOP to 2ms (to display a full cycle). Select voltage markers from the toolbar and place as shown. (This causes PSpice to automatically create the photos.) Run the simulation. When the simulation is complete, the waveforms of Figure 15–75 should appear.

FIGURE 15–74 Studying phase relationships using OrCAD PSpice.

FIGURE 15–75

Problems

595

You can verify the angle between the waveforms using cursors. First, note that the period T 2 ms 2000 ms. (This corresponds to 360°.) Now using the cursors (see Appendix A if you need help), measure the time between crossover points as indicated in Figure 15–73. You should get 333 ms. This yields an angular displacement of 333 ms v 360° 60° 2000 ms

which agrees with the given sources.

PROBLEMS

15.1 Introduction 1. What do we mean by “ac voltage”? By “ac current”? 15.2 Generating AC Voltages 2. The waveform of Figure 15–8 is created by a 600-rpm generator. If the speed of the generator changes so that its cycle time is 50 ms, what is its new speed? 3. a. What do we mean by instantaneous value? b. For Figure 15–76, determine instantaneous voltages at t 0, 1, 2, 3, 4, 5, 6, 7, and 8 ms. 15.3 Voltage and Current Conventions for ac 4. For Figure 15–77, what is I when the switch is in position 1? When in position 2? Include sign.

e (V) 20 10 0 10 20

1 2 3 45 6 7 8

FIGURE 15–76 1 2 40 V

70 V

FIGURE 15–77

e(t)

i(t) R

4 k

FIGURE 15–78

5. The source of Figure 15–78 has the waveform of Figure 15–76. Determine the current at t 0, 1, 2, 3, 4, 5, 6, 7, and 8 ms. Include sign. 15.4 Frequency, Period, Amplitude, and Peak Value 6. For each of the following, determine the period: a. f 100 Hz b. f 40 kHz c. f 200 MHz 7. For each of the following, determine the frequency: a. T 0.5 s b. T 100 ms c. 5T 80 ms 8. For a triangular wave, f 1.25 MHz. What is its period? How long does it take to go through 8 107 cycles?

t (ms)

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AC Fundamentals 9. Determine the period and frequency for the waveform of Figure 15–79.

v i 9V 3 mA

t (ms)

6 V

0 1

9 10 11 12 13 14

t (µs)

3 mA

FIGURE 15–80

FIGURE 15–79

10. Determine the period and frequency for the waveform of Figure 15–80. How many cycles are shown? 11. What is the peak-to-peak voltage for Figure 15–79? What is the peak-topeak current of Figure 15–80? 12. For a certain waveform, 625T 12.5 ms. What is the waveform’s period and frequency? 13. A square wave with a frequency of 847 Hz goes through how many cycles in 2 minutes and 57 seconds? 14. For the waveform of Figure 15–81, determine a. period b. frequency c. peak-to-peak value 15. Two waveforms have periods of T1 and T2 respectively. If T1 0.25 T2 and f1 10 kHz, what are T1, T2, and f2? 16. Two waveforms have frequencies f1 and f2 respectively. If T1 4 T2 and waveform 1 is as shown in Figure 15–79, what is f2?

525 s 75 V

FIGURE 15–81

i (A) 50 0

200° 40° 120°

50 FIGURE 15–82

310°

15.5 Angular and Graphic Relationships for Sine Waves 17. Given voltage v Vm sin a. If Vm 240 V, what is v at a 37°? 18. For the sinusoidal waveform of Figure 15–82, a. Determine the equation for i. b. Determine current at all points marked. 19. A sinusoidal voltage has a value of 50 V at a 150°. What is Vm? 20. Convert the following angles from radians to degrees: a. p/12 b. p/1.5 c. 3p/2 d. 1.43 e. 17 f. 32p 21. Convert the following angles from degrees to radians: a. 10° b. 25° c. 80° d. 150° e. 350° f. 620° 22. A 50-kHz sine wave has an amplitude of 150 V. Sketch the waveform with its axis scaled in microseconds. 23. If the period of the waveform in Figure 15–82 is 180 ms, compute current at t 30, 75, 140, and 315 ms. 24. A sinusoidal waveform has a period of 60 ms and Vm 80 V. Sketch the waveform. What is its voltage at 4 ms?

597

Problems 25. A 20-kHz sine wave has a value of 50 volts at t 5 ms. Determine Vm and sketch the waveform. 26. For the waveform of Figure 15–83, determine v2.

v (V) 57 330°

15.6 Voltages and Currents as Functions of Time 27. Calculate q in radians per second for each of the following: a. T 100 ns b. f 30 Hz c. 100 cycles in 4 s d. period 20 ms e. 5 periods in 20 ms 28. For each of the following values of q, compute f and T: a. 100 rad/s b. 40 rad in 20 ms c. 34 103 rad/s 29. Determine equations for sine waves with the following: a. Vm 170 V, f 60 Hz b. Im 40 mA, T 10 ms

v2 FIGURE 15–83

c. T 120 ms, v 10 V at t 12 ms 30. Determine f, T, and amplitude for each of the following: a. v 75 sin 200pt b. i 8 sin 300t 31. A sine wave has a peak-to-peak voltage of 40 V and T 50 ms. Determine its equation. 32. Sketch the following waveforms with the horizontal axis scaled in degrees, radians, and seconds: a. v 100 sin 200pt V b. i 90 sin qt mA, T 80 ms 33. Given i 47 sin 8260t mA, determine current at t 0 s, 80 ms, 410 ms, and 1200 ms. 34. Given v 100 sin a. Sketch one cycle. a. Determine at which two angles v 86.6 V. b. If q 100p/60 rad/s, at which times do these occur? 35. Write equations for the waveforms of Figure 15–84. Express the phase angle in degrees. i

10 A

40 V

5 (mA)

ωt

ωt π rad 5

(a) ω 1000 rad/s

ωt

60° π rad 4

(b) t 50 ms

0 60°

FIGURE 15–84

36. Sketch the following waveforms with the horizontal axis scaled in degrees and seconds: a. v 100 sin(232.7t 40°) V b. i 20 sin(qt 60°) mA, f 200 Hz 37. Given v 5 sin(qt 45°). If q 20p rad/s, what is v at t 20, 75, and 90 ms?

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AC Fundamentals 38. Repeat Problem 35 for the waveforms of Figure 15–85. v v 80 V 100 V

ωt

160 µs

320° (a) Period 10 µs

(b) f 833.3 Hz

FIGURE 15–85

39. Determine the equation for the waveform shown in Figure 15–86. i 54° v

i1 = 15.6 A

100 V

ωt

350°

ωt

180° i2 Im sin ( 60°)

1620 µs

FIGURE 15–87

FIGURE 15–86

40. For the waveform of Figure 15–87, determine i2. 41. Given v 30 sin(qt 45°) where q 40p rad/s. Sketch the waveform. At what time does v reach 0 V? At what time does it reach 23 V and 23 V? 15.7 Introduction to Phasors 42. For each of the phasors of Figure 15–88, determine the equation for v(t) or i(t) as applicable, and sketch the waveform. j

j Vm = 100 V

70°

Im = 20 A

(a)

40°

35°

(b)

Im = 10 mA (c)

FIGURE 15–88

43. With the aid of phasors, sketch the waveforms for each of the following pairs and determine the phase difference and which waveform leads: a. v 100 sin qt b. v1 200 sin(qt 30°) i 80 sin(qt 20°) v2 150 sin(qt 30°) c. i1 40 sin(qt 30°) d. v 100 sin(qt 140°) i2 50 sin(qt 20°) i 80 sin(qt 160°)

599

Problems 44. Repeat Problem 43 for the following. a. i 40 sin(qt 80°) b. v 20 cos(qt 10°) v 30 sin(qt 70°) i 15 sin(qt 10°) c. v 20 cos(qt 10°) d. v 80 cos(qt 30°) i 15 sin(qt 120°) i 10 cos(qt 15°) 45. For the waveforms in Figure 15–89, determine the phase differences. Which waveform leads? 46. Draw phasors for the waveforms of Figure 15–89. 15.8 AC Waveforms and Average Value 47. What is the average value of each of the following over an integral number of cycles? a. i 5 sin qt b. i 40 cos qt c. v 400 sin(qt 30°) d. v 20 cos 2qt 48. Using Equation 15–20, compute the area under the half-cycle of Figure 15– 54 using increments of p/12 rad. 49. Compute Iavg for the waveforms of Figure 15–90.

A B 300° 30°

ωt

(a)

A 60°

ωt

180°

(b) FIGURE 15–89

v (V)

i (A)

i (A) 3 2 1 0 1 2 3

1 2 3 4 5 6 7 8 9 10

t (s)

100

t (ms)

200

(b)

50. For the waveform of Figure 15–91, compute Im. 51. For the circuit of Figure 15–92, e 25 sin qt V and period T 120 ms. a. Sketch voltage v(t) with the axis scaled in milliseconds. b. Determine the peak and minimum voltages. c. Compute v at t 10, 20, 70, and 100 ms. d. Determine Vavg.

Iavg = 10 A

1 2 3 4 5 6 1 cycle

FIGURE 15–91

t (s)

3 2 2 (c)

FIGURE 15–90

i (A)

Sine wave

1 0

20 (a)

1 cycle

v(t)

15 V

e = 25 sin t FIGURE 15–92

ωt

600

Chapter 15

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AC Fundamentals 52. Using numerical methods for the curved part of the waveform (with increment size t 0.25 s), determine the area and the average value for the waveform of Figure 15–93. 53. ∫ Using calculus, find the average value for Figure 15–93. v 12 V

3 t2

0 1

t (s)

6 V

FIGURE 15–93

15.9 Effective Values 54. Determine the effective values of each of the following: a. v 100 sin qt V b. i 8 sin 377t A c. v 40 sin(qt 40°) V d. i 120 cos qt mA 55. Determine the rms values of each for the following. a. A 12 V battery b. 24 sin(qt 73°) mA c. 10 24 sin qt V d. 45 27 cos 2 qt V 56. For a sine wave, Veff 9 V. What is its amplitude? i 4A 0 2

t (µs)

−12 A

(a)

v 30 V

1 cycle

0 5V

10 V (b)

FIGURE 15–94

t (ms)

Problems 57. Determine the root mean square values for a. i 3 兹2 苶(4) sin(qt 44°) mA b. Voltage v of Figure 15–92 with e 25 sin qt V 58. Compute the rms values for Figures 15–90(a), and 15–91. For Figure 15– 91, Im 30 A. 59. Compute the rms values for the waveforms of Figure 15–94. 60. Compute the effective value for Figure 15–95. v 2V 0

−10 V

FIGURE 15–95

61. Determine the rms value of the waveform of Figure 15–96. Why is it the same as that of a 24-V battery? v 24 V

t (ms)

24 V

FIGURE 15–96

62. Compute the rms value of the waveform of Figure 15–52(c). To handle the triangular portion, use Equation 15–20. Use a time interval t 1s. 63. ∫ Repeat Problem 62, using calculus to handle the triangular portion. 15.11 AC Voltage and Current Measurement 64. Determine the reading of an average responding AC meter for each of the following cases. (Note: Meaningless is a valid answer if applicable.) Assume the frequency is within the range of the instrument. a. v 153 sin qt V b. v 兹2苶(120) sin(qt 30°) V c. The waveform of Figure 15–61 d. v 597 cos qt V 65. Repeat Problem 64 using a true rms meter. 15.12 Circuit Analysis Using Computers Use Electronics Workbench or PSpice for the following. 66. EWB PSpice Plot the waveform of Problem 37 and, using the cursor, determine voltage at the times indicated. Don’t forget to convert the frequency to Hz.

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AC Fundamentals 67. EWB PSpice Plot the waveform of Problem 41. Using the cursor, determine the time at which v reaches 0 V. Don’t forget to convert the frequency to Hz. 68. EWB PSpice Assume the equations of Problem 43 all represent voltages. For each case, plot the waveforms, then use the cursor to determine the phase difference between waveforms.

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 1. 16.7 ms 2. Frequency doubles, period halves 3. 50 Hz; 20 ms 4. 20 V; 0.5 ms and 2.5 ms; 35 V: 4 ms and 5 ms 5. (c) and (d); Since current is directly proportional to the voltage, it will have the same waveshape. 6. 250 Hz 7. f1 100 Hz; f2 33.3 Hz 8. 50 kHz and 1 MHz 9. 22.5 Hz 10. At 12 ms, direction →; at 37 ms, direction ←; at 60 ms, → 11. At 75 ms, i 5 A In-Process Learning Check 2 1. a (deg) i (mA)

0 0

45 10.6

90 15

135 10.6

180 0

225 10.6

270 15

315 10.6

2. a. 0.349 b. 0.873 c. 2.09 d. 4.36 3. 30°; 120°; 450° 4. Same as Figure 15–27 with T 10 s and amplitude 50 mA. 5. 1.508 106 rad/s 6. i 8 sin 157t A 7. i 6 sin 69.81t A 8. a. i 250 sin(251t 30°) A b. i 20 sin(62.8t 45°) A c. v 40 sin(628t 30°) V d. v 80 sin(314 103t 36°) V 9. a. 2.95 ms; 7.05 ms; 22.95 ms; 27.05 ms b. 11.67 ms; 18.33 ms; 31.67 ms; 38.33 ms

360 0

Answers to In-Process Learning Checks 10.

80 V 40 V ωt

45°

135°

ωt

315°

80°

(a)

260° (b)

i2 5 mA

10 mA ωt

90°

360°

70°

(c)

11. a. 2.25 s 12. a.

110° (d)

b. 1.33 s j

b. 125°

c. v1 leads

b. 20°

c. i1 leads

Vm1 80 V

45°

80° Vm2 40 V

13. a.

j Im1 10 mA Im2 5 mA 70°

290°

440°

603

R, L, and C Elements and the Impedance Concept OBJECTIVES

KEY TERMS

After studying this chapter, you will be able to • express complex numbers in rectangular and polar forms, • represent ac voltage and current phasors as complex numbers, • represent voltage and current sources in transformed form, • add and subtract currents and voltages using phasors, • compute inductive and capacitive reactance, • determine voltages and currents in simple ac circuits, • explain the impedance concept, • determine impedance for R, L, and C circuit elements, • determine voltages and currents in simple ac circuits using the impedance concept, • use Electronics Workbench and PSpice to solve simple ac circuit problems.

Capacitive Reactance Complex Number Impedance Inductive Reactance j 兹苶1苶 Phasor Domain Polar Form Rectangular Form Time Domain

OUTLINE Complex Number Review Complex Numbers in AC Analysis R, L, and C Circuits with Sinusoidal Excitation Resistance and Sinusoidal AC Inductance and Sinusoidal AC Capacitance and Sinusoidal AC The Impedance Concept Computer Analysis of AC Circuits

n Chapter 15, you learned how to analyze a few simple ac circuits in the time domain using voltages and currents expressed as functions of time. However, this is not a very practical approach. A more practical approach is to represent ac voltages and currents as phasors, circuit elements as impedances, and analyze circuits in the phasor domain using complex algebra. With this approach, ac circuit analysis is handled much like dc circuit analysis, and all basic relationships and theorems—Ohm’s law, Kirchhoff’s laws, mesh and nodal analysis, superposition and so on—apply. The major difference is that ac quantities are complex rather than real as with dc. While this complicates computational details, it does not alter basic circuit principles. This is the approach used in practice. The basic ideas are developed in this chapter. Since phasor analysis and the impedance concept require a familiarity with complex numbers, we begin with a short review.

Charles Proteus Steinmetz CHARLES STEINMETZ WAS BORN IN Breslau, Germany in 1865 and emigrated to the United States in 1889. In 1892, he began working for the General Electric Company in Schenectady, New York, where he stayed until his death in 1923, and it was there that his work revolutionized ac circuit analysis. Prior to his time, this analysis had to be carried out using calculus, a difficult and time-consuming process. By 1893, however, Steinmetz had reduced the very complex alternatingcurrent theory to, in his words, “a simple problem in algebra.” The key concept in this simplification was the phasor—a representation based on complex numbers. By representing voltages and currents as phasors, Steinmetz was able to define a quantity called impedance and then use it to determine voltage and current magnitude and phase relationships in one algebraic operation. Steinmetz wrote the seminal textbook on ac analysis based on his method, but at the time he introduced it he was practically the only person who understood it. Now, however, it is common knowledge and one of the basic tools of the electrical engineer and technologist. In this chapter, we learn the method and illustrate its application to the solution of basic ac circuit problems. In addition to his work for GE, Charles Steinmetz was a professor of electrical engineering (1902–1913) and electrophysics (1913–1923) at Union University (now Union College) in Schenectady.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

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16.1

Complex Number Review

A complex number is a number of the form C a jb, where a and b are 苶1苶. The number a is called the real part of C and b real numbers and j 兹 is called its imaginary part. (In circuit theory, j is used to denote the imaginary component rather than i to avoid confusion with current i.)

Imaginary Axis

C = 6 j8

8 6 4 2 0 2 4 6 j

8 10

Real Axis

FIGURE 16–1 A complex number in rectangular form.

Geometrical Representation Complex numbers may be represented geometrically, either in rectangular form or in polar form as points on a two-dimensional plane called the complex plane (Figure 16–1). The complex number C 6 j8, for example, represents a point whose coordinate on the real axis is 6 and whose coordinate on the imaginary axis is 8. This form of representation is called the rectangular form. Complex numbers may also be represented in polar form by magnitude and angle. Thus, C 10∠53.13° (Figure 16–2) is a complex number with magnitude 10 and angle 53.13°. This magnitude and angle representation is just an alternate way of specifying the location of the point represented by C a jb. Conversion between Rectangular and Polar Forms To convert between forms, note from Figure 16–3 that C a jb

(rectangular form)

C C∠v (polar form)

C = 10∠53.13°

(16–1) (16–2)

where C is the magnitude of C. From the geometry of the triangle, 53.13°

Real Axis

a C cos v

(16–3a)

b C sin v

(16–3b)

C 兹a苶2苶苶苶b2苶

(16–4a)

b v tan 1 a

(16–4b)

where FIGURE 16–2 A complex number in polar form.

and j C = a jb = C∠θ C θ

b Real Axis

FIGURE 16–3 Polar and rectangular equivalence.

Equations 16–3 and 16–4 permit conversion between forms. When using Equation 16–4b, however, be careful when the number to be converted is in the second or third quadrant, as the angle obtained is the supplementary angle rather than the actual angle in these two quadrants. This is illustrated in Example 16–1 for the complex number W.

Section 16.1

EXAMPLE 16–1 Determine rectangular and polar forms for the complex numbers C, D, V, and W of Figure 16–4(a) j W

4 3 2

1 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1 2 3

⫺2 V ⫺3

36.87°

⫺4

(a) Complex numbers

(b) In polar form, C = 5∠36.87° j

j 45°

5. 66

45°

135°

D (c) In polar form, D = 5.66∠ 45°

(d) In polar form, W = 5.66∠135°

FIGURE 16–4

Solution Point C: Real part 4; imaginary part 3. Thus, C 4 j3. In polar form, C 兹苶42苶苶苶32苶 5 and vC tan 1 (3/4) 36.87°. Thus, C 5∠36.87° as indicated in (b). Point D: In rectangular form, D 4 j4. Thus, D 兹4苶2苶苶苶42苶 5.66 and 1 vD tan ( 4/4) 45°. Therefore, D 5.66∠ 45°, as shown in (c). Point V: In rectangular form, V j2. In polar form, V 2∠ 90°. Point W: In rectangular form, W 4 j4. Thus, W 兹4苶2苶苶苶42苶 5.66 1 and tan ( 4/4) 45°. Inspection of Figure 16–4(d) shows, however, that this 45° angle is the supplementary angle. The actual angle (measured from the positive horizontal axis) is 135°. Thus, W 5.66∠135°.

In practice (because of the large amount of complex number work that you will do), a more efficient conversion process is needed than that described previously. As discussed later in this section, inexpensive calculators are available that perform such conversions directly—you simply enter

■

Complex Number Review

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R, L, and C Elements and the Impedance Concept

the complex number components and press the conversion key. With these, the problem of determining angles for numbers such as W in Example 16–1 does not occur; you just enter 4 j4 and the calculator returns 5.66∠135°.

Powers of j Powers of j are frequently required in calculations. Here are some useful powers: j2 (兹苶1苶)(兹苶1苶) 1 3 2 j j j j j4 j2j2 ( 1) ( 1) 1 ( j)j 1

(16–5)

1 1 j j 2 j j j j j

Addition and Subtraction of Complex Numbers Addition and subtraction of complex numbers can be performed analytically or graphically. Analytic addition and subtraction is most easily illustrated in rectangular form, while graphical addition and subtraction is best illustrated in polar form. For analytic addition, add real and imaginary parts separately. Similarly for subtraction. For graphical addition, add vectorially as in Figure 16–5(a); for subtraction, change the sign of the subtrahend, then add, as in Figure 16–5(b).

Given A 2 j1 and B 1 j3. Determine their sum and difference analytically and graphically.

EXAMPLE 16–2

Solution A B (2 j1) (1 j3) (2 1) j(1 3) 3 j4. A B (2 j1) (1 j3) (2 1) j(1 3) 1 j2. Graphical addition and subtraction are shown in Figure 16–5. j j 4 3 2

3 2

A B = 3 j4

B A 1 2 3

A 1 2 (a)

FIGURE 16–5

3 4

A ( B) = A B = 1 j2 B (b)

Section 16.1

■

Complex Number Review

Multiplication and Division of Complex Numbers These operations are usually performed in polar form. For multiplication, multiply magnitudes and add angles algebraically. For division, divide the magnitude of the denominator into the magnitude of the numerator, then subtract algebraically the angle of the denominator from that of the numerator. Thus, given A A∠vA and B B∠vB,

Ⲑ A/B A/BⲐv

A B AB vA vB

(16–6)

(16–7)

EXAMPLE 16–3 Given A 3∠35° and B 2∠ 20°, determine the product A B and the quotient A/B. Solution A B (3∠35°)(2∠ 20°) (3)(2)Ⲑ 35° 20° 6∠15° A (3∠35°) 3 Ⲑ 35° ( 20°) 1.5∠55° B (2∠ 20°) 2

EXAMPLE 16–4 For computations involving purely real, purely imaginary, or small integer numbers, it is sometimes easier to multiply directly in rectangular form than it is to convert to polar. Compute the following directly: a. ( j3)(2 j4). b. (2 j3)(1 j5). Solution a. ( j3)(2 j4) ( j3)(2) ( j3)( j4) j6 j212 12 j6 b. (2 j3)(1 j5) (2)(1) (2)( j5) ( j3)(1) ( j3)( j5) 2 j10 j3 j215 2 j13 15 13 j13

1. Polar numbers with the same angle can be added or subtracted directly without conversion to rectangular form. For example, the sum of 6 ∠36.87° and 4 ∠36.87° is 10 ∠36.87°, while the difference is 6 ∠36.87° 4 ∠36.87° 2 ∠36.87°. By means of sketches, indicate why this procedure is valid. 2. To compare methods of multiplication with small integer values, convert the numbers of Example 16–4 to polar form, multiply them, then convert the answers back to rectangular form. Answers: 1. Since the numbers have the same angle, their sum also has the same angle and thus, their magnitudes simply add (or subtract).

PRACTICE PROBLEMS 1

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R, L, and C Elements and the Impedance Concept

Reciprocals The reciprocal of a complex number C C∠v is 1 1 ∠ v C∠v C

(16–8)

Thus, 1 0.05∠ 30° 20∠30°

Complex Conjugates The conjugate of a complex number (denoted by an asterisk *) is a complex number with the same real part but the opposite imaginary part. Thus, the conjugate of C C∠v a jb is C* C∠ v a jb. For example, if C 3 j4 5∠53.13°, then C* 3 j4 5∠ 53.13°. Calculators for AC Analysis The analysis of ac circuits involves a considerable amount of complex number arithmetic; thus, you will need a calculator that can work easily with complex numbers. There are several inexpensive calculators on the market that are suitable for this purpose in that they can perform all required calculations (addition, subtraction, multiplication, and division) in either rectangular or polar form without the need for conversion. This is important, because it saves you a great deal of time and cuts down on errors. To illustrate, consider Example 16–5. Using a calculator with only basic complex number conversion capabilities requires that you convert between forms as illustrated. On the other hand, a calculator with more sophisticated complex number capabilities (such as that shown in Figure 16–6) allows you to perform the calculation without going through all the intermediate conversion steps. You need to acquire an appropriate calculator and learn to use it proficiently. FIGURE 16–6 This calculator displays complex numbers in standard mathematical notation.

EXAMPLE 16–5

The following illustrates the type of calculations that you will encounter. Use your calculator to reduce the following to rectangular form: (3 j4)(10∠40°) (6 j5) 6 30∠53.13° Solution Using a calculator with basic capabilities requires a number of intermediate steps, some of which are shown below. (5∠ 53.13)(10∠40) answer (6 j5) 6 (18 j24) (5∠ 53.13)(10∠40) (6 j5) 24 j24

Section 16.2

■

611

Complex Numbers in AC Analysis

(5∠ 53.13)(10∠40) (6 j5) 33.94∠45 (6 j5) 1.473∠ 58.13 (6 j5) (0.778 j1.251) 6.778 j3.749 Using a calculator such as that shown in Figure 16–6 saves steps, as it multiplies (3 j4)(10∠40°) and adds 6 30∠53.13°, etc., directly, without your having to convert forms.

16.2

Complex Numbers in AC Analysis

Representing AC Voltages and Currents by Complex Numbers As you learned in Chapter 15, ac voltages and currents can be represented as phasors. Since phasors have magnitude and angle, they can be viewed as complex numbers. To get at the idea, consider the voltage source of Figure 16–7(a). Its phasor equivalent (b) has magnitude Em and angle v. It therefore can be viewed as the complex number E Em∠v

(16–9)

j E

(a) e(t) = Em sin (ωt θ)

⫹ e(t) ⫺

(b) E = Em∠θ

FIGURE 16–7 Representation of a sinusoidal source voltage as a complex number.

From this point of view, the sinusoidal voltage e(t) 200 sin(qt 40°) of Figure 16–8(a) and (b) can be represented by its phasor equivalent, E 200 V∠40°, as in (c).

j e ⫹ e ⫺ (a) e = 200 sin (ωt 40°) V

E = 200 V∠40°

200 ωt

40° (b) Waveform

FIGURE 16–8 Transforming e 200 sin(qt 40°) V to E 200 V∠40°.

0V 40°

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R, L, and C Elements and the Impedance Concept

⫹ E = 200 V∠40° ⫺ Transformed source FIGURE 16–9 Direct transformation of the source.

We can take advantage of this equivalence. Rather than show a source as a time-varying voltage e(t) that we subsequently convert to a phasor, we can represent the source by its phasor equivalent right from the start. This viewpoint is illustrated in Figure 16–9. Since E 200 V∠40°, this representation retains all the original information of Figure 16–8 with the exception of the sinusoidal time variation. However, since the sinusoidal waveform and time variation is implicit in the definition of a phasor, you can easily restore this information if it is needed. The idea illustrated in Figure 16–9 is of fundamental importance to circuit theory. By replacing the time function e(t) with its phasor equivalent E, we have transformed the source from the time domain to the phasor domain. The value of this approach is illustrated next. Before we move on, we should note that both Kirchhoff’s voltage law and Kirchhoff’s current law apply in the time domain (i.e., when voltages and currents are expressed as functions of time) and in the phasor domain (i.e., when voltages and currents are represented as phasors). For example, e v1 v2 in the time domain can be transformed to E V1 V2 in the phasor domain and vice versa. Similarly for currents.

Summing AC Voltages and Currents Sinusoidal quantites must sometimes be added or subtracted as in Figure 16–10. Here, we want the sum of e1 and e2, where e1 10 sin qt and e2 15 sin(qt 60°). The sum of e1 and e2 can be found by adding waveforms point by point as in (b). For example, at qt 0°, e1 10 sin 0° 0 and e2 15 sin(0° 60°) 13 V, and their sum is 13 V. Similarly, at qt 90°, e1 10 sin 90° 10 V and e2 15 sin(90° 60°) 15 sin 150° 7.5, and their sum is 17.5 V. Continuing in this manner, the sum of e1 e2 (the green waveform) is obtained.

17.5 V

⫹ e1 ⫺ ⫹ e2 ⫺

⫹ v (t) ⫺

v = e1 e 2

13.0 V

e1 = 10 sin ωt 0 60°

90°

180°

270°

ωt

360°

e2 = 15 sin (ωt 60°) (a) e1 = 10 sin ωt e2 = 15 sin (ωt 60°)

(b) Waveforms

FIGURE 16–10 Summing waveforms point by point.

As you can see, the process is tedious and provides no analytic expression for the resulting voltage. A better way is to transform the sources and use complex numbers to perform the addition. This is shown in Figure 16–11.

Section 16.2

■

613

Complex Numbers in AC Analysis

Here, we have replaced voltages e1 and e2 with their phasor equivalents, E1 and E2, and v with its phasor equivalent, V. Since v e1 e2, replacing v, e1, and e2 with their phasor equivalents yields V E1 E2. Now V can be found by adding E1 and E2 as complex numbers. Once V is known, its corresponding time equation and companion waveform can be determined.

EXAMPLE 16–6 Given e1 10 sin qt V and e2 15 sin(qt 60°) V as before, determine v and sketch it.

⫹ e1 ⫺ ⫹ e2 ⫺

⫹ v ⫺

(a) Original network. v(t) = e1(t) e2(t)

Solution e1 10 sin qt V. Thus, E1 10 V∠0°. e2 15 sin(qt 60°) V. Thus, E2 15 V∠60°. Transformed sources are shown in Figure 16–12(a) and phasors in (b).

Thus,

V E1 E2 10∠0° 15∠60° (10 j0) (7.5 j13) (17.5 j13) 21.8 V∠36.6° v 21.8 sin(qt 36.6°) V

Waveforms are shown in (c).

⫹ FIGURE 16–11 Transformed circuit. This is one of the key ideas of sinusoidal circuit analysis.

V = 21.8 V ∠36.6°

⫹ 15 V∠60° ⫺

⫺

(a) Phasor summation

21.8 V v(t) V e1(t)

60°

e2(t)

36.6°

ωt

E1 60°

(b) Phasors FIGURE 16–12

⫹ E2 ⫺

⫹ V ⫺

(b) Transformed network. V = E1 E 2

⫹ 10 V∠0° ⫺

⫹ E1 ⫺

36.6° (c) Waveforms

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R, L, and C Elements and the Impedance Concept

Verify by direct substitution that v 21.8 sin(qt 36.6°) V, as in Figure 16–12, is the sum of e1 and e2. To do this, compute e1 and e2 at a point, add them, then compare the sum to 21.8 sin(qt 36.6°) V computed at the same point. Perform this computation at qt 30° intervals over the complete cycle to satisfy yourself that the result is true everywhere. (For example, at qt 0°, v 21.8 sin(qt 36.6°) 21.8 sin(36.6°) 13 V, as we saw earlier in Figure 16–10.) Answer: Here are the points on the graph at 30° intervals: ␻t 0° 30° 60° 90° 120° 150° 180° 210° 240°

270°

300° 330° 360°

v 13 20 21.7 17.5 8.66 2.5 13 20 21.7 17.5 8.66 2.5

IMPORTANT NOTES... 1. To this point, we have used peak values such as Vm and Im to represent the magnitudes of phasor voltages and currents, as this has been most convenient for our purposes. In practice, however, rms values are used instead. Accordingly, we will now change to rms. Thus, from here on, the phasor V 120 V ∠0° will be taken to mean a voltage of 120 volts rms at an angle of 0°. If you need to convert this to a time function, first multiply the rms value by 兹2苶, then follow the usual procedure. Thus, v 兹2 苶 (120) sin qt 170 sin qt. 2. To add or subtract sinusoidal voltages or currents, follow the three steps outlined in Example 16–6. That is, • convert sine waves to phasors and express them in complex number form, • add or subtract the complex numbers, • convert back to time functions if desired. 3. Although we use phasors to represent sinusoidal waveforms, it should be noted that sine waves and phasors are not the same thing. Sinusoidal voltages and currents are real—they are the actual quantities that you measure with meters and whose waveforms you see on oscilloscopes. Phasors, on the other hand, are mathematical abstractions that we use to help visualize relationships and solve problems. 4. Quantities expressed as time functions are said to be in the time domain, while quantities expressed as phasors are said to be in the phasor (or frequency) domain. Thus, e 170 sin qt V is in the time domain, while V 120 V ∠0° is in the phasor domain.

Section 16.2

■

Complex Numbers in AC Analysis

EXAMPLE 16–7

Express the voltages and currents of Figure 16–13 in both the time and the phasor domains. v (V)

100

40 mA ωt

ωt

80°

25° (a)

(b)

FIGURE 16–13

Solution a. Time domain: v 100 sin(qt 80°) volts. Phasor domain: V (0.707)(100 V ∠80°) 70.7 V∠80°. b. Time domain: i 40 sin(qt 25°) mA. Phasor domain: I (0.707)(40 mA∠ 25°) 28.3 mA∠ 25°.

If i1 14.14 sin(qt 55°) A and i2 4 sin(qt 15°) A, determine their sum, i. Work with rms values.

EXAMPLE 16–8 Solution

I1 (0.707)(14.14 A)∠ 55° 10 A∠ 55° I2 (0.707)(4 A)∠15° 2.828 A∠15° I I1 I2 10 A∠ 55° 2.828 A∠15° (5.74 A j8.19 A) (2.73 A j0.732 A) 8.47 A j7.46 A 11.3 A∠ 41.4 i(t) 兹2 苶(11.3) sin(qt 41.4°) 16 sin(qt 41.4°) A

While it may seem silly to convert peak values to rms and then convert rms back to peak as we did here, we did it for a reason. The reason is that very soon, we will stop working in the time domain entirely and work only with phasors. At that point, the solution will be complete when we have the answer in the form I 11.3∠ 41.4°. (To help focus on rms, voltages and currents in the next two examples (and in other examples to come) are expressed as an rms value times 兹2苶.)

615

616

Chapter 16

■

R, L, and C Elements and the Impedance Concept

EXAMPLE 16–9 For Figure 16–14, v1 兹2苶(16) sin qt V, v2 兹2苶(24) sin(qt 90°) and v3 兹2 苶(15) sin(qt 90°) V. Determine source voltage e. v1

FIGURE 16–14

Solution The answer can be obtained by KVL. First, convert to phasors. Thus, V1 16 V∠0°, V2 24 V∠90°, and V3 15 V∠ 90°. KVL yields E V1 V2 V3 16 V∠0° 24 V∠90° 15 V∠ 90° 18.4 V∠29.4°. Converting back to a function of time yields e 兹2苶(18.4) sin(qt 29.4°) V.

EXAMPLE 16–10 For Figure 16–15, i1 兹2苶(23) sin qt mA, i2 兹2苶(0.29) sin(qt 63°)A and i3 兹2 苶(127) 10 3 sin(qt 72°)A. Determine current iT. iT

FIGURE 16–15

Solution Convert to phasors. Thus, I1 23 mA∠0°, I2 0.29 A∠63°, and I3 127 10 3 A∠ 72°. KCL yields IT I1 I2 I3 23 mA∠0° 290 mA∠63° 127 mA∠ 72° 238 mA∠35.4°. Converting back to a function of time yields iT 兹2 苶(238) sin(qt 35.4°) mA. PRACTICE PROBLEMS 3

1. Convert the following to time functions. Values are rms. a. E 500 mV∠ 20° b. I 80 A∠40° 2. For the circuit of Figure 16–16, determine voltage e1. FIGURE 16–16

e2 = 141.4 sin (ωt 30°) V ⫹ ⫺ ⫹ e1 ⫺

Answers: 1. a. e 707 sin(qt 20°) mV 2. e1 221 sin(qt 99.8°) V

⫹ v = 170 sin (ωt 60°) V ⫺

b. i 113 sin(qt 40°) A

Section 16.4

■

1. Convert the following to polar form: a. j6 b. j4 c. 3 j3 d. 4 j6 e. 5 j8 f. 1 j2 g. 2 j3 2. Convert the following to rectangular form: a. 4∠90° b. 3∠0° c. 2∠ 90° d. 5∠40° e. 6∠120° f. 2.5∠ 20° g. 1.75∠ 160° 3. If C 12∠ 140°, what is C? 4. Given: C1 36 j4 and C2 52 j11. Determine C1 C2 and C1 C2. Express in rectangular form. 5. Given: C1 24∠25° and C2 12∠ 125°. Determine C1 C2 and C1/C2. 6. Compute the following and express answers in rectangular form: (12∠0°)(14 j2) 6 j4 a. (14 j2) b. (1 j6) 2 6 (10∠20°)(2∠ 10°) 10∠20° 7. For Figure 16–17, determine iT where i1 10 sin qt, i2 20 sin(qt 90°), and i3 5 sin(qt 90°).

冤

FIGURE 16–17

IN-PROCESS

LEARNING CHECK 1

冥

iT = i1 i2 i3 i1

(Answers are at the end of the chapter.)

16.3

R, L, and C Circuits with Sinusoidal Excitation

R, L, and C circuit elements each have quite different electrical properties. Resistance, for example, opposes current, while inductance opposes changes in current, and capacitance opposes changes in voltage. These differences result in quite different voltage-current relationships as you saw earlier. We now investigate these relationships for the case of sinusoidal ac. Sine waves have several important characteristics that you will discover from this investigation: 1. When a circuit consisting of linear circuit elements R, L, and C is connected to a sinusoidal source, all currents and voltages in the circuit will be sinusoidal. 2. These sine waves have the same frequency as the source and differ from it only in terms of their magnitudes and phase angles.

16.4

Resistance and Sinusoidal AC

We begin with a purely resistive circuit. Here, Ohm’s law applies and thus, current is directly proportional to voltage. Current variations therefore follow voltage variations, reaching their peak when voltage reaches its peak,

617

618

Chapter 16

■

R, L, and C Elements and the Impedance Concept

changing direction when voltage changes polarity, and so on (Figure 16–18). From this, we conclude that for a purely resistive circuit, current and voltage are in phase. Since voltage and current waveforms coincide, their phasors also coincide (Figure 16–19). vR ⫹ e ⫺

⫹ vR ⫺

iR t

(a) Source voltage is a sine wave. Therefore, vR is a sine wave.

(b) iR = vR/R. Therefore iR is a sine wave.

FIGURE 16–18 Ohm’s law applies to resistors. Note that current and voltage are in phase.

j IR

The relationship illustrated in Figure 16–18 may be stated mathematically as

VR FIGURE 16–19 For a resistor, voltage and current phasors are in phase.

v Vm sin qt Vm iR R sin qt Im sin qt R R R

(16–10)

Im Vm/R

(16–11)

Vm ImR

(16–12)

where

Transposing, The in-phase relationship is true regardless of reference. Thus, if vR Vm sin(qt v), then iR Im sin(qt v).

EXAMPLE 16–11 For the circuit of Figure 16–18(a), if R 5 ⍀ and iR 12 sin(qt 18) A, determine vR.

Solution vR RiR 5 12 sin(qt 18°) 60 sin(qt 18°) V. The waveforms are shown in Figure 16–20. 60 V

12 A

3␲ 2 2␲

iR ␲ 2 FIGURE 16–20

␲ 18°

ωt

Section 16.5

■

1. If vR 150 cos qt V and R 25 k⍀, determine iR and sketch both waveforms. 2. If vR 100 sin(qt 30°) V and R 0.2 M⍀, determine iR and sketch both waveforms.

Inductance and Sinusoidal AC

619

PRACTICE PROBLEMS 4

Answers: 1. iR 6 cos qt mA. vR and iR are in phase. 2. iR 0.5 sin(qt 30°) mA. vR and iR are in phase.

16.5

Inductance and Sinusoidal AC

Phase Lag in an Inductive Circuit As you saw in Chapter 13, for an ideal inductor, voltage vL is proportional to the rate of change of current. Because of this, voltage and current are not in phase as they are for a resistive circuit. This can be shown with a bit of calculus. From Figure 16–21, vL LdiL/dt. For a sine wave of current, you get when you differentiate di d vL L L L (Im sin qt) qLIm cos qt Vm cos qt dt dt

Utilizing the trigonometric identity cos qt sin(qt 90°), you can write this as vL Vm sin(qt 90°)

(16–13)

Vm qLIm

(16–14)

iL L

⫹ di vL = L L dt ⫺

FIGURE 16–21 Voltage vL is proportional to the rate of change of current iL.

where

Voltage and current waveforms are shown in Figure 16–22, and phasors in Figure 16–23. As you can see, for a purely inductive circuit, current lags voltage by 90° (i.e., 1⁄ 4 cycle). Alternatively you can say that voltage leads current by 90°.

Vm Im

vL iL 2π π

3π

j ωt

FIGURE 16–22 For inductance, current lags voltage by 90°. Here iL is reference.

Although we have shown that current lags voltage by 90° for the case of Figure 16–22, this relationship is true in general, that is, current always lags voltage by 90° regardless of the choice of reference. This is illustrated in Figure 16–24. Here, VL is at 0° and IL at 90°. Thus, voltage vL will be a sine wave and current iL a negative cosine wave, i.e., iL Im cos qt. Since

VL IL

FIGURE 16–23 Phasors for the waveforms of Fig. 16–22 showing the 90° lag of current.

620

Chapter 16

■

R, L, and C Elements and the Impedance Concept

iL is a negative cosine wave, it can also be expressed as iL Im sin(qt 90°). The waveforms are shown in (b). Since current always lags voltage by 90° for a pure inductance, you can, if you know the phase of the voltage, determine the phase of the current, and vice versa. Thus, if vL is known, iL must lag it by 90°, while if iL is known, vL must lead it by 90°.

Inductive Reactance From Equation 16–14, we see that the ratio Vm to Im is

(a) Current IL always lags voltage VL by 90°

2␲ ␲ 2

␲

3␲ 2

(16–15)

This ratio is defined as inductive reactance and is given the symbol XL. Since the ratio of volts to amps is ohms, reactance has units of ohms. Thus,

Vm qL Im

Vm XL Im

ωt

(⍀)

(16–16)

Combining Equations 16–15 and 16–16 yields XL qL (⍀)

(b) Waveforms FIGURE 16–24 Phasors and waveforms when VL is used as reference.

(16–17)

where q is in radians per second and L is in henries. Reactance XL represents the opposition that inductance presents to current for the sinusoidal ac case. We now have everything that we need to solve simple inductive circuits with sinusoidal excitation, that is, we know that current lags voltage by 90° and that their amplitudes are related by Vm Im XL

(16–18)

Vm ImXL

(16–19)

and

EXAMPLE 16–12 The voltage across a 0.2-H inductance is vL 100 sin (400t 70°) V. Determine iL and sketch it. Solution

q 400 rad/s. Therefore, XL qL (400)(0.2) 80 ⍀. Vm 100 V Im 1.25 A XL 80 ⍀

The current lags the voltage by 90°. Therefore iL 1.25 sin(400t 20°) A as indicated in Figure 16–25.

Section 16.5 vL = 100 sin (400t 70°)V

100 V

j VL = 70.7 V∠70°

1.25 A

70°

3␲ 2 ␲ 2 20°

70°

IL = 0.884 A∠ 20° 20°

■

Inductance and Sinusoidal AC

NOTE… Remember to show phasors as rms values from now on.

2␲ ωt

␲

iL = 1.25 sin (400t 20°)A

90°

(b)

(a)

FIGURE 16–25 With voltage VL at 70°, current IL will be 90° later at 20°.

The current through a 0.01-H inductance is iL 20 sin(qt 50°) A and f 60 Hz. Determine vL.

EXAMPLE 16–13 Solution

q 2pf 2p(60) 377 rad/s XL qL (377)(0.01) 3.77 ⍀ Vm Im XL (20 A)(3.77 ⍀) 75.4 V Voltage leads current by 90°. Thus, vL 75.4 sin(377t 40°) V as shown in Figure 16–26.

vL = 75.4 sin (ωt 40°)

75.4 V

20 A VL = 53.3 V 40°

90°

50°

40°

␲ 2

40°

3␲ 2

␲

2␲ ωt

50° IL = 14.1 A 50°

iL = 20 sin (ωt 50°)

(a)

(b)

FIGURE 16–26

1. Two inductances are connected in series (Figure 16–27). If e 100 sin qt and f 10 kHz, determine the current. Sketch voltage and current waveforms. FIGURE 16–27

0.6 H ⫹ e(t) ⫺

0.2 H

621

PRACTICE PROBLEMS 5

622

Chapter 16

■

R, L, and C Elements and the Impedance Concept 2. The current through a 0.5-H inductance is iL 100 sin(2400t 45°) mA. Determine vL and sketch voltage and current phasors and waveforms. Answers: 1. iL 1.99 sin(qt 90°) mA. Waveforms same as Figure 16–24. 2. vL 120 sin(2400t 135° V). See following art for waveforms. j

iL 135° 45°

45°

3π

2π

ωt

135°

Variation of Inductive Reactance with Frequency Since XL qL 2pfL, inductive reactance is directly proportional to frequency (Figure 16–28). Thus, if frequency is doubled, reactance doubles, while if frequency is halved, reactance halves, and so on. In addition, XL is directly proportional to inductance. Thus, if inductance is doubled, XL is doubled, and so on. Note also that at f 0, XL 0 ⍀. This means that inductance looks like a short circuit to dc. (We already concluded this earlier in Chapter 13.)

XL (⍀) L2 7500

6000 4500 3000 1500 0

f (Hz) 100 200 300 400 500

FIGURE 16–28 Variation of XL with frequency. Note that L2 L1.

PRACTICE PROBLEMS 6

A circuit has 50 ohms inductive reactance. If both the inductance and the frequency are doubled, what is the new XL? Answer: 200 ⍀

Section 16.6

16.6

■

Capacitance and Sinusoidal AC

623

Capacitance and Sinusoidal AC

Phase Lead in a Capacitive Circuit For capacitance, current is proportional to the rate of change of voltage, i.e., iC C dvC /dt [Figure 16–29(a)]. Thus if vC is a sine wave, you get upon substitution dvC d iC C C (Vm sin qt) qCVm cos qt Im cos qt dt dt

Vm Im iC

(a) iC = C

dvC dt

ωt

3π

IC IC

2π

⫹ vC ⫺

60º VC

(b) Waveforms with vC as reference

VC (d) VC at −30º

FIGURE 16–29 For capacitance, current always leads voltage by 90°.

Using the appropriate trigonometric identity, this can be written as iC Im sin(qt 90°)

(16–20)

Im qCVm

(16–21)

where

Waveforms are shown in Figure 16–29(b) and phasors in (c). As indicated, for a purely capacitive circuit, current leads voltage by 90°, or alternatively, voltage lags current by 90°. This relationship is true regardless of reference. Thus, if the voltage is known, the current must lead by 90° while if the current is known, the voltage must lag by 90°. For example, if IC is at 60° as in (d), VC must be at 30°.

1. The current source of Figure 16–30(a) is a sine wave. Sketch phasors and capacitor voltage vC.

iC ig

(a) FIGURE 16–30

iC = ig ⫹ vC ⫺

3␲ 2 2␲

Im 0

␲ 2

␲ (b)

ωt

30º

PRACTICE PROBLEMS 7

624

Chapter 16

■

R, L, and C Elements and the Impedance Concept 2. Refer to the circuit of Figure 16–31(a): a. Sketch the phasors. b. Sketch capacitor current iC. e ⫹ e ⫺

Em ωt

45°

(b)

(a) FIGURE 16–31

Answers: 1. IC is at 0°; VC is at 90°; vC is a negative cosine wave. 2. a. VC is at 45° and IC is at 135°. b. Waveforms are the same as for Problem 2, Practice Problem 5, except that voltage and current waveforms are interchanged.

Capacitive Reactance Now consider the relationship between maximum capacitor voltage and current magnitudes. As we saw in Equation 16–21, they are related by Im qCVm. Rearranging, we get Vm /Im 1/qC. The ratio of Vm to Im is defined as capacitive reactance and is given the symbol XC. That is, Vm XC (⍀) Im

Since Vm /Im 1/qC, we also get 1 XC (⍀) qC

(16–22)

where q is in radians per second and C is in farads. Reactance XC represents the opposition that capacitance presents to current for the sinusoidal ac case. We now have everything that we need to solve simple capacitive circuits with sinusoidal excitation, i.e., we know that current leads voltage by 90° and that Vm Im XC

(16–23)

Vm Im XC

(16–24)

and

Section 16.6

■

The voltage across a 10-mF capacitance is vC 100 sin(qt 40°) V and f 1000 Hz. Determine iC and sketch its waveform.

EXAMPLE 16–14 Solution

q 2pf 2p(1000 Hz) 6283 rad/s 1 1 XC 15.92 ⍀ qC (6283)(10 10 6) Vm 100 V Im 6.28 A XC 15.92 ⍀ Since current leads voltage by 90°, iC 6.28 sin(6283t 50°) A as indicated in Figure 16–32. 100 V j

6.28 A

3␲ 2

IC 50°

␲ ␲ 2 40° iC

50°

40° VC

2␲

ωt

(b)

(a)

FIGURE 16–32 Phasors are not to scale with waveform.

The current through a 0.1-mF capacitance is iC 5 sin(1000t 120°) mA. Determine vC.

EXAMPLE 16–15 Solution

1 1 XC 10 k⍀ (1000 rad/s)(0.1 10 6 F) qC Thus, Vm Im XC (5 mA)(10 k⍀) 50 V. Since voltage lags current by 90°, vC 50 sin(1000t 30°) V. Waveforms and phasors are shown in Figure 16–33. vC

50 V 5 mA

j IC 30°

(a)

ωt

120°

30° 120° iC (b)

FIGURE 16–33 Phasors are not to scale with waveform.

Capacitance and Sinusoidal AC

625

626

Chapter 16

■

PRACTICE PROBLEMS 8

R, L, and C Elements and the Impedance Concept

Two capacitances are connected in parallel (Figure 16–34). If e 100 sin qt V and f 10 Hz, determine the source current. Sketch current and voltage phasors and waveforms. i

FIGURE 16–34 ⫹ e(t) ⫺

100 ␮F

50 ␮F

Answer: i 0.942 sin(62.8t 90°) 0.942 cos 62.8t A See Figure 16–29(b) and (c).

Variation of Capacitive Reactance with Frequency Since XC 1/qC 1/2pfC, the opposition that capacitance presents varies inversely with frequency. This means that the higher the frequency, the lower the reactance, and vice versa (Figure 16–35). At f 0 (i.e., dc), capacitive reactance is infinite. This means that a capacitance looks like an open circuit to dc. (We already concluded this earlier in Chapter 10.) Note that XC is also inversely proportional to capacitance. Thus, if capacitance is doubled, XC is halved, and so on. XC (⍀) 3183 ⍀

3000 2000 1000 0

1591 ⍀

1061 ⍀ 797 ⍀

637 ⍀ 5

f (kHz)

FIGURE 16–35 XC varies inversely with frequency. Values shown are for C 0.05 mF.

IN-PROCESS

LEARNING CHECK 2

1. For a pure resistance, vR 100 sin(qt 30°) V. If R 2 ⍀, what is the expression for iR? 2. For a pure inductance, vL 100 sin(qt 30°) V. If XL 2 ⍀, what is the expression for iL? 3. For a pure capacitance, vC 100 sin(qt 30°) V. If XC 2 ⍀, what is the expression for iC? 4. If f 100 Hz and XL 400 ⍀, what is L? 5. If f 100 Hz and XC 400 ⍀, what is C? 6. For each of the phasor sets of Figure 16–36, identify whether the circuit is resistive, inductive, or capacitive. Justify your answers.

Section 16.7 j j

(a)

627

The Impedance Concept

■

(b)

(c)

(d)

FIGURE 16–36 (Answers are at the end of the chapter.)

16.7

The Impedance Concept

In Sections 16.5 and 16.6, we handled magnitude and phase analysis separately. However, this is not the way it is done in practice. In practice, we represent circuit elements by their impedance, and determine magnitude and phase relationships in one step. Before we do this, however, we need to learn how to represent circuit elements as impedances.

Impedance The opposition that a circuit element presents to current in the phasor domain is defined as its impedance. The impedance of the element of Figure 16–37, for example, is the ratio of its voltage phasor to its current phasor. Impedance is denoted by the boldface, uppercase letter Z. Thus, V Z (ohms) I

(16–25)

(This equation is sometimes referred to as Ohm’s law for ac circuits.) Since phasor voltages and currents are complex, Z is also complex. That is, V V Z ∠v I I

⫹ V ⫺

V ohms I

FIGURE 16–37 Impedance concept.

(16–26)

where V and I are the rms magnitudes of V and I respectively, and v is the angle between them. From Equation 16–26, Z Z∠v

(16–27)

where Z V/I. Since V 0.707Vm and I 0.707Im, Z can also be expressed as Vm/Im. Once the impedance of a circuit is known, the current and voltage can be determined using

NOTES...

V I Z

(16–28)

V IZ

(16–29)

and Let us now determine impedance for the basic circuit elements R, L, and C.

Although Z is a complex number, it is not a phasor since it does not represent a sinusoidally varying quantity.

628

Chapter 16

■

R, L, and C Elements and the Impedance Concept Resistance

I ⫹ VR ⫺

(a) Voltage and current

ZR = R ⍀

For a pure resistance (Figure 16–38), voltage and current are in phase. Thus, if voltage has an angle v, current will have the same angle. For example, if VR VR∠v, then I I∠v. Substituting into Equation 16–25 yields: V VR ∠v VR ZR R ∠0° R I I∠v I

(b) Impedance

Thus the impedance of a resistor is just its resistance. That is,

FIGURE 16–38 Impedance of a pure resistance.

ZR R

(16–30)

This agrees with what we know about resistive circuits, i.e., that the ratio of voltage to current is R, and that the angle between them is 0°. Inductance I

+ VL

ZL = jωL

For a pure inductance, current lags voltage by 90°. Assuming a 0° angle for voltage (we can assume any reference we want because we are interested only in the angle between VL and I), we can write VL VL∠0° and I I∠ 90°. The impedance of a pure inductance (Figure 16–39) is therefore

−

(a) Voltage and current

V VL VL∠0° ZL L ∠90° qL∠90° jqL I I I∠ 90° (b) Impedance

where we have used the fact that VL/IL qL. Thus, ZL jqL jXL

FIGURE 16–39 Impedance of a pure inductance.

(16–31)

since qL is equal to XL.

Consider again Example 16–12. Given v L 100 sin(400t 70°) and L 0.2 H, determine iL using the impedance concept.

EXAMPLE 16–16 Solution

See Figure 16–40. VL 70.7 V∠70° and q 400 rad/s ZL jqL j(400)(0.2) j80 ⍀ VL 70.7∠70° 70.7∠70° IL 0.884 A∠ 20° ZL j80 80∠90°

In the time domain, iL 兹2苶(0.884) sin(400t 20°) 1.25 sin(400t 20°) A, which agrees with our previous solution. IL

ZL = j80 ⍀ FIGURE 16–40

⫹ VL = 70.7 V∠0° ⫺

Section 16.7

■

The Impedance Concept

629

Capacitance

For a pure capacitance, current leads voltage by 90°. Its impedance (Figure 16–41) is therefore VC VC∠0° VC 1 1 ZC ∠ 90° ∠ 90° j (ohms) I I∠90° I qC qC

ZC = j 1 ωC

Thus, 1 ZC j jXC qC

(ohms)

(16–32)

FIGURE 16–41 Impedance of a pure capacitance.

since 1/qC is equal to XC.

EXAMPLE 16–17 Given vC 100 sin(qt 40°), f 1000 Hz, and C 10 mF, determine iC in Figure 16–42. Solution q 2pf 2p(1000 Hz) 6283 rads/s VC 70.7 V∠ 40°

冢

冣

1 1 ZC j j j15.92 ⍀. qC 6283 10 10 6 VC 70.7∠ 40° 70.7∠ 40° IC 4.442 A∠50° ZC j15.92 15.92∠ 90° In the time domain, iC 兹2 苶(4.442) sin(6283t 50°) 6.28 sin(6283t 50°) A, which agrees with our previous solution, in Example 16–14. IC

ZC = j15.92 ⍀

⫹ VC = 70.7 V∠ 40° ⫺

FIGURE 16–42

1. If IL 5 mA∠ 60°, L 2 mH, and f 10 kHz, what is VL? 2. A capacitor has a reactance of 50 ⍀ at 1200 Hz. If vC 80 sin 800t V, what is iC? Answers: 1. 628 mV∠30° 2. 0.170 sin(800t 90°) A

A Final Note The real power of the impedance method becomes apparent when you consider complex circuits with elements in series, parallel, and so on. This we do

PRACTICE PROBLEMS 9

630

Chapter 16

■

R, L, and C Elements and the Impedance Concept

later, beginning in Chapter 18. Before we do this, however, there are some ideas on power that you need to know. These are considered in Chapter 17.

16.8

ELECTRONICS WORKBENCH

PSpice

Computer Analysis of AC Circuits

In Chapter 15, you saw how to represent sinusoidal waveforms using PSpice and Electronics Workbench. Let us now apply these tools to the ideas of this chapter. To illustrate, recall that in Example 16–6, we summed voltages e1 10 sin qt V and e2 15 sin(qt 60°) V using phasor methods to obtain their sum v 21.8 sin(qt 36.6°) V. We will now verify this summation numerically. Since the process is independent of frequency, let us choose f 500 Hz. This yields a period of 2 ms; thus, 1⁄2 cycle is 1 ms, 1⁄4 cycle (90°) is 500 ms, 45° is 250 ms, etc.

Electronics Workbench As noted in Chapter 15, Electronics Workbench requires that you enter rms values for waveforms. Thus, to represent 10 sin qt V, enter (0.7071)(10 V) 7.071 V as its magnitude, and to represent 15 sin(qt 60°) V, enter 10.607 V. You must also enter its angle. Procedure: Create the circuit of Figure 16–43(a) on the screen. Set Source 1 to 7.071 V, 0 deg, and 500 Hz using the procedure of Chapter 15. Similarly, set Source 2 to 10.607 V, 60 deg, and 500 Hz. Click Analysis, Transient, set TSTOP to 0.002 (to run the solution out to 2 ms so that you display a full cycle) and TMAX to 2e-06 (to avoid getting a choppy waveform.) Highlight Node 1 (to display e1) and click Add. Repeat for Node 2 (to display v) Click Simulate. Following simulation, graphs e1 (the red curve, left-hand scale) and v (the blue curve, right-hand scale) appear. Expand the Analysis Graph window to full screen, then click the cursor icon and drag a cursor to 500 ms or as close as you can get it and read values. (You should get about 10 V for e1 and 17.5 V for v.) Scale values from the graph at 200-ms increments and tabulate. Now replace the sources with a single source of 21.8

(a) The circuit

(b) Waveforms e1 and v

FIGURE 16–43 Summing sinusoidal waveforms with Electronics Workbench.

Section 16.8

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Computer Analysis of AC Circuits

631

sin(qt 36.6°) V. (Don’t forget to convert to rms). Run a simulation. The resulting graph should be identical to v (the blue curve) of Figure 16–43(b).

OrCAD PSpice With PSpice, you can plot all three waveforms of Fig. 16–10 simultaneously. Procedure: Create the circuit of Fig. 16–44 using source VSIN. Double click Source 1 and select Parts in the Property Editor. Set VOFF to OV, VAMPL to 10V, FREQ to 500Hz, and PHASE to 0deg. Click Apply, then close the Property Editor. Set up source 2 similarly. Now place markers as shown (Note 2) so that PSpice will automatically create the plots. Click the New Simulation Profile icon, choose Transient, set TSTOP to 2ms (to display a full cycle), set Maximum Step Size to 1us (to yield a smooth plot), then click OK. Run the simulation and the waveforms of Figure 16–45 should

FIGURE 16–44 Summing sinusoidal waveforms with PSpice. Source 2 is displayed using a differential marker.

FIGURE 16–45 PSpice waveforms. Compare to Figure 16–10.

NOTES… 1. Make sure that the polarities of the sources are as indicated. (You will have to rotate V2 three times to get it into the position shown.) 2. To display V2, use differential markers (indicated as and on the toolbar).

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appear. Using the cursor, scale voltages at 500 ms. You should get 10 V for e1, 7.5 V for e2 and 17.5 V for v. Read values at 200-ms intervals and tabulate. Now replace the sources of Figure 16–44 with a single source of 21.8 sin(qt 36.6°) V, run a simulation, scale values, and compare results. They should agree. Another Example

PSpice makes it easy to study the response of circuits over a range of frequencies. This is illustrated in Example 16–18.

EXAMPLE 16–18 Compute and plot the reactance of a 12-␮F capacitor over the range 10 Hz to 1000 Hz. Solution PSpice has no command to compute reactance; however, we can calculate voltage and current over the desired frequency range, then plot their ratio. This gives reactance. Procedure: Create the circuit of Figure 16–46 on the screen. (Use source VAC here as it is the source to use for phasor analyses—see Appendix A). Note its default of 0V. Double click the default value (not the symbol) and in the dialog box, enter 120V, then click OK. Click the New Simulations Profile icon, enter fig 16-46 for a name and then in the dialog box that opens, select AC Sweep/Noise. For the Start Frequency, key in 10Hz; for the End Frequency, key 1kHz; set AC Sweep type to Logarithmic, select Decade and type 100 into the Pts/Decade (points per decade) box. Run the simultation and a set of empty axes appears. Click Trace, Add Trace and in the dialog box, click V1(C1), press the / key on the keyboard, then click I(C1) to yield the ratio V1(C1)/I(C1) (which is the capacitor’s reactance). Click OK and PSpice will compute and plot the capacitor’s reactance versus frequency, Figure 16–47. Compare its shape to Figure 16–35. Use the cursor to scale some values off the screen and verify each point using XC 1/qC.

FIGURE 16–46 Circuit for plotting reactance.

Section 16.8

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Computer Analysis of AC Circuits

FIGURE 16–47 Reactance for a 12-mF capacitor versus frequency.

Phasor Analysis

As a last example, we will show how to use PSpice to perform phasor analysis—i.e., to solve problems with voltages and currents expressed in phasor form. To illustrate, consider again Example 16–17. Recall, VC 70.7 V∠ 40°, C 10 mF, and f 1000 Hz. Procedure: Create the circuit on the screen (Figure 16–48) using source VAC and component IPRINT (Note 1). Double click the VAC symbol and in the Property Editor, set ACMAG to 70.7V and ACPHASE to ⫺40deg. (See Note 2). Double click IPRINT and in the Property Editor, type yes into cells AC, MAG, and PHASE. Click Apply and close the editor. Click the New Simulation Profile icon, select AC Sweep/Noise, Linear, set both Start Frequency and End Frequency to

FIGURE 16–48 Phasor analysis using PSpice. Component IPRINT is a software ammeter.

633

NOTES… 1. Component IPRINT is a software ammeter, found in the SPECIAL parts library. In this example, we configure it to display ac current in magnitude and phase angle format. Make sure that it is connected as shown in Figure 16–48, since if it is reversed, the phase angle of the measured current will be in error by 180°. 2. If you want to display the phase of the source voltage on the schematic as in Figure 16–48, double click the source symbol and in the Property Editor, click ACPHASE, Display, then select Value Only. 3. The results displayed by IPRINT are expressed in exponential format. Thus, frequency (Figure 16–49) is shown as 1.000E 03, which is 1.000 103 1000 Hz, etc.

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1000Hz and Total Points to 1. Run the simulation. When the simulation window opens, click View, Output File, then scroll until you find the answers (Figure 16–49 and Note 3). The first number is the frequency (1000 Hz), the second number (IM) is the magnitude of the current (4.442 A), and the third (IP) is its phase (50 degrees). Thus, IC 4.442 A∠50° as we determined earlier in Example 16–17.

FIGURE 16–49 Current for the circuit of Figure 16–48. I 4.442 A ∠50°.

PRACTICE PROBLEMS 10

PROBLEMS

Modify Example 16–18 to plot both capacitor current and reactance on the same graph. You will need to add a second Y-axis for the capacitor current. (See Appendix A if you need help.)

16.1 Complex Number Review 1. Convert each of the following to polar form: a. 5 j12 b. 9 j6 c. 8 j15 d. 10 j4 2. Convert each of the following to rectangular form: a. 6∠30° b. 14∠90° c. 16∠0° d. 6∠150° e. 20∠ 140° f. 12∠30° g. 15∠ 150° 3. Plot each of the following on the complex plane: a. 4 j6 b. j4 c. 6∠ 90° d. 10∠135° 4. Simplify the following using powers of j: a. j(1 j1) b. ( j)(2 j5) c. j[ j(1 j6)] d. ( j4)( j2 4) e. (2 j3)(3 j4) 5. Add or subtract as indicated. Express your answer in rectangular form. a. (4 j8) (3 j2) b. (4 j8) (3 j2) c. (4.1 j7.6) 12∠20° d. 2.9∠25° 7.3∠ 5° e. 9.2∠ 120° (2.6 j4.1) 6. Multiply or divide as indicated. Express your answer in polar form. a. (37 j9.8)(3.6 j12.3) b. (41.9∠ 80°)(16 j2) 42 j18.6 c. 19.1 j4.8

42.6 j187.5 d. 11.2∠38°

Problems 7. Reduce each of the following to polar form: 18∠40° (12 j8) a. 15 j6 11 j11 21∠20° j41 b. 36∠0° (1 j12) 11∠40° 16 j17 21∠ 60° 18∠40° 18∠ 40° c. 7 j12 4

冤

冥

16.2 Complex Numbers in AC Analysis 8. In the manner of Figure 16–9, represent each of the following as transformed sources. a. e 100 sin(qt 30°) V b. e 15 sin(qt 20°) V c. e 50 sin(qt 90°) V d. e 50 cos qt V e. e 40 sin(qt 120°) V f. e 80 sin(qt 70°) V 9. Determine the sinusoidal equivalent for each of the transformed sources of Figure 16–50. 10. Given: e1 10 sin(qt 30°) V and e2 15 sin(qt 20°) V. Determine their sum v e1 e2 in the manner of Example 16–6, i.e., a. Convert e1 and e2 to phasor form. b. Determine V E1 E2. c. Convert V to the time domain. d. Sketch e1, e2, and v as per Figure 16–12. 11. Repeat Problem 10 for v e1 e2. Note: For the remaining problems and throughout the remainder of the book, express phasor quantities as rms values rather than as peak values.

⫹ 10 V∠30° ⫺

(a)

⫹ 15 V∠ 10° ⫺

12. Express the voltages and currents of Figure 16–51 as time domain and phasor domain quantities. (b) v (V)

v (V)

FIGURE 16–50

70.7

150° ωt

ωt

(b)

(a) i (mA)

v (V)

100

10 ωt

ωt

200° (c) FIGURE 16–51

(d)

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FIGURE 16–52

13. For Figure 16–52, i1 25 sin (qt 36°) mA and i2 40 cos (qt 10°) mA. a. Determine phasors I1, I2 and IT. b. Determine the equation for iT in the time domain. 14. For Figure 16–52, iT 50 sin(qt 60°) A and i2 20 sin(qt 30°) A. a. Determine phasors IT and I2. b. Determine I1. c. From (b), determine the equation for i1. 15. For Figure 16–17, i1 7 sin qt mA, i2 4 sin (qt 90°) mA, and i3 6 sin (qt 90°) mA. a. Determine phasors I1, I2, I3 and IT. b. Determine the equation for iT in the time domain. 16. For Figure 16–17, iT 38.08 sin(qt 21.8°) A, i1 35.36 sin qt A, and i2 28.28 sin(qt 90°) A. Determine the equation for i3.

iL ⫹ e ⫺

FIGURE 16–53

+ vL −

16.4 to 16.6 17. For Figure 16–18(a), R 12 ⍀. For each of the following, determine the current or voltage and sketch. a. v 120 sin qt V, i b. v 120 sin (qt 27°) V, i c. i 17 sin (qt 56°) mA, v d. i 17 cos(qt 67°) mA, v 18. Given v 120 sin (qt 52°) V and i 15 sin (qt 52°) mA, what is R? 19. Two resistors R1 10 k⍀ and R2 12.5 k⍀ are in series. If i 14.7 sin(qt 39°) mA, a. What are vR1 and vR2? b. Compute vT vR1 vR2 and compare to vT calculated from vT i RT. 20. The voltage across a certain component is v 120 sin(qt 55°) V and its current is 18 cos(qt 145°) mA. Show that the component is a resistor and determine its value. 21. For Figure 16–53, Vm 10 V and Im 5 A. For each of the following, determine the missing quantity: a. vL 10 sin(qt 60°) V, iL b. vL 10 sin(qt 15°) V, iL c. iL 5 cos(qt 60°) A, vL d. iL 5 sin(qt 10°) A, vL 22. What is the reactance of a 0.5-H inductor at a. 60 Hz b. 1000 Hz c. 500 rad/s 23. For Figure 16–53, e 100 sin qt and L 0.5 H. Determine iL at a. 60 Hz b. 1000 Hz c. 500 rad/s 24. For Figure 16–53, let L 200 mH. a. If vL 100 sin377t V, what is iL? b. If iL 10 sin(2p 400t 60°) mA, what is vL?

Problems 25. For Figure 16–53, if a. vL 40 sin(qt 30°) V, iL 364 sin(qt 60°) mA, and L 2 mH, what is f? b. iL 250 sin(qt 40°) mA, vL 40 sin(qt v) V, and f 500 kHz, what are L and v? 26. Repeat Problem 21 if the given voltages and currents are for a capacitor instead of an inductor. 27. What is the reactance of a 5-mF capacitor at a. 60 Hz b. 1000 Hz c. 500 rad/s 28. For Figure 16–54, e 100 sin qt and C 5 mF. Determine iC at a. 60 Hz b. 1000 Hz c. 500 rad/s 29. For Figure 16–54, let C 50 mF. a. If vC 100 sin377t V, what is iC? b. If iC 10 sin(2p 400t 60°) mA, what is vC? 30. For Figure 16–54, if a. vC 362 sin(qt 33°) V, iC 94 sin(qt 57°) mA, and C 2.2 mF, what is f? b. iC 350 sin(qt 40°) mA, vC 3.6 sin(qt v) V, and f 12 kHz, what are C and v? 16.7 The Impedance Concept 31. Determine the impedance of each circuit element of Figure 16–55.

48 ⍀

(a)

(b) 0.1 H, 60 Hz

FIGURE 16–55

32. If E 100V∠0° is applied across each of the circuit elements of Figure 16–56: a. Determine each current in phasor form. b. Express each current in time domain form.

IL IC

50 ⍀

(a) EWB

FIGURE 16–56

j25 ⍀

(b)

j10 ⍀

(c)

⫹ e ⫺

FIGURE 16–54

iC + vC −

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IL = 2 A∠ 90°

100 V∠0°

33. If the current through each circuit element of Figure 16–56 is 0.5 A∠0°: a. Determine each voltage in phasor form. b. Express each voltage in time domain form. 34. For each of the following, determine the impedance of the circuit element and state whether it is resistive, inductive, or capacitive. a. V 240 V∠ 30°, I 4 A∠ 30°. b. V 40 V∠30°, I 4 A∠ 60°. c. V 60 V∠ 30°, I 4 A∠60°. d. V 140 V∠ 30°, I 14 mA∠ 120°. 35. For each circuit of Figure 16–57, determine the unknown. 36. a. If VL 120 V ∠67°, L 600 mH, and f 10 kHz, what is IL? b. If IL 48 mA ∠ 43°, L 550 mH, and f 700 Hz, what is VL? c. If VC 50 V ∠ 36°, C 390 pF, and f 470 kHz, what is IC?

d. If IC 95 mA ∠87°, C 6.5 nF, and f 1.2 MHz, what is VC? (a) L = 0.2 H. Determine f. IC = 0.4 A∠90°

100 V∠0°

16.8 Computer Analysis of AC Circuits The version of Electronics Workbench current at the time of writing of this book is unable to measure phase angles. Thus, in the problems that follow, we ask only for magnitudes of voltages and currents. 37. EWB Create the circuit of Figure 16–58 on the screen. (Use the ac source from the Sources Parts bin and the ammeter from the Indicators Parts bin.) Double click the ammeter symbol and set Mode to AC. Click the ON/OFF switch at the top right hand corner of the screen to energize the circuit. Compare the measured reading against the theoretical value.

(b) f = 100 Hz. Determine C. FIGURE 16–57

FIGURE 16–58

38. EWB Replace the capacitor of Figure 16–58 with a 200-mH inductor and repeat Problem 37. 39. PSpice Create the circuit of Figure 16–53 on the screen. Use a source of 100 V∠0°, L 0.2 H, and f 50 Hz. Solve for current IL (magnitude and angle). See note below.

Answers to In-Process Learning Checks

639

40. PSpice Plot the reactance of a 2.39-H inductor versus frequency from 1 Hz to 500 Hz and compare to Figure 16–28. Change the x-axis scale to linear. 41. PSpice For the circuit of Problem 39, plot current magnitude versus frequency from f 1 Hz to f 20 Hz. Measure the current at 10 Hz and verify with your calculator. Note: PSpice does not permit source/inductor loops. To get around this, add a very small resistor in series, for example, R 0.00001 ⍀.

In-Process Learning Check 1 1. a. 6∠90° d. b. 4∠ 90° e. c. 4.24∠45° f. 2. a. j4 d. b. 3 j0 e. c. j2 f. 3. 12∠40° 4. 88 j7; 16 j15 5. 288∠ 100°; 2∠150° 6. a. 14.70 j2.17 7. 18.0 sin(qt 56.3°)

7.21∠ 56.3° 9.43∠122.0° 2.24∠ 63.4° 3.83 j3.21 3 j5.20 2.35 j0.855

g. 3.61∠ 123.7°

g. 1.64 j0.599

b. 8.94 j7.28

In-Process Learning Check 2 1. 50 sin(qt 30°) A 2. 50 sin(qt 60°) A 3. 50 sin(qt 120°) A 4. 0.637 H 5. 3.98 mF 6. a. Voltage and current are in phase. Therefore, R b. Current leads by 90°. Therefore, C c. Current leads by 90°. Therefore, C d. Current lags by 90°. Therefore, L

ANSWERS TO IN-PROCESS LEARNING CHECKS

Power in AC Circuits OBJECTIVES After studying this chapter, you will be able to • explain what is meant by active, reactive, and apparent power, • compute the active power to a load, • compute the reactive power to a load, • compute the apparent power to a load, • construct and use the power triangle to analyze power to complex loads, • compute power factor, • explain why equipment is rated in VA instead of watts, • measure power in single-phase circuits, • describe why effective resistance differs from geometric resistance, • describe energy relations in ac circuits, • use PSpice to study instantaneous power.

KEY TERMS Active Power Apparent Power Average Power Effective Resistance Fp Instantaneous Power

Power Factor Correction Power Factor Power Triangle Q Reactive Power S Skin Effect VA VAR Wattless Power Wattmeter

OUTLINE Introduction Power to a Resistive Load Power to an Inductive Load Power to a Capacitive Load Power in More Complex Circuits Apparent Power The Relationship Between P, Q, and S Power Factor AC Power Measurement Effective Resistance Energy Relationships for AC Circuit Analysis Using Computers

n Chapter 4, you studied power in dc circuits. In this chapter, we turn our attention to power in ac circuits. In ac circuits, there are additional considerations that are not present with dc. In dc circuits, for example, the only power relationship you encounter is P ⫽ VI watts. This is referred to as real power or active power and is the power that does useful work such as light a lamp, power a heater, run an electric motor, and so on. In ac circuits, you also encounter this type of power. For ac circuits that contain reactive elements however, (i.e., inductance or capacitance), a second component of power also exists. This component, termed reactive power, represents energy that oscillates back and forth throughout the system. For example, during the buildup of current in an inductance, energy flows from the power source to the inductance to create its magnetic field. When the magnetic field collapses, this energy is returned to the circuit. This movement of energy in and out of the inductance constitutes a flow of power. However, since it flows first in one direction, then in the other, it contributes nothing to the average flow of power from the source to the load. For this reason, reactive power is sometimes referred to as wattless power. (A similar situation exists regarding power flow to and from the electric field of a capacitor.) For a circuit that contains resistive as well as reactive elements, some energy is dissipated while the remainder is shuttled back and forth as described above; thus, both active and reactive components of power are present. This combination of real and reactive power is termed apparent power. In this chapter, we look at all three components of power. New ideas that emerge include the concept of power factor, the power triangle, the measurement of power in ac circuits, and the concept of effective resistance.

Henry Cavendish CAVENDISH, AN ENGLISH CHEMIST and physicist born in 1731, is included here not for what he did for the emerging electrical field, but for what he didn’t do. A brilliant man, Cavendish was 50 years ahead of his time, and his experiments in electricity preceded and anticipated almost all the major discoveries that came about over the next half century (e.g., he discovered Coulomb’s law before Coulomb did). However, Cavendish was interested in research and knowledge purely for its own sake and never bothered to publish most of what he learned, in effect depriving the world of his findings and holding back the development of the field of electricity by many years. Cavendish’s work lay unknown for nearly a century before another great scientist, James Clerk Maxwell, had it published. Nowadays, Cavendish is better known for his work in the gravitational field than for his work in the electrical field. One of the amazing things he did was to determine the mass of the earth using the rather primitive technology of his day.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

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Power in AC Circuits

17.1 i(t) p(t)

Load

⫹ v(t) ⫺

p = vi FIGURE 17–1 Voltage, current, and power references. When p is positive, power is in the direction of the reference arrow.

Introduction

At any given instant, the power to a load is equal to the product of voltage times current (Figure 17–1). This means that if voltage and current vary with time, so will power. This time-varying power is referred to as instantaneous power and is given the symbol p(t) or just p. Thus, p vi

(watts)

(17–1)

Now consider the case of sinusoidal ac. Since voltage and current are positive at various times during their cycle and negative at others, instantaneous power may also be positive at some times and negative at others. This is illustrated in Figure 17–2, where we have multiplied voltage times current point by point to get the power waveform. For example, from t 0 s to t t1, v and i are both positive; therefore, power is positive. At t t1, v 0 V and thus p 0 W. From t1 to t2, i is positive and v is negative; therefore, p is negative. From t2 to t3, both v and i are negative; therefore power is positive, and so on. As discussed in Chapter 4, a positive value for p means that power transfer is in the direction of the reference arrow, while a negative value means that it is in the opposite direction. Thus, during positive parts of the power cycle, power flows from the source to the load, while during negative parts, it flows out of the load back into the circuit. The waveform of Figure 17–2 is the actual power waveform. We will now show that the key aspects of power flow embodied in this waveform can be described in terms of active power, reactive power, and apparent power. 1 cycle Ⳮ

0 ⳮ

p(t) i(t)

v(t)

Energy Energy Energy Energy stored released stored released

FIGURE 17–2 Instantaneous power in an ac circuit. Positive p represents power to the load; negative p represents power returned from the load.

Active Power Since p represents the power flowing to the load, its average will be the average power to the load. Denote this average by the letter P. If P is positive, then, on average, more power flows to the load than is returned from it. (If P is zero, all power sent to the load is returned.) Thus, if P has a positive value, it represents the power that is really dissipated by the load. For this reason, P is called real power. In modern terminology, real power is also called active power. Thus, active power is the average value of the instantaneous power, and the terms real power, active power, and average power mean the same

Section 17.2

thing. (We usually refer to it simply as power.) In this book, we use the terms interchangeably.

Reactive Power Consider again Figure 17–2. During the intervals that p is negative, power is being returned from the load. (This can only happen if the load contains reactive elements: L or C.) The portion of power that flows into the load then back out is called reactive power. Since it first flows one way then the other, its average value is zero; thus, reactive power contributes nothing to the average power to the load. Although reactive power does no useful work, it cannot be ignored. Extra current is required to create reactive power, and this current must be supplied by the source; this also means that conductors, circuit breakers, switches, transformers, and other equipment must be made physically larger to handle the extra current. This increases the cost of a system. At this point, it should be noted that real power and reactive power do not exist as separate entities. Rather, they are components of the power waveform shown in Figure 17–2. However, as you will see, we are able to conceptually separate them for purposes of analysis.

17.2

Power to a Resistive Load

First consider power to a purely resistive load (Figure 17–3). Here, current is in phase with voltage. Assume i Imsin qt and v Vmsin qt. Then, p vi (Vmsin qt)(Imsin qt) Vm Imsin2qt i Heat ⫹ e

p R

⫺

⫹ v ⫺

Load (a)

Ⳮ

1 cycle

V mIm

p(t)

Average power VmIm 2

0 ⳮ

T 2

i(t)

v(t) (b)

FIGURE 17–3

Power to a purely resistive load. The peak value of p is Vm Im.

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Power to a Resistive Load

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Therefore, V Im p m (1 cos 2 qt) 2

(17–2)

where we have used the trigonometric relationship sin2qt 1⁄ 2(1 cos 2 qt) from inside the front cover of the book. A sketch of p versus time is shown in (b). Note that p is always positive (except where it is momentarily zero). This means that power flows only from the source to the load. Since none is ever returned, all power delivered by the source is absorbed by the load. We therefore conclude that power to a pure resistance consists of active power only. Note also that the frequency of the power waveform is double that of the voltage and current waveforms. (This is confirmed by the 2q in Equation 17–2.)

Average Power Inspection of the power waveform of Figure 17–3 shows that its average value lies half way between zero and its peak value of Vm Im. That is, P Vm Im /2

(You can also get the same result by averaging Equation 17–2 as we did in Chapter 15.) Since V (the magnitude of the rms value of voltage) is Vm /兹2苶 and I (the magnitude of the rms value of current) is Im /兹2苶, this can be written as P VI. Thus, average power to a purely resistive load is P VI

(watts)

(17–3)

Alternate forms are obtained by substituting V IR and I V/R into Equation 17–3. They are P I 2R (watts) V 2/R (watts)

(17–4) (17–5)

Thus the active power relationships for resistive circuits are the same for ac as for dc.

17.3

Power to an Inductive Load

For a purely inductive load as in Figure 17–4(a), current lags voltage by 90°. If we select current as reference, i Imsin qt and v Vmsin(qt 90°). A sketch of p versus time (obtained by multiplying v times i) then looks as shown in (b). Note that during the first quarter-cycle, p is positive and hence power flows to the inductance, while during the second quarter-cycle, p is negative and all power transferred to the inductance during the first quarter-cycle flows back out. Similarly for the third and fourth quarter-cycles. Thus, the average power to an inductance over a full cycle is zero, i.e., there are no power losses associated with a pure inductance. Consequently, PL 0 W and the only power flowing in the circuit is reactive power. This is true in general, that is, the power that flows into and out of a pure inductance is reactive power only. To determine this power, consider again equation 17–1.

Section 17.3 i

⫹ v ⫺

(a) Let i = Im sin ωt v = Vm sin (ωt Ⳮ 90°) Ⳮ

1 cycle pL(t)

v(t)

i(t)

3T 4

T 4

T 2

ⳮVI Energy Energy Energy Energy stored released stored released (b) FIGURE 17–4 Power to a purely inductive load. Energy stored during each quartercycle is returned during the next quarter-cycle. Average power is zero.

With v Vmsin(qt 90°) and i Imsin qt, pL vi becomes pL Vm Imsin(qt 90°)sin qt

After some trigonometric manipulation, this reduces to pL VI sin 2 qt

(17–6)

where V and I are the magnitudes of the rms values of the voltage and current respectively. The product VI in Equation 17–6 is defined as reactive power and is given the symbol QL. Because it represents “power” that alternately flows into, then out of the inductance, QL contributes nothing to the average power to the load and, as noted earlier, is sometimes referred to as wattless power. As you will soon see, however, reactive power is of major concern in the operation of electrical power systems. Since QL is the product of voltage times current, its unit is the volt-amp (VA). To indicate that QL represents reactive volt-amps, an “R” is appended to yield a new unit, the VAR (volt-amps reactive). Thus, QL VI (VAR)

(17–7)

Substituting V IXL and I V/XL yields the following alternate forms: V2 QL I 2XL (VAR) XL

(17–8)

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Power in AC Circuits

By convention, QL is taken to be positive. Thus, if I 4 A and XL 2 ⍀, QL (4 A)2(2 ⍀) 32 VAR. Note that the VAR (like the watt) is a scalar quantity with magnitude only and no angle.

17.4

Power to a Capacitive Load

For a purely capacitive load, current leads voltage by 90°. Taking current as reference, i Imsin qt and v Vmsin(qt 90°). Multiplication of v times i yields the power curve of Figure 17–5. Note that negative and positive loops of the power wave are identical; thus, over a cycle, the power returned to the circuit by the capacitance is exactly equal to that delivered to it by the source. This means that the average power to a capacitance over a full cycle is zero, i.e., there are no power losses associated with a pure capacitance. Consequently, PC 0 W and the only power flowing in the circuit is reactive power. This is true in general, that is, the power that flows into and out of a pure capacitance is reactive power only. This reactive power is given by pC vi Vm Imsin qt sin(qt 90°)

which reduces to pC VI sin 2 qt

i ⫹ v ⫺

(a) Let i = Im sin ωt v = Vm sin (ωt ⳮ 90°)

Ⳮ VI

1 cycle

v(t)

i(t) p(t)

ⳮVI Energy Energy Energy Energy released stored released stored (b) FIGURE 17–5

Power to a purely capacitive load. Average power is zero.

(17–9)

Section 17.4

where V and I are the magnitudes of the rms values of the voltage and current respectively. Now define the product VI as QC. This product represents reactive power. That is, QC VI

(VAR)

(17–10)

Since V IXC and I V/XC, QC can also be expressed as V2 QC I 2XC (VAR) XC

(17–11)

By convention, reactive power to capacitance is defined as negative. Thus, if I 4 A and XC 2 ⍀, then I 2XC (4 A)2(2 ⍀) 32 VAR. We can either explicitly show the minus sign as QC 32 VAR or imply it by stating that Q represents capacitive vars, i.e. QC 32 VAR (cap.).

EXAMPLE 17–1

For each circuit of Figure 17–6, determine real and reac-

tive power.

100 V∠0°

(a) R 25 Ω

(b) XL 20 Ω

100 V∠0°

Solution Only voltage and current magnitudes are needed. a. I 100 V/25 ⍀ 4 A. P VI (100 V)(4 A) 400 W. Q 0 VAR b. I 100 V/20 ⍀ 5 A. Q VI (100 V)(5 A) 500 VAR (ind.). P 0 W c. I 100V/40 ⍀ 2.5A. Q VI (100V)(2.5A) 250VAR (cap.). P 0 W The answer for (c) can also be expressed as Q 250 VAR.

■

Power to a Capacitive Load

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PRACTICE PROBLEMS 1

■

Power in AC Circuits

1. If the power at some instant in Figure 17–1 is p 27 W, in what direction is the power at that instant? 2. For a purely resistive load, v and i are in phase. Given v 10 sin qt V and i 5 sin qt A. Using graph paper, carefully plot v and i at 30° intervals. Now multiply the values of v and i at these points and plot the power. (The result should look like Figure 17–3(b). a. From the graph, determine the peak power and average power. b. Compute power using P VI and compare to the average value determined in (a). 3. Repeat Example 17–1 using equations 17–4, 17–5, 17–8, and 17–11. Answers: 1. From the load to the source. 2. a. 50 W; 25 W b. Same

17.5

Power in More Complex Circuits

The relationships described above were developed using the load of Figure 17–1. However, they hold true for every element in a circuit, no matter how complex the circuit or how its elements are interconnected. Further, in any circuit, total real power PT is found by summing real power to all circuit elements, while total reactive power QT is found by summing reactive power, taking into account that inductive Q is positive and capacitive Q is negative. It is sometimes convenient to show power to circuit elements symbolically as illustrated in the next example.

EXAMPLE 17–2

For the RL circuit of Figure 17–7(a), I 5 A. Determine

P and Q. XL = 4 ⍀

R=3⍀

75 W P Q

(a) FIGURE 17–7

P Q

100 VAR (ind.)

(b) Symbolic representation

From the terminals, P and Q are the same for both (a) and (b).

Solution P I 2R (5 A)2(3 ⍀) 75 W Q QL I 2XL (5 A)2(4 ⍀) 100 VAR (ind.) These can be represented symbolically as in Figure 17–7(b).

Section 17.5

■

Power in More Complex Circuits

EXAMPLE 17–3 For the RC circuit of Figure 17–8(a), determine P and Q. ⫹ 40 V ⫺

⫹ V 20 ⍀ 80 ⍀ ⫺

⫹ V ⫺

P Q

80 W

(a) FIGURE 17–8

20 VAR (cap.)

(b)

From the terminals, P and Q are the same for both (a) and (b).

Solution P V 2/R (40 V)2/(20 ⍀) 80 W Q QC V 2/XC (40 V)2/(80 ⍀) 20 VAR (cap.) These can be represented symbolically as in Figure 17–8(b).

In terms of determining total P and Q, it does not matter how the circuit or system is connected or what electrical elements it contains. Elements can be connected in series, in parallel, or in series-parallel, for example, and the system can contain electric motors and the like, and total P is still found by summing the power to individual elements, while total Q is found by algebraically summing their reactive powers.

EXAMPLE 17–4 a. For Figure 17–9(a), compute PT and QT. b. Reduce the circuit to its simplest form. R 3Ω 20 A

XC1 6 Ω XC2

10 Ω

XL 5Ω

200 V

(a)

1200 W

2400 VAR (cap.) 4000 VAR (cap.)

PT QT

(b) FIGURE 17–9

8000 VAR (ind.)

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Power in AC Circuits

3Ω

1200 W 20 A

1600 VAR (ind.)

Xeq

j4Ω

(d)

Continued.

Solution a. P I 2R (20 A)2(3 ⍀) 1200 W QC1 I 2XC1 (20 A)2(6 ⍀) 2400 VAR (cap.) V 22 (200 V)2 QC2 4000 VAR (cap.) (10 ⍀) XC2 V2 (200 V)2 QL 2 8000 VAR (ind.) XL 5⍀ These are represented symbolically in part (b). PT 1200 W and QT 2400 VAR 4000 VAR 8000 VAR 1600 VAR. Thus, the load is net inductive as shown in (c). b. QT I 2Xeq. Thus, Xeq QT/I 2 (1600 VAR)/(20 A)2 4 ⍀. Circuit resistance remains unchanged. Thus, the equivalent is as shown in (d).

PRACTICE PROBLEMS 2

For the circuit of Figure 17–10, PT 1.9 kW and QT 900 VAR (ind.). Determine P2 and Q2.

Load 1 200 W 500 VAR (ind.)

PT QT

Load 3 1.4 kW 800 VAR (ind.) P2 = ? Q2 = ? Load 2

FIGURE 17–10 Answer: 300 W

17.6

400 VAR (cap.)

Apparent Power

When a load has voltage V across it and current I through it as in Figure 17–11, the power that appears to flow to it is VI. However, if the load contains both resistance and reactance, this product represents neither real power nor reactive

Section 17.7

■

The Relationship Between P, Q, and S I

power. Since it appears to represent power, it is called apparent power. Apparent power is given the symbol S and has units of volt-amperes (VA). Thus, S VI (VA)

(17–12)

where V and I are the magnitudes of the rms voltage and current respectively. Since V IZ and I V/Z, S can also be written as S I 2Z V 2/Z (VA)

(17–13)

For small equipment (such as found in electronics), VA is a convenient unit. However, for heavy power apparatus (Figure 17–12), it is too small and kVA (kilovolt-amps) is frequently used, where VI S (kVA) 1000

(17–14)

In addition to its VA rating, it is common practice to rate electrical apparatus in terms of its operating voltage. Once you know these two, it is easy to determine rated current. For example, a piece of equipment rated at 250 kVA, 4.16 kV has a rated current of I S/V (250 103 VA)/(4.16 103 V) 60.1 A.

FIGURE 17–12 Power apparatus is rated in apparent power. The transformer shown is a 167-kVA unit. (Courtesy Carte International Ltd.)

17.7

The Relationship Between P, Q, and S

Until now, we have treated real, reactive, and apparent power separately. However, they are related by a very simple relationship through the power triangle.

FIGURE 17–11 S VI.

Load

⫹ V ⫺

Apparent power

651

652

Chapter 17

Power in AC Circuits

■

I ⫹ ⫹ E ⫺

V ⫺

⫹ VR ⫺ ⫹ VL ⫺

(a)

The Power Triangle Consider the series circuit of Figure 17–13(a). Let the current through the circuit be I I∠0°, with phasor representation (b). The voltages across the resistor and inductance are VR and VL respectively. As noted in Chapter 16, VR is in phase with I, while VL leads it by 90°. Kirchhoff’s voltage law applies for ac voltages in phasor form. Thus, V VR VL as indicated in (c). The voltage triangle of (c) may be redrawn as in Figure 17–14(a) with magnitudes of VR and VL replaced by IR and IXL respectively. Now multiply all quantities by I. This yields sides of I 2R, I 2XL, and hypotenuse VI as indicated in (b). Note that these represent P, Q, and S respectively as indicated in (c). This is called the power triangle. From the geometry of this triangle, you can see that S 兹P 苶2苶苶苶 Q2L苶

Alternatively, the relationship between P, Q, and S may be expressed as a complex number:

(b)

S P jQL

(17–16a)

S S∠v

(17–16b)

If the circuit is capacitive instead of inductive, Equation 17–16a becomes

(c)

S P jQC

FIGURE 17–13 Steps in the development of the power triangle.

(17–15)

(17–17)

The power triangle in this case has a negative imaginary part as indicated in Figure 17–15.

IXL

I2XL

I2R

IR (a) Magnitudes only shown

(b) Multiplied by I

FIGURE 17–14 Steps in the development of the power triangle (continued).

The power relationships may be written in generalized forms as

P θ

QC S FIGURE 17–15 Power triangle for capacitive case.

S P Q

(17–18)

S VI*

(17–19)

and where P P∠0°, QL jQL , QC jQC, and I* is the conjugate of current I. These relationships hold true for all networks regardless of what they contain or how they are configured. When solving problems involving power, remember that P values can be added to get PT, and Q values to get QT (where Q is positive for inductive elements and negative for capacitive). However, apparent power values cannot be added to get ST, i.e., ST S1 S2 … SN. Instead, determine PT and QT, then use the power triangle to obtain ST.

Section 17.7

EXAMPLE 17–5

■

The Relationship Between P, Q, and S

The P and Q values for a circuit are shown in Figure

17–16(a). a. Determine the power triangle. b. Determine the magnitude of the current supplied by the source. I

⫹ ⫺

700 W 800 W 1300 VAR 600 VAR (cap.) (ind.) 100 VAR (cap.) 120 V (cap.) 120 W 80 W 1200 VAR

PT = 1700 W 19.4°

θT ST = 1803 VA

(a)

QT = 600 VAR (cap.)

(b)

FIGURE 17–16

Solution a. PT 700 800 80 120 1700 W QT 1300 600 100 1200 600 VAR 600 VAR (cap.) ST PT jQT 1700 j600 1803∠ 19.4° VA The power triangle is as shown. The load is net capacitive. b. I ST/E 1803 VA/120 V 15.0 A

EXAMPLE 17–6 A generator supplies power to an electric heater, an inductive element, and a capacitor as in Figure 17–17(a). a. Find P and Q for each load. b. Find total active and reactive power supplied by the generator. c. Draw the power triangle for the combined loads and determine total apparent power. d. Find the current supplied by the generator.

120 V∠0°

6Ω

2.5 kW heater (a) FIGURE 17–17

24 Ω XC

653

654

Chapter 17

■

Power in AC Circuits

1 08

1.8 kVAR QT

35.8° PT 2.5 kW (b) FIGURE 17–17 Continued.

Solution a. The components of power are as follows: Heater:

PH 2.5 kW

Inductor:

PL 0 W

Capacitor:

PC 0 W

QH 0 VAR V2 (120 V)2 QL 2.4 kVAR (ind.) XL 6⍀ (120 V)2 V2 QC 600 VAR (cap.) 24 ⍀ XC

b. PT 2.5 kW 0 W 0 W 2.5 kW QT 0 VAR 2.4 kVAR 600 VAR 1.8 kVAR (ind.) c. The power triangle is sketched as Figure 17–7(b). Both the hypotenuse and the angle can be obtained easily using rectangular to polar conversion. ST PT jQT 2500 j1800 3081∠35.8°. Thus, apparent power is ST 3081 VA. S 3081 VA d. I T 25.7 A E 120 V

Active and Reactive Power Equations An examination of the power triangle of Figures 17–14 and 17–15 shows that P and Q may be expressed respectively as P VI cos v S cos v (W)

(17–20)

Q VI sin v S sin v (VAR)

(17–21)

and

where V and I are the magnitudes of the rms values of the voltage and current respectively and v is the angle between them. P is always positive, while Q is positive for inductive circuits and negative for capacitive circuits. Thus, if V 120 volts, I 50 A, and v 30°, P (120)(50)cos 30° 5196 W and Q (120)(50)sin 30° 3000 VAR.

Section 17.8

A 208-V generator supplies power to a group of three loads. Load 1 has an apparent power of 500 VA with v 36.87° (i.e., it is net inductive). Load 2 has an apparent power of 1000 VA and is net capacitive with a power triangle angle of 53.13°. Load 3 is purely resistive with power P3 200 W. Determine the power triangle for the combined loads and the generator current. Answers: ST 1300 VA, vT 22.6°, I 6.25 A

17.8

Power Factor

The quantity cos v in Equation 17–20 is defined as power factor and is given the symbol Fp. Thus, Fp cos v

(17–22)

From Equation 17–20, we see that Fp may be computed as the ratio of real power to apparent power. Thus, cos v P/S

(17–23)

Power factor is expressed as a number or as a percent. From Equation 17–23, it is apparent that power factor cannot exceed 1.0 (or 100% if expressed in percent). The power factor angle v is of interest. It can be found as v cos 1(P/S)

(17–24)

Angle v is the angle between voltage and current. For a pure resistance, therefore, v 0°. For a pure inductance, v 90°; for a pure capacitance, v 90°. For a circuit containing both resistance and inductance, v will be somewhere between 0° and 90°; for a circuit containing both resistance and capacitance, v will be somewhere between 0° and 90°.

Unity, Lagging, and Leading Power Factor As indicated by Equation 17–23, a load’s power factor shows how much of its apparent power is actually real power. For example, for a purely resistive circuit, v 0° and Fp cos 0° 1.0. Therefore, P VI (watts) and all the load’s apparent power is real power. This case (Fp 1) is referred to as unity power factor. For a load containing only resistance and inductance, the load current lags voltage. The power factor in this case is described as lagging. On the other hand, for a load containing only resistance and capacitance, current leads voltage and the power factor is described as leading. Thus, an inductive circuit has a lagging power factor, while a capacitive circuit has a leading power factor. A load with a very poor power factor can draw excessive current. This is discussed next.

■

Power Factor

PRACTICE PROBLEMS 3

655

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Chapter 17

■

Power in AC Circuits

I = 200 A ⫹ 600 V ⫺

120 kW P = 120 kW Q=0 (a) S = 120 kVA

Poof! I = 333 A ⫹ 600 V ⫺

120 kW P = 120 kW QL = 160 kVAR

(b) S = (120)2 Ⳮ (160)2 = 200 kVA The generator is overloaded. FIGURE 17–18 Illustrating why electrical apparatus is rated in VA instead of watts. Both loads dissipate 120 kW, but the current rating of generator (b) is exceeded because of the power factor of its load.

Why Equipment Is Rated in VA We now examine why electrical apparatus is rated in VA instead of watts. Consider Figure 17–18. Assume that the generator is rated at 600 V, 120 kVA. This means that it is capable of supplying I 120 kVA/600 V 200 A. In (a), the generator is supplying a purely resistive load with 120 kW. Since S P for a purely resistive load, S 120 kVA and the generator is supplying its rated kVA. In (b), the generator is supplying a load with P 120 kW as before, but Q 160 kVAR. Its apparent power is therefore S 200 kVA, which means that the generator current is I 200 kVA/600 V 333.3 A. Even though it is supplying the same power as in (a), the generator is now greatly overloaded, and damage may result as indicated in (b). This example illustrates clearly that rating a load or device in terms of power is a poor choice, as its current-carrying capability can be greatly exceeded (even though its power rating is not). Thus, the size of electrical apparatus (generators, interconnecting wires, transformers, etc.) required to supply a load is governed, not by the load’s power requirements, but rather by its VA requirements. Power Factor Correction The problem shown in Figure 17–18 can be alleviated by cancelling some or all of the reactive component of power by adding reactance of the opposite type to the circuit. This is referred to as power factor correction. If you completely cancel the reactive component, the power factor angle is 0° and Fp 1. This is referred to as unity power factor correction. Residential customers are charged solely on the basis of energy used. This is because all residential power factors are essentially the same, and the

EXAMPLE 17–7 For the circuit of Figure 17–18(b), a capacitance with QC 160 kVAR is added in parallel with the load as in Figure 17–19(a). Determine generator current I. I ⫹ 600 V ⫺

120 kW 160 kVAR (ind.)

Power factor corrected load (a) Let QC = 160 kVAR

⫹ 600 V ⫺

120 kW 0 VAR Corrected load

(b) Load corrected to unity power factor

FIGURE 17–19 Power factor correction. The parallel capacitor greatly reduces source current.

Solution QT 160 kVAR 160 kVAR 0. Therefore, ST 120 kW j0 kVAR. Thus, ST 120 kVA, and I 120 kVA/600 V 200 A. Thus, the generator is no longer overloaded.

Section 17.8

power factor effect is simply built into the tariff. Industrial customers, on the other hand, have widely different power factors, and the electrical utility may have to take the power factors of these customers into account. To illustrate, assume that the loads of Figures 17–18(a) and (b) are two small industrial plants. If the utility based its charge solely on power, both customers would pay the same amount. However, it costs the utility more to supply customer (b) since larger conductors, larger transformers, larger switchgear, and so on are required to handle the larger current. For this reason, industrial customers may pay a penalty if their power factor drops below a prescribed value.

EXAMPLE 17–8

An industrial client is charged a penalty if the plant power factor drops below 0.85. The equivalent plant loads are as shown in Figure 17–20. The frequency is 60 Hz. a. Determine PT and QT. b. Determine what value of capacitance (in microfarads) is required to bring the power factor up to 0.85. c. Determine generator current before and after correction. Solution a. The components of power are as follows: Lights: P 12 kW, Q 0 kVAR 2 2 Furnace: P I R (150) (2.4) 54 kW Q I 2X (150)2(3.2) 72 kVAR (ind.) Motor: vm cos 1(0.8) 36.9°. Thus, from the motor power triangle, Qm Pm tan vm 80 tan 36.9° 60 kVAR (ind.) Total: PT 12 kW 54 kW 80 kW 146 kW QT 0 72 kVAR 60 kVAR 132 kVAR

⫹ 600 V ⫺

150 A

Lights Electric Furnace Motor loads 80 kW 12 kW 2.4 Ⳮ j3.2⍀ 0.8 Fp (lag) Plant loads (a) EWB

FIGURE 17–20

Qm θm

Pm = 80 kW b) Power triangle for motor.

■

Power Factor

657

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Chapter 17

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Power in AC Circuits

b. The power triangle for the plant is shown in Figure 17–21(a). However, we must correct the power factor to 0.85. Thus we need v⬘ cos 1(0.85) 31.8°, where v⬘ is the power factor angle of the corrected load as indicated in Figure 17–21(b). The maximum reactive power that we can tolerate is thus Q⬘T PT tan v⬘ 146 tan 31.8° 90.5 kVAR. A

8 6.

19 42.1° PT = 146 kW

QT = 132 kVAR

(a) Power triangle for the plant

Q'T = PT tan θ' 1.8 17 = 90.5 kVAR θ' = 31.8° PT = 146 kW (b) Power triangle after correction

FIGURE 17–21 Initial and final power triangles. Note that PT does not change when we correct the power factor.

Now consider Figure 17–22. Q⬘T QC 132 kVAR, where Q⬘T 90.5 kVAR. Therefore, QC 41.5 kVAR 41.5 kVAR (cap.). But QC V 2/XC. Therefore, XC V 2/QC (600)2/41.5 kVAR 8.67 ⍀. But XC 1/qC. Thus a capacitor of 1 1 C 306 ␮F qXC (2p)(60)(8.67) will provide the required correction.

⫹

600 V ⫺

QC Q'T

146 kW 132 kVAR (ind.) Plant

Corrected load FIGURE 17–22

c. For the original circuit Figure 17–21(a), ST 196.8 kVA. Thus, S 196.8 kVA I T 328 A E 600 V For the corrected circuit 17–21(b), S⬘T 171.8 kVA and 171.8 kVA I 286 A 600 V Thus, power factor correction has dropped the current by 42 A.

Section 17.9

1. Repeat Example 17–8 except correct the power factor to unity. 2. Due to plant expansion, 102 kW of purely resistive load is added to the plant of Figure 17–20. Determine whether power factor correction is needed to correct the expanded plant to 0.85 Fp, or better.

■

AC Power Measurement

PRACTICE PROBLEMS 4

Answers:1. 973 mF, 243 A. Other answers remain unchanged. 2. Fp 0.88. No correction needed.

In practice, almost all loads (industrial, residential, and commercial) are inductive due to the presence of motors, fluorescent lamp ballasts, and the like. Consequently, you will likely never run into capacitive loads that need power factor correcting.

1. Sketch the power triangle for Figure 17–9(c). Using this triangle, determine the magnitude of the applied voltage. 2. For Figure 17–10, assume a source of E 240 volts, P2 300 W, and Q2 400 VAR (cap.). What is the magnitude of the source current I? 3. What is the power factor of each of the circuits of Figure 17–7, 17–8, and 17–9? Indicate whether they are leading or lagging. 4. Consider the circuit of Figure 17–18(b). If P 100 kW and QL 80 kVAR, is the source overloaded, assuming it is capable of handling a 120-kVA load? (Answers are at the end of the chapter.)

17.9

AC Power Measurement

To measure power in an ac circuit, you need a wattmeter (since the product of voltage times current is not sufficient to determine ac power). Figure 17–23 shows such a meter. It is a digital device that monitors voltage and current and from these, computes and displays power. (You may also encounter older electrodynamometer type wattmeters, i.e., electromechanical analog instruments that use a pivoted pointer to indicate the power reading on a scale, much like the analog meters of Chapter 2. Although their details differ dramatically from the electronic types, the manner in which they are connected in a circuit to measure power is the same. Thus, the measurement techniques described below apply to them as well.) To help understand power measurement, consider Figure 17–1. Instantaneous load power is the product of load voltage times load current, and average power is the average of this product. One way to implement power measurement is therefore to create a meter with a current sensing circuit, a voltage sensing circuit, a multiplier circuit, and an averaging circuit. Figure 17–24 shows a simplified symbolic representation of such an instrument. Current is passed through its current coil (CC) to create a magnetic field proportional to the current, and a sensor circuit connected across the load voltage reacts with this field to produce an output voltage proportional to the product of instantaneous voltage and current (i.e., proportional to

IN-PROCESS

LEARNING CHECK 1

659

660

Chapter 17

■

Power in AC Circuits

FIGURE 17–23 Multifunction power/energy meter. It can measure active power (W), reactive power (VARs), apparent power (VA), power factor, energy, and more. Load current CC

Vload

Voltage sensing circuit

CC Current coil

FIGURE 17–24 Conceptual representation of a wattmeter.

instantaneous power). An averaging circuit averages this voltage and drives a display to indicate average power. (The scheme used by the meter of Figure 17–23 is actually considerably more sophisticated than this because it measures many things besides power—e.g., it measures, VARs, VA, energy, etc. However, the basic idea is conceptually correct.) Figure 17–25 shows how to connect a wattmeter (whether electronic or electromechanical) into a circuit. Load current passes through its current coil circuit, and load voltage is impressed across its voltage sensing circuit. With this connection, the wattmeter computes and displays the product of the magnitude of the load voltage, the magnitude of the load current, and the cosine of the angle between them, i.e. Vload Iload cos vload. Thus, it measures load power. Note the marking on the terminals. You usually connect the meter so that load current enters the current terminal and the higher potential end of the load is connected to the voltage terminal. On many meters, the voltage terminal is internally connected so that only three terminals are brought out as in Figure 17–26.

Section 17.9

I load

W E

Load

Vload

FIGURE 17–25 Connection of wattmeter.

When power is to be measured in a low power factor circuit, a low power factor wattmeter must be used. This is because, for low power factor loads, currents can be very high, even though the power is low. Thus, you can easily exceed the current rating of a standard wattmeter and damage it, even though the power indication on the meter is small.

EXAMPLE 17–9

For the circuit of Figure 17–25, what does the wattmeter

indicate if a. Vload 100 V∠0° and Iload 15 A∠60°, b. Vload 100 V∠10° and Iload 15 A∠30°? Solution a. vload 60°. Thus, P (100)(15)cos 60° 750 W, b. vload 10° 30° 20°. Thus, P (100)(15)cos( 20°) 1410 W. Note: For (b), since cos( 20°) cos( 20°), it does not matter whether we include the minus sign.

EXAMPLE 17–10 For Figure 17–26, determine the wattmeter reading. W ⫹ E ⫺

600 W 700 VAR

FIGURE 17–26 This wattmeter has its voltage side terminals connected internally.

Solution

A wattmeter reads only active power. Thus, it indicates 600 W.

It should be noted that the wattmeter reads power only for circuit elements on the load side of the meter. In addition, if the load consists of several elements, it reads the sum of the powers.

■

AC Power Measurement

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Chapter 17

■

PRACTICE PROBLEMS 5

Power in AC Circuits

Determine the wattmeter reading for Figure 17–27.

⫹ E ⫺

100 W 70 VAR (ind.)

40 W

300 W 80 VAR (cap.)

10 W

700 W 60 VAR (cap.)

FIGURE 17–27 Answer: 750 W

17.10 Effective Resistance Up to now, we have assumed that resistance is constant, independent of frequency. However, this is not entirely true. For a number of reasons, the resistance of a circuit to ac is greater than its resistance to dc. While this effect is small at low frequencies, it is very pronounced at high frequencies. AC resistance is known as effective resistance. Before looking at why ac resistance is greater than dc resistance, we need to reexamine the concept of resistance itself. Recall from Chapter 3 that resistance was originally defined as opposition to current, that is, R V/I. (This is ohmic resistance.) Building on this, you learned in Chapter 4 that P I 2R. It is this latter viewpoint that allows us to give meaning to ac resistance. That is, we define ac or effective resistance as P Reff 2 (⍀) I

(17–25)

where P is dissipated power (as determined by a wattmeter). From this, you can see that anything that affects dissipated power affects resistance. For dc and low-frequency ac, both definitions for R, i.e., R V/I and R P/I 2 yield the same value. However, as frequency increases, other factors cause an increase in resistance. We will now consider some of these.

Eddy Currents and Hysteresis The magnetic field surrounding a coil or other circuit carrying ac current varies with time and thus induces voltages in nearby conductive material such as metal equipment cabinets, transformer cores, and so on. The resulting currents (called eddy currents because they flow in circular patterns like eddies in a brook) are unwanted and create power losses called eddy current losses. Since additional power must be supplied to make up for these losses, P in Equation 17–25 increases, increasing the effective resistance of the coil. If ferromagnetic material is also present, an additional power loss occurs due to hysteresis effects caused by the magnetic field alternately magnetizing the material in one direction, then the other. Hysteresis and eddy current losses are important even at low frequencies, such as the 60-Hz power system frequency. This is discussed in Chapter 24.

Section 17.11

■

Energy Relationships for AC

663

Skin Effect Magnetically induced voltages created inside a conductor by its own changing magnetic field force electrons to the periphery of the conductor (Figure 17–28), resulting in a nonuniform distribution of current, with current density greatest near the periphery and smallest in the center. This phenomenon is known as skin effect. Because the center of the wire carries little current, its cross-sectional area has effectively been reduced, thus increasing resistance. While skin effect is generally negligible at power line frequencies (except for conductors larger than several hundred thousand circular mils), it is so pronounced at microwave frequencies that the center of a wire carries almost no current. For this reason, hollow conductors are often used instead of solid wires, as shown in Figure 17–28(c). Current density varies across the wire.

Wire (a) Varying magnetic field creates forces on electrons within the conductor

(b) This force pushes electrons outward, leaving few free electrons in the center of the conductor

FIGURE 17–28 Skin effect in ac circuits.

Radiation Resistance At high frequencies some of the energy supplied to a circuit escapes as radiated energy. For example, a radio transmitter supplies power to an antenna, where it is converted into radio waves and radiated into space. The resistance effect here is known as radiation resistance. This resistance is much higher than simple dc resistance. For example, a TV transmitting antenna may have a resistance of a fraction of an ohm to dc but several hundred ohms effective resistance at its operating frequency.

17.11 Energy Relationships for AC Recall, power and energy are related by the equation p dw/dt. Thus, energy can be found by integration as

冕

W pdt vidt

(17–26)

Inductance For an inductance, v Ldi/dt. Substituting this into Equation 17–26, cancelling dt, and rearranging terms yields WL

冕冢L dd it 冣idt L冕idi

(17–27)

FINAL NOTES... 1. The resistance measured by an ohmmeter is dc resistance. 2. Many of the effects noted above will be treated in detail in your various electronics courses. We will not pursue them further here.

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Power in AC Circuits

Recall from Figure 17–4(b), energy flows into an inductor during time interval 0 to T/4 and is released during time interval T/4 to T/2. The process then repeats itself. The energy stored (and subsequently released) can thus be found by integrating power from t 0 to t T/4. Current at t 0 is 0 and current at t T/4 is Im. Using these as our limits of integration, we find

冕

WL L

1 idi LI 2m LI 2 2

(J)

(17–28)

where we have used I Im /兹2苶 to express energy in terms of effective current.

Capacitance For a capacitance, i Cdv/dt. Substituting this into Equation 17–26 yields

冕冢冣

冕

dv WC v C dt C v dv dt

(17–29)

Consider Figure 17–5(b). Energy stored can be found by integrating power from T/4 to T/2. The corresponding limits for voltage are 0 to Vm. Thus,

冕

WC C

1 v dv CV 2m CV 2 2

(J)

(17–30)

where we have used V Vm /兹2苶. You will use these relationships later.

17.12 Circuit Analysis Using Computers

NOTES 1. As of this writing, Electronics Workbench has no simple way to plot current and power; thus, no example is included here. If this changes when EWB is updated, we will add examples and problems to our web site. 2. If you want to include the identification VSIN, 1000Hz, etc. on your schematic as in Figure 17–29, do it from the Property Editor as described in Chapter 16.

The time-varying relationships between voltage, current and power described earlier in this chapter can be investigated easily using PSpice. To illustrate, consider the circuit of Figure 17–3 with v 1.2 sin qt, R 0.8 ⍀ and f 1000 Hz. Create the circuit on the screen, including voltage and current markers as in Figure 17–29, then set VSIN parameters to VOFF 0V, VAMPL 1.2V and FREQ 1000Hz as you did in Chapter 16. Click the New Simulation Profile, type fig17-29 for a name, choose Transient, set TSTOP to 1ms, then click OK. Run the simulation and the voltage and current waveforms of Figure 17–30 should appear. To plot power (i.e., the product of vi), click Trace, then Add Trace, and when the dialog box opens, use the asterisk to create the product

FIGURE 17–29 Using PSpice to investigate instantaneous power in an ac circuit.

665

Problems

V(R1:1)*I(R1), then click OK. The blue power curve should now appear. Compare to Figure 17–3. Note that all curves agree exactly.

FIGURE 17–30 Voltage, current, and power waveforms for Figure 17–29.

17.1–17.5 1. Note that the power curve of Figure 17–4 is sometimes positive and sometimes negative. What is the significance of this? Between t T/4 and t T/2, what is the direction of power flow? 2. What is real power? What is reactive power? Which power, real or reactive, has an average value of zero? 3. A pair of electric heating elements is shown in Figure 17–31. a. Determine the active and reactive power to each. b. Determine the active and reactive power delivered by the source. 4. For the circuit of Figure 17–32, determine the active and reactive power to the inductor. 5. If the inductor of Figure 17–32 is replaced by a 40-␮F capacitor and source frequency is 60 Hz, what is QC? 6. Find R and XL for Figure 17–33. 12.5 A

625 W

PROBLEMS

⫹ 100 V ⫺

10 ⍀ R2

20 ⍀

FIGURE 17–31

⫹ 100 V ⫺

j8 ⍀

FIGURE 17–32

2500 VAR

R 5A 250 W 150 VAR

FIGURE 17–33

7. For the circuit of Figure 17–34, f 100 Hz. Find a. R b. XC

c. C

FIGURE 17–34

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Chapter 17

P 40 VAR

■

Power in AC Circuits

⫹ 60 V ⫺

8⍀

8. For the circuit of Figure 17–35, f 10 Hz. Find a. P b. XL 9. For Figure 17–36, find XC.

40 Ω

360 W

FIGURE 17–35

c. L

480 VAR

FIGURE 17–36

10. For Figure 17–37, XC 42.5 ⍀. Find R, P, and Q. 11. Find the total average power and the total reactive power supplied by the source for Figure 17–38.

380.1 VA

R XC

85 V

10 Ω

40 Ω

4 A ∠0°

FIGURE 17–37

15 Ω

FIGURE 17–38

12. If the source of Figure 17–38 is reversed, what is PT and QT? What conclusion can you draw from this? 13. Refer to Figure 17–39. Find P2 and Q3. Is the element in Load 3 inductive or capacitive? P2 600 VAR (cap.) 2.9 kW

1200 W 1400 VAR

1.1 kVAR (ind.)

800 W Q3

(ind.)

FIGURE 17–39

14. For Figure 17–40, determine PT and QT. 10 Ω

PT QT

FIGURE 17–40

22 Ω

j6 Ω

2.5 A∠0°

Problems 15. For Figure 17–41, q 10 rad/s. Determine a. RT b. R2 c. XC I=6A

j20

d. Leq

ⳮjXC

10 ⍀

720 W

720 W 432 VAR (ind)

432 VAR (ind.)

Leq

R3 4⍀ (b)

(a) FIGURE 17–41

16. For Figure 17–42, determine the total PT and QT. PT QT

⫹ 600 V ⫺

1152 W 633.6 VAR (cap.)

R1 20 ⍀

2.4 A R2 40 ⍀

XC 50 ⍀

XL 40 ⍀

FIGURE 17–42

17.7 The Relationship Between P, Q, and S 17. For the circuit of Figure 17–7, draw the power triangle and determine the apparent power. 18. Repeat Problem 17 for Figure 17–8. 19. Ignoring the wattmeter of Figure 17–27, determine the power triangle for the circuit as seen by the source. 20. For the circuit of Figure 17–43, what is the source current? 21. For Figure 17–44, the generator supplies 30 A. What is R? 30 A

I ⫹ ⫺

1200 W 0W 800 VAR 200 VAR (cap.) (ind.) 200 W 120 V 0 VAR

FIGURE 17–43

R ⫹ ⫺

600 V

FIGURE 17–44

1377 W 7450 W 2750 VAR 17 880 VAR (ind.) (cap.) 6915 W 12 960 VAR (ind.)

(c)

667

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Power in AC Circuits

1014 W 4608 VAR (ind.)

⫹ 600 V

0W 1912 VAR (cap.)

⫺

22. Suppose V 100 V∠60° and I 10A∠40°: a. What is v, the angle between V and I? b. Determine P from P VI cos v. c. Determine Q from Q VI sin v. d. Sketch the power triangle and from it, determine S. e. Show that S VI* gives the same answer as (d). 23. For Figure 17–45, Sgen 4835 VA. What is R? 24. Refer to the circuit of Figure 17–16: a. Determine the apparent power for each box. b. Sum the apparent powers that you just computed. Why does the sum not equal ST 1803 VA as obtained in Example 17–5? 17.8 Power Factor 25. Refer to the circuit of Figure 17–46: a. Determine PT, QT, and ST.

FIGURE 17–45

b. Determine whether the fuse will blow. Ig

⫹

⫺

⫹ 600 V 60 Hz ⫺

Motor

Po = 10 hp η = 0.87 Fp = 0.65 (lag)

FIGURE 17–47 ⫹ 600 V ⫺

Motors

Electric utility 300 kVA capacity

Lights

Other loads

Factory loads

(a)

k 00

36° P (b) Factory power triangle FIGURE 17–48

15-A fuse 120 V

182 VA Fp = 0.8 (lag)

772 VA Fp = 0.385 (lead)

278 W 521 VAR (ind.)

FIGURE 17–46

26. A motor with an efficiency of 87% supplies 10 hp to a load (Figure 17–47). Its power factor is 0.65 (lag). a. What is the power input to the motor? b. What is the reactive power to the motor? c. Draw the motor power triangle. What is the apparent power to the motor? 27. To correct the circuit power factor of Figure 17–47 to unity, a power factor correction capacitor is added. a. Show where the capacitor is connected. b. Determine its value in microfarads. 28. Consider Figure 17–20. The motor is replaced with a new unit requiring Sm (120 j35) kVA. Everything else remains the same. Find the following: a. PT b. QT c. ST d. Determine how much kVAR capacitive correction is needed to correct to unity Fp. 29. A small electrical utility has a 600-V, 300-kVA capacity. It supplies a factory (Figure 17–48) with the power triangle shown in (b). This fully loads the utility. If a power factor correcting capacitor corrects the load to unity

Problems

669

power factor, how much more power (at unity power factor) can the utility sell to other customers? 17.9 AC Power Measurement 30. a. Why does the wattmeter of Figure 17–49 indicate only 1200 watts? b. Where would the wattmeter have to be placed to measure power delivered by the source? Sketch the modified circuit. c. What would the wattmeter indicate in (b)?

700 W 600 VAR ⫹ E ⫺

W 300 W 0 VAR

120 V∠50°

1200 W 400 VAR

25 A∠20°

FIGURE 17–50

FIGURE 17–49

31. Determine the wattmeter reading for Figure 17–50. 32. Determine the wattmeter reading for Figure 17–51.

W 120 V

40 Ω XC 60 Ω

FIGURE 17–51

17.10 Effective Resistance 33. Measurements on an iron-core solenoid coil yield the following values: V 80 V, I 400 mA, P 25.6 W, and R 140 ⍀. (The last measurement was taken with an ohmmeter.) What is the ac resistance of the solenoid coil? 17.12 Circuit Analysis Using Computers 34. PSpice An inductance L 1 mH has current i 4 sin (2p 1000)t. Use PSpice to investigate the power waveform and compare to Figure 17–4. Use current source ISIN (see Note). 35. PSpice A 10-mF capacitor has voltage v 10 sin(qt 90°) V. Use PSpice to investigate the power waveform and compare to Figure 17–5. Use voltage source VSIN with f 1000 Hz.

NOTE... PSpice represents current into devices. Thus, when you double click a current source symbol (ISIN, IPWL, etc.) and specify a current waveform, you are specifying the current into the source.

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■

Power in AC Circuits 36. PSpice The voltage waveform of Figure 17–52 is applied to a 200-mF capacitor. v 10 V

t (ms)

0 1

10 V FIGURE 17–52

a. Using the principles of Chapter 10, determine the current through the capacitor and sketch. (Also sketch the voltage waveform on your graph.) Multiply the two waveforms to obtain a plot of p(t). Compute power at its max and min points. b. Use PSpice to verify the results. Use voltage source VPWL. You have to describe the waveform to the source. It has a value of 0 V at t 0, 10 V at t 1 ms, 10 V at t 3 ms, and 0 V at t 4 ms. To set these, double click the source symbol and enter values via the Property Editor as follows: 0 for T1, 0V for V1, 1ms for T2, 10V for V2, etc. Run the simulation and plot voltage, current, and power using the procedure we used to create Figure 17–30. Results should agree with those of (a). 37. PSpice Repeat Question 36 for a current waveform identical to Figure 17–52 except that it oscillates between 2 A and 2 A applied to a 2-mH inductor. Use current source IPWL (see Note).

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 1. 100 V

2000 VA

1600 VAR

53.13° 1200 W FIGURE 17–53

2. 8.76 A 3. Fig. 17–7: 0.6 (lag); Fig. 17–8: 0.97 (lead); Fig. 17–9: 0.6 (lag) 4. Yes. (S 128 kVA)

AC Series-Parallel Circuits OBJECTIVES After studying this chapter, you will be able to • apply Ohm’s law to analyze simple series circuits, • apply the voltage divider rule to determine the voltage across any element in a series circuit, • apply Kirchhoff’s voltage law to verify that the summation of voltages around a closed loop is equal to zero, • apply Kirchhoff’s current law to verify that the summation of currents entering a node is equal to the summation of currents leaving the same node, • determine unknown voltage, current, and power for any series/parallel circuit, • determine the series or parallel equivalent of any network consisting of a combination of resistors, inductors, and capacitors.

KEY TERMS AC Parallel Circuits AC Series Circuits Current Divider Rule

Frequency Effects Impedance Kirchhoff’s Current Law Kirchhoff’s Voltage Law Voltage Divider Rule

OUTLINE Ohm’s Law for AC Circuits AC Series Circuits Kirchhoff’s Voltage Law and the Voltage Divider Rule AC Parallel Circuits Kirchhoff’s Current Law and the Current Divider Rule Series-Parallel Circuits Frequency Effects Applications Circuit Analysis Using Computers

n this chapter we examine how simple circuits containing resistors, inductors, and capacitors behave when subjected to sinusoidal voltages and currents. Principally, we find that the rules and laws which were developed for dc circuits will apply equally well for ac circuits. The major difference between solving dc and ac circuits is that analysis of ac circuits requires using vector algebra. In order to proceed successfully, it is suggested that the student spend time reviewing the important topics covered in dc analysis. These include Ohm’s law, the voltage divider rule, Kirchhoff’s voltage law, Kirchhoff’s current law, and the current divider rule. You will also find that a brief review of vector algebra will make your understanding of this chapter more productive. In particular, you should be able to add and subtract any number of vector quantities.

Heinrich Rudolph Hertz HEINRICH HERTZ WAS BORN IN HAMBURG, Germany, on February 22, 1857. He is known mainly for his research into the transmission of electromagnetic waves. Hertz began his career as an assistant to Hermann von Helmholtz in the Berlin Institute physics laboratory. In 1885, he was appointed Professor of Physics at Karlsruhe Polytechnic, where he did much to verify James Clerk Maxwell’s theories of electromagnetic waves. In one of his experiments, Hertz discharged an induction coil with a rectangular loop of wire having a very small gap. When the coil discharged, a spark jumped across the gap. He then placed a second, identical coil close to the first, but with no electrical connection. When the spark jumped across the gap of the first coil, a smaller spark was also induced across the second coil. Today, more elaborate antennas use similar principles to transmit radio signals over vast distances. Through further research, Hertz was able to prove that electromagnetic waves have many of the characteristics of light: they have the same speed as light; they travel in straight lines; they can be reflected and refracted; and they can be polarized. Hertz’s experiments ultimately led to the development of radio communication by such electrical engineers as Guglielmo Marconi and Reginald Fessenden. Heinrich Hertz died at the age of 36 on January 1, 1894.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

673

674

Chapter 18

■

AC Series-Parallel Circuits

18.1

Ohm’s Law for AC Circuits

This section is a brief review of the relationship between voltage and current for resistors, inductors, and capacitors. Unlike Chapter 16, all phasors are given as rms rather than as peak values. As you saw in Chapter 17, this approach simplifies the calculation of power.

NOTES... Although currents and voltages may be shown in either time domain (as sinusoidal quantities) or in phasor domain (as vectors), resistance and reactance are never shown as sinusoidal quantities. The reason for this is that whereas currents and voltages vary as functions of time, resistance and reactance do not.

Resistors In Chapter 16, we saw that when a resistor is subjected to a sinusoidal voltage as shown in Figure 18–1, the resulting current is also sinusoidal and in phase with the voltage. The sinusoidal voltage v Vmsin(qt v) may be written in phasor form as V V∠v. Whereas the sinusoidal expression gives the instantaneous value of voltage for a waveform having an amplitude of Vm (volts peak), the phasor form has a magnitude which is the effective (or rms) value. The relationship between the magnitude of the phasor and the peak of the sinusoidal voltage is given as Vm V 兹2苶

Because the resistance vector may be expressed as ZR R∠0°, we evaluate the current phasor as follows: V V∠v V I ∠v I∠v ZR R∠0° R

i R

(a)

v = Vm sin (ωt + θ) i = Im sin (ωt + θ)

ωt (rad)

2 radians (b) FIGURE 18–1

Sinusoidal voltage and current for a resistor.

Section 18.1

■

675

Ohm’s Law for AC Circuits

If we wish to convert the current from phasor form to its sinusoidal equivalent in the time domain, we would have i Imsin(qt v). Again, the relationship between the magnitude of the phasor and the peak value of the sinusoidal equivalent is given as I I m 兹2苶

The voltage and current phasors may be shown on a phasor diagram as in Figure 18–2. Because one phasor is a current and the other is a voltage, the relative lengths of these phasors are purely arbitrary. Regardless of the angle v, we see that the voltage across and the current through a resistor will always be in phase.

j V V ∠θ

I I ∠θ

j FIGURE 18–2 Voltage and current phasors for a resistor.

EXAMPLE 18–1

Refer to the resistor shown in Figure 18–3: i v 72 sin ωt

FIGURE 18–3

a. Find the sinusoidal current i using phasors. b. Sketch the sinusoidal waveforms for v and i. c. Sketch the phasor diagram of V and I. Solution: a. The phasor form of the voltage is determined as follows: v 72 sin qt ⇔ V 50.9 V∠0° From Ohm’s law, the current phasor is determined to be 50.9 V∠0° V I 2.83 A∠0° 18 ∠0° ZR which results in the sinusoidal current waveform having an amplitude of Im (兹2 苶)(2.83 A) 4.0 A Therefore, the current i will be written as i 4 sin qt

676

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AC Series-Parallel Circuits

b. The voltage and current waveforms are shown in Figure 18–4. v = 72 sin ωt i = 4 sin ωt

3 2

ωt

FIGURE 18–4

c. Figure 18–5 shows the voltage and current phasors. j

V 50.9 V∠0 I 2.83 A∠0

FIGURE 18–5

EXAMPLE 18–2

Refer to the resistor of Figure 18–6: i 3 × 10 −3 sin(ωt 40 ) v

2 k

FIGURE 18–6

a. Use phasor algebra to find the sinusoidal voltage, v. b. Sketch the sinusoidal waveforms for v and i. c. Sketch a phasor diagram showing V and I. Solution a. The sinusoidal current has a phasor form as follows: i 3 10 3sin(qt 40°) ⇔ I 2.12 mA∠ 40°

Section 18.1

From Ohm’s law, the voltage across the 2-k resistor is determined as the phasor product V IZR (2.12 mA∠ 40°)(2 k ∠0°) 4.24 V∠ 40° The amplitude of the sinusoidal voltage is Vm (兹2 苶)(4.24 V) 6.0 V The voltage may now be written as v 6.0 sin(qt 40°) b. Figure 18–7 shows the sinusoidal waveforms for v and i. v = 6 sin (ωt 40 )

ωt

i 3 × 10 −3 sin(ωt 40 ) FIGURE 18–7

c. The corresponding phasors for the voltage and current are shown in Figure 18–8. j

I 2.12 mA∠ 40

V 4.24 V∠ 40

FIGURE 18–8

Inductors When an inductor is subjected to a sinusoidal current, a sinusoidal voltage is induced across the inductor such that the voltage across the inductor leads the current waveform by exactly 90°. If we know the reactance of an inductor, then from Ohm’s law the current in the inductor may be expressed in phasor form as V V∠v V I ∠(v 90°) ZL XL∠90° XL

■

Ohm’s Law for AC Circuits

677

678

Chapter 18

■

AC Series-Parallel Circuits

In vector form, the reactance of the inductor is given as ZL XL∠90°

where XL qL 2pfL.

EXAMPLE 18–3

Consider the inductor shown in Figure 18–9: i

v 1.05 sin(ωt 120 )

FIGURE 18–9

a. Determine the sinusoidal expression for the current i using phasors. b. Sketch the sinusoidal waveforms for v and i. c. Sketch the phasor diagram showing V and I. Solution: a. The phasor form of the voltage is determined as follows: v 1.05 sin(qt 120°) ⇔ V 0.742 V∠120° From Ohm’s law, the current phasor is determined to be 0.742 V∠120° V I 29.7 mA∠30° 25 ∠90° ZL The amplitude of the sinusoidal current is Im (兹2 苶)(29.7 mA) 42 mA The current i is now written as i 0.042 sin(qt 30°) b. Figure 18–10 shows the sinusoidal waveforms of the voltage and current. v 1.05 sin(ωt 120°)

30°

ωt

120° i 0.042 sin(ωt 30°)

FIGURE 18–10 Sinusoidal voltage and current for an inductor.

Section 18.1

c. The voltage and current phasors are shown in Figure 18–11. V 0.742 V∠120 j

120

Voltage leads current by 90° I 29.7 mA∠30 30

j FIGURE 18–11 Voltage and current phasors for an inductor.

Capacitors When a capacitor is subjected to a sinusoidal voltage, a sinusoidal current results. The current through the capacitor leads the voltage by exactly 90°. If we know the reactance of a capacitor, then from Ohm’s law the current in the capacitor expressed in phasor form is V V∠v V I ∠(v 90°) ZC XL∠ 90° XL

In vector form, the reactance of the capacitor is given as ZC XC∠ 90°

where 1 1 XC qC 2pfC

EXAMPLE 18–4

Consider the capacitor of Figure 18–12. i 2.4 × 10−3 sin(ωt 62 )

1.2 k v

FIGURE 18–12

a. Find the voltage v across the capacitor. b. Sketch the sinusoidal waveforms for v and i. c. Sketch the phasor diagram showing V and I. Solution a. Converting the sinusoidal current into its equivalent phasor form gives i 2.4 10 3sin(qt 62°) ⇔ I 1.70 mA∠62°

■

Ohm’s Law for AC Circuits

679

680

Chapter 18

■

AC Series-Parallel Circuits

From Ohm’s law, the phasor voltage across the capacitor must be V IZC (1.70 mA∠62°)(1.2 k ∠ 90°) 2.04 V∠ 28° The amplitude of the sinusoidal voltage is Vm (兹2 苶)(2.04 V) 2.88 V The voltage v is now written as v 2.88 sin(qt 28°) b. Figure 18–13 shows the waveforms for v and i. v = 2.88 sin(ωt 28 )

i = 2.4 10−3 sin(ωt 62 )

2π

3π

ωt

FIGURE 18–13 Sinusoidal voltage and current for a capacitor.

c. The corresponding phasor diagram for V and I is shown in Figure 18–14. j I 1.70 mA∠62

Current leads voltage by 90°

62 28 V 2.04 V∠ 28 j FIGURE 18–14 Voltage and current phasors for a capacitor.

The relationships between voltage and current, as illustrated in the previous three examples, will always hold for resistors, inductors, and capacitors.

IN-PROCESS

LEARNING CHECK 1

1. What is the phase relationship between current and voltage for a resistor? 2. What is the phase relationship between current and voltage for a capacitor? 3. What is the phase relationship between current and voltage for an inductor? (Answers are at the end of the chapter.)

Section 18.2 A voltage source, E 10 V∠30°, is applied to an inductive impedance of 50 . a. Solve for the phasor current, I. b. Sketch the phasor diagram for E and I. c. Write the sinusoidal expressions for e and i. d. Sketch the sinusoidal expressions for e and i.

■

AC Series Circuits

PRACTICE PROBLEMS 1

Answers: a. I 0.2 A∠ 60° c. e 14.1 sin(qt 30°) i 0.283 sin(qt 60°)

A voltage source, E 10 V∠30°, is applied to a capacitive impedance of 20 . a. Solve for the phasor current, I. b. Sketch the phasor diagram for E and I. c. Write the sinusoidal expressions for e and i. d. Sketch the sinusoidal expressions for e and i. Answers: a. I 0.5 A∠120° c. e 14.1 sin(qt 30°) i 0.707 sin(qt 120°)

18.2

AC Series Circuits

When we examined dc circuits we saw that the current everywhere in a series circuit is always constant. This same applies when we have series elements with an ac source. Further, we had seen that the total resistance of a dc series circuit consisting of n resistors was determined as the summation RT R1 R2 … Rn

When working with ac circuits we no longer work with only resistance but also with capacitive and inductive reactance. Impedance is a term used to collectively determine how the resistance, capacitance, and inductance “impede” the current in a circuit. The symbol for impedance is the letter Z and the unit is the ohm ( ). Because impedance may be made up of any combination of resistances and reactances, it is written as a vector quantity Z, where Z Z∠v

( )

Each impedance may be represented as a vector on the complex plane, such that the length of the vector is representative of the magnitude of the impedance. The diagram showing one or more impedances is referred to as an impedance diagram. Resistive impedance ZR is a vector having a magnitude of R along the positive real axis. Inductive reactance ZL is a vector having a magnitude of

PRACTICE PROBLEMS 2

681

682

Chapter 18

■

AC Series-Parallel Circuits

XL along the positive imaginary axis, while the capacitive reactance ZC is a vector having a magnitude of XC along the negative imaginary axis. Mathematically, each of the vector impedances is written as follows:

ZR R∠0° R j0 R ZL XL∠90° 0 jXL jXL ZC XC∠90° 0 jXC jXC

ZL XL∠90 j XL ZR R∠0

ZC XC∠ 90 j XC j

An impedance diagram showing each of the above impedances is shown in Figure 18–15. All impedance vectors will appear in either the first or the fourth quadrants, since the resistive impedance vector is always positive. For a series ac circuit consisting of n impedances, as shown in Figure 18–16, the total impedance of the circuit is found as the vector sum ZT Z1 Z2 … Zn

FIGURE 18–15

(18–1)

Consider the branch of Figure 18–17. By applying Equation 18–1, we may determine the total impedance of the circuit as ZT (3 j0) (0 j4 ) 3 j4 5 ∠53.13°

The above quantities are shown on an impedance diagram as in Figure 18–18. 4 ∠90

ZT 3 j 4 5 ∠53.13

53.13

3 ∠0

3 FIGURE 18–16

FIGURE 18–17

FIGURE 18–18

From Figure 18–18 we see that the total impedance of the series elements consists of a real component and an imaginary component. The corresponding total impedance vector may be written in either polar or rectangular form. The rectangular form of an impedance is written as Z R jX

If we are given the polar form of the impedance, then we may determine the equivalent rectangular expression from R Z cos v

(18–2)

X Z sin v

(18–3)

and

Section 18.2

In the rectangular representation for impedance, the resistance term, R, is the total of all resistance looking into the network. The reactance term, X, is the difference between the total capacitive and inductive reactances. The sign for the imaginary term will be positive if the inductive reactance is greater than the capacitive reactance. In such a case, the impedance vector will appear in the first quadrant of the impedance diagram and is referred to as being an inductive impedance. If the capacitive reactance is larger, then the sign for the imaginary term will be negative. In such a case, the impedance vector will appear in the fourth quadrant of the impedance diagram and the impedance is said to be capacitive. The polar form of any impedance will be written in the form Z Z∠v

The value Z is the magnitude (in ohms) of the impedance vector Z and is determined as follows: Z 兹R 苶2苶苶苶 X2苶

( )

(18–4)

The corresponding angle of the impedance vector is determined as

冢冣

X v tan 1 R

(18–5)

Whenever a capacitor and an inductor having equal reactances are placed in series, as shown in Figure 18–19, the equivalent circuit of the two components is a short circuit since the inductive reactance will be exactly balanced by the capacitive reactance. Any ac circuit having a total impedance with only a real component, is referred to as a resistive circuit. In such a case, the impedance vector ZT will be located along the positive real axis of the impedance diagram and the angle of the vector will be 0°. The condition under which series reactances are equal is referred to as “series resonance” and is examined in greater detail in a later chapter. If the impedance Z is written in polar form, then the angle v will be positive for an inductive impedance and negative for a capacitive impedance. In the event that the circuit is purely reactive, the resulting angle v will be either 90° (inductive) or 90° (capacitive). If we reexamine the impedance diagram of Figure 18–18, we conclude that the original circuit is inductive.

XC X ZT 0 XL X

ZT jXL jXC jX jX 0 FIGURE 18–19

■

AC Series Circuits

683

684

Chapter 18

■

AC Series-Parallel Circuits

EXAMPLE 18–5

Consider the network of Figure 18–20.

VR R 25

10 V∠0 ZT

FIGURE 18–20

200

XC 225 VC

a. Find ZT. b. Sketch the impedance diagram for the network and indicate whether the total impedance of the circuit is inductive, capacitive, or resistive. c. Use Ohm’s law to determine I, VR , and VC . Solution a. The total impedance is the vector sum ZT 25 j200 ( j225 ) 25 j25 35.36 ∠ 45° b. The corresponding impedance diagram is shown in Figure 18–21. j j200

45 j200 j225 j25

ZT 25 j 25 35.36 ∠ 45

j225 FIGURE 18–21

Because the total impedance has a negative reactance term ( j25 ), ZT is capacitive. c.

10 V∠0° I 0.283 A∠45° 35.36 ∠ 45° VR (282.8 mA∠45°)(25 ∠0°) 7.07 V∠45° VC (282.8 mA∠45°)(225 ∠ 90°) 63.6 V∠ 45°

Section 18.2

Notice that the magnitude of the voltage across the capacitor is many times larger than the source voltage applied to the circuit. This example illustrates that the voltages across reactive elements must be calculated to ensure that maximum ratings for the components are not exceeded.

EXAMPLE 18–6 Determine the impedance Z which must be within the indicated block of Figure 18–22 if the total impedance of the network is 13 ∠22.62°. R 2 ZT 13 ∠22.62

XL 10

FIGURE 18–22

Solution we get

Converting the total impedance from polar to rectangular form, ZT 13 ∠22.62° ⇔ 12 j5

Now, we know that the total impedance is determined from the summation of the individual impedance vectors, namely ZT 2 j10 Z 12 j5 Therefore, the impedance Z is found as Z 12 j5 (2 j10 ) 10 j5 11.18 ∠ 26.57° In its most simple form, the impedance Z will consist of a series combination of a 10- resistor and a capacitor having a reactance of 5 . Figure 18–23 shows the elements which may be contained within Z to satisfy the given conditions.

FIGURE 18–23

Z 10 j 5 11.18 ∠ 26.57

■

AC Series Circuits

685

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Chapter 18

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AC Series-Parallel Circuits

EXAMPLE 18–7

Find the total impedance for the network of Figure 18–24. Sketch the impedance diagram showing Z1, Z2, and ZT.

Z1 50 j 90 ZT Z2 30 j 30 FIGURE 18–24

Solution: ZT Z1 Z2 (50 j90 ) (30 j30 ) (80 j60 ) 100 ∠36.87° The polar forms of the vectors Z1 and Z2 are as follows: Z1 50 j90 102.96 ∠60.95° Z2 30 j30 42.43 ∠ 45° The resulting impedance diagram is shown in Figure 18–25. j Z1 102.96 ∠60.95

60.95

ZT 100 ∠36.87

36.87 45 Z2 42.43 ∠ 45

FIGURE 18–25

The phase angle v for the impedance vector Z Z∠v provides the phase angle between the voltage V across Z and the current I through the impedance. For an inductive impedance the voltage will lead the current by v. If the impedance is capacitive, then the voltage will lag the current by an amount equal to the magnitude of v.

Section 18.2

The phase angle v is also useful for determining the average power dissipated by the circuit. In the simple series circuit shown in Figure 18–26, we know that only the resistor will dissipate power. The average power dissipated by the resistor may be determined as follows: V2 P VR I R I 2R R

■

687

AC Series Circuits I

R VR

V ZT

(18–6)

Notice that Equation 18–6 uses only the magnitudes of the voltage, current, and impedance vectors. Power is never determined by using phasor products. Ohm’s law provides the magnitude of the current phasor as

FIGURE 18–26

V I Z

Substituting this expression into Equation 18–6, we obtain the expression for power as

冢冣

V2 V2 R P R 2 Z Z Z

(18–7) j

From the impedance diagram of Figure 18–27, we see that R cos v Z

The previous chapter had defined the power factor as Fp cos v, where v is the angle between the voltage and current phasors. We now see that for a series circuit, the power factor of the circuit can be determined from the magnitudes of resistance and total impedance. R Fp cos v Z

(18–8) FIGURE 18–27

The power factor, Fp , is said to be leading if the current leads the voltage (capacitive circuit) and lagging if the current lags the voltage (inductive circuit). Now substituting the expression for the power factor into Equation 18–7, we express power delivered to the circuit as P VI cos v

Since V IZ, power may be expressed as V2 P VI cos v I 2Z cos v cos v Z

Z Z∠θ

(18–9)

688

Chapter 18

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AC Series-Parallel Circuits

EXAMPLE 18–8

Refer to the circuit of Figure 18–28. I

e 20 2 sin ωt

FIGURE 18–28

a. Find the impedance ZT. b. c. d. e. f.

Calculate the power factor of the circuit. Determine I. Sketch the phasor diagram for E and I. Find the average power delivered to the circuit by the voltage source. Calculate the average power dissipated by both the resistor and the capacitor.

Solution a. ZT 3 j4 5 ∠ 53.13° b. Fp cos v 3 \5 0.6 (leading) c. The phasor form of the applied voltage is (兹2 苶)(20 V) E ∠0° 20 V∠0° 兹2苶 which gives a current of 20 V∠0° I 4.0 A∠53.13° 5 ∠ 53.13° d. The phasor diagram is shown in Figure 18–29. j

I 4.0 A∠53.13

53.13

E 20 V∠0

FIGURE 18–29

From this phasor diagram, we see that the current phasor for the capacitive circuit leads the voltage phasor by 53.13°.

Section 18.3

■

Kirchhoff’s Voltage Law and the Voltage Divider Rule

e. The average power delivered to the circuit by the voltage source is P (20 V)(4 A) cos 53.13° 48.0 W f. The average power dissipated by the resistor and capacitor will be PR (4 A)2(3 )cos 0° 48 W PC (4 A)2(4 )cos 90° 0 W (as expected!) Notice that the power factor used in determining the power dissipated by each of the elements is the power factor for that element and not the total power factor for the circuit. As expected, the summation of powers dissipated by the resistor and capacitor is equal to the total power delivered by the voltage source.

A circuit consists of a voltage source E 50 V∠25° in series with L 20 mH, C 50 mF, and R 25 . The circuit operates at an angular frequency of 2 krad/s. a. Determine the current phasor, I. b. Solve for the power factor of the circuit. c. Calculate the average power dissipated by the circuit and verify that this is equal to the average power delivered by the source. d. Use Ohm’s law to find VR , VL, and VC . Answers: a. I 1.28 A∠ 25.19° b. Fp 0.6402 c. P 41.0 W d. VR 32.0 V∠ 25.19° VC 12.8 V∠ 115.19° VL 51.2 V∠64.81°

18.3

Kirchhoff’s Voltage Law and the Voltage Divider Rule

When a voltage is applied to impedances in series, as shown in Figure 18–30, Ohm’s law may be used to determine the voltage across any impedance as Vx IZx

The current in the circuit is E I ZT

Now, by substitution we arrive at the voltage divider rule for any series combination of elements as Zx Vx E ZT

(18–10)

PRACTICE PROBLEMS 3

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AC Series-Parallel Circuits

Equation 18–10 is very similar to the equation for the voltage divider rule in dc circuits. The fundamental differences in solving ac circuits are that we use impedances rather than resistances and that the voltages found are phasors. Because the voltage divider rule involves solving products and quotients of phasors, we generally use the polar form rather than the rectangular form of phasors. Kirchhoff’s voltage law must apply for all circuits whether they are dc or ac circuits. However, because ac circuits have voltages expressed in either sinusoidal or phasor form, Kirchhoff’s voltage law for ac circuits may be stated as follows:

The phasor sum of voltage drops and voltage rises around a closed loop is equal to zero.

E Vx

When adding phasor voltages, we find that the summation is generally done more easily in rectangular form rather than the polar form.

EXAMPLE 18–9

Consider the circuit of Figure 18–31. I

FIGURE 18–30

5 k

E 26 V∠0 XL

12 k VL

FIGURE 18–31

a. Find ZT. b. Determine the voltages VR and VL using the voltage divider rule. c. Verify Kirchhoff’s voltage law around the closed loop. Solution a. ZT 5 k j12 k 13 k ∠67.38° 5 k ∠0° b. VR (26 V∠0°) 10 V∠ 67.38° 13 k ∠67.38° 12 k ∠90° VL (26 V∠0°) 24 V∠22.62° 13 k ∠67.38° c. Kirchhoff’s voltage law around the closed loop will give

冢冢

冣冣

26 V∠0° 10 V∠ 67.38° 24 V∠22.62° 0 (26 j0) (3.846 j9.231) (22.154 j9.231) 0 (26 3.846 22.154) j(0 9.231 9.231) 0 0 j0 0

Section 18.3

■

Kirchhoff’s Voltage Law and the Voltage Divider Rule

EXAMPLE 18–10 Consider the circuit of Figure 18–32:

Z1 v 40 j80 1 e 100 sin ωt

Z2 30 j40

FIGURE 18–32

a. Calculate the sinusoidal voltages v1 and v2 using phasors and the voltage divider rule. b. Sketch the phasor diagram showing E, V1, and V2. c. Sketch the sinusoidal waveforms of e, v1, and v2. Solution a. The phasor form of the voltage source is determined as e 100 sin qt ⇔ E 70.71∠V 0° Applying VDR, we get 40 j80 V1 (70.71 V∠0°) (40 j80 ) (30 j40 )

冢冣 89.44 ∠ 63.43° 冢冣(70.71 V∠0°) 80.62 ∠ 29.74° 78.4 V∠ 33.69°

and 30 j40 V2 (70.71 V∠0°) (40 j80 ) (30 j40 )

冢冣 50.00 ∠53.13° 冢冣(70.71 V∠0°) 80.62 ∠ 29.74° 43.9 V∠82.87°

The sinusoidal voltages are determined to be v1 (兹2 苶)(78.4)sin(qt 33.69°) 111 sin(qt 33.69°) and v2 (兹2 苶)(43.9)sin(qt 82.87°) 62.0 sin(qt 82.87°) b. The phasor diagram is shown in Figure 18–33.

691

692

Chapter 18

■

AC Series-Parallel Circuits j V2 43.9 V∠82.87 82.87

E 70.71 V∠0

33.69

V1 78.4 V∠ 33.69

FIGURE 18–33

c. The corresponding sinusoidal voltages are shown in Figure 18–34. v1 111 V 100 V 62.0 V

v2 e

82.9°

ωt (rad)

33.7°

FIGURE 18–34

IN-PROCESS

LEARNING CHECK 2

1. Express Kirchhoff’s voltage law as it applies to ac circuits. 2. What is the fundamental difference between how Kirchhoff’s voltage law is used in ac circuits as compared with dc circuits? (Answers are at the end of the chapter.)

PRACTICE PROBLEMS 4

A circuit consists of a voltage source E 50 V∠25° in series with L 20 mH, C 50 mF, and R 25 . The circuit operates at an angular frequency of 2 krad/s. a. Use the voltage divider rule to determine the voltage across each element in the circuit. b. Verify that Kirchhoff’s voltage law applies for the circuit. Answers: a. VL 51.2 V∠64.81°, VC 12.8 V∠ 115.19° VR 32.0 V∠ 25.19° b. 51.2 V∠64.81° 12.8 V∠ 115.19° 32.0 V∠ 25.19° 50 V∠25°

Section 18.4

18.4

■

AC Parallel Circuits

693

AC Parallel Circuits

The admittance Y of any impedance is defined as a vector quantity which is the reciprocal of the impedance Z. Mathematically, admittance is expressed as

冢冣

1 1 1 YT ∠ v YT∠ v (S) ZT ZT∠v ZT

(18–11)

where the unit of admittance is the siemens (S). In particular, we have seen that the admittance of a resistor R is called conductance and is given the symbol YR. If we consider resistance as a vector quantity, then the corresponding vector form of the conductance is 1 1 YR ∠0° G∠0° G j0 R∠0° R

(S)

(18–12)

If we determine the admittance of a purely reactive component X, the resultant admittance is called the susceptance of the component and is assigned the symbol B. The unit for susceptance is siemens (S). In order to distinguish between inductive susceptance and capacitive susceptance, we use the subscripts L and C respectively. The vector forms of reactive admittance are given as follows: 1 1 YL ∠ 90° BL∠ 90° 0 jBL (S) XL∠90° XL

(18–13)

1 1 YC ∠90° BC ∠90° 0 jBC XC∠ 90° XC

(18–14)

(S)

In a manner similar to impedances, admittances may be represented on the complex plane in an admittance diagram as shown in Figure 18–35. The lengths of the various vectors are proportional to the magnitudes of the corresponding admittances. The resistive admittance vector G is shown on the positive real axis, whereas the inductive and capacitive admittance vectors YL and YC are shown on the negative and positive imaginary axes respectively.

EXAMPLE 18–11

Determine the admittances of the following impedances. Sketch the corresponding admittance diagram. a. R 10 b. XL 20 c. XC 40 Solutions

1 1 a. YR 100 mS∠0° R 10 ∠0° 1 1 b. YL 50 mS∠ 90° XL 20 ∠90° 1 1 c. YC 25 mS∠90° XC 40 ∠ 90° The admittance diagram is shown in Figure 18–36.

j YC BC ∠90 j BC

YR G ∠0

YL BL ∠ 90 j BL j FIGURE 18–35 Admittance diagram showing conductance (YR) and susceptance (YL and YC).

694

Chapter 18

■

AC Series-Parallel Circuits j YC 25 mS∠90 YR 100 mS∠0

FIGURE 18–36

YL 50 mS∠ 90

For any network of n admittances as shown in Figure 18–37, the total admittance is the vector sum of the admittances of the network. Mathematically, the total admittance of a network is given as YT Y1 Y2 … Yn

(S)

(18–15)

Z1 Y1

Zn Y2

FIGURE 18–37

The resultant impedance of a parallel network of n impedances is determined to be 1 1 ZT Y1 Y2 … Yn YT 1 ZT 1 1 1 … Z1 Z2 Zn

( )

(18–16)

EXAMPLE 18–12 Find the equivalent admittance and impedance of the network of Figure 18–38. Sketch the admittance diagram.

FIGURE 18–38

Section 18.4

Solution The admittances of the various parallel elements are 1 Y1 25.0 mS∠0° 25.0 mS j0 40 ∠0° 1 苶 mS∠90° 0 j16.6苶 mS Y2 16.6 60 ∠ 90° 1 Y3 33.3 苶 mS∠ 90° 0 j33.3 苶 mS 30 ∠90° The total admittance is determined as YT Y1 Y2 Y3 25.0 mS j16.6苶 mS ( j33.3 苶 mS) 25.0 mS j16.6苶 mS 30.0 mS∠ 33.69° This results in a total impedance for the network of 1 ZT YT 1 30.0 mS∠ 33.69° 33.3 ∠33.69° The admittance diagram is shown in Figure 18–39. j YC 16.7 mS∠90 YR 25.0 mS∠0 33.70

YT 30.0 mS∠ 33.7 YL 33.3 mS∠ 90 FIGURE 18–39

Two Impedances in Parallel By applying Equation 18–14 for two impedances, we determine the equivalent impedance of two impedances as Z Z2 ZT 1 Z1 Z2

( )

(18–17)

From the above expression, we see that for two impedances in parallel, the equivalent impedance is determined as the product of the impedances over the sum. Although the expression for two impedances is very similar to the expression for two resistors in parallel, the difference is that the calculation of impedance involves the use of complex algebra.

■

AC Parallel Circuits

695

696

Chapter 18

■

AC Series-Parallel Circuits

EXAMPLE 18–13

Find the total impedance for the network shown in Fig-

ure 18–40.

200 XL

250

FIGURE 18–40

Solution (200 ∠ 90°)(250 ∠90°) ZT j200 j250 50 k ∠0° 1 k ∠ 90° 50∠90°

XC X

XL X

The previous example illustrates that unlike total parallel resistance, the total impedance of a combination of parallel reactances may be much larger that either of the individual impedances. Indeed, if we are given a parallel combination of equal inductive and capacitive reactances, the total impedance of the combination is equal to infinity (namely an open circuit). Consider the network of Figure 18–41. The total impedance ZT is found as (XL∠90°)(XC∠ 90°) X 2∠0° ZT ∠0° jXL jXC 0∠0°

Because the denominator of the above expression is equal to zero, the magnitude of the total impedance will be undefined (Z ). The magnitude is undefined and the algebra yields a phase angle v 0°, which indicates that the vector lies on the positive real axis of the impedance diagram. Whenever a capacitor and an inductor having equal reactances are placed in parallel, the equivalent circuit of the two components is an open circuit. ZT

(open circuit)

FIGURE 18–41

The principle of equal parallel reactances will be studied in a later chapter dealing with “resonance.”

Three Impedances in Parallel Equation 18–16 may be solved for three impedances to give the equivalent impedance as Z1Z2Z3 ZT Z1Z2 Z1Z3 Z2Z3

( )

although this is less useful than the general equation.

(18–18)

Section 18.4

EXAMPLE 18–14

■

AC Parallel Circuits

Find the equivalent impedance of the network of Figure

18–42.

2 k

3 k

2 k

FIGURE 18–42

Solution (2 k ∠0°)(3 k ∠90°)(2 k ∠ 90°) ZT (2 k ∠0°)(3 k ∠90°) (2 k ∠0°)(2 k ∠ 90°) (3 k ∠90°)(2 k ∠ 90°) 12 109 ∠0° 6 6 10 ∠90° 4 106∠ 90° 6 106∠0° 12 109 ∠0° 12 109 ∠0° 6 6 6 10 j2 10 6.325 106∠18.43° 1.90 k ∠ 18.43°

And so the equivalent impedance of the network is ZT 1.80 k j0.6 k

A circuit consists of a current source, i 0.030 sin 500t, in parallel with L 20 mH, C 50 mF, and R 25 . a. Determine the voltage V across the circuit. b. Solve for the power factor of the circuit. c. Calculate the average power dissipated by the circuit and verify that this is equal to the power delivered by the source. d. Use Ohm’s law to find the phasor quantities, IR, IL, and IC.

PRACTICE PROBLEMS 5

Answers: a. V 0.250 V∠61.93° b. Fp 0.4705 c. PR 41.0 W PT d. IR 9.98 mA∠61.98° IC 6.24 mA∠151.93° IL 25.0 mA∠ 28.07°

A circuit consists of a 2.5-Arms current source connected in parallel with a resistor, an inductor, and a capacitor. The resistor has a value of 10 and dissipates 40 W of power.

PRACTICE PROBLEMS 6

697

698

Chapter 18

■

AC Series-Parallel Circuits a. Calculate the values of XL and XC if XL 3XC. b. Determine the magnitudes of current through the inductor and the capacitor. Answers: a. XL 80 , XC 26.7 b. IL 0.25 mA, IC 0.75 mA

18.5

Kirchhoff’s Current Law and the Current Divider Rule

The current divider rule for ac circuits has the same form as for dc circuits with the notable exception that currents are expressed as phasors. For a parallel network as shown in Figure 18–43, the current in any branch of the network may be determined using either admittance or impedance. Yx Ix I YT

ZT Ix I Zx

(18–19)

I I1

FIGURE 18–43

For two branches in parallel the current in either branch is determined from the impedances as Z2 I1 I Z1 Z2

(18–20)

Also, as one would expect, Kirchhoff’s current law must apply to any node within an ac circuit. For such circuits, KCL may be stated as follows: The summation of current phasors entering and leaving a node is equal to zero.

EXAMPLE 18–15 Calculate the current in each of the branches in the network of Figure 18–44.

Section 18.5

■

Kirchoff’s Current Law and the Current Divider Rule

I 2 A∠0 I1

200

250

FIGURE 18–44

Solution 250 ∠ 90° I1 (2 A∠0°) j200 j250 250 ∠ 90° (2 A∠0°) 10 A∠0° 50 ∠ 90°

冢冢

冣

and 200 ∠90° I2 (2 A∠0°) j200 j250 200 ∠90° (2 A∠0°) 8 A∠180° 50 ∠ 90°

冢冢

冣

The above results illustrate that the currents in parallel reactive components may be significantly larger than the applied current. If the current through the component exceeds the maximum current rating of the element, severe damage may occur.

EXAMPLE 18–16 Refer to the circuit of Figure 18–45: a

I1 E 5 V∠0

FIGURE 18–45

20 k

I2 XL

1 k

I3 XC

1 k

699

700

Chapter 18

■

AC Series-Parallel Circuits

a. b. c. d.

Find the total impedance, ZT. Determine the supply current, IT. Calculate I1, I2, and I3 using the current divider rule. Verify Kirchhoff’s current law at node a.

Solution a. Because the inductive and capacitive reactances are in parallel and have the same value, we may replace the combination by an open circuit. Consequently, only the resistor R needs to be considered. As a result ZT 20 k ∠0° 5 V∠0° IT 250 mA∠0° 20 k ∠0° 20 k ∠0° I1 (250 mA∠0°) 250 mA∠0° 20 k ∠0° 20 k ∠0° I2 (250 mA∠0°) 5.0 mA∠ 90° 1 k ∠90° 20 k ∠0° I3 (250 mA∠0°) 5.0 mA∠90° 1 k ∠ 90°

冢冢冢

冣冣

冣

d. Notice that the currents through the inductor and capacitor are 180° out of phase. By adding the current phasors in rectangular form, we have IT 250 mA j5.0 A j5.0 A 250 mA j0 250 mA∠0° The above result verifies Kirchhoff’s current law at the node.

IN-PROCESS

LEARNING CHECK 3

1. Express Kirchhoff’s current law as it applies to ac circuits. 2. What is the fundamental difference between how Kirchhoff’s current law is applied to ac circuits as compared with dc circuits? (Answers are at the end of the chapter.)

PRACTICE PROBLEMS 7

a. Use the current divider rule to determine current through each branch in the circuit of Figure 18–46.

IL 250 µA∠0

FIGURE 18–46

IR R

IC XC

Section 18.6

■

Series-Parallel Circuits

701

b. Verify that Kirchhoff’s current law applies to the circuit of Figure 18–46. Answers: a. IL 176 mA∠ 69.44° IR 234 mA∠20.56° IC 86.8 mA∠110.56° b. Iout Iin 250 mA

18.6

Series-Parallel Circuits

We may now apply the analysis techniques of series and parallel circuits in solving more complicated circuits. As in dc circuits, the analysis of such circuits is simplified by starting with easily recognized combinations. If necessary, the original circuit may be redrawn to make further simplification more apparent. Regardless of the complexity of the circuits, we find that the fundamental rules and laws of circuit analysis must apply in all cases. Consider the network of Figure 18–47. We see that the impedances Z2 and Z3 are in series. The branch containing this combination is then seen to be in parallel with the impedance Z1. The total impedance of the network is expressed as ZT Z1 㥋 (Z2 Z3)

Z2 2 j5

Solving for ZT gives the following:

Z1 2 j8

ZT (2 j8 )㥋(2 j5 6 j7 ) (2 j8 ) 㥋 (8 j2 ) (2 j8 )(8 j2 ) 2 j8 8 j2 ) (8.246 ∠ 75.96°)(8.246 ∠14.04°) 11.66 ∠ 30.96° 5.832 ∠ 30.96° 5.0 j3.0

EXAMPLE 18–17

Determine the total impedance of the network of Figure 18–48. Express the impedance in both polar form and rectangular form.

X C1

XC 2

8 XL

FIGURE 18–48

Solution After redrawing and labelling the given circuit, we have the circuit shown in Figure 18–49.

Z3 6 j7 FIGURE 18–47

702

Chapter 18

■

AC Series-Parallel Circuits

Z1 j18 Z2 j12 12 ∠90 Z3 4 j8 8.94 ∠ 63.43 FIGURE 18–49

The total impedance is given as ZT Z1 Z2 㥋 Z3 where Z1 j18 18 ∠ 90° Z2 j12 12 ∠90° Z3 4 j8 8.94 ∠ 63.43° We determine the total impedance as (12 ∠90°)(8.94 ∠ 63.43°) ZT j18 j12 4 j8

冤

冥

107.3 ∠26.57° j18 5.66∠45° j18 19.0 ∠ 18.43° j18 18 j6 18 j24 30 ∠ 53.13°

冢

EXAMPLE 18–18

冣

Consider the circuit of Figure 18–50: Z1 R1 I1

50 V∠0

1 k R2

1 k XC

Z2 XL

FIGURE 18–50

2 k

2 k Z3

Section 18.6

a. b. c. d.

Find ZT. Determine the currents I1, I2, and I3. Calculate the total power provided by the voltage source. Determine the average powers P1, P2, and P3 dissipated by each of the impedances. Verify that the average power delivered to the circuit is the same as the power dissipated by the impedances.

Solution a. The total impedance is determined by the combination ZT Z1 Z2 㥋 Z3 For the parallel combination we have (1 k j2 k )( j2 k ) Z2 㥋 Z3 1 k j2 k j2 k (2.236 k ∠63.43°)(2 k ∠ 90°) 1 k ∠0° 4.472 k ∠ 26.57° 4.0 k j2.0 k And so the total impedance is ZT 5 k j2 k 5.385 k ∠ 21.80° 50 V∠0° I1 5.385 k ∠ 21.80° 9.285 mA∠21.80° Applying the current divider rule, we get (2 k ∠ 90°)(9.285 mA∠21.80°) I2 1 k j2 k j2 k 18.57 mA∠ 68.20° and (1 k j2 k )(9.285 mA∠21.80°) I3 1 k j2 k j2 k (2.236 k ∠63.43°)(9.285 mA∠21.80°) 1 k ∠0° 20.761 mA∠85.23° c.

PT (50 V)(9.285 mA)cos 21.80° 431.0 mW

d. Because only the resistors will dissipate power, we may use P I 2R: P1 (9.285 mA)2(1 k ) 86.2 mW P2 (18.57 mA)2(1 k ) 344.8 mW Alternatively, the power dissipated by Z2 may have been determined as P I 2Z cos v: P2 (18.57 mA)2(2.236 k )cos 63.43° 344.8 mW

■

Series-Parallel Circuits

703

704

Chapter 18

■

AC Series-Parallel Circuits

Since Z3 is purely capacitive, it will not dissipate any power: P3 0 By combining these powers, the total power dissipated is found: PT 86.2 mW 344.8 mW 0 431.0 mW (checks!)

PRACTICE PROBLEMS 8

I E 20 V∠0

Z1 20 j40 Z2 10 j10 FIGURE 18–51

Refer to the circuit of Figure 18–51: a. Calculate the total impedance, ZT. b. Find the current I. c. Use the current divider rule to find I1 and I2. d. Determine the power factor for each impedance, Z1 and Z2. e. Determine the power factor for the circuit. f. Verify that the total power dissipated by impedances Z1 and Z2 is equal to the power delivered by the voltage source. Answers: a. ZT 18.9 ∠ 45° b. I 1.06 A∠45° c. I1 0.354 A∠135°, I2 1.12 A∠26.57° d. FP(1) 0.4472 leading, FP(2) 0.7071 lagging e. FP 0.7071 leading f. PT 15.0 W, P1 2.50 W, P2 12.5 W P1 P2 15.0 W PT

18.7

Frequency Effects

As we have already seen, the reactance of inductors and capacitors depends on frequency. Consequently, the total impedance of any network having reactive elements is also frequency dependent. Any such circuit would need to be analyzed separately at each frequency of interest. We will examine several

Section 18.7

■

Frequency Effects

fairly simple combinations of resistors, capacitors, and inductors to see how the various circuits operate at different frequencies. Some of the more important combinations will be examined in greater detail in later chapters which deal with resonance and filters.

RC Circuits As the name implies, RC circuits consist of a resistor and a capacitor. The components of an RC circuit may be connected either in series or in parallel as shown in Figure 18–52.

R R

(a) Series RC circuit

(b) Parallel RC circuit

FIGURE 18–52

Consider the RC series circuit of Figure 18–53. Recall that the capacitive reactance, XC , is given as

R 1 k

1 1 XC qC 2pfC

The total impedance of the circuit is a vector quantity expressed as 1 1 ZT R j R qC jqC 1 jqRC ZT jqC

FIGURE 18–53

(18–21)

If we define the cutoff or corner frequency for an RC circuit as 1 1 qc t RC

(rad/s)

(18–22)

(Hz)

(18–23)

or equivalently as 1 fc 2pRC

then several important points become evident. For q qc /10 (or f fc /10) Equation 18–21 can be expressed as 1 j0 1 ZT ⯝ jqC jqC and for q 10qc, the expression of (18–21) can be simplified as 0 jqRC ZT ⯝ R jqC

1 F

705

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AC Series-Parallel Circuits

TABLE 18–1 Angular Frequency, ␻ (Rad/s)

XC (⍀)

ZT (⍀)

0 1 10 100 200 500 1000 2000 5000 10 k 100 k

1M 100 k 10 k 5k 2k 1k 500 200 100 10

1M 100 k 10.05 k 5.099 k 2.236 k 1.414 k 1118 1019 1005 1000

Solving for the magnitude of the impedance at several angular frequencies, we have the results shown in Table 18–1. If the magnitude of the impedance ZT is plotted as a function of angular frequency q, we get the graph of Figure 18–54. Notice that the abscissa and ordinate of the graph are not scaled linearly, but rather logarithmically. This allows for the the display of results over a wide range of frequencies. ZT (Ω) 100 k

10 k

100

200

500

1000

10 k

20 k

50 k

100 k

ω (rad/s)

FIGURE 18–54 Impedance versus angular frequency for the network of Figure 18–53.

The graph illustrates that the reactance of a capacitor is very high (effectively an open circuit) at low frequencies. Consequently, the total impedance of the series circuit will also be very high at low frequencies. Secondly, we notice that as the frequency increases, the reactance decreases. Therefore, as the frequency gets higher, the capacitive reactance has a diminished effect in the circuit. At very high frequencies (typically for q 10qc), the impedance of the circuit will effectively be R 1 k . Consider the parallel RC circuit of Figure 18–55. The total impedance, ZT, of the circuit is determined as

1 F

Z ZC ZT R ZR ZC 1 R jqC 1 R jqC R jqC 1 jqRC jqC

1 k

冢冣

FIGURE 18–55

which may be simplified as R ZT 1 jqRC

(18–24)

Section 18.7

As before, the cutoff frequency is given by Equation 18–22. Now, by examining the expression of (18–24) for q qc /10, we have the following result: R ZT ⯝ R 1 j0

For q 10qc, we have R 1 ZT ⯝ 0 jqRC jqC

If we solve for the impedance of the circuit in Figure 18–55 at various angular frequencies, we obtain the results of Table 18–2. Plotting the magnitude of the impedance ZT as a function of angular frequency q, we get the graph of Figure 18–56. Notice that the abscissa and ordinate of the graph are again scaled logarithmically, allowing for the display of results over a wide range of frequencies.

■

Frequency Effects

707

TABLE 18–2 Angular Frequency, ␻ (Rad/s)

XC (⍀)

ZT (⍀)

0 1 10 100 200 500 1k 2k 5k 10 k 100 k

1M 100 k 10 k 5k 2k 1k 500 200 100 10

1000 1000 1000 995 981 894 707 447 196 99.5 10

ZT ( ) 1000 500 200 100 50 20 10 5 2 1

10 20

50 100 200 500 1 k 2 k 5 k 10 k 20 k 50 k 100 k

ω (rad/s)

ωC

FIGURE 18–56 Impedance versus angular frequency for the network of Figure 18–55.

The results indicate that at dc ( f 0 Hz) the capacitor, which behaves as an open circuit, will result in a circuit impedance of R 1 k . As the frequency increases, the capacitor reactance approaches 0 , resulting in a corresponding decrease in circuit impedance. R

RL Circuits RL circuits may be analyzed in a manner similar to the analysis of RC circuits. Consider the parallel RL circuit of Figure 18–57.

FIGURE 18–57

1 k

XL 1 H

708

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AC Series-Parallel Circuits

The total impedance of the parallel circuit is found as follows: Z ZL ZT R ZR ZL R( jqL) R jqL jqL ZT L 1 jq R

(18–25)

If we define the cutoff or corner frequency for an RL circuit as R 1 qc L t

(rad/s)

(18–26)

(Hz)

(18–27)

or equivalently as R fc 2p L

then several important points become evident. For q qc /10 (or f fc /10) Equation 18–25 can be expressed as jqL ZT ⯝ jqL 1 j0

The above result indicates that for low frequencies, the inductor has a very small reactance, resulting in a total impedance which is essentially equal to the inductive reactance. For q 10qc, the expression of (18–25) can be simplified as jqL ZT ⯝ R L 0 jq R

TABLE 18–3 Angular Frequency, ␻ (Rad/s)

XL (⍀)

ZT (⍀)

0 1 10 100 200 500 1k 2k 5k 10 k 100 k

0 1 10 99.5 196 447 707 894 981 995 1000

The above results indicate that for high frequencies, the impedance of the circuit is essentially equal to the resistance, due to the very high impedance of the inductor. Evaluating the impedance at several angular frequencies, we have the results of Table 18–3. When the magnitude of the impedance ZT is plotted as a function of angular frequency q, we get the graph of Figure 18–58.

RLC Circuits When numerous capacitive and inductive components are combined with resistors in series-parallel circuits, the total impedance ZT of the circuit may rise and fall several times over the full range of frequencies. The analysis of such complex circuits is outside the scope of this textbook. However, for illustrative purposes we examine the simple series RLC circuit of Figure 18–59.

Section 18.7

■

709

Frequency Effects

ZT ( ) 1000 500 200 100 50 20 10 5 2 ω (rad/s)

1 10

100

10 k

100 k

FIGURE 18–58 Impedance versus angular frequency for the network of Figure 18–57. R

The impedance ZT at any frequency will be determined as ZT R jXL jXC R j(XL XC)

At very low frequencies, the inductor will appear as a very low impedance (effectively a short circuit), while the capacitor will appear as a very high impedance (effectively an open circuit). Because the capacitive reactance will be much larger than the inductive reactance, the circuit will have a very large capacitive reactance. This results in a very high circuit impedance, ZT. As the frequency increases, the inductive reactance increases, while the capacitive reactance decreases. At some frequency, f0, the inductor and the capacitor will have the same magnitude of reactance. At this frequency, the reactances cancel, resulting in a circuit impedance which is equal to the resistance value. As the frequency increases still further, the inductive reactance becomes larger than the capacitive reactance. The circuit becomes inductive and the magnitude of the total impedance of the circuit again rises. Figure 18–60 shows how the impedance of a series RLC circuit varies with frequency. The complete analysis of the series RLC circuit and the parallel RLC circuit is left until we examine the principle of resonance in a later chapter. 1. For a series network consisting of a resistor and a capacitor, what will be the impedance of the network at a frequency of 0 Hz (dc)? What will be the impedance of the network as the frequency approaches infinity? 2. For a parallel network consisting of a resistor and an inductor, what will be the impedance of the network at a frequency of 0 Hz (dc)? What will be the impedance of the network as the frequency approaches infinity? (Answers are at the end of the chapter.)

FIGURE 18–59 ZT ( )

R fo FIGURE 18–60

IN-PROCESS

LEARNING CHECK 4

f (Hz)

710

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■

AC Series-Parallel Circuits

PRACTICE PROBLEMS 9

FIGURE 18–61

47 k C 0.22 F

Given the series RC network of Figure 18–61, calculate the cutoff frequency in hertz and in radians per second. Sketch the frequency response of ZT (magnitude) versus angular frequency q for the network. Show the magnitude ZT at qc/10, qc, and 10qc. Answers: qc 96.7 rad/s At 0.1 qc: ZT 472 k

18.8

FIGURE 18–62

ZT YT

At 10 qc: ZT 47.2 k

Applications

As we have seen, we may determine the impedance of any ac circuit as a vector Z R jX. This means that any ac circuit may now be simplified as a series circuit having a resistance and a reactance, as shown in Figure 18–62. Additionally, an ac circuit may be represented as an equivalent parallel circuit consisting of a single resistor and a single reactance as shown in Figure 18–63. Any equivalent circuit will be valid only at the given frequency of operation. We will now examine the technique used to convert any series impedance into its parallel equivalent. Suppose that the two circuits of Figure 18–62 and Figure 18–63 are exactly equivalent at some frequency. These circuits can be equivalent only if both circuits have the same total impedance, ZT, and the same total admittance, YT. From the circuit of Figure 18–62, the total impedance is written as

RS ZT YT

fc 15.4 Hz At qc: ZT 66.5 k

ZT RS jXS

Therefore, the total admittance of the circuit is 1 1 YT ZT RS jXS

FIGURE 18–63

Multiplying the numerator and denominator by the complex conjugate, we obtain the following: RS jXS YT (RS jXS)(RS jXS) RS jXS RS2 XS2 RS XS j YT RS2 XS2 RS2 XS2

(18–28)

Now, from the circuit of Figure 18–63, the total admittance of the parallel circuit may be found from the parallel combination of RP and XP as 1 1 YT RP jXP

Section 18.8

which gives 1 1 YT j RP XP

(18–29)

Two vectors can only be equal if both the real components are equal and the imaginary components are equal. Therefore the circuits of Figure 18–62 and Figure 18–63 can only be equivalent if the following conditions are met: RS2 XS2 RP RS

(18–30)

RS2 XS2 XP XS

(18–31)

and

In a similar manner, we have the following conversion from a parallel circuit to an equivalent series circuit: RPXP2 RS RP2 XP2

(18–32)

RP2XP XS RP2 XP2

(18–33)

and

EXAMPLE 18–19 A circuit has a total impedance of ZT 10 j50 . Sketch the equivalent series and parallel circuits. Solution The series circuit will be an inductive circuit having RS 10 and XLS 50 . The equivalent parallel circuit will also be an inductive circuit having the following values: (10 )2 (50 )2 RP 260 10 (10 )2 (50 )2 XLP 52 50 The equivalent series and parallel circuits are shown in Figure 18–64.

10 RP

XLS 50

FIGURE 18–64

260 XLP 52

■

Applications

711

712

Chapter 18

■

AC Series-Parallel Circuits A circuit has a total admittance of YT 0.559 mS∠63.43°. Sketch the equivalent series and parallel circuits.

EXAMPLE 18–20

Solution Because the admittance is written in polar form, we first convert to the rectangular form of the admittance. GP (0.559 mS) cos 63.43° 0.250 mS ⇔ RP 4.0 k BCP (0.559 mS) sin 63.43° 0.500 mS ⇔ XCP 2.0 k The equivalent series circuit is found as (4 k )(2 k )2 RS 0.8 k (4 k )2 (2 k )2 and (4 k )2(2 k ) XCS 1.6 k (4 k )2 (2 k )2 The equivalent circuits are shown in Figure 18–65.

RS RP

4 k

XCP

0.8 k

6 k XCS

1.6 k

FIGURE 18–65

EXAMPLE 18–21

Refer to the circuit of Figure 18–66. R1 IT

200 V∠0

20 k XC 10 k ZT

FIGURE 18–66

a. Find ZT. b. Sketch the equivalent series circuit. c. Determine IT.

R2 30 k

10 k

Section 18.8

Solution a. The circuit consists of two parallel networks in series. We apply Equations 18–32 and 18–33 to arrive at equivalent series elements for each of the parallel networks as follows: (20 k )(10 k )2 RS1 4 k (20 k )2 (10 k )2 (20 k )2(10 k ) XCS 8 k (20 k )2 (10 k )2 and (30 k )(10 k )2 RS2 3 k (30 k )2 (10 k )2 (30 k )2(10 k ) XLS 9 k (30 k )2 (10 k )2 The equivalent circuits are shown in Figure 18–67. R1 RS1

XCS

4 k

8 k

200 k XC 10 k (a)

RS2 3 k R2

30 k

XL 10 k XLS 9 k

(b) FIGURE 18–67

The total impedance of the circuit is found to be ZT (4 k j8 k ) (3 k j9 k ) 7 k j1 k 7.071 k ∠8.13° b. Figure 18–68 shows the equivalent series circuit. IT

200 V∠0 FIGURE 18–68

7 k

1 k

■

Applications

713

714

Chapter 18

■

AC Series-Parallel Circuits 200 V∠0° IT 28.3 mA∠ 8.13° 7.071 k ∠8.13°

An inductor of 10 mH has a series resistance of 5 . a. Determine the parallel equivalent of the inductor at a frequency of 1 kHz. Sketch the equivalent showing the values of LP (in henries) and RP. b. Determine the parallel equivalent of the inductor at a frequency of 1 MHz. Sketch the equivalent showing the values of LP (in henries) and RP. c. If the frequency were increased still further, predict what would happen to the values of LP and RP

IN-PROCESS

LEARNING CHECK 5

(Answers are at the end of the chapter.)

A network has an impedance of ZT 50 k ∠75° at a frequency of 5 kHz. a. Determine the most simple equivalent series circuit (L and R). b. Determine the most simple equivalent parallel circuit.

PRACTICE PROBLEMS 10

Answers: a. RS 12.9 k , LS 1.54 H b. RP 193 k , LP 1.65 H

18.9

ELECTRONICS WORKBENCH

PSpice

Circuit Analysis Using Computers

Electronics Workbench In this section we will use Electronics Workbench to simulate how sinusoidal ac measurements are taken with an oscilloscope. The “measurements” are then interpreted to verify the ac operation of circuits. You will use some of the display features of the software to simplify your work. The following example provides a guide through each step of the procedure. EXAMPLE 18–22

Given the circuit of Figure 18–69. L 6.366 mH

I 10 V∠0° 1000 Hz

FIGURE 18–69

Section 18.9

■

Circuit Analysis Using Computers

a. Determine the current I and the voltage VR. b. Use Electronics Workbench to display the resistor voltage vR and the source voltage e. Use the results to verify the results of part (a). Solution a. XL 2p(1000 Hz)(6.366 10 3 H) 40 Z 30 j40 50 ∠53.13° 10 V∠0° I 0.200 A∠ 53.13° 50 ∠53.13° VR (0.200 A∠ 53.13°)(30 ) 6.00 V∠ 53.13° b. Use the schematic editor to input the circuit shown in Figure 18–70.

EWB

FIGURE 18–70

The ac voltage source is obtained from the Sources parts bin. The properties of the voltage source are changed by double clicking on the symbol and then selecting the Value tab. Change the values as follows: Voltage (V): Frequency: Phase:

10 V 1 kHz 0 Deg

The oscilloscope is selected from the Instruments parts bin and the settings are changed as follows: Time base: Channel A: Channel B:

0.2 ms/div 5 V/Div 5 V/Div

715

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AC Series-Parallel Circuits

At this point, the circuit could be simulated and the display will resemble actual lab results. However, we can refine the display to provide information that is more useful. First, click on Analysis/Options. Next click on the Instruments tab and change the following settings for the oscilloscope: ✓Pause after each screen •Minimum number of time points

1000

After returning to the main window, click on the Power switch. It is necessary to press the Pause/Resume button once, since the display on the oscilloscope will not immediately show the circuit steady state values, but rather begins from t 0. Consequently the display will show the transient (which in this case will last approximately 5 L/R 1 ms). Now, we can obtain a more detailed display by clicking on the Display Graphs button. We will use cursors and a grid to help in the analysis. The display will show several cycles of selected voltages. However, we are really only interested in viewing one complete cycle. This is done as follows: Click on the Properties button. Select the General tab. We enable the cursors by selecting • All traces and ✓Cursors On from the Cursors box. Next, we enable the grid display by selecting Grid On from the Grid box. The abscissa (time axis) is adjusted to show one period from t 4 ms to t 5 ms by selecting the Bottom Axis tab and adjusting the values as follows: Range Divisions

Minimum Maximum Number

0.004 0.005 20

The resulting display is shown in Figure 18–71.

FIGURE 18–71

Section 18.9

■

Circuit Analysis Using Computers

After positioning the cursors and using the display window, we are able to obtain various measurements for the circuit. The phase angle of the resistor voltage with respect to the source voltage is found by using the difference between the cursors in the bottom axis, dx 146.88 ms. Now we have 146.88 ms v 360° 52.88° 1000 ms The amplitude of the resistor voltage is 8.49 V, which results in an rms value of 6.00 V. As a result of the measurements we have VR 6.00 V∠ 52.88° and 6 V∠ 52.88° I 0.200 A∠ 52.88° 30 These values correspond very closely to the theoretical results calculated in part (a) of the example.

OrCAD PSpice In the following example we will use the Probe postprocessor of PSpice to show how the impedance of an RC circuit changes as a function of frequency. The Probe output will provide a graphical result that is very similar to the frequency responses determined in previous sections of this chapter.

EXAMPLE 18–23 Refer to the network of Figure 18–72. Use the OrCAD Capture CIS Demo to input the circuit. Run the Probe postprocessor to procide a graphical display of network impedance as a function of frequency from 50 Hz to 500 Hz. R1 1 k ZT

1 F

FIGURE 18–72

Solution Since PSpice is unable to analyze an incomplete circuit, it is necessary to provide a voltage source (and ground) for the circuit of Figure 18–72. The input impedance is not dependent on the actual voltage used, and so we may use any ac voltage source. In this example we arbitrarily select a voltage of 10 V. • Open the CIS Demo software. • Open a new project and call it Ch 18 PSpice 1. Ensure that the Analog or Mixed-Signal Circuit Wizard is activated.

717

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AC Series-Parallel Circuits

• Enter the circuit as show in Figure 18–73. Simply click on the voltage value and change its value to 10V from the default value of 0V. (There must be no spaces between the magnitude and the units.

FIGURE 18–73

• Click on PSpice, New Simulation Profile and give a name such as Example 18-23 to the new simulation. The Simulation Settings dialog box will open. • Click on the Analysis tab and select AC Sweep/Noise as the Analysis type. Select General Settings from the Options box. • Select Linear AC Sweep Type (Quite often, logarithmic frequency sweeps such as Decade, are used). Enter the following values into the appropriate dialog boxes. Start Frequency; 50, End Frequency: 500, Total Points: 1001. Click OK. • Click on PSpice and Run. The Probe postprocessor will appear on the screen. • Click on Trace and Add Trace. You may simply click on the appropriate values from the list of variables and use the division symbol to result in impedance. Enter the following expression into the Trace Expression box: V(R1:1)/I(R1) Notice that the above expression is nothing other than an application of Ohm’s Law. The resulting display is shown in Figure 18–74.

Section 18.9

■

Circuit Analysis Using Computers

FIGURE 18–74

EWB

PRACTICE PROBLEMS 11

Given the circuit of Figure 18–75. C 0.5305 µF

I 10 V∠0° 100 Hz

4k VR

FIGURE 18–75

a. Determine the current I and the voltage VR. b. Use Electronics Workbench to display the resistor voltage vR and the source voltage, e. Use the results to verify the results of part (a). Answers: a. I 2.00 mA∠36.87°, VR 8.00 V∠36.87° b. vR 11.3 sin(qt 36.87°), e 14.1 sin qt

719

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PRACTICE PROBLEMS 12

AC Series-Parallel Circuits

Use OrCAD PSpice to input the circuit of Figure 18–55. Use the Probe postprocessor to obtain a graphical display of the network impedance as a function of frequency from 100 Hz to 2000 Hz.

PSpice

PUTTING IT INTO PRACTICE

ou are working in a small industrial plant where several small motors are powered by a 60-Hz ac line voltage of 120 Vac. Your supervisor tells you that one of the 2-Hp motors which was recently installed draws too much current when the motor is under full load. You take a current reading and find that the current is 14.4 A. After doing some calculations, you determine that even if the motor is under full load, it shouldn’t require that much current. However, you have an idea. You remember that a motor can be represented as a resistor in series with an inductor. If you could reduce the effect of the inductive reactance of the motor by placing a capacitor across the motor, you should be able to reduce the current since the capacitive reactance will cancel the inductive reactance. While keeping the motor under load, you place a capacitor into the circuit. Just as you suspected, the current goes down. After using several different values, you observe that the current goes to a minimum of 12.4 A. It is at this value that you have determined that the reactive impedances are exactly balanced. Sketch the complete circuit and determine the value of capacitance that was added into the circuit. Use the information to determine the value of the motor’s inductance. (Assume that the motor has an efficiency of 100%.)

PROBLEMS

18.1 Ohm’s Law for AC Circuits 1. For the resistor shown in Figure 18–76: a. Find the sinusoidal current i using phasors. b. Sketch the sinusoidal waveforms for v and i. c. Sketch the phasor diagram for V and I. I 30 mA∠ 40

200 v 25 sinωt

FIGURE 18–76

6 k

FIGURE 18–77

2. Repeat Problem 1 for the resistor of Figure 18–77. 3. Repeat Problem 1 for the resistor of Figure 18–78.

Problems i 10 sin(ωt 60 )

47 k V 62 V∠30

FIGURE 18–79

FIGURE 18–78

4. Repeat Problem 1 for the resistor of Figure 18–79. 5. For the component shown in Figure 18–80: a. Find the sinusoidal voltage v using phasors. b. Sketch the sinusoidal waveforms for v and i. c. Sketch the phasor diagram for V and I. i 2.0 10–3 sinωt

5 k v 2.3 sin(ωt 120 )

680 v

FIGURE 18–81

FIGURE 18–80

6. Repeat Problem 5 for the component shown in Figure 18–81. 7. Repeat Problem 5 for the component shown in Figure 18–82. I 5 A∠ 60 L 30 mH f 1 kHz

C 0.47 F V 200 mV∠ 90 f 1 kHz

FIGURE 18–82

FIGURE 18–83

8. Repeat Problem 5 for the component shown in Figure 18–83. 9. Repeat Problem 5 for the component shown in Figure 18–84. i 6.25 10–3 sin 10 000 t v

FIGURE 18–84

0.01 F

721

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Chapter 18

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AC Series-Parallel Circuits 10. Repeat Problem 5 for the component shown in Figure 18–85.

240 H

i 22.5 10–3 cos(20 000 t 20 )

v 8 cos(200t 30 )

150 mH

FIGURE 18–86

FIGURE 18–85

11. Repeat Problem 5 for the component shown in Figure 18–86. 12. Repeat Problem 5 for the component shown in Figure 18–87. i

0.0033 F v 170 sin(377t 40 )

FIGURE 18–87

18.2 AC Series Circuits 13. Find the total impedance of each of the networks shown in Figure 18–88. R1

3 k

30 XC

ZT XL

35 (a)

XC1 1.5 k

XL XC2

5.9 k

4.2 k

3.3 k

(b)

FIGURE 18–88

14. Repeat Problem 13 for the networks of Figure 18–89. 15. Refer to the network of Figure 18–90. a. Determine the series impedance Z which will result in the given total impedance, ZT. Express your answer in rectangular and polar form. b. Sketch an impedance diagram showing ZT and Z. 16. Repeat Problem 15 for the network of Figure 18–91. 17. A circuit consisting of two elements has a total impedance of ZT 2 k ∠15° at a frequency of 18 kHz. Determine the values in ohms, henries, or farads of the unknown elements.

Problems

723

Z1 30 ∠20°

Z1 30 j40

Z2 55 ∠−30°

Z3 20 ∠−90

Z2 80 ∠60

(b)

(a) FIGURE 18–89

Z1 16 ∠ 30

ZT 100 ∠30

ZT 90 ∠25

R Z 47 FIGURE 18–90

FIGURE 18–91

18. A network has a total impedance of ZT 24.0 k ∠ 30° at a frequency of 2 kHz. If the network consists of two series elements, determine the values in ohms, henries, or farads of the unknown elements. 19. Given that the network of Figure 18–92 is to operate at a frequency of 1 kHz, what series components R and L (in henries) or C (in farads) must be in the indicated block to result in a total circuit impedance of ZT 50 ∠60°? 20. Repeat Problem 19 for a frequency of 2 kHz. 21. Refer to the circuit of Figure 18–93. a. Find ZT, I, VR, VL, and VC. b. Sketch the phasor diagram showing I, VR, VL, and VC.

E 120 V∠0

XC R

40 VR FIGURE 18–93

FIGURE 18–92

ZT 50 ∠60

20 mH

VL XL I

724

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AC Series-Parallel Circuits c. Determine the average power dissipated by the resistor. d. Calculate the average power delivered by the voltage source. Compare the result to (c). 22. Consider the circuit of Figure 18–94. a. Find ZT, I, VR, VL, and VC. b. Sketch the phasor diagram showing I, VR, VL, and VC. c. Write the sinusoidal expressions for the current i and the voltages e, vR, vC, and vL. d. Sketch the sinusoidal current and voltages found in (c). e. Determine the average power dissipated by the resistor. f. Calculate the average power delivered by the voltage source. Compare the result to (e). VR

vC XC

R 4 k E 10 V∠0

XL XC

6 k VL

e 24 sin(ωt)

P 200 mW R i 100 XC

50 FIGURE 18–96

36 vR

20 vL FIGURE 18–95

FIGURE 18–94

3 k VC

e 10 sin(ωt)

23. Refer to the circuit of Figure 18–95. a. Determine the circuit impedance, ZT. b. Use phasors to solve for i, vR, vC, and vL. c. Sketch the phasor diagram showing I, VR, VL, and VC. d. Sketch the sinusoidal expressions for the current and voltages found in (b). e. Determine the average power dissipated by the resistor. f. Calculate the average power delivered by the voltage source. Compare the result to (e). 24. Refer to the circuit of Figure 18–96. a. Determine the value of the capacitor reactance, XC, needed so that the resistor in the circuit dissipates a power of 200 mW. b. Using the value of XC from (a), determine the sinusoidal expression for the current i in the circuit. 18.3 Kirchhoff’s Voltage Law and the Voltage Divider Rule 25. a. Suppose a voltage of 10 V∠0° is applied across the network in Figure 18–88a. Use the voltage divider rule to find the voltage appearing across each element. b. Verify Kirchhoff’s voltage law for each network.

Problems 26. a. Suppose a voltage of 240 V∠30° is applied across the network in Figure 18–89a. Use the voltage divider rule to find the voltage appearing across each impedance. b. Verify Kirchhoff’s voltage law for each network. 27. Given the circuit of Figure 18–97: a. Find the voltages VC and VL. b. Determine the value of R. VC XC

E 250 V∠0

VL X L 400

63.3 R

VR 125 V∠60

E 120 V∠0

VC 118.8 V∠ 45

20 VL

50 VR FIGURE 18–98

FIGURE 18–97

28. Refer to the circuit of Figure 18–98. a. Find the voltages VR and VL. b. Determine the value of XC. 29. Refer to the circuit of Figure 18–99: a. Find the voltage across XC. b. Use Kirchhoff’s voltage law to find the voltage across the unknown impedance. c. Calculate the value of the unknown impedance Z. d. Determine the average power dissipated by the circuit. VR 4 V∠ 20 R

I 2 A∠θ

20 VZ

Z XC

E 600 V∠30

Z XL

30 VC FIGURE 18–99

PT 500 W

100

E 10 V∠30

VR R1

200 VL FIGURE 18–100

30. Given that the circuit of Figure 18–100 has a current with a magnitude of 2.0 A and dissipates a total power of 500 W: a. Calculate the value of the unknown impedance Z. (Hint: Two solutions are possible.)

725

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AC Series-Parallel Circuits b. Calculate the phase angle v of the current I. c. Find the voltages VR, VL, and VZ. 18.4 AC Parallel Circuits 31. Determine the input impedance, ZT, for each of the networks of Figure 18–101.

200

500

460

500

(a)

3 k

6 k

3 k

6 k

(b)

FIGURE 18–101

32. Repeat Problem 31 for Figure 18–102.

600

900

1800

50 k

(a)

(b)

FIGURE 18–102

33. Given the circuit of Figure 18–103. a. Find ZT, IT, I1, I2, and I3.

IT E 10 V∠0

FIGURE 18–103

20 k

1 k

0.8 k

Problems b. c. d. e.

Sketch the admittance diagram showing each of the admittances. Sketch the phasor diagram showing E, IT, I1, I2, and I3. Determine the average power dissipated by the resistor. Find the power factor of the circuit and calculate the average power delivered by the voltage source. Compare the answer with the result obtained in (d). 34. Refer to the circuit of Figure 18–104. a. Find ZT, IT, I1, I2, and I3. b. c. d. e. f. g.

Sketch the admittance diagram for each of the admittances. Sketch the phasor diagram showing E, IT, I1, I2, and I3. Determine the expressions for the sinusoidal currents iT, i1, i2, and i3. Sketch the sinusoidal voltage e and current iT. Determine the average power dissipated by the resistor. Find the power factor of the circuit and calculate the average power delivered by the voltage source. Compare the answer with the result obtained in (f).

IT e 50 sin(10 000t)

25 k

1 F

20 mH

FIGURE 18–104

35. Refer to the network of Figure 18–105. a. Determine ZT. b. Given the indicated current, use Ohm’s law to find the voltage, V, across the network. I 30 mA∠0 ZT

Z1 3 k j4 k

Z1 V 3 k j8 k

FIGURE 18–105

727

728

Chapter 18

■

AC Series-Parallel Circuits 36. Consider the network of Figure 18–106. a. Determine ZT. b. Given the indicated current, use Ohm’s law to find the voltage, V, across the network. c. Solve for I2 and I. I I1 100 mA∠30 ZT

Z1 30 j60

Z2 V 10 j40

FIGURE 18–106

37. Determine the impedance, Z2, which will result in the total impedance shown in Figure 18–107.

ZT 4 ∠60

FIGURE 18–107

Z1 2 j5

Z1 ZT 1.2 k ∠0 3 k j1 k

FIGURE 18–108

38. Determine the impedance, Z2, which will result in the total impedance shown in Figure 18–108. 18.5 Kirchhoff’s Current Law and the Current Divider Rule 39. Solve for the current in each element of the networks in Figure 18–101 if the current applied to each network is 10 mA∠ 30°. 40. Repeat Problem 39 for Figure 18–102. 41. Use the current divider rule to find the current in each of the elements in Figure 18–109. Verify that Kirchhoff’s current law applies. 42. Given that IL 4 A∠30° in the circuit of Figure 18–110, find the currents I, IC, and IR. Verify that Kirchhoff’s current law appplies for this circuit. 43. Suppose that the circuit of Figure 18–111 has a current I with a magnitude of 8 A: a. Determine the current IR through the resistor. b. Calculate the value of resistance, R. c. What is the phase angle of the current I?

Problems

1 mA∠0

5 k

4 k

20 k

IR a

FIGURE 18–109

IC I

IL 4 A∠30

FIGURE 18–110 I I IL 5 A∠0

IR 50

IR IC 2 A∠0

FIGURE 18–111

FIGURE 18–112

44. Assume that the circuit of Figure 18–112 has a current I with a magnitude of 3 A: a. Determine the current IR through the resistor. b. Calculate the value of capacitive reactance XC. c. What is the phase angle of the current I? 18.6 Series-Parallel Circuits 45. Refer to the circuit of Figure 18–113. a. Find ZT, IL , IC, and IR. b. Sketch the phasor diagram showing E, IL, IC, and IR. c. Calculate the average power dissipated by the resistor. d. Use the circuit power factor to calculate the average power delivered by the voltage source. Compare the answer with the results obtained in (c).

729

730

Chapter 18

■

AC Series-Parallel Circuits XL 5

IL E 120 V∠0

FIGURE 18–113

46. Refer to the circuit of Figure 18–114. a. Find ZT, I1, I2, and I3. b. Sketch the phasor diagram showing E, I1, I2, and I3. c. Calculate the average power dissipated by each of the resistors. d. Use the circuit power factor to calculate the average power delivered by the voltage source. Compare the answer with the results obtained in (c). 8 k

4 k

I1 20 V∠0

3 k

30 k V

ZT 9 k

FIGURE 18–114

47. Refer to the circuit of Figure 18–115. a. Find ZT, IT, I1, and I2. b. Determine the voltage Vab. 3 IT 20 V∠0

I1 ZT

6 a

Vab 2

FIGURE 18–115

Problems 48. Consider the circuit of Figure 18–116. a. Find ZT, IT, I1, and I2. b. Determine the voltage V. 30

5 V∠0

40 V

10 FIGURE 18–116

49. Refer to the circuit of Figure 18–117: a. Find ZT, I1, I2, and I3. b. Determine the voltage V. I3 I2 XL 1 mA∠0

10 k

9 k

10 k R2

3 k

I1 FIGURE 18–117

50. Refer to the circuit of Figure 18–118: a. Find ZT, I1, I2, and I3. b. Determine the voltage V. R2 = 20 k

1 A∠0

400 k

I1 FIGURE 18–118

50 k

40 k

731

732

Chapter 18

■

AC Series-Parallel Circuits 18.7 Frequency Effects 51. A 50-k resistor is placed in series with a 0.01-mF capacitor. Determine the cutoff frequency qC (in rad/s) and sketch the frequency response (ZT vs. q) of the network. 52. A 2-mH inductor is placed in parallel with a 2-k resistor. Determine the cutoff frequency qC (in rad/s) and sketch the frequency response (ZT vs. q) of the network. 53. A 100-k resistor is placed in parallel with a 0.47-mF capacitor. Determine the cutoff frequency fC (in Hz) and sketch the frequency response (ZT vs. f ) of the network. 54. A 2.7-k resistor is placed in parallel with a 20-mH inductor. Determine the cutoff frequency fC (in Hz) and sketch the frequency response (ZT vs. f ) of the network. 18.8 Applications 55. Convert each each of the networks of Figure 18–119 into an equivalent series network consisting of two elements.

2.5 k

30 k

270 300

10 k 90

(a)

(b)

FIGURE 18–119

56. Convert each each of the networks of Figure 18–119 into an equivalent parallel network consisting of two elements. 57. Show that the networks of Figure 18–120 have the same input impedance at frequencies of 1 krad/s and 10 krad/s. (It can be shown that these networks are equivalent at all frequencies.) R

100

10 F

100

100 mH (a)

FIGURE 18–120

R 100

(b)

Problems 58. Show that the networks of Figure 18–121 have the same input impedance at frequencies of 5 rad/s and 10 rad/s. 150

100 100 ZT

100

25 H

100 H

(b)

(a) FIGURE 18–121

18.9 Circuit Analysis Using Computers 59. EWB Given the circuit of Figure 18–122: a. Use Electronics Workbench to simultaneously display the capacitor voltage vC and e. Record your results and determine the phasor voltage VC. b. Interchange the positions of the resistor and the capacitor relative to the ground. Use Electronics Workbench to simultaneously display the resistor voltage vR and e. Record your results and determine the phasor voltage VR. c. Compare your results to those obtained in Example 18–8.

20 V∠0° 100 Hz

EWB

398 µF

VR VC

FIGURE 18–122

60. EWB Given the circuit of Figure 18–123: a. Use Electronics Workbench to display the inductor voltage vL and e. Record your results and determine the phasor voltage VL. b. Interchange the positions of the resistor and the inductor relative to the ground. Use Electronics Workbench to display the inductor voltage vR and e. Record your results and determine the phasor voltage VR. c. Compare your results to those obtained in Example 18–9.

733

734

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AC Series-Parallel Circuits

26 V∠0° 2 kHz

EWB

5 k

0.955 mH

FIGURE 18–123

61. PSpice A 50-k resistor is placed in series with a 0.01-mF capacitor. Use OrCAD PSpice to input these components into a circuit. Run the Probe postprocessor to provide a graphical display of the network impedance as a function of frequency from 50 Hz to 500 Hz. Let the frequency sweep logarithmically in octaves. 62. PSpice A 2-mH inductor is placed in parallel with a 2-k resistor. Use OrCAD PSpice to input these components into a circuit. Run the Probe postprocessor to provide a graphical display of the network impedance as a function of frequency from 50 kHz to 500 kHz. Let the frequency sweep logarithmically in octaves. 63. PSpice A 100-k resistor is placed in parallel with a 0.47-mF capacitor. Use OrCAD PSpice to input these components into a circuit. Run the Probe postprocessor to provide a graphical display of the network impedance as a function of frequency from 0.1 Hz to 10 Hz. Let the frequency sweep logarithmically in octaves. 64. PSpice A 2.7-k resistor is placed in parallel with a 20-mH inductor. Use OrCAD PSpice to input these components into a circuit. Run the Probe postprocessor to provide a graphical display of the network impedance as a function of frequency from 100 kHz to 1 MHz. Let the frequency sweep logarithmically in octaves.

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 1. Current and voltage are in phase. 2. Current leads voltage by 90°. 3. Voltage leads current by 90°. In-Process Learning Check 2 1. The phasor sum of voltage drops and rises around a closed loop is equal to zero.

Answers to In-Process Learning Checks 2. All voltages must be expressed as phasors, i.e., V V∠v V cos v jV sin v. In-Process Learning Check 3 1. The phasor sum of currents entering a node is equal to the phasor sum of currents leaving the same node. 2. All currents must be expressed as phasors, i.e., I I∠v I cos v jI sin v. In-Process Learning Check 4 1. At f 0 Hz, Z (open circuit) 2. As f → , Z R In-Process Learning Check 5 1. RP 795 , LP 10.1 mH 2. RP 790 M , LP 10.0 mH

735

Methods of AC Analysis OBJECTIVES

KEY TERMS

After studying this chapter, you will be able to • convert an ac voltage source into its equivalent current source, and conversely, convert a current source into an equivalent voltage source, • solve for the current or voltage in a circuit having either a dependent current source or a dependent voltage source, • set up simultaneous linear equations to solve an ac circuit using mesh analysis, • use complex determinants to find the solutions for a given set of linear equations, • set up simultaneous linear equations to solve an ac circuit using nodal analysis, • perform delta-to-wye and wye-to-delta conversions for circuits having reactive elements, • solve for the balanced condition in a given ac bridge circuit. In particular, you will examine the Maxwell, Hay, and Schering bridges, • use Electronics Workbench to analyze bridge circuits, • use PSpice to calculate current and voltage in an ac circuit.

Balanced Bridges Controlling Elements Delta-Wye Conversion Dependent Sources Hay Bridge Maxwell Bridge Mesh Analysis Nodal Analysis Schering Bridge Source Conversion

OUTLINE Dependent Sources Source Conversion Mesh (Loop) Analysis Nodal Analysis Delta-to-Wye and Wye-to-Delta Conversions Bridge Networks Circuit Analysis Using Computers

o this point, we have examined only circuits having a single ac source. In this chapter, we continue our study by analyzing multisource circuits and bridge networks. You will find that most of the techniques used in analyzing ac circuits parallel those of dc circuit analysis. Consequently, a review of Chapter 8 will help you understand the topics of this chapter. Near the end of the chapter, we examine how computer techniques are used to analyze even the most complex ac circuits. It must be emphasized that although computer techniques are much simpler than using a pencil and calculator, there is virtually no knowledge to be gained by mindlessly entering data into a computer. We use the computer merely as a tool to verify our results and to provide a greater dimension to the analysis of circuits.

Hermann Ludwig Ferdinand von Helmholtz HERMANN HELMHOLTZ WAS BORN IN POTSDAM (near Berlin, Germany) on August 31, 1821. Helmholtz was a leading scientist of the nineteenth century, whose legacy includes contributions in the fields of acoustics, chemistry, mathematics, magnetism, electricity, mechanics, optics, and physiology. Helmholtz graduated from the Medical Institute in Berlin in 1843 and practiced medicine for five years as a surgeon in the Prussian army. From 1849 to 1871, he served as a professor of physiology at universities in Königsberg, Bonn, and Heidelberg. In 1871, Helmholtz was appointed Professor of Physics at the University of Berlin. Helmholtz’ greatest contributions were as a mathematical physicist, where his work in theoretical and practical physics led to the proof of the Law of Conservation of Energy in his paper “Über die Erhaltung der Kraft,” published in 1847. He showed that mechanics, heat, light, electricity, and magnetism were simply manifestations of the same force. His work led to the understanding of electrodynamics (the motion of charge in conductors), and his theory of the electromagnetic properties of light set the groundwork for later scientists to understand how radio waves are propagated. For his work, the German emperor Kaiser Wilhelm I made Helmholtz a noble in 1883. This great scientist died on September 8, 1894, at the age of 73.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

737

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Chapter 19

■

Methods of AC Analysis

19.1

Dependent Sources

The voltage and current sources we have worked with up to now have been independent sources, meaning that the voltage or current of the supply was not in any way dependent upon any voltage or current elsewhere in the circuit. In many amplifier circuits, particularly those involving transistors, it is possible to explain the operation of the circuits by replacing the device with an equivalent electronic model. These models often use voltage and current sources which have values dependent upon some internal voltage or current. Such sources are called dependent sources. Figure 19–1 compares the symbols for both independent and dependent sources. E

(a) Independent sources

k1 V

k2V

k 3I

k4I

(b) Dependent sources FIGURE 19–1

Although the diamond is the accepted symbol for representing dependent sources, many articles and textbooks still use a circle. In this textbook we use both forms of the dependent source to familiarize the student with the various notations. The dependent source has a magnitude and phase angle determined by voltage or current at some internal element multiplied by a constant, k. The magnitude of the constant is determined by parameters within the particular model, and the units of the constant correspond to the required quantities in the equation.

EXAMPLE 19–1 Refer to the resistor shown in Figure 19–2. Determine the voltage VR across the resistor given that the controlling voltage has the following values: a. V ⫽ 0 V. b. V ⫽ 5 V∠30°. c. V ⫽ 3 V∠⫺150°.

Section 19.2

R I gmV

■

Source Conversion

2 k VR

Given: gm 4 mS

FIGURE 19–2

Solution Notice that the dependent source of this example has a constant, gm, called the transconductance. Here, gm ⫽ 4 mS. I ⫽ (4 mS)(0 V) ⫽ 0 VR ⫽ 0 V b. I ⫽ (4 mS)(5 V∠30°) ⫽ 20 mA∠30° VR ⫽ (20 mA∠30°)(2 k ) ⫽ 40 V∠30° c. I ⫽ (4 mS)(3 V∠⫺150°) ⫽ 12 mA∠⫺150° VR ⫽ (12 mA∠⫺150°)(2 k ) ⫽ 24 V∠⫺150° a.

The circuit of Figure 19–3 represents a simplified model of a transistor amplifier. I

hie

2 k

hfe I

4 VR

hfe 120 FIGURE 19–3

Determine the voltage VR for each of the following applied voltages: a. V ⫽ 10 mV∠0°. b. V ⫽ 2 mV∠180°. c. V ⫽ 0.03 V∠90°. Answers: a. 2.4 mV∠180°

19.2

b. 0.48 mV∠0°

c. 7.2 mV∠⫺90°

Source Conversion

When working with dc circuits, the analysis of a circuit is often simplified by replacing the source (whether a voltage source or a current source) with its equivalent. The conversion of any ac source is similar to the method used in dc circuit analysis.

PRACTICE PROBLEMS 1

739

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Chapter 19

■

Methods of AC Analysis

A voltage source E in series with an impedance Z is equivalent to a current source I having the same impedance Z in parallel. Figure 19–4 shows the equivalent sources. a

Z E I Z

E IZ

FIGURE 19–4

From Ohm’s law, we perform the source conversion as follows: E I ⫽ ᎏᎏ Z

and E ⫽ IZ

It is important to realize that the two circuits of Figure 19–4 are equivalent between points a and b. This means that any network connected to points a and b will behave exactly the same regardless of which type of source is used. However, the voltages or currents within the sources will seldom be the same. In order to determine the current through or the voltage across the source impedance, the circuit must be returned to its original state.

EXAMPLE 19–2

Convert the voltage source of Figure 19–5 into an equiva-

lent current source. R

10 V∠0

FIGURE 19–5

Section 19.2

Solution ZT ⫽ 3 ⫹ j4 ⫽ 5 ∠53.13° 10 V∠0° I ⫽ ᎏᎏ ⫽ 2 A∠⫺53.13° 5 ∠53.13° The equivalent current source is shown in Figure 19–6. a

2 A∠ 53.13

FIGURE 19–6

EXAMPLE 19–3

Convert the current source of Figure 19–7 into an equiva-

lent voltage source. a

240 mA∠30

40 XC

b FIGURE 19–7

Solution The impedance of the parallel combination is determined to be (40 ∠0°)(20 ∠⫺90°) Z ⫽ ᎏᎏᎏ 40 ⫺ j20 800 ∠⫺90° ⫽ ᎏᎏ 44.72∠⫺26.57° ⫽ 17.89 ∠⫺63.43° ⫽ 8 ⫺ j16 and so E ⫽ (240 mA∠30°)(17.89 ∠⫺63.43°) ⫽ 4.29 V∠⫺33.43°

■

Source Conversion

741

742

Chapter 19

■

Methods of AC Analysis

The resulting equivalent circuit is shown in Figure 19–8. Z 8 j16 17.89 ∠ 63.43

8 a

4.29V∠ 33.43

16 a

4.29V∠146.57

FIGURE 19–8

It is possible to use the same procedure to convert a dependent source into its equivalent provided that the controlling element is external to the circuit in which the source appears. If the controlling element is in the same circuit as the dependent source, this procedure cannot be used.

EXAMPLE 19–4

Convert the current source of Figure 19–9 into an equiva-

lent voltage source. a I I∠0 R1 100I

25 k

b FIGURE 19–9

Solution In the circuit of Figure 19–9 the controlling element, R1, is in a separate circuit. Therefore, the current source is converted into an equivalent voltage source as follows: E ⫽ (100I1)(Z) ⫽ (100I1)(25 k ∠0°) ⫽ (2.5 ⫻ 106 )I1 The resulting voltage source is shown in Figure 19–10. Notice that the equivalent voltage source is dependent on the current, I, just as the original current source.

Section 19.2

I I∠0

■

Source Conversion

25 k

R1 (2.5 106 ) I

b FIGURE 19–10

Convert the voltage sources of Figure 19–11 into equivalent current sources. 16 k ∠ 30 Z

5 V∠30 b (a)

0.03 mA∠30

2 k a 0.05 F

V 8 k

b ω 5 krad/s

(b) FIGURE 19–11

Answers: a. I ⫽ 0.3125 mA∠60° (from b to a) in parallel with Z ⫽ 16 k ∠⫺30° b. I ⫽ 0.161 mA∠93.43° (from a to b) in parallel with Z ⫽ 2 k ⫺ j4 k

PRACTICE PROBLEMS 2

743

744

Chapter 19

IN-PROCESS

LEARNING CHECK 1

■

Methods of AC Analysis

Given a 40-mA∠0° current source in parallel with an impedance, Z. Determine the equivalent voltage source for each of the following impedances: a. Z ⫽ 25 k ∠30°. b. Z ⫽ 100 ∠⫺90°. c. Z ⫽ 20 k ⫺ j16 k . (Answers are at the end of the chapter.)

19.3

Mesh (Loop) Analysis

Mesh analysis allows us to determine each loop current within a circuit, regardless of the number of sources within the circuit. The following steps provide a format which simplifies the process of using mesh analysis: 1. Convert all sinusoidal expressions into equivalent phasor notation. Where necessary, convert current sources into equivalent voltage sources. 2. Redraw the given circuit, simplifying the given impedances wherever possible and labelling the impedances (Z1, Z2, etc.). 3. Arbitrarily assign clockwise loop currents to each interior closed loop within a circuit. Show the polarities of all impedances using the assumed current directions. If an impedance is common to two loops, it may be thought to have two simultaneous currents. Although in fact two currents will not occur simultaneously, this maneuver makes the algebraic calculations fairly simple. The actual current through a common impedance is the vector sum of the individual loop currents. 4. Apply Kirchhoff’s voltage law to each closed loop in the circuit, writing each equation as follows: (ZI) ⫽ E

If the current directions are originally assigned in a clockwise direction, then the resulting linear equations may be simplified to the following format: Loop 1: ⫹( Z1)I1 ⫺ ( Z1⫺2)I2 ⫺ … ⫺( Z1⫺n)In ⫽ ( E1) Loop 2: ⫺( Z2⫺1)I1 ⫹ ( Z2)I2 ⫺ … ⫺( Z2⫺n)In ⫽ ( E2) . . . . . . … Loop n: ⫺( Zn⫺1)I1 ⫺ ( Zn⫺2)I2 ⫺ ⫹( Zn)In ⫽ ( En)

In the above format, Zx is the summation of all impedances around loop x. The sign in front of all loop impedances will be positive. Zx⫺y is the summation of impedances which are common between loop x and loop y. If there is no common impedance between two loops, this term is simply set to zero. All common impedance terms in the linear equations are given negative signs.

Section 19.3

Ex is the summation of voltage rises in the direction of the assumed current Ix. If a voltage source has a polarity such that it appears as a voltage drop in the assumed current direction, then the voltage is given a negative sign. 5. Solve the resulting simultaneous linear equations using substitution or determinants. If necessary, refer to Appendix B for a review of solving simultaneous linear equations.

EXAMPLE 19–5

Solve for the loop equations in the circuit of Figure

19–12. E1 5 V∠0

R1 2

XL 4

1.25 A∠ 90

R3 3 FIGURE 19–12

Solution Step 1: The current source is first converted into an equivalent voltage source as shown in Figure 19–13.

E1 5 V∠0

R1 2

4 E2

5 V∠0

R3 3 E2 (1.25 A∠ 90 ) (4 ∠90 ) 5 V∠0 ) FIGURE 19–13

Steps 2 and 3: Next the circuit is redrawn as shown in Figure 19–14. The impedances have been simplified and the loop currents are drawn in a clockwise direction.

■

Mesh (Loop) Analysis

745

746

Chapter 19

■

Methods of AC Analysis

E1 5 V∠0

5 V∠0

Z1 3 j3 Z2 1 j0 Z3 2 j4 FIGURE 19–14

Step 4: The loop equations are written as Loop 1: (Z1 ⫹ Z2)I1 ⫺ (Z2)I2 ⫽ ⫺E1 Loop 2: ⫺(Z2)I1 ⫹ (Z2 ⫹ Z3)I2 ⫽ ⫺E2 The solution for the currents may be found by using determinants: ⫺E ⫺Z 冨 ⫺E Z ⫹ Z 冨 I ⫽ 冨Z⫺Z⫹ Z ⫺Z Z ⫹Z冨 1

(⫺E1)(Z2 ⫹ Z3) ⫺ E2Z2 ⫽ ᎏᎏᎏ (Z1 ⫹ Z2)(Z2 ⫹ Z3) ⫺ Z2Z2 (⫺E1)(Z2 ⫹ Z3) ⫺ E2Z2 ⫽ ᎏᎏᎏ Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 and Z ⫹Z 冨 ⫺Z I ⫽ 冨Z⫺Z⫹ Z 1

冨

⫺E1 ⫺E2 ⫺Z2 Z2 ⫹ Z3

冨

⫺E2(Z1 ⫹ Z2) ⫺ E1Z2 ⫽ ᎏᎏᎏ (Z1 ⫹ Z2)(Z2 ⫹ Z3) ⫺ Z2Z2 ⫺E2(Z1 ⫹ Z2) ⫺ E1Z2 ⫽ ᎏᎏᎏ Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 Solving these equations using the actual values of impedances and voltages, we have ⫺(5)(3 ⫹ j4) ⫺ (5)(1) I1 ⫽ ᎏᎏᎏᎏᎏ (3 ⫺ j3)(1) ⫹ (3 ⫺ j3)(2 ⫹ j4) ⫹ (1)(2 ⫹ j4) (⫺15 ⫺ j20) ⫺ 5 ⫽ ᎏᎏᎏᎏ (3 ⫺ j3) ⫹ (6 ⫹ j6 ⫺ j212) ⫹ (2 ⫹ j4) ⫺20 ⫺ j20 ⫽ ᎏᎏ 23 ⫹ j7

Section 19.3 28.28∠⫺135° ⫽ ᎏᎏ 24.04∠16.93° ⫽ 1.18 A∠⫺151.93° and ⫺(5)(4 ⫺ j3) ⫺ (5)(1) I2 ⫽ ᎏᎏᎏ 23 ⫹ j7 (⫺20 ⫹ j15) ⫺ (5) ⫽ ᎏᎏ 23 ⫹ j7 ⫺25 ⫹ j15 ⫽ ᎏᎏ 23 ⫹ j7 29.15∠149.04° ⫽ ᎏᎏ 24.04∠16.93° ⫽ 1.21 A∠132.11°

EXAMPLE 19–6 Given the circuit of Figure 19–15, write the loop equations and solve for the loop currents. Determine the voltage, V. XC

2 I 1 A∠40

6 V 4 E2

12 V∠0

FIGURE 19–15

Solution Step 1: Converting the current source into an equivalent voltage source gives us the circuit of of Figure 19–16. 3

XC 2

E1 3 V∠40

6 V 4 E2

12 V∠0

FIGURE 19–16

Steps 2 and 3: After simplifying the impedances and assigning clockwise loop currents, we have the circuit of Figure 19–17.

■

Mesh (Loop) Analysis

747

748

Chapter 19

■

Methods of AC Analysis

E1 3 V∠40

12 V∠0

Z1 3 j2 Z2 0 j4 Z3 6 FIGURE 19–17

Step 4: The loop equations for the circuit of Figure 19–17 are as follows: Loop 1: (Z1 ⫹ Z2)I1 ⫺ (Z2)I2 ⫽ E1 Loop 2: ⫺(Z2)I1 ⫹ (Z2 ⫹ Z3)I2 ⫽ ⫺E2 which, after substituting the impedance values into the expressions, become Loop 1: (3 ⫹ j2 ) ⫺ ( j4 ) ⫽ 3 V∠40° Loop 2: ⫺( j4 ) ⫹ (6 ⫹ j4 ) ⫽ ⫺12 V∠0° Step 5: Solve for the currents using determinants, where the elements of the determinants are expressed as vectors. Since the determinant of the denominator is common to both terms we find this value first. 3 ⫹ j2 ⫺j4 D⫽ ⫺ j4 6 ⫹ j4 ⫽ (3 ⫹ j2)(6 ⫹ j4) ⫺ (⫺j4)(⫺j4) ⫽ 18 ⫹ j12 ⫹ j12 ⫺ 8 ⫹ 16 ⫽ 26 ⫹ j24 ⫽ 35.38∠42.71°

冨

Now, the currents are found as 3∠40° 冨 ⫺12∠0° I ⫽ 1

⫺j4 6 ⫹ j4

冨

D (3∠40°)(7.211∠33.69°) ⫺ (12∠0°)(4∠90°) ⫽ ᎏᎏᎏᎏᎏ 35.38∠42.71° 21.633∠73.69° ⫺ 48∠90° ⫽ ᎏᎏᎏ 35.38∠42.71° 27.91∠⫺77.43° ⫽ ᎏᎏ 35.38∠42.71° ⫽ 0.7887 A∠⫺120.14°

Section 19.3

and 3 ⫹ j2 冨 ⫺j4 3 I ⫽

冨

3∠40°⫺ ⫺12∠0° 2 D (3.606∠33.69°)(⫺12∠0°) ⫹ (3∠40°)(4∠90°) I2 ⫽ ᎏᎏᎏᎏᎏ 35.38∠42.71° ⫺43.27∠33.69° ⫹ 12∠130° ⫽ ᎏᎏᎏ 35.38∠42.71° 46.15∠⫺161.29° ⫽ ᎏᎏ 35.38∠42.71° ⫽ 1.304 A∠156.00° The current through the 4- inductive reactance is I ⫽ I1 ⫺ I2 ⫽ (0.7887 A∠⫺120.14°) ⫺ (1.304 A∠156.00°) ⫽ (⫺0.3960 A ⫺ j0.6821 A) ⫺ (⫺1.1913 A ⫹ j0.5304 A) ⫽ 0.795 A ⫺ j1.213 A ⫽ 1.45 A∠⫺56.75° The voltage is now easily found from Ohm’s law as V ⫽ IZL ⫽ (1.45 A∠⫺56.75°)(4 ∠90°) ⫽ 5.80 V∠33.25°

EXAMPLE 19–7 Given the circuit of of Figure 19–18, write the loop equations and show the determinant of the coefficients for the loop equations. Do not solve them. R2 2

E1 3 V∠0

3 R1

E2 6 V∠0

FIGURE 19–18

Solution The circuit is redrawn in Figure 19–19, showing loop currents and impedances together with the appropriate voltage polarities.

■

Mesh (Loop) Analysis

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Methods of AC Analysis

Z1 = 1

Z2 = 2 Z3 = j 2 Z4 = −j3

E1 3 V∠0

E2 6 V∠0

FIGURE 19–19

The loop equations may now be written as Loop 1: (Z1 ⫹ Z3)I1 ⫺ (Z3)I2 ⫺ (Z1)I3 ⫽ E1 Loop 2: ⫺(Z3)I1 ⫹ (Z2 ⫹ Z3 ⫹ Z4)I2 ⫺ (Z4)I3 ⫽ 0 Loop 3: ⫺(Z1)I1 ⫺ (Z4)I2 ⫹ (Z1 ⫹ Z4)I3 ⫽ E2 Using the given impedance values, we have Loop 1: (1 ⫹ j2 )I1 ⫺ ( j2 )I2 ⫺ (1 )I3 ⫽ 3 V Loop 2: ⫺( j2 )I1 ⫹ (2 ⫺ j1 )I2 ⫺ (⫺j3 )I3 ⫽ 0 Loop 3: ⫺(1 )I1 ⫺ (⫺j3 )I2 ⫹ (1 ⫺ j3 )I3 ⫽ 6 V Notice that in the above equations, the phase angles (v ⫽ 0°) for the voltages have been omitted. This is because 3 V∠0° ⫽ 3 V ⫹ j0 V ⫽ 3 V. The determinant for the coefficients of the loop equations is written as D⫽

冨

1 ⫹ j2 ⫺j2 ⫺1

⫺j2 2 ⫺ j1 j3

⫺1 j3 1 ⫺ j3

冨

Notice that the above determinant is symmetrical about the principle diagonal. The coefficients in the loop equations were given the required signs (positive for the loop impedance and negative for all common impedances). However, since the coefficients in the determinant may contain imaginary numbers, it is no longer possible to generalize about the signs (⫹/⫺) of the various coefficients.

PRACTICE PROBLEMS 3

Given the circuit of Figure 19–20, write the mesh equations and solve for the loop currents. Use the results to determine the current I.

Section 19.4

■

Nodal Analysis

4 6

I 2 A∠30

4 V∠0

FIGURE 19–20 Answers: I1 ⫽ 1.19 A∠1.58°, I2 ⫽ 1.28 A∠⫺46.50°, I ⫽ 1.01 A∠72.15°

Briefly list the steps followed in using mesh analysis to solve for the loop currents of a circuit. (Answers are at the end of the chapter.)

19.4

Nodal Analysis

Nodal analysis allows us to calculate all node voltages with respect to an arbitrary reference point in a circuit. The following steps provide a simple format to apply nodal analysis. 1. Convert all sinusoidal expressions into equivalent phasor notation. If necessary, convert voltage sources into equivalent current sources. 2. Redraw the given circuit, simplifying the given impedances wherever possible and relabelling the impedances as admittances (Y1, Y2, etc.). 3. Select and label an appropriate reference node. Arbitrarily assign subscripted voltages (V1, V2, etc.) to each of the remaining n nodes within the circuit. 4. Indicate assumed current directions through all admittances in the circuit. If an admittance is common to two nodes, it is considered in each of the two node equations. 5. Apply Kirchhoff’s current law to each of the n nodes in the circuit, writing each equation as follows: (YV) ⫽ Isources

The resulting linear equations may be simplified to the following format: Node 1: ⫹( Y1)V1 ⫺ ( Y1⫺2)V2 ⫺ … ⫺( Y1⫺n)Vn ⫽ ( I1) Node 2: ⫺( Y2⫺1)I1 ⫹ ( Y2)I2 ⫺ … ⫺( Y2⫺n)In ⫽ ( I2) . . . . . . Node n: ⫺( Yn⫺1)I1 ⫺ ( Yn⫺2)I2 ⫺ … ⫹ ( Yn)In ⫽ ( In)

IN-PROCESS

LEARNING CHECK 2

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In the above format, Yx is the summation of all admittances connected to node x. The sign in front of all node admittances will be positive. Yx⫺y is the summation of common admittances between node x and node y. If there are no common admittances between two nodes, this term is simply set to zero. All common admittance terms in the linear equations will have a negative sign. Ix is the summation of current sources entering node x. If a current source leaves the node, the current is given a negative sign. 6. Solve the resulting simultaneous linear equations using substitution or determinants.

EXAMPLE 19–8

Given the circuit of Figure 19–21, write the nodal equations and solve for the node voltages. XL 2 I1 1 A∠0

I2 2 A∠0

FIGURE 19–21

Solution The circuit is redrawn in Figure 19–22, showing the nodes and a simplified representation of the admittances. The nodal equations are written as Node 1: (Y1 ⫹ Y2)V1 ⫺ (Y2)V2 ⫽ I1 Node 2: ⫺(Y2)V1 ⫹ (Y2 ⫹ Y3)V2 ⫽ ⫺I2 Using determinants, the following expressions for nodal voltages are obtained: IB V1

V2 Y2

I1 1 A∠0

Y1 0.5 S Y2 j 0.5 S Y3 j 0.25 S FIGURE 19–22

I2 2 A∠0

Section 19.4

冨I⫺I

冨

⫺Y2 Y2 ⫹ Y3 2 V1 ⫽ Y1 ⫹ Y2 ⫺Y2 ⫺Y2 Y2 ⫹ Y3 I1(Y2 ⫹ Y3) ⫺ I2Y2 ⫽ ᎏᎏᎏ Y1Y2 ⫹ Y1Y3 ⫹ Y2Y3 1

冨

and

冨Y⫺Y⫹ Y

冨

I1 ⫺I2 V2 ⫽ Y1 ⫹ Y2 ⫺Y2 ⫺Y2 Y2 ⫹ Y3 ⫺I2(Y1 ⫹ Y2) ⫹ I1Y2 ⫽ ᎏᎏᎏ Y1Y2 ⫹ Y1Y3 ⫹ Y2Y3 1

冨

Substituting the appropriate values into the above expressions we find the nodal voltages as 1(⫺j0.5 ⫹ j0.25) ⫺ (2)(⫺j0.5) V1 ⫽ ᎏᎏᎏᎏᎏ (0.5)(⫺j0.5) ⫹ (0.5)( j0.25) ⫹ (⫺j0.5)( j0.25) j0.75 ⫽ ᎏᎏ 0.125 ⫺ j 0.125 0.75∠90° ⫽ ᎏᎏ 0.1768∠⫺45° ⫽ 4.243 V∠135° and ⫺2(0.5 ⫺ j0.5) ⫹ (1)(⫺j0.5) V2 ⫽ ᎏᎏᎏᎏᎏ (0.5)(⫺j0.5) ⫹ (0.5)( j0.25) ⫹ (⫺j0.5)( j0.25) ⫺1 ⫹ j 0.5 ⫽ ᎏᎏ 0.125 ⫺ j 0.125 1.118∠153.43° ⫽ ᎏᎏ 0.1768∠⫺45° ⫽ 6.324 V∠198.43° ⫽ 6.324 V∠⫺161.57°

EXAMPLE 19–9 Use nodal analysis to determine the voltage V for the circuit of Figure 19–23. Compare the results to those obtained when the circuit was analyzed using mesh analysis in Example 19–6.

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Nodal Analysis

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Methods of AC Analysis

I1 1 A∠40

V XL 4 E2

12 V∠0

FIGURE 19–23

Solution Step 1: Convert the voltage source into an equivalent current source as illustrated in Figure 19–24. XC 2 I1

1 A∠40 R1

2 A∠0

E2 12 V∠0 2 A∠0 6 6

FIGURE 19–24

Steps 2, 3, and 4: The reference node is selected to be at the bottom of the circuit and the admittances are simplified as shown in Figure 19–25. IB V1

V2 Y2

I1 1 A∠40

I2 2 A∠0

Y1 1 S j 0 3

Y2 0 j 1 S 2 Y3 1 S j 1 S 6

FIGURE 19–25

Step 5: Applying Kirchhoff’s current law to each node, we have the following:

Section 19.4

Node 1: IA ⫹ IB ⫽ I1 Y1V1 ⫹ Y2(V1 ⫺ V2) ⫽ I1 (Y1 ⫹ Y2)V1 ⫺ Y2V2 ⫽ I1 Node 2: IC ⫽ IB ⫹ I2 Y3V2 ⫽ Y2(V1 ⫺ V2) ⫹ I2 ⫺Y2V1 ⫹ (Y22 ⫹ Y3)V2 ⫽ I2 After substituting the appropriate values into the above equations, we have the following simultaneous linear equations: (0.3333 S ⫹ j0.5 S)V1 ⫺ (j0.5 S)V2 ⫽ 1 A∠40° ⫺( j0.5 S)V1 ⫹ (0.16667 S ⫹ j0.25 S)V2 ⫽ 2 A∠0° Step 6: We begin by solving for the determinant for the denominator: D⫽

⫹ j0.5 冨0.3333 ⫺j0.5

⫺j0.5 0.1667 ⫹ j0.25

冨

⫽ (0.3333 ⫹ j0.5)(0.1667 ⫹ j0.25) ⫺ ( j0.5)( j0.5) ⫽ 0.0556 ⫹ j0.0833 ⫹ j0.0833 ⫺ 0.125 ⫹ 0.25 ⫽ 0.1806 ⫹ j0.1667 ⫽ 0.2457∠42.71° Next we solve the node voltages as 1∠40° 冨 2∠0° V ⫽

冨

⫺j0.5 0.1667 ⫹ j0.25 1 D (1∠40°)(0.3005∠56.31°) ⫺ (2∠0°)(0.5∠⫺90°) ⫽ ᎏᎏᎏᎏᎏ D 0.3005∠96.31° ⫺ 1.0∠⫺90° ⫽ ᎏᎏᎏᎏ 0.2457∠42.71° 1.299∠91.46° ⫽ ᎏᎏᎏᎏ 0.2457∠42.71° ⫽ 5.29 V∠48.75°

and 0.3333 ⫹ j0.5 冨 ⫺j0.5 V ⫽ 2

冨

1∠40° 2∠0°

D (0.6009∠56.31°)(2∠0°) ⫺ (0.5∠⫺90°)(1∠40°) ⫽ ᎏᎏᎏᎏᎏ D 1.2019∠56.31° ⫺ 0.5∠⫺50° ⫽ ᎏᎏᎏᎏ 0.2457∠42.71° 1.426∠75.98° ⫽ ᎏᎏ ⫽ 5.80 V∠33.27° 0.2457∠42.71°

Examining the circuit of Figure 19–23, we see that the voltage V is the same as the node voltage V2. Therefore V ⫽ 5.80 V∠33.27°, which is the same result obtained in Example 19–6. (The slight difference in phase angle is the result of rounding error.)

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Nodal Analysis

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EXAMPLE 19–10 Given the circuit of Figure 19–26, write the nodal equations expressing all coefficients in rectangular form. Do not solve the equations. 2

2 A∠90

V1 2 2

I1 1 A∠0 V4

2 V∠0

FIGURE 19–26

Solution As in the previous example, we need to first convert the voltage source into its equivalent current source. The current source will be a phasor I4, where V 2 V∠0° I4 ⫽ ᎏᎏ4 ⫽ ᎏᎏ ⫽ 1.0 A∠90° XC 2 ∠⫺90° Figure 19–27 shows the circuit as it appears after the source conversion. Notice that the direction of the current source is downward to correspond with the polarity of the voltage source V4. 2

2 A∠90

I4 1 A∠90

I1 1 A∠0

2 V∠0 1 ∠90 2 ∠ 90

FIGURE 19–27

Now, by labelling the nodes and admittances, the circuit may be simplified as shown in Figure 19–28.

Section 19.4

Y1 0 j 1 S

Y3 1 S j 0

Y2 0

Y4 0 + j 1 S

2 j1 2

Y5 1 S j 1 S

V1 Y2

2 A∠90

I1 1 A∠0

I4 1 A∠90

FIGURE 19–28

The admittances of Figure 19–28 are as follows: Y1 ⫽ 0 ⫺ j0.5 S Y2 ⫽ 0 ⫹ j0.5 S Y3 ⫽ 0.5 S ⫹ j0 Y4 ⫽ 0 ⫹ j0.5 S Y5 ⫽ 1.0 S ⫺ j0.5 S Using the assigned admittances, the nodal equations are written as follows: Node 1: Node 2: Node 3:

(Y1 ⫹ Y2)V1 ⫺ (Y2)V2 ⫺ (0)V3 ⫽ ⫺I1 ⫹ I2 ⫺(Y2)V1 ⫹ (Y2 ⫹ Y3 ⫹ Y5)V2 ⫺ (Y3)V3 ⫽ ⫺I2 ⫺ I3 ⫺(0)V1 ⫺ (Y3)V2 ⫹ (Y3 ⫹ Y4)V3 ⫽ I3 ⫺ I4

By substituting the rectangular form of the admittances and current into the above linear equations, the equations are rewritten as Node 1: Node 2: Node 3:

(⫺j0.5 ⫹ j0.5)V1 ⫺ ( j0.5)V2 ⫺ (0)V3 ⫽ ⫺1 ⫹ j2 ⫺( j0.5)V1 ⫹ ( j0.5 ⫹ 0.5 ⫹ 1 ⫺ j0.5)V2 ⫺ (0.5)V3 ⫽ ⫺j2 ⫺ j2 ⫺(0)V1 ⫺ (0.5)V2 ⫹ (0.5 ⫹ j0.5)V3 ⫽ j2 ⫺ j1

Finally, the nodal equations are simplified as follows: Node 1: Node 2: Node 3:

(0)V1 ⫺ ( j0.5)V2 ⫺ (0)V3 ⫽ ⫺1 ⫹ j2 ⫺( j0.5)V1 ⫹ (1.5)V2 ⫺ (0.5)V3 ⫽ ⫺j4 (0)V1 ⫺ (0.5)V2 ⫹ (0.5 ⫹ j0.5)V3 ⫽ j1

■

Nodal Analysis

757

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PRACTICE PROBLEMS 4

Methods of AC Analysis

Given the circuit of Figure 19–29, use nodal analysis to find the voltages V1 and V2. Use your results to find the current I. I XL

2 E1 4 V∠0

I2 1 A∠0

FIGURE 19–29 Answers: V1 ⫽ 4.22 V∠⫺56.89°, V2 ⫽ 2.19 V∠1.01°, I ⫽ 1.19 A∠1.85°

IN-PROCESS

LEARNING CHECK 3

Briefly list the steps followed in using nodal analysis to solve for the node voltages of a circuit. (Answers are at the end of the chapter.)

19.5

Delta-to-Wye and Wye-to-Delta Conversions

In Chapter 8, we derived the relationships showing the equivalence of “delta” (or “pi”) connected resistance to a “wye” (or “tee”) configuration. In a similar manner, impedances connected in a configuration are equivalent to a unique Y configuration. Figure 19–30 shows the equivalent circuits. a

c Z1

Z3 Zc

b b FIGURE 19–30 Delta-wye equivalence.

A configuration is converted to a Y equivalent by using the following: ZbZc Z1 ⫽ ᎏᎏ Za ⫹ Zb ⫹ Zc

(19–1)

Section 19.5

■

Delta-to-Wye and Wye-to-Delta Conversions

Z Zc Z2 ⫽ ᎏaᎏ Za ⫹ Zb ⫹ Zc

(19–2)

Z Zb Z3 ⫽ ᎏaᎏ Za ⫹ Zb ⫹ Zc

(19–3)

The above conversion indicates that the impedance in any arm of a Y circuit is determined by taking the product of the two adjacent impedances at this arm and dividing by the summation of the impedances. If the impedances of the sides of the network are all equal (magnitude and phase angle), the equivalent Y network will have identical impedances, where each impedance is determined as Z ZY ⫽ ᎏᎏ 3

(19–4)

A Y configuration is converted to a equivalent by using the following: Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 Za ⫽ ᎏᎏᎏ Z1

(19–5)

Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 Zb ⫽ ᎏᎏᎏ Z2

(19–6)

Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 Zc ⫽ ᎏᎏᎏ Z3

(19–7)

Any impedance in a “ ” is determined by summing the possible twoimpedance product combinations of the “Y” and then dividing by the impedance found in the opposite branch of the “Y.” If the arms of a “Y” have identical impedances, the equivalent “ ” will have impedances given as Z ⫽ 3ZY

(19–8)

PRACTICE PROBLEMS 5 Determine the Y equivalent of the network shown in

EXAMPLE 19–11 Figure 19–31.

a 6

3 b FIGURE 19–31

c 9

Solution (3 )(⫺j6 ) ⫺j18 18 ∠⫺90° Z1 ⫽ ᎏᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ 3 ⫺ j6 ⫹ j9 3 ⫹ j3 4.242∠45° ⫽ 4.242 ∠⫺135° ⫽ ⫺3.0 ⫺ j3.0

759

760

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■

Methods of AC Analysis (3 )( j9 ) j27 27 ∠90° Z2 ⫽ ᎏᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ 3 ⫺ j6 ⫹ j9 3 ⫹ j3 4.242∠45° ⫽ 6.364 ∠45° ⫽ 4.5 ⫹ j4.5 ( j9 )(⫺j6 ) 54 54 ∠0° Z3 ⫽ ᎏᎏᎏ ⫽ ᎏᎏ ⫽ ᎏᎏ 3 ⫺ j6 ⫹ j9 3 ⫹ j3 4.242∠45° ⫽ 12.73 ∠⫺45° ⫽ 9.0 ⫺ j9.0 In the above solution, we see that the given network has an equivalent Y network with one arm having a negative resistance. This result indicates that although the circuit has an equivalent Y circuit, the Y circuit cannot actually be constructed from real components since negative resistors do not exist (although some active components may demonstrate negative resistance characteristics). If the given conversion is used to simplify a circuit we would treat the impedance Z1 ⫽ ⫺3 ⫺ j3 as if the resistance actually were a negative value. Figure 19–32 shows the equivalent Y circuit. a 3

j 3

j 4.5

4.5 FIGURE 19–32

j 9

It is left to the student to show that the Y of Figure 19–32 is equivalent to the of Figure 19–31.

EXAMPLE 19–12 Find the total impedance of the network in Figure 19–33. a

6 3

3 c FIGURE 19–33

Section 19.5

■

Delta-to-Wye and Wye-to-Delta Conversions

Solution If we take a moment to examine this network, we see that the circuit contains both a and a Y. In calculating the total impedance, the solution is easier when we convert the Y to a . The conversion is shown in Figure 19–34. a

a Z1 3Ω Zc 3Ω b

3Ω Z3 c

FIGURE 19–34

Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 Za ⫽ ᎏᎏᎏ Z1 (3 )( j3 ) ⫹ (3 )(⫺j3 ) ⫹ ( j3 )(⫺j3 ) ⫽ ᎏᎏᎏᎏᎏ 3 ⫺j 2 9 ⫽ ᎏᎏ ⫽ 3 3 Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 9 Zb ⫽ ᎏᎏᎏ ⫽ ᎏᎏ ⫽ j3 Z2 ⫺j3 Z1Z2 ⫹ Z1Z3 ⫹ Z2Z3 9 Zc ⫽ ᎏᎏᎏ ⫽ ᎏᎏ ⫽ ⫺j3 Z3 j3 Now, substituting the equivalent into the original network, we have the revised network of Figure 19–35. a

3 3

FIGURE 19–35

The network of Figure 19–35 shows that the corresponding sides of the are parallel. Because the inductor and the capacitor in the left side of the have the same values, we may replace the parallel combination of these two components with an open circuit. The resulting impedance of the network is now easily determined as ZT ⫽ 3 㥋6 ⫹ ( j3 )㥋(⫺j6 ) ⫽ 2 ⫹ j6

761

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Methods of AC Analysis

A Y network consists of a 60-

capacitor, a 180- inductor, and a 540- resistor. Determine the corresponding network. Answer: Za ⫽ ⫺1080 ⫹ j180 , Zb ⫽ 20 ⫹ j120 , Zc ⫽ 360 ⫺ j60

A network consists of a resistor, inductor, and capacitor, each having an impedance of 150 . Determine the corrresponding Y network.

IN-PROCESS

LEARNING CHECK 4

(Answers are at the end of the chapter.)

19.6 a Z2

Z1 b

Z5 Z3

Z4 d

Bridge Networks

Bridge circuits, similar to the network of Figure 19–36, are used extensively in electronics to measure the values of unknown components. Recall from Chapter 8 that any bridge circuit is said to be balanced when the current through the branch between the two arms is zero. In a practical circuit, component values of very precise resistors are adjusted until the current through the central element (usually a sensitive galvanometer) is exactly equal to zero. For ac circuits, the condition of a balanced bridge occurs when the impedance vectors of the various arms satisfy the following condition: Z Z ᎏᎏ1 ⫽ ᎏᎏ2 Z3 Z4

FIGURE 19–36

(19–9)

When a balanced bridge occurs in a circuit, the equivalent impedance of the bridge network is easily determined by removing the central impedance and replacing it by either an open or a short circuit. The resulting impedance of the bridge circuit is then found as either of the following: ZT ⫽ Z1㥋Z2 ⫹ Z3㥋Z4

or ZT ⫽ (Z1 ⫹ Z3)㥋(Z2 ⫹ Z4)

If, on the other hand, the bridge is not balanced, then the total impedance must be determined by performing a -to-Y conversion. Alternatively, the circuit may be analyzed by using either mesh analysis or nodal analysis.

Section 19.6

EXAMPLE 19–13

Given that the circuit of Figure 19–37 is a balanced

bridge. a. Calculate the unknown impedance, Z x. b. Determine the values of Lx and Rx if the circuit operates at a frequency of 1 kHz.

Z1 E f 1 kHz

IG 0

Rx Lx

Z1 30 k ∠ 20 Z2 10 k ∠0 Z3 100 ∠0 FIGURE 19–37

Solution a. The expression for the unknown impedance is determined from Equation 19–9 as Z Z3 Z x ⫽ ᎏ2ᎏ Z1 (10 k )(100 ) ⫽ ᎏᎏ 30 k ∠⫺20° ⫽ 33.3 ∠20° ⫽ 31.3 ⫹ j11.4 b. From the above result, we have R x ⫽ 31.3 and XL 11.4 L x ⫽ ᎏᎏ ⫽ ᎏᎏ ⫽ 1.81 mH 2pf 2p(1000 Hz)

We will now consider various forms of bridge circuits which are used in electronic circuits to determine the values of unknown inductors and capacitors. As in resistor bridges, the circuits use variable resistors together with very sensitive galvanometer movements to ensure a balanced condition for the bridge. However, rather than using a dc source to provide current in the circuit, the bridge circuits use ac sources operating at a known frequency

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Bridge Networks

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(usually 1 kHz). Once the bridge is balanced, the value of unknown inductance or capacitance may be easily determined by obtaining the reading directly from the instrument. Most instruments using bridge circuitry will incorporate several different bridges to enable the measurement of various types of unknown impedances.

Maxwell Bridge The Maxwell bridge, shown in Figure 19–38, is used to determine the inductance and series resistance of an inductor having a relatively large series resistance (in comparison to XL ⫽ qL).

C R1

Z1 E ω

IG 0

Z2 R2

R3 Lx

Rx Z4

FIGURE 19–38 Maxwell bridge.

Resistors R1 and R3 are adjusted to provide the balanced condition (when the current through the galvanometer is zero: IG ⫽ 0). When the bridge is balanced, we know that the following condition must apply: Z Z ᎏᎏ1 ⫽ ᎏᎏ3 Z2 Z4

If we write the impedances using the rectangular forms, we obtain

冤

冢

冣

冥

1 (R1) ⫺j ᎏᎏ qC ᎏᎏ 1 R1 ⫺ j ᎏᎏ 3 qC ⫽ ᎏRᎏ R2 R x ⫹ jqL x R1 ⫺j ᎏᎏ qC R R3 ᎏᎏ ⫽ ᎏ2ᎏ qR1C ⫺ j1 Rx ⫹ jqL x ᎏᎏ qC

冢

冣

R R3 ⫺jR1 ᎏᎏ ⫺ j ⫽ ᎏ2ᎏ qCR1 R x ⫹ jqL x (⫺jR1)(R x ⫹ jqL x) ⫽ R2 R3(qCR1 ⫺ j) qLxR1 ⫺ jR1R x ⫽ qR1R2 R3C ⫺ jR2 R3

Section 19.6

Now, since two complex numbers can be equal only if their real parts are equal and if their imaginary parts are equal, we must have the following: qL x R1 ⫽ qR1R2 R3C

and R1R x ⫽ R2 R3

Simplifying these expressions, we get the following equations for a Maxwell bridge: L x ⫽ R2 R3C

(19–10)

R R3 Rx ⫽ ᎏ2ᎏ R1

(19–11)

and

EXAMPLE 19–14 I C 0.01 F R2

10 k

IG 0

E 10 V∠0 f 1000 Hz

G R3

Lx 16 mH

50 FIGURE 19–39

a. Determine the values of R1 and R3 so that the bridge of Figure 19–39 is balanced. b. Calculate the current I when the bridge is balanced. Solution a. Rewriting Equations 19–10 and 19–11 and solving for the unknowns, we have Lx 16 mH R3 ⫽ ᎏᎏ ⫽ ᎏᎏ ⫽ 160 R2C (10 k )(0.01 mF) and R R3 (10 k )(160 ) R1 ⫽ ᎏ2ᎏ ⫽ ᎏᎏ ⫽ 32 k Rx 50

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Methods of AC Analysis

b. The total impedance is found as ZT ⫽ (ZC 㥋R1㥋R2) ⫹ [R3㥋(R x ⫹ ZLx)] ZT ⫽ (⫺j15.915 k )㥋32 k 㥋10 k ⫹ [160 㥋(50 ⫹ j100.5 )] ⫽ 6.87 k ∠⫺25.6° ⫹ 77.2 ∠38.0° ⫽ 6.91 k ∠⫺25.0° The resulting circuit current is 10 V∠0° I ⫽ ᎏᎏ ⫽ 1.45 mA∠25.0° 6.91 k ∠⫺25°

Hay Bridge In order to measure the inductance and series resistance of an inductor having a small series resistance, a Hay bridge is generally used. The Hay bridge is shown in Figure 19–40. By applying a method similar to that used to determine the values of the unknown inductance and resistance of the Maxwell bridge, it may be shown that the following equations for the Hay bridge apply: R2R3C Lx ⫽ ᎏ ᎏ q2R12C2 ⫹ 1

(19–12)

q2R1R2R3C2 Rx ⫽ ᎏ ᎏ q2R12C2 ⫹ 1

(19–13)

and

Z2 R1

Z1 E ω

IG 0

G Z3

Z1 E ω

IG 0

G Z3

FIGURE 19–40 Hay bridge.

Z2 R2

Cx Rx

FIGURE 19–41 Schering bridge.

Schering Bridge The Schering bridge, shown in Figure 19–41, is a circuit used to determine the value of unknown capacitance. By solving for the balanced bridge condition, we have the following equations for the unknown quantities of the circuit: RC Cx ⫽ ᎏ1ᎏ3 R2

(19–14)

Section 19.6 C R2 R x ⫽ ᎏ1ᎏ C3

■

Bridge Networks

(19–15)

EXAMPLE 19–15

Determine the values of C1 and C3 which will result in a balanced bridge for the circuit of Figure 19–42.

C1 R1 E f 1 kHz

IG 0

G Cx

EWB

10 k

R2 5 M

1 F

200

FIGURE 19–42

Solution Rewriting Equations 19–14 and 19–15, we solve for the unknown capacitances as R Cx (10 k )(1 mF) C3 ⫽ ᎏ2ᎏ ⫽ ᎏᎏ ⫽ 0.002 mF R1 5 M and C Rx (0.002 mF)(200 ) C1 ⫽ ᎏ3ᎏ ⫽ ᎏᎏ ⫽ 40 pF R2 10 k

Determine the values of R1 and R3 so that the bridge of Figure 19–43 is balanced.

R1 C 0.01 F

IG 0

E f 1 kHz

10 k

G Lx R3

16 mH 50

FIGURE 19–43 Answers:

R1 ⫽ 7916 , R3 ⫽ 199.6

PRACTICE PROBLEMS 6

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Methods of AC Analysis

19.7 ELECTRONICS WORKBENCH

PSpice

Circuit Analysis Using Computers

In some of the examples in this chapter, we analyzed circuits that resulted in as many as three simultaneous linear equations. You have no doubt wondered if there is a less complicated way to solve these circuits without the need for using complex algebra. Computer programs are particularly useful for solving such ac circuits. Both Electronics Workbench and PSpice have individual strengths in the solution of ac circuits. As in previous examples, Electronics Workbench provides an excellent simulation of how measurements are taken in a lab. PSpice, on the other hand, provides voltage and current readings, complete with magnitude and phase angle. The following examples show how these programs are useful for examining the circuits in this chapter.

EXAMPLE 19–16

Use Electronics Workbench to show that the bridge circuit of Figure 19–44 is balanced.

0.01 F 10 k 32 k

⫹

E ⫽ 10 V∠0 f ⫽ 1000 Hz ⫺

G 16 mH 160 k

FIGURE 19–44

Solution Recall that a bridge circuit is balanced when the current through the branch between the two arms of the bridge is equal to zero. In this example, we will use a multimeter set on its ac ammeter range to verify the condition of the circuit. The ammeter is selected by clicking on A and it is set to its ac range by clicking on the sinusoidal button. Figure 19–45 shows the circuit connections and the ammeter reading. The results correspond to the conditions that were previously analyzed in Example 19–14. (Note: When using Electronics Workbench, the ammeter may not show exactly zero current in the balanced condition. This is due to the way the program does the calculations. Any current less than 5 mA is considered to be effectively zero.)

Section 19.7

EWB

■

Circuit Analysis Using Computers

FIGURE 19–45

Use Electronics Workbench to verify that the results obtained in Example 19–15 result in a balanced bridge circuit. (Assume that the bridge is balanced if the galvanometer current is less than 5 mA.)

OrCAD PSpice

EXAMPLE 19–17 Use OrCAD Capture CIS to input the circuit of Figure 19–15. Assume that the circuit operates at a frequency of q ⫽ 50 rad/s ( f ⫽ 7.958 Hz). Use PSpice to obtain a printout showing the currents through XC, R2, and XL. Compare the results to those obtained in Example 19–6. Solution Since the reactive components in Figure 19–15 were given as impedance, it is necessary to first determine the corresponding values in henries and farads. 4 L ⫽ ᎏᎏ ⫽ 80 mH 50 rad/s and 1 C ⫽ ᎏᎏ ⫽ 10 mF (2 )(50 rad/s) Now we are ready to use OrCAD Capture to input the circuit as shown in Figure 19–46. The basic steps are reviewed for you. Use the ac current source,

PRACTICE PROBLEMS 7

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FIGURE 19–46

ISRC from the SOURCE library and place one IPRINT part from the SPECIAL library. The resistor, inductor, and capacitor are selected from the ANALOG library and the ground symbol is selected by using the Place ground tool. Change the value of the current source by double clicking on the part and moving the horizontal scroll bar until you find the field titled AC. Type 1A 40Deg into this field. A space must be placed between the magnitude and phase angle. Click on Apply. In order for these values to be displayed on the schematic, you must click on the Display button and then Value Only. Click on OK to return to the properties editor and then close the editor by clicking on X. The IPRINT part is similar to an ammeter and provides a printout of the current magnitude and phase angle. The properties of the IPRINT part are changed by double clicking on the part and scrolling across to show the appropriate fields. Type OK in the AC, MAG, and PHASE fields. In order to display the selected fields on the schematic, you must click on the Display button and then select Name and Value after changing each field. Since we need to measure three currents in the circuit, we could follow this procedure two more times. However, an easier method is to click on the IPRINT part and copy the part by using ⬍Ctrl⬎⬍C⬎ and ⬍Ctrl⬎⬍V⬎. Each IPRINT will then have the same properties. Once the rest of the circuit is completed and wired, click on the New Simulation Profile tool. Give the simulation a name (such as ac Branch Currents). Click on the Analysis tab and select AC Sweep/ Noise as the analysis type. Type the following values: Start Frequency: 7.958Hz End Frequency: 7.958Hz Total Points: 1

Problems

Since we do not need the Probe postprocessor to run, it is disabled by selecting the Probe Window tab (from the Simulation Settings dialog box). Click on Display Probe window and exit the simulation settings by clicking on OK. Click on the Run tool. Once PSpice has successfully run, click on the View menu and select the Output File menu item. Scroll through the file until the currents are shown as follows: FREQ 7.958E+00 FREQ 7.958E+00 FREQ 7.958E+00

IM(V_PRINT1)IP(V_PRINT1) 7.887E-01 -1.201E+02 IM(V_PRINT2)IP(V_PRINT2) 1.304E+00 1.560E+02 IM(V_PRINT3)IP(V_PRINT3) 1.450E+00 -5.673E+01

The above printout provides: I1 ⫽ 0.7887 A∠⫺120.1°, I2 ⫽ 1.304 A∠156.0°, and I3 ⫽ 1.450 A∠⫺56.73°. These results are consistent with those calculated in Example 19–6.

Use OrCAD PSpice to evaluate the node voltages for the circuit of Figure 19–23. Assume that the circuit operates at an angular frequency of q ⫽ 1000 rad/s ( f ⫽ 159.15 Hz).

PRACTICE PROBLEMS 8

PUTTING IT INTO PRACTICE

he Schering bridge of Figure 19–74 (p. 779) is balanced. In this chapter, you have learned several methods that allow you to find the current anywhere in a circuit. Using any method, determine the current through the galvanometer if the value of Cx ⫽ 0.07 mF (All other values remain unchanged.) Repeat the calculations for a value of Cx ⫽ 0.09 mF. Can you make a general statement for current through the galvanometer if Cx is smaller that the value required to balance the bridge? What general statement can be made if the value of Cx is larger than the value in the balanced bridge?

19.1 Dependent Sources 1. Refer to the circuit of Figure 19–47. Find V when the controlling current I is the following: a. 20 mA∠0° b. 50 mA∠⫺180° c. 60 mA∠60°

PROBLEMS

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Methods of AC Analysis

R 1 k

60 I

25 k

5 k V

FIGURE 19–47

2. Refer to the circuit of Figure 19–48. Find I when the controlling voltage, V, is the following: a. 30 mV∠0° b. 60 mV∠⫺180° c. 100 mV∠⫺30°

100 k V

R 2 k

200 V

20 k

FIGURE 19–48

3. Repeat Problem 1 for the circuit of Figure 19–49. 40 k

R 1 k

10 k V

(800 )I

FIGURE 19–49

4. Repeat Problem 2 for the circuit of Figure 19–50.

FIGURE 19–50

I R 2 k

(6.0 mS)V

60 k

30 k

Problems 5. Find the output voltage, Vout, for the circuit of Figure 19–51. I 10 mV∠30

2 k 130 I

Vout 20 k

30 k R3

FIGURE 19–51

6. Repeat Problem 5 for the circuit of Figure 19–52. (5 mS) V1 XC

2 k 20 mV∠0 R1

V1 1.5 k

Vout 5 k

20 k R3

FIGURE 19–52

19.2 Source Conversion 7. Given the circuits of Figure 19–53, convert each of the current sources into an equivalent voltage source. Use the resulting circuit to find VL. 30 mA∠0

1800

600 VL

2 mA∠30

225

(a)

(b)

FIGURE 19–53

8. Convert each voltage source of Figure 19–54 into an equivalent current source. 40

9 82 V∠70

5 V∠ 25

(a)

FIGURE 19–54

9. Refer to the circuit of Figure 19–55. a. Solve for the voltage, V.

1 k

(b)

773

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■

Methods of AC Analysis b. Convert the current source into an equivalent voltage source and again solve for V. Compare to the result obtained in (a). 6 k

30 mV∠0

2 k

50 k V

150 I

10 k

FIGURE 19–55

10. Refer to the circuit of Figure 19–56. a. Solve for the voltage, VL. b. Convert the current source into an equivalent voltage source and again solve for VL. c. If I ⫽ 5 mA∠90°, what is VL? I 3 A∠0

10 k (5 mS)V

RL 40 k VL

120 k

FIGURE 19–56

19.3 Mesh (Loop) Analysis 11. Consider the circuit of Figure 19–57. a. Write the mesh equations for the circuit. b. Solve for the loop currents. c. Determine the current I through the 4- resistor. 2 3

20 V∠0

4 30 V∠30

30 4 V∠20 12 8

I 5

2 V∠ 40

12 FIGURE 19–57

FIGURE 19–58

12. Refer to the circuit of Figure 19–58. a. Write the mesh equations for the circuit. b. Solve for the loop currents. c. Determine the current through the 25- inductor.

Problems 13. Refer to the circuit of Figure 19–59. a. Simplify the circuit and write the mesh equations. b. Solve for the loop currents. c. Determine the voltage V across the 15- capacitor. 25

9 V

12 10 V∠0 10

15 2 A∠0

15 V∠60

FIGURE 19–59

14. Consider the circuit of Figure 19–60. a. Simplify the circuit and write the mesh equations. b. Solve for the loop currents. c. Determine the voltage V across the 2- resistor. 15. Use mesh analysis to find the current I and the voltage V in the circuit of Figure 19–61. 2 A∠45

4 A∠0

2 V

3 V∠ 90

FIGURE 19–60 18 V∠0

XL 4 k

8 k R1 4 k

5 mA∠30 R2 3 k

FIGURE 19–61

30 V∠53.13

775

776

Chapter 19

■

Methods of AC Analysis 16. Repeat Problem 15 for the circuit of Figure 19–62. R2 4 k

R3 3 k

I 12 k XC2

10 k

1 mA∠0

XC1

16 k

9 k V

30 V∠30

FIGURE 19–62

19.4 Nodal Analysis 17. Consider the circuit of Figure 19–63. a. Write the nodal equations. b. Solve for the node voltages. c. Determine the current I through the 4- capacitor. 3 A∠90 6 4 4 A∠ 90 3

I 2 A∠0

FIGURE 19–63

2 A∠0

V 2

3 2 A∠90

3 A∠ 90

FIGURE 19–64

18. Refer to the circuit of Figure 19–64. a. Write the nodal equations. b. Solve for the node voltages. c. Determine the voltage V across the 3- capacitor. 19. a. Simplify the circuit of Figure 19–59, and write the nodal equations. b. Solve for the node voltages. c. Determine the voltage across the 15- capacitor. 20. a. Simplify the circuit of Figure 19–60, and write the nodal equations. b.Solve for the node voltages. c. Determine the current through the 2- resistor. 21. Use nodal analysis to determine the node voltages in the circuit of Figure 19–61. Use the results to find the current I and the voltage V. Compare your answers to those obtained using mesh analysis in Problem 15. 22. Use nodal analysis to determine the node voltages in the circuit of Figure 19–62. Use the results to find the current I and the voltage V. Compare your answers to those obtained using mesh analysis in Problem 16.

Problems 19.5 Delta-to-Wye and Wye-to-Delta Conversions 23. Convert each of the networks of Figure 19–65 into an equivalent Y network. 24. Convert each of the Y networks of Figure 19–66 into an equivalent network.

25 75

(a) 270

810

36 k

18 k 9 k

90 b

2 k

(b)

(a)

3 k

FIGURE 19–65

25. Using →Y or Y→ conversion, calculate I for the circuit of Figure 19–67. 26. Using →Y or Y→ conversion, calculate I for the circuit of Figure 19–68.

10 I

10 V∠0

6 2 15

18 10

3 3

25 V∠30 10

FIGURE 19–67

FIGURE 19–68

27. Refer to the circuit of Figure 19–69: a. Determine the equivalent impedance, ZT, of the circuit. b. Find the currents I and I1. I

16 V∠0

9 k ZT

3 k

I1 2 k 6 k 3 k 6 k

FIGURE 19–69

2 k

(b) FIGURE 19–66

777

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■

Methods of AC Analysis 28. Refer to the circuit of Figure 19–70: a. Determine the equivalent impedance, ZT, of the circuit. b. Find the voltages V and V1. 5H

0.2 F 10 mA∠0 1 krad/s

8 k

V1 2 k

10 k

0.1 F

FIGURE 19–70

19.6 Bridge Networks 29. Given that the bridge circuit of Figure 19–71 is balanced: a. Determine the value of the unknown impedance. b. Solve for the current I.

I Z2 2

500 mV∠0

IG 0 G

IG 0

20 V∠0

G 50 k 200

3 FIGURE 19–71

5 k

25 k FIGURE 19–72

30. Given that the bridge circuit of Figure 19–72 is balanced: a. Determine the value of the unknown impedance. b. Solve for the current I. 31. Show that the bridge circuit of Figure 19–73 is balanced. 32. Show that the bridge circuit of Figure 19–74 is balanced. 33. Derive Equations 19–14 and 19–15 for the balanced Schering bridge. 34. Derive Equations 19–12 and 19–13 for the balanced Hay bridge.

Problems

0.2 F 1 k 100 k

10 V sin 2000 t

G IG 0

Rx 5

500

EWB

Lx 100 mH

FIGURE 19â&#x20AC;&#x201C;73

0.05 F 2 k 800

10 V sin 2000 t

G IG 0

Cx 0.08 F

0.2 F

EWB

Rx 500

FIGURE 19â&#x20AC;&#x201C;74

R1 1 k 0.01 F

IG 0

E f 1 kHz

G Rx R3

50 500 mH

FIGURE 19â&#x20AC;&#x201C;75

35. Determine the values of the unknown resistors which will result in a balanced bridge for the circuit of Figure 19â&#x20AC;&#x201C;75. 36. Determine the values of the unknown capacitors which will result in a balanced bridge for the circuit of Figure 19â&#x20AC;&#x201C;76.

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Methods of AC Analysis

C1 5 k 1 M E f 1 kHz

IG 0 G

Cx C3

0.1 F 1 k

FIGURE 19–76

19.7 Circuit Analysis Using Computers 37. EWB Use Electronics Workbench to show that the bridge circuit of Figure 19–73 is balanced. (Assume that the bridge is balanced if the galvanometer current is less than 5 mA.) 38. EWB Repeat Problem 37 for the bridge circuit of Figure 19–74. 39. PSpice Use the OrCAD Capture to input the file for the circuit of Figure 19–21. Assume that the circuit operates at a frequency q ⫽ 2 krad/s. Use IPRINT and VPRINT to obtain a printout of the node voltages and the current through each element of the circuit. 40. PSpice Use the OrCAD Capture to input the file for the circuit of Figure 19–29. Assume that the circuit operates at a frequency q ⫽ 1 krad/s. Use IPRINT and VPRINT to obtain a printout of the node voltages and the current through each element of the circuit. 41. PSpice Use the OrCAD Capture to input the file for the circuit of Figure 19–68. Assume that the circuit operates at a frequency q ⫽ 20 rad/s. Use IPRINT to obtain a printout of the current I. 42. PSpice Use the OrCAD Capture to input the file for the circuit of Figure 19–69. Assume that the circuit operates at a frequency q ⫽ 3 krad/s. Use IPRINT to obtain a printout of the current I.

ANSWERS TO IN-PROCESS LEARNING CHECKS

In-Process Learning Check 1 a. E ⫽ 1000 V∠30° b. E ⫽ 4 V∠⫺90° c. E ⫽ 1024 V∠⫺38.66° In-Process Learning Check 2 1. Convert current sources to voltage sources. 2. Redraw the circuit. 3. Assign a clockwise current to each loop. 4. Write loop equations using Kirchhoff’s voltage law. 5. Solve the resulting simultaneous linear equations to find the loop currents.

Answers to In-Process Learning Checks In-Process Learning Check 3 1. Convert voltage sources to current sources. 2. Redraw the circuit. 3. Label all nodes, including the reference node. 4. Write nodal equations using Kirchhoff’s current law. 5. Solve the resulting simultaneous linear equations to find the node voltages. In-Process Learning Check 4 Z1 ⫽ 150 ∠90° Z2 ⫽ 150 ∠⫺90°

Z3 ⫽ 150 ∠0°

781

AC Network Theorems OBJECTIVES

KEY TERMS

After studying this chapter you will be able to • apply the superposition theorem to determine the voltage across or current through any component in a given circuit, • determine the Thévenin equivalent of circuits having independent and/or dependent sources, • determine the Norton equivalent of circuits having independent and/or dependent sources, • apply the maximum power transfer theorem to determine the load impedance for which maximum power is transferred to the load from a given circuit, • use PSpice to find the Thévenin and Norton equivalents of circuits having either independent or dependent sources, • use Electronics Workbench to verify the operation of ac circuits.

Absolute Maximum Power Maximum Power Transfer Norton’s Theorem Relative Maximum Power Superposition Theorem

OUTLINE Superposition Theorem—Independent Sources Superposition Theorem—Dependent Sources Thévenin’s Theorem—Independent Sources Norton’s Theorem—Independent Sources Thévenin’s and Norton’s Theorems for Dependent Sources Maximum Power Transfer Theorem Circuit Analysis Using Computers

n this chapter we apply the superposition, ThĂŠvenin, Norton, and maximum power transfer theorems in the analysis of ac circuits. Although the Millman and reciprocity theorems apply to ac circuits as well as to dc circuits, they are omitted since the applications are virtually identical with those used in analyzing dc circuits. Many of the techniques used in this chapter are similar to those used in Chapter 9, and as a result, most students will find a brief review of dc theorems useful. This chapter examines the application of the network theorems by considering both independent and dependent sources. In order to show the distinctions between the methods used in analyzing the various types of sources, the sections are labelled according to the types of sources involved. An understanding of dependent sources is particularly useful when working with transistor circuits and operational amplifiers. Sections 20.2 and 20.5 are intended to provide the background for analyzing the operation of feedback amplifiers. Your instructor may find that these topics are best left until you cover this topic in a course dealing with such amplifiers. Consequently, the omission of sections 20.2 and 20.5 will not in any way detract from the continuity of the important ideas presented in this chapter.

William Bradford Shockley SHOCKLEY WAS BORN the son of a mining engineer in London, England on February 13, 1910. After graduating from the California Institute of Technology and Massachusetts Institute of Technology, Shockley joined Bell Telephone Laboratories. With his co-workers, John Bardeen and Walter Brattain, Shockley developed an improved solid-state rectifier using a germanium crystal which had been injected with minute amounts of impurities. Unlike vacuum tubes, the resulting diodes were able to operate at much lower voltages without the need for inefficient heater elements. In 1948, Shockley combined three layers of germanium to produce a device which was able to not only rectify a signal but to amplify it. Thus was developed the first transistor. Since its humble beginning, the transistor has been improved and decreased in size to the point where now a circuit containing thousands of transistors can easily fit into an area not much bigger than the head of a pin. The advent of the transistor has permitted the construction of elaborate spacecraft, unprecedented communication, and new forms of energy generation. Shockley, Bardeen, and Brattain received the 1956 Nobel Prize in Physics for their discovery of the transistor.

CHAPTER PREVIEW

PUTTING IT IN PERSPECTIVE

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784

Chapter 20

■

AC Network Theorems

20.1

Superposition Theorem—Independent Sources

The superposition theorem states the following: The voltage across (or current through) an element is determined by summing the voltage (or current) due to each independent source.

NOTES... As in dc circuits, the superposition theorem can be applied only to voltage and current; it cannot be used to solve for the total power dissipated by an element. This is because power is not a linear quantity, but rather follows a square-law relationship (P V 2/R I 2R).

In order to apply this theorem, all sources other than the one being considered are eliminated. As in dc circuits, this is done by replacing current sources with open circuits and by replacing voltage sources with short circuits. The process is repeated until the effects due to all sources have been determined. Although we generally work with circuits having all sources at the same frequency, occasionally a circuit may operate at more than one frequency at a time. This is particularly true in diode and transistor circuits which use a dc source to set a “bias” (or operating) point and an ac source to provide the signal to be conditioned or amplified. In such cases, the resulting voltages or currents are still determined by applying the superposition theorem. The topic of how to solve circuits operating at several different frequencies simultaneously is covered in Chapter 25.

EXAMPLE 20–1

Determine the current I in Figure 20–1 by using the superposition theorem. 4

5 V∠0

2 A∠0

FIGURE 20–1

Solution Current due to the 5 V∠0° voltage source: Eliminating the current source, we obtain the circuit shown in Figure 20–2. 4

5 V∠0 I(1)

Current source is replaced with an open circuit. FIGURE 20–2

Section 20.1

■

Superposition Theorem—Independent Sources

Applying Ohm’s law, we have 5 V∠0° 5 V∠0° I(1) 4 j2 4.472 ∠ 26.57° 1.118 A∠26.57° Current due to the 2 A∠0° current source: Eliminating the voltage source, we obtain the circuit shown in Figure 20–3. 4

2 A∠0 I(2)

Voltage source is replaced with a short circuit.

FIGURE 20–3

The current I(2) due to this source is determined by applying the current divider rule: 4 ∠0° I(2) (2 A∠0°) 4 j2 8 V∠0° 4.472 ∠ 26.57° 1.789 A∠26.57° The total current is determined as the summation of currents I(1) and I(2): I I(1) I(2) 1.118 A∠26.57° 1.789 A∠26.57° (1.0 A j0.5 A) (1.6 A j0.8 A) 2.6 j1.3 A 2.91 A∠26.57°

785

786

Chapter 20

■

AC Network Theorems

EXAMPLE 20–2

Consider the circuit of Figure 20–4: I 2 A∠0

VC XC

15 20 VR E

20 V∠0

FIGURE 20–4

Find the following: a. VR and VC using the superposition theorem. b. Power dissipated by the circuit. c. Power delivered to the circuit by each of the sources. Solution a. The superposition theorem may be employed as follows: Voltages due to the current source: Eliminating the voltage source, we obtain the circuit shown in Figure 20–5. I 2 A∠0

VC(1) XC

15 20 VR(1)

FIGURE 20–5

The impedance “seen” by the current source will be the parallel combination of R㥋ZC. (20 )( j 15 ) 300 ∠ 90° Z1 12 ∠ 53.13° 20 j 15 25 ∠ 36.87° The voltage VR(1) is the same as the voltage across the capacitor, VC(1). Hence, VR(1) VC(1) (2 A∠0°)(12 ∠ 53.13°) 24 V∠ 53.13°

Section 20.1

■

Superposition Theorem—Independent Sources

Voltages due to the voltage source: Eliminating the current source, we have the circuit shown in Figure 20–6.

VC(2) XC

15 20 VR(2) E

20 V∠0

FIGURE 20–6

The voltages VR(2) and VC(2) are determined by applying the voltage divider rule, 20 ∠0° VR(2) (20 V∠0°) 20 j15 400 V∠0° 16 V∠ 36.87° 25 ∠ 36.87° and 15 ∠ 90° VC(2) (20 V∠0°) 20 j 15 300 V∠90° 12 V∠126.87° 25 ∠ 36.87° Notice that VC(2) is assigned to be negative relative to the originally assumed polarity. The negative sign is eliminated from the calculation by adding (or subtracting) 180° from the corresponding calculation. By applying superposition, we get VR VR(1) VR(2) 24 V∠ 53.13° 16 V∠36.87° (14.4 V j19.2 V) (12.8 V j9.6 V) 27.2 V j9.6 V 28.84 V∠ 19.44° and VC VC(1) VC(2) 24 V∠ 53.13° 12 V∠126.87° (14.4 V j19.2 V) ( 7.2 V j9.6 V) 7.2 V j9.6 V 12 V∠ 53.13° b. Since only the resistor will dissipate power, the total power dissipated by the circuit is found as (28.84 V)2 PT 41.60 W 20

787

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AC Network Theorems

c. The power delivered to the circuit by the current source is P1 V1I cos v1 where V1 VC 12 V∠ 53.13° is the voltage across the current source and v1 is the phase angle between V1 and I. The power delivered by the current source is P1 (12 V)(2 A) cos 53.13° 14.4 W The power delivered to the circuit by the voltage source is similarly determined as P2 EI2 cos v2 where I2 is the current through the voltage source and v2 is the phase angle between E and I2.

冢

冣

28.84 V P2 (20 V) cos 19.44° 27.2 W 20 As expected, the total power delivered to the circuit must be the summation PT P1 P2 41.6 W

PRACTICE PROBLEMS 1

Use superposition to find V and I for the circuit of Figure 20–7. FIGURE 20–7

4 V 2 A∠0 I

5 V∠30

Answers: I 2.52 A∠ 25.41°, V 4.45 V∠104.18°

IN-PROCESS

LEARNING CHECK 1

A 20- resistor is in a circuit having three sinusoidal sources. After analyzing the circuit, it is found that the current through the resistor due to each of the sources is as follows: I1 1.5 A∠20° I2 1.0 A∠110° I3 2.0 A∠0° a. Use superposition to calculate the resultant current through the resistor. b. Calculate the power dissipated by the resistor. c. Show that the power dissipated by the resistor cannot be found by applying superposition, namely, PT I 21 R I 22 R I 23 R. (Answers are at the end of the chapter.)

Section 20.2

20.2

■

Superposition Theorem—Dependent Sources

Chapter 19 introduced the concept of dependent sources. We now examine ac circuits which are powered by dependent sources. In order to analyze circuits having dependent sources, it is first necessary to determine whether the dependent source is conditional upon a controlling element in its own circuit or whether the controlling element is located in some other circuit. If the controlling element is external to the circuit under consideration, the method of analysis is the same as for an independent source. However, if the controlling element is in the same circuit, the analysis follows a slightly different stategy. The next two examples show the techniques used to analyze circuits having dependent sources.

EXAMPLE 20–3

Consider the circuit of Figure 20–8.

a. Determine the general expression for V in terms of I. b. Calculate V if I 1.0 A∠0°. c. Calculate V if I 0.3 A∠90°. V R2 I

6 V∠0

FIGURE 20–8

Solution a. Since the current source in the circuit is dependent on current through an element which is located outside of the circuit of interest, the circuit may be analyzed in the same manner as for independent sources. Voltage due to the voltage source: Eliminating the current source, we obtain the circuit shown in Figure 20–9. V(1) R2 8 R1

6 V∠0

FIGURE 20–9

8 V(1) (6 V∠0°) 4.8 V∠0° 10

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Voltage due to the current source: Eliminating the voltage source, we have the circuit shown in Figure 20–10. V(2) R2 8 5I

FIGURE 20–10

Z T 2 㥋8 1.6 ∠0° V(2) VZT (5I)(1.6 ∠0°) 8.0 I From superposition, the general expression for voltage is determined to be V V(1) V(2) 4.8 V∠0° 8.0 I b.

If I 1.0 A∠0°, V 4.8 V∠0° (8.0 )(1.0 A∠0°) 3.2 V 3.2 V∠180°

If I 0.3 A∠90°, V 4.8 V∠0° (8.0 )(0.3 A∠90°) 4.8 V j2.4 V 5.367 V∠ 26.57°

EXAMPLE 20–4

Given the circuit of Figure 20–11, calculate the voltage across the 40- resistor. I2 R2 40

2 A∠0

R1 10

0.2 V

FIGURE 20–11

Solution In the circuit of Figure 20–11, the dependent source is controlled by an element located in the circuit. Unlike the sources in the previous examples, the dependent source cannot be eliminated from the circuit since doing so would contradict Kirchhoff’s voltage law and/or Kirchhoff’s current law. The circuit must be analyzed by considering all effects simultaneously. Applying Kirchhoff’s current law, we have I1 I2 2 A∠0°

Section 20.3

■

791

Thévenin’s Theorem—Independent Sources

From Kirchhoff’s voltage law, we have (10 ) I1 V 0.2 V 1.2 V I1 0.12 V and, V I2 0.025 V 40 Combining the above expressions, we have 0.12 V 0.025 V 2.0 A∠0° 0.145 V 2.0 A∠0° V 13.79 V∠0°

Determine the voltage V in the circuit of Figure 20–12. FIGURE 20–12

PRACTICE PROBLEMS 2

10 5 A∠0 I

2I 2

Answer: V 2.73 V∠180°

20.3

Thévenin’s Theorem—Independent Sources

Thévenin’s theorem is a method which converts any linear bilateral ac circuit into a single ac voltage source in series with an equivalent impedance as shown in Figure 20–13. The resulting two-terminal network will be equivalent when it is connected to any external branch or component. If the original circuit contains reactive elements, the Thévenin equivalent circuit will be valid only at the frequency at which the reactances were determined. The following method may be used to determine the Thévenin equivalent of an ac circuit having either independent sources or sources which are dependent upon voltage or current in some other circuit. The outlined method may not be used in circuits having dependent sources controlled by voltage or current in the same circuit. 1. Remove the branch across which the Thévenin equivalent circuit is to be found. Label the resulting two terminals. Although any designation will do, we will use the notations a and b. 2. Set all sources to zero. As in dc circuits, this is achieved by replacing voltage sources with short circuits and current sources with open circuits.

ZTh

ETh b FIGURE 20–13 Thévenin equivalent circuit.

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AC Network Theorems

3. Determine the Thévenin equivalent impedance, ZTh by calculating the impedance seen between the open terminals a and b. Occasionally it may be necessary to redraw the circuit to simplify this process. 4. Replace the sources removed in Step 3 and determine the open-circuit voltage across the terminals a and b. If any of the sources are expressed in sinusoidal form, it is first necessary to convert these sources into an equivalent phasor form. For circuits having more than one source, it may be necessary to apply the superposition theorem to calculate the open-circuit voltage. Since all voltages will be phasors, the resultant is found by using vector algebra. The open-circuit voltage is the Thévenin voltage, ETh. 5. Sketch the resulting Thévenin equivalent circuit by including that portion of the circuit removed in Step 1.

EXAMPLE 20–5

Find the Thévenin equivalent circuit external to ZL for the circuit of Figure 20–14.

80 20 V∠0

XC b

FIGURE 20–14

Solution Steps 1 and 2: Removing the load impedance ZL and setting the voltage source to zero, we have the circuit of Figure 20–15. XL

80 R

ZTh

XC b 60 Voltage source is replaced with a short circuit. FIGURE 20–15

Section 20.3

■

Thévenin’s Theorem—Independent Sources

Step 3: The Thévenin impedance between terminals a and b is found as ZTh R㥋(ZL ZC) (40 ∠0°)(20 ∠90°) 40 j20 800 ∠90° 44.72 ∠26.57° 17.89 ∠63.43° 8 j16 Step 4: The Thévenin voltage is found by using the voltage divider rule as shown in the circuit of Figure 20–16. XL

80 20 V∠0

XC b 60

FIGURE 20–16

40 ∠0° ETh Vab (20 V∠0°) 40 j80 j60 800 V∠0° 44.72 ∠26.57° 17.89 V∠ 26.57° Step 5: The resultant Thévenin equivalent circuit is shown in Figure 20–17. 8 j16 ZTh ETh 17.89 V∠ 26.6

FIGURE 20–17

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EXAMPLE 20–6

Determine the Thévenin equivalent circuit external to ZL in the circuit in Figure 20–18.

30 50 V∠20

2 A∠0

b FIGURE 20–18

Solution Step 1: Removing the branch containing ZL, we have the circuit of Figure 20–19. XC a 30 50 V∠20

2 A∠0 b

FIGURE 20–19

Step 2: After setting the voltage and current sources to zero, we have the circuit of Figure 20–20. XC a 30 R

ZTh b

Voltage source replaced with a short circuit.

FIGURE 20–20

Current source replaced with an open circuit.

Section 20.3

■

Thévenin’s Theorem—Independent Sources

Step 3: The Thévenin impedance is determined as ZTh ZC㥋ZR (30 ∠ 90°)(60 ∠0°) 60 j30 1800 ∠ 90° 67.08 ∠ 26.57° 26.83 ∠ 63.43° Step 4: Because the given network consists of two independent sources, we consider the individual effects of each upon the open-circuit voltage. The total effect is then easily determined by applying the superposition theorem. Reinserting only the voltage source into the original circuit, as shown in Figure 20–21, allows us to find the open-circuit voltage, Vab(1), by applying the voltage divider rule: 60 Vab(1) (50 V∠20°) 60 j30 3000 V∠20° 67.08∠ 26.57° 44.72 V∠46.57° XC a 30

50 V∠20

Vab(1) b

FIGURE 20–21

Now, considering only the current source as shown in Figure 20–22, we determine Vab(2) by Ohm’s law: XC a 30 2 A∠0

Vab(2)

FIGURE 20–22

(2 A∠0°)(30 ∠ 90°)(60 ∠0°) Vab(2) 60 j30 (2 A∠0°)(26.83 ∠ 63.43°) 53.67 V∠ 63.43°

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From the superposition theorem, the Thévenin voltage is determined as ETh Vab(1) Vab(2) 44.72 V∠46.57° 53.67 V∠ 63.43° (30.74 V j32.48 V) (24.00 V j48.00 V) (54.74 V j15.52 V) 56.90 V∠ 15.83° Step 5: The resulting Thévenin equivalent circuit is shown in Figure 20–23. 26.8 ∠ 63.43 a

ZTh ETh 56.9 V∠ 15.8

b FIGURE 20–23

IN-PROCESS

LEARNING CHECK 2

Refer to the circuit shown in Figure 20–24 of Practice Problem 3. List the steps that you would use to find the Thévenin equivalent circuit. (Answers are at the end of the chapter.)

R1 5 k

PRACTICE PROBLEMS 3 10 sin 2000t

R2 0.5 k a

0.2 F

500 mH VL b

FIGURE 20–24

a. Find the Thévenin equivalent circuit external to the inductor in the circuit in Figure 20–24. (Notice that the voltage source is shown as sinusoidal.) b. Use the Thévenin equivalent circuit to find the phasor output voltage, VL. c. Convert the answer of (b) into the equivalent sinusoidal voltage. Answers: a. ZTh 1.5 k j2.0 k 2.5 k ∠ 53.13°, ETh 3.16 V∠ 63.43° b. VL 1.75 V∠60.26° c. vL 2.48 sin(2000t 60.26°)

Section 20.4

20.4

■

Norton’s Theorem—Independent Sources

797

Norton’s Theorem—Independent Sources

Norton’s theorem converts any linear bilateral network into an equivalent circuit consisting of a single current source and a parallel impedance as shown in Figure 20–25. Although Norton’s equivalent circuit may be determined by first finding the Thévenin equivalent circuit and then performing a source conversion, we generally use the more direct method outlined below. The steps to find the Norton equivalent cirucuit are as follows: 1. Remove the branch across which the Norton equivalent circuit is to be found. Label the resulting two terminals a and b. 2. Set all sources to zero. 3. Determine the Norton equivalent impedance, ZN, by calculating the impedance seen between the open terminals a and b. NOTE: Since the previous steps are identical with those followed for finding the Thévenin equivalent circuit, we conclude that the Norton impedance must be the same as the Thévenin impedance. 4. Replace the sources removed in Step 3 and determine the current that would occur between terminals a and b if these terminals were shorted. Any voltages and currents that are given in sinusoidal notation must first be expressed in equivalent phasor notation. If the circuit has more than one source it may be necessary to apply the superposition theorem to calculate the total short-circuit current. Since all currents will be in phasor form, any addition must be done using vector algebra. The resulting current is the Norton current, IN. 5. Sketch the resulting Norton equivalent circuit by inserting that portion of the circuit removed in Step 1. As mentioned previously, it is possible to find the Norton equivalent circuit from the Thévenin equivalent by simply performing a source conversion. We have already determined that both the Thévenin and Norton impedances are determined in the same way. Consequently, the impedances must be equivalent, and so we have ZN ZTh

(20–1)

Now, applying Ohm’s law, we determine the Norton current source from the Thévenin voltage and impedance, namely, ETh IN ZTh

Figure 20–26 shows the equivalent circuits.

(20–2)

b FIGURE 20–25 Norton equivalent circuit.

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AC Network Theorems

ZTh

ETh

b ZN ZTh E IN Th ZTh

FIGURE 20–26

EXAMPLE 20–7

Given the circuit of Figure 20–27, find the Norton equiv-

alent.

80 20 V∠0

XC b

FIGURE 20–27

Solution Steps 1 and 2: By removing the load impedance, ZL, and setting the voltage source to zero, we have the network of Figure 20–28. XL

80 R

XC b 60 FIGURE 20–28

Section 20.4

■

Norton’s Theorem—Independent Sources

Step 3: The Norton impedance may now be determined by evaluating the impedance between terminals a and b. Hence, we have (40 ∠0°)(20 ∠90°) ZN 40 j20 800 ∠90° 44.72∠26.57° 17.89 ∠63.43° 8 j16 Step 4: Reinserting the voltage source, as in Figure 20–29, we find the Norton current by calculating the current between the shorted terminals, a and b. XL

80 20 V∠0

Iab

XC b

60 FIGURE 20–29

Because the resistor R 40 is shorted, the current is determined by the impedances XL and XC as 20 V∠0° IN Iab j80 j 60 20 V∠0° 20 ∠ 90° 1.00 A∠ 90° Step 5: The resultant Norton equivalent circuit is shown in Figure 20–30. ZN 8 j16 17.89 ∠63.43 a

1 A∠ 90

b FIGURE 20–30

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AC Network Theorems

EXAMPLE 20–8

Find the Norton equivalent circuit external to RL in the circuit of Figure 20–31. Use the equivalent circuit to calculate the current IL when RL 0 , 400 , and 2 k . R

400 400 50 V∠45

400

0→2 k

RL 100 mA∠0

IL b

FIGURE 20–31

Solution Steps 1 and 2: Removing the load resistor and setting the sources to zero, we obtain the network shown in Figure 20–32. XC

400 400 XL

400

b FIGURE 20–32

Step 3: The Norton impedance is determined as (400 ∠90°)(400 j400 ) ZN j400 400 j400 (400 ∠90°)(565.69 ∠ 45°) 400 ∠0° 565.69 ∠ 45° Step 4: Because the network consists of two sources, we determine the effects due to each source separately and then apply superposition to evaluate the Norton current source. Reinserting the voltage source into the original network, we see from Figure 20–33 that the short-circuit current between the terminals a and b is easily found by using Ohm’s law.