Fundamental Electrical and Electronic Principles, Third Edition

Fundamental Electrical and Electronic Principles

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Fundamental Electricaland ElectronicPrinciplesThird EditionChristopher R RobertsonAMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORDPARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYONewnes is an imprint of Elsevier

Newnes is an imprint of ElsevierLinacre House, Jordan Hill, Oxford OX2 8DP, UK30 Corporate Drive, Suite 400, Burlington, MA 01803, USAFirst published 1993 as Electrical and Electronic Principles 1 by Edward ArnoldSecond edition 2001Third edition 2008Copyright © C. R. Robertson 1993, 2001Copyright © 2008 Elsevier Ltd. All rights reservedThe right of Christopher R. Robertson to be identified as the author of this work hasbeen asserted in accordance with the Copyright, Designs and Patents Act 1988No part of this publication may be reproduced, stored in a retrieval system ortransmitted in any form or by any means electronic, mechanical, photocopying,recording or otherwise without the prior written permission of the publisherPermissions may be sought directly from Elsevier ’ s Science & TechnologyRights Department in Oxford, UK: phone ( 44) (0) 1865 843830; fax ( 44) (0)1865 853333; email: permissions@elsevier.com . Alternatively you can submityour request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions , and selecting Obtaining permission to use Elsevier materialNoticeNo responsibility is assumed by the publisher for any injury and/or damage topersons or property as a matter of products liability, negligence or otherwise, or fromany use or operation of any methods, products, instructions or ideas contained in thematerial herein.British Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryLibrary of Congress Cataloguing in Publication DataA catalogue record for this book is available from the Library of CongressISBN: 978-0-7506-8737-9For information on all Newnes publications visit ourweb site at http://books.elsevier.comTypeset by Charon Tec Ltd., A Macmillan Company.(www.macmillansolutions.com)Printed and bound in Slovenia08 09 10 11 12 10 9 8 7 6 5 4 3 2 1

ContentsPreface ........................................................................................................................................................ixIntroduction ............................................................................................................................................xi1 Fundamentals ...........................................................................11.1 Units .................................................................................................................................................11.2 Standard Form Notation ......................................................................................................21.3 ‘ Scientific ’ Notation ..................................................................................................................21.4 Conversion of Areas and Volumes .................................................................................41.5 Graphs .............................................................................................................................................51.6 Basic Electrical Concepts .....................................................................................................71.7 Communication .....................................................................................................................26Summary of Equations ......................................................................................................29Assignment Questions ......................................................................................................302 D.C. Circuits ............................................................................................. 312.1 Resistors in Series ..................................................................................................................312.2 Resistors in Parallel ...............................................................................................................352.3 Potential Divider ....................................................................................................................402.4 Current Divider .......................................................................................................................412.5 Series/Parallel Combinations .........................................................................................432.6 Kirchhoff ’ s Current Law .....................................................................................................482.7 Kirchhoff ’ s Voltage Law .....................................................................................................492.8 The Wheatstone Bridge Network ................................................................................552.9 The Wheatstone Bridge Instrument ..........................................................................632.10 The Slidewire Potentiometer .........................................................................................65Summary of Equations ......................................................................................................68Assignment Questions ......................................................................................................69Suggested Practical Assignments ..............................................................................723 Electric Fields and Capacitors .................................................... 753.1 Coulomb ’ s Law .......................................................................................................................753.2 Electric Fields ...........................................................................................................................763.3 Electric Field Strength (E) .................................................................................................783.4 Electric Flux (ψ) and Flux Density (D) .......................................................................793.5 The Charging Process and Potential Gradient ....................................................803.6 Capacitance (C) ......................................................................................................................833.7 Capacitors ..................................................................................................................................843.8 Permittivity of Free Space (ε 0 ) .......................................................................................843.9 Relative Permittivity (ε r ) .....................................................................................................843.10 Absolute Permittivity (ε) ...................................................................................................85v

viContents3.11 Calculating Capacitor Values ..........................................................................................853.12 Capacitors in Parallel ...........................................................................................................873.13 Capacitors in Series ..............................................................................................................893.14 Series/Parallel Combinations .........................................................................................923.15 Multiplate Capacitors .........................................................................................................953.16 Energy Stored ..........................................................................................................................973.17 Dielectric Strength and Working Voltage ...........................................................1013.18 Capacitor Types ...................................................................................................................102Summary of Equations ...................................................................................................105Assignment Questions ...................................................................................................107Suggested Practical Assignment .............................................................................1104 Magnetic Fields and Circuits .....................................................1114.1 Magnetic Materials ............................................................................................................1114.2 Magnetic Fields ...................................................................................................................1114.3 The Magnetic Circuit ........................................................................................................1144.4 Magnetic Flux and Flux Density ................................................................................1154.5 Magnetomotive Force (mmf ) ....................................................................................1164.6 Magnetic Field Strength ................................................................................................1174.7 Permeability of Free Space (μ 0 ) .................................................................................1184.8 Relative Permeability (μ r ) ...............................................................................................1194.9 Absolute Permeability (μ) .............................................................................................1194.10 Magnetisation (B/H) Curve ...........................................................................................1224.11 Composite Series Magnetic Circuits ......................................................................1264.12 Reluctance (S) .......................................................................................................................1284.13 Comparison of Electrical, Magnetic and Electrostatic Quantities ........1314.14 Magnetic Hysteresis .........................................................................................................1324.15 Parallel Magnetic Circuits ..............................................................................................134Summary of Equations ...................................................................................................135Assignment Questions ...................................................................................................136Suggested Practical Assignments ...........................................................................1385 Electromagnetism .......................................................................... 1415.1 Faraday ’ s Law of Electromagnetic Induction ....................................................1415.2 Lenz ’ s Law ...............................................................................................................................1445.3 Fleming ’ s Righthand Rule .............................................................................................1445.4 EMF Induced in a Single Straight Conductor ...................................................1475.5 Force on a Current-Carrying Conductor ..............................................................1515.6 The Motor Principle ..........................................................................................................1535.7 Force between Parallel Conductors .......................................................................1565.8 The Moving Coil Meter ...................................................................................................1585.9 Shunts and Multipliers ....................................................................................................1625.10 Shunts .......................................................................................................................................1625.11 Multipliers ...............................................................................................................................163

Contents vii5.12 Figure of Merit and Loading Effect .........................................................................1665.13 The Ohmmeter ....................................................................................................................1705.14 Wattmeter ...............................................................................................................................1715.15 Eddy Currents .......................................................................................................................1725.16 Self and Mutual Inductance ........................................................................................1745.17 Self-Inductance ...................................................................................................................1755.18 Self-Inductance and Flux Linkages .........................................................................1765.19 Factors Affecting Inductance .....................................................................................1795.20 Mutual Inductance ............................................................................................................1805.21 Relationship between Self- and Mutual-Inductance ...................................1825.22 Energy Stored .......................................................................................................................1845.23 The Transformer Principle .............................................................................................1865.24 Transformer Voltage and Current Ratios ..............................................................188Summary of Equations ...................................................................................................191Assignment Questions ...................................................................................................192Suggested Practical Assignments ...........................................................................1956 Alternating Quantities ................................................................ 1976.1 Production of an Alternating Waveform .............................................................1976.2 Angular Velocity and Frequency ..............................................................................2006.3 Standard Expression for an Alternating Quantity ..........................................2006.4 Average Value .......................................................................................................................2036.5 r.m.s. Value ..............................................................................................................................2056.6 Peak Factor .............................................................................................................................2066.7 Form Factor ............................................................................................................................2076.8 Rectifiers ..................................................................................................................................2086.9 Half-wave Rectifier ............................................................................................................2096.10 Full-wave Bridge Rectifier .............................................................................................2106.11 Rectifier Moving Coil Meter .........................................................................................2126.12 Phase and Phase Angle ..................................................................................................2136.13 Phasor Representation ....................................................................................................2166.14 Addition of Alternating Quantities ..........................................................................2196.15 The Cathode Ray Oscilloscope ..................................................................................2246.16 Operation of the Oscilloscope ...................................................................................2266.17 Dual Beam Oscilloscopes ..............................................................................................228Summary of Equations ...................................................................................................229Assignment Questions ...................................................................................................230Suggested Practical Assignments ...........................................................................2327 D.C. Machines .................................................................................... 2337.1 Motor/Generator Duality ..............................................................................................2337.2 The Generation of d.c. Voltage ...................................................................................2357.3 Construction of d.c. Machines ................................................................................... 2387.4 Classification of Generators .........................................................................................238

viiiContents7.5 Separately Excited Generator .....................................................................................2397.6 Shunt Generator .................................................................................................................2407.7 Series Generator .................................................................................................................2427.8 D. C. Motors ...........................................................................................................................2447.9 Shunt Motor ..........................................................................................................................2447.10 Series Motor ..........................................................................................................................245Summary of Equations ...................................................................................................247Assignment Questions ...................................................................................................2488D.C. Transients ................................................................................... 2498.1 Capacitor-Resistor Series Circuit (Charging) ......................................................2498.2 Capacitor-Resistor Series Circuit (Discharging) ...............................................2538.3 Inductor-Resistor Series Circuit (Connection to Supply) ...........................2568.4 Inductor-Resistor Series Circuit (Disconnection) ............................................259Summary of Equations ...................................................................................................260Assignment Questions ...................................................................................................261Suggested Practical Assignments ...........................................................................2629Semiconductor Theory and Devices ................................... 2639.1 Atomic Structure ................................................................................................................2639.2 Intrinsic (Pure) Semiconductors ................................................................................2649.3 Electron-Hole Pair Generation and Recombination .....................................2669.4 Conduction in Intrinsic Semiconductors ............................................................2679.5 Extrinsic (Impure) Semiconductors ........................................................................2689.6 n-type Semiconductor ...................................................................................................2689.7 p-type Semiconductor ...................................................................................................2709.8 The p-n Junction ................................................................................................................2719.9 The p-n Junction Diode .................................................................................................2729.10 Forward-biased Diode ....................................................................................................2739.11 Reverse-biased Diode .....................................................................................................2739.12 Diode Characteristics .......................................................................................................2749.13 The Zener Diode .................................................................................................................276Assignment Questions ...................................................................................................281Suggested Practical Assignments ...........................................................................282Appendix A: SI Units and Quantities ..................................................................................283Answers to Assignment Questions .....................................................................................285Index .......................................................................................................................................................289

PrefaceThis Textbook supersedes the second edition of FundamentalElectrical and Electronic Principles. In response to commentsfrom colleges requesting that the contents more closely match theobjectives of the BTEC unit Electrical and Electronic Principles,some chapters have been removed and some exchanged with thecompanion book Further Electrical and Electronic Principles, ISBN9780750687478. Also, in order to encourage students to use otherreference sources, those chapters that have been totally removedmay be accessed on the website address http://books.elsevier.com/companions/9780750687379. The previous edition includedSupplementary Worked Examples at the end of each chapter. Themajority of these have now been included within each chapter asWorked Examples, and those that have been removed may be accessedon the above website.This book continues with the philosophy of the previous editionsin that it may be used as a complete set of course notes for studentsundertaking the study of Electrical and Electronic Principles in thefirst year of a BTEC National Diploma/Certificate course. It alsoprovides coverage for some other courses, including foundation/bridging courses which require the study of Electrical and ElectronicEngineering.Fundamental Electrical and Electronic Principles contains 349illustrations, 112 worked examples, 26 suggested practical assignmentsand 234 assignment questions. The answers to the latter are to be foundtowards the end of the book.The order of the chapters does not necessarily follow the orderset out in any syllabus, but rather follows a logical step-by-stepsequence through the subject matter. Some topic areas may extendbeyond current syllabus requirements, but do so both for the sake ofcompleteness and to encourage those students wishing to extend theirknowledge.Coverage of the second year BTEC National Diploma/Certificateunit, Further Electrical Principles, is found in the third edition of thecompanion book Further Electrical and Electronic Principles.C. RobertsonTonbridgeMarch 2008ix


IntroductionThe chapters follow a sequence that I consider to be a logicalprogression through the subject matter, and in the main, followthe order of objectives stated in the BTEC unit of Electrical andElectronic Principles. The major exception to this is that the topics ofinstrumentation and measurements do not appear in a specific chapterof that title. Instead, the various instruments and measurement methodsare integrated within those chapters where the relevant theory iscovered.Occasionally a word or phrase will appear in bold blue type, and closeby will be a box with a blue background. These emphasised words orphrases may be ones that are not familiar to students, and within thebox will be an explanation of the words used in the text.Throughout the book, Worked Examples appear as Q questionsin bold type, followed by A answers. In all chapters, AssignmentQuestions are provided for students to solve.The first chapter deals with the basic concepts of electricity; the use ofstandard form and its adaptation to scientific notation; SI and derivedunits; and the plotting of graphs. This chapter is intended to providea means of ensuring that all students on a given course start with thesame background knowledge. Also included in this chapter are notesregarding communication. In particular, emphasis is placed on logicaland thorough presentation of information, etc. in the solution ofAssignment Questions and Practical Assignment reports.xi


Chapter 1Fundamentals1.1 UnitsWherever measurements are performed there is a need for a coherentand practical system of units. In science and engineering theInternational System of units (SI units) form the basis of all units used.There are seven ‘ base ’ units from which all the other units are derived,called derived units.Table 1.1 The SI base unitsQuantity Unit Unit symbolMass kilogram kgLength metre mTime second sElectric current ampere ATemperature kelvin KLuminous intensity candela cdAmount of substance mole molA few examples of derived units are shown in Table 1.2 , and it is worthnoting that different symbols are used to represent the quantity and itsassociated unit in each case.Table 1.2 Some SI derived unitsQuantityUnitName Symbol Name SymbolForce F Newton NPower P Watt WEnergy W Joule JResistance R Ohm ΩFor a more comprehensive list of SI units see Appendix A at the backof the book.1

2 Fundamental Electrical and Electronic Principles1.2 Standard Form NotationStandard form is a method of writing large and small numbers in aform that is more convenient than writing a large number of trailing orleading zeroes.For example the speed of light is approximately 300 000 000 m/s.When written in standard form this figure would appears as3. 0 108 m/s, where 108represents100 000 000Similarly, if the wavelength of ‘ red ’ light is approximately0.000 000 767 m, it is more convenient to write it in standard form as7. 67107 m, where 1071/10 000 000It should be noted that whenever a ‘ multiplying ’ factor is required,the base 10 is raised to a positive power. When a ‘ dividing ’ factor isrequired, a negative power is used. This is illustrated below:10 101 1 / 10 0.1 101100 102 1 / 100 0.01 1021000 1031/ 1000 0.001 10 etc.etc.One restriction that is applied when using standard form is that onlythe first non-zero digit must appear before the decimal point.Thus, 46 500 is written asSimilarly, 0.002 69 is written as465 . 10 and not as 465 . 104 32. 6910 and not as 26.910 or 269103 4 51.3 ‘Scientific ’ NotationThis notation has the advantage of using the base 10 raised to a powerbut it is not restricted to the placement of the decimal point. It has theadded advantage that the base 10 raised to certain powers have uniquesymbols assigned.For example if a body has a mass m 500 000 g.In standard form this would be written asm 50 . 10 5 g.Using scientific notation it would appear asm 500 kg ( 500 kilogram)

Fundamentals 3where the ‘ k ’ in front of the g for gram represents 10 3 .Not only is the latter notation much neater but it gives a better ‘ feel ’ tothe meaning and relevance of the quantity.See Table 1.3 for the symbols (prefixes) used to represent the variouspowers of 10. It should be noted that these prefixes are arranged inmultiples of 10 3 . It is also a general rule that the positive powers of10 are represented by capital letters, with the negative powers beingrepresented by lower case (small) letters. The exception to this rule isthe ‘ k ’ used for kilo.Table 1.3 Unit prefixes used in ‘scientific ’ notationMultiplying factor Prefix name Symbol10 12 tera T10 9 giga G10 6 mega M10 3 kilo k10 3 milli m10 6 micro μ10 9 nano n10 12 pico pWorked Example 1.1Q Write the following quantities in a concise form using (a) standard form, and (b) scientific notation(i) 0.000 018 A (ii) 15 000 V (iii) 250 000 000 WA(a)(b)(i) 0.000 018 A 1.8 10 5 A(ii) 15 000 V 1.5 10 4 V(iii) 250 000 000 W 2.5 10 8 W(i) 0.000 018 A 18 μ A(ii) 15 000 V 15 kV(iii) 250 000 000 W 250 MW AnsThe above example illustrates the neatness and convenience of the scientific orengineering notation.Worked Example 1.2Q Write the following quantities in scientific (engineering) notation. (a) 25 10 5 A, (b) 3 10 4 W,(c) 850 000 J, (d) 0.0016 V.A(a) 2 5 10 5 25 10 10 6 250 10 6and since 10 6 is represented by μ (micro)

4 Fundamental Electrical and Electronic Principlesthen 25 10 5 A 250 μ A AnsAlternatively, 25 10 5 25 10 2 10 3 0.25 10 3so 25 10 5 A 0.25 mA Ans(b) 3 10 4 3 10 1 10 3 or 300 10 6so 3 10 4 W 0.3 mW or 300 μ W Ans(c) 850 000 850 10 3 or 0.85 10 6so 850 000 J 850 kJ or 0.85 MJ Ans(d) 0.0016 1.6 10 3so 0.0016 V 1.6 mV Ans1.4 Conversion of Areas and VolumesConsider a square having sides of 1 m as shown in Fig. 1.1 . In this caseeach side can also be said to be 100 cm or 1000 mm. Hence the area Aenclosed could be stated as:A 111m2or A 100 100 102 102 104 cm2or A 1000 1000 103 103 10 6 mm2.1m1mFig. 1.1From the above it may be seen that1m2 104 cm2and that 1m2 106mm2.Similarly, if the square had sides of 1 cm the area would beA 1cm2 102 102 104 m2.Again if the sides were of length 1 mm the area would beA 1mm2 103 103 106 m2.Thus 1 cm 2 4 10 m 2 2 6and 1 mm 10 m 2 .

Fundamentals 5Since the basic unit for area is m 2 , then areas quoted in other unitsshould firstly be converted into square metres before calculationsproceed. This procedure applies to all the derived units, and it isgood practice to convert all quantities into their ‘ basic ’ units beforeproceeding with calculations.It is left to the reader to confirm that the following conversions forvolumes are correct:1mm3 109m31cm3 106 m3.Worked Example 1.3Q A mass m of 750 g is acted upon by a force F of 2 N. Calculate the resulting acceleration given that thethree quantities are related by the equationF ma newtonAm 750 g 0.75 kg; F 2 NSince Fma newton, thenFa metre/second2m2075 .so a 2. 667m/s2 Ans1.5 GraphsA graph is simply a pictorial representation of how one quantity orvariable relates to another. One of these is known as the dependentvariable and the other as the independent variable. It is generalpractice to plot the dependent variable along the vertical axis andthe independent variable along the horizontal axis of the graph. Toillustrate the difference between these two types of variable considerthe case of a vehicle that is travelling between two points. If a graphof the distance travelled versus the time elapsed is plotted, then thedistance travelled would be the dependent variable. This is because thedistance travelled depends on the time that has elapsed. But the timeis independent of the distance travelled, since the time will continue toincrease regardless of whether the vehicle is moving or not.Such a graph is shown in Fig. 1.2 , from which it can be seen that overthe first three hours the distance travelled was 30 km. Over the nexttwo hours a further 10 km was travelled, and subsequently no furtherdistance was travelled. Since distance travelled divided by the time

6 Fundamental Electrical and Electronic Principless (km)4030201001 2 3 4 5δ t 1δ s 1δ t 2δ s 2t (h)Fig. 1.2taken is velocity, then the graph may be used to determine the speed ofthe vehicle at any time. Another point to note about this graph is thatit consists of straight lines. This tells us that the vehicle was travellingat different but constant velocities at different times. It should beapparent that the steepest part of the graph occurs when the vehicle wastravelling fastest. To be more precise, we refer to the slope or gradientof the graph. In order to calculate the velocity over the first three hours,the slope can be determined as follows:Change in time, δ t 1 3 0 3 hchange in distance δ s 1 30 0 30 kmδs130slope or gradient velocity v 10km/hδt13Similarly, for the second section of the graph:δs2( 40 30)v 5 km/hδt2( 53)Considering the final section of the graph, it can be seen that there is nochange in distance (the vehicle is stationary). This is confirmed, sinceif δ s is zero then the velocity must be zero.In some ways this last example is a special case, since it involved astraight line graph. In this case we can say that the distance is directlyproportional to time. In many cases a non-linear graph is produced, butthe technique for determining the slope at any given point is similar.Such a graph is shown in Fig. 1.3 , which represents the displacementof a mass when subjected to simple harmonic motion. The resultinggraph is a sinewave. To determine the slope at any given instant in timewe would have to determine the slope of the tangent to the curve atthat point on the graph. If this is done then the figure obtained in eachcase would be the velocity of the mass at that instant. Notice that theslope is steepest at the instants that the curve passes through the zerodisplacement axis (maximum velocity). It is zero at the ‘ peaks ’ of thegraph (zero velocity). Also note that if the graph is sloping upwards

Fundamentals 7S (mm)positiveslopenegativeslope0t (s)zero slopeFig. 1.3as you trace its path from left to right it is called a positive slope. If itslopes downwards it is called a negative slope.1.6 Basic Electrical ConceptsAll matter is made up of atoms, and there are a number of ‘ models ’used to explain physical effects that have been both predicted andsubsequently observed. One of the oldest and simplest of these is theBohr model. This describes the atom as consisting of a central nucleuscontaining minute particles called protons and neutrons. Surroundingthe nucleus are a number of electrons in various orbits. This model isillustrated in Fig. 1.4 . The possible presence of neutrons in the nucleushas been ignored, since these particles play no part in the electricalconcepts to be described. It should be noted that this atomic model isgreatly over-simplified. It is this very simplicity that makes it ideal forthe beginner to achieve an understanding of what electricity is and howmany electrical devices operate.The model shown in Fig. 1.4 is not drawn to scale since a proton isapproximately 2000 times more massive than an electron. Due tothis relatively large mass the proton does not play an active part inelectrical current flow. It is the behaviour of the electrons that is moreimportant. However, protons and electrons do share one thing in Fig. 1.4

8 Fundamental Electrical and Electronic Principlescommon; they both possess a property known as electric charge. Theunit of charge is called the coulomb (C). Since charge is consideredas the quantity of electricity it is given the symbol Q. An electron andproton have exactly the same amount of charge. The electron has anegative charge, whereas the proton has a positive charge. Any atomin its ‘ normal ’ state is electrically neutral (has no net charge). So, inthis state the atom must possess as many orbiting electrons as thereare protons in its nucleus. If one or more of the orbiting electronscan somehow be persuaded to leave the parent atom then this chargebalance is upset. In this case the atom acquires a net positive charge,and is then known as a positive ion. On the other hand, if ‘ extra ’electrons can be made to orbit the nucleus then the atom acquires a netnegative charge. It then becomes a negative ion.An analogy is a technique where the behaviour of one system is compared to thebehaviour of another system. The system chosen for this comparison will be one that ismore familiar and so more easily understood. HOWEVER, it must be borne in mind that ananalogy should not be extended too far. Since the two systems are usually very differentphysically there will come a point where comparisons are no longer validYou may now be wondering why the electrons remain in orbit aroundthe nucleus anyway. This can best be explained by considering ananalogy . Thus, an electron orbiting the nucleus may be comparedto a satellite orbiting the Earth. The satellite remains in orbit due toa balance of forces. The gravitational force of attraction towards theEarth is balanced by the centrifugal force on the satellite due to its highvelocity. This high velocity means that the satellite has high kineticenergy. If the satellite is required to move into a higher orbit, then itsmotor must be fired to speed it up. This will increase its energy. Indeed,if its velocity is increased sufficiently, it can be made to leave Earthorbit and travel out into space. In the case of the electron there is alsoa balance of forces involved. Since both electrons and protons havemass, there will be a gravitational force of attraction between them.However the masses involved are so minute that the gravitational forceis negligible. So, what force of attraction does apply here? Rememberthat electrons and protons are oppositely charged particles, andoppositely charged bodies experience a force of attraction. Comparethis to two simple magnets, whereby opposite polarities attract and like(the same) polarities repel each other. The same rule applies to chargedbodies. Thus it is the balance between this force of electrostaticattraction and the kinetic energy of the electron that maintains the orbit.It may now occur to you to wonder why the nucleus remains intact,since the protons within it are all positively charged particles! It isbeyond the scope of this book (and of the course of study on which youare now embarked) to give a comprehensive answer. Suffice to say thatthere is a force within the nucleus far stronger than the electrostaticrepulsion between the protons that binds the nucleus together.

Fundamentals 9All materials may be classified into one of three major groups—conductors, insulators and semiconductors. In simple terms, the groupinto which a material falls depends on how many ‘ free ’ electrons it has.The term ‘ free ’ refers to those electrons that have acquired sufficientenergy to leave their orbits around their parent atoms. In general wecan say that conductors have many free electrons which will be driftingin a random manner within the material. Insulators have very few freeelectrons (ideally none), and semiconductors fall somewhere betweenthese two extremes.Electric current This is the rate at which free electrons can be madeto drift through a material in a particular direction. In other words, itis the rate at which charge is moved around a circuit. Since charge ismeasured in coulombs and time in seconds then logically the unit forelectric current would be the coulomb/second. In fact, the amount ofcurrent flowing through a circuit may be calculated by dividing theamount of charge passing a given point by the time taken. The unithowever is given a special name, the ampere (often abbreviated toamp). This is fairly common practice with SI units, whereby the nameschosen are those of famous scientists whose pioneering work is thuscommemorated. The relationship between current, charge and time canbe expressed as a mathematical equation as follows:I Qtamp, or QIt coulomb (1.1)Worked Example 1.4QA charge of 35 mC is transferred between two points in a circuit in a time of 20 ms. Calculate the valueof current flowing.AQ 35 10 3 C; t 20 10 3 sQI tamp35 10201033I 1. 75 A AnsWorked Example 1.5QIf a current of 120 μ A flows for a time of 15 s, determine the amount of charge transferred.AI 120 10 6 A; t 15 sQI t coulomb12010615Q 1. 8 mC Ans

10 Fundamental Electrical and Electronic PrinciplesWorked Example 1.6Q80 coulombs of charge was transferred by a current of 0.5 A. Calculate the time for which the currentflowed.AQ 80 C; I 0.5 AQt secondsI8005 .t 160s AnsElectromotive Force (emf) The random movement of electrons withina material does not constitute an electrical current. This is because itdoes not result in a drift in one particular direction. In order to cause the‘ free ’ electrons to drift in a given direction an electromotive force mustbe applied. Thus the emf is the ‘ driving ’ force in an electrical circuit.The symbol for emf is E and the unit of measurement is the volt (V).Typical sources of emf are cells, batteries and generators.The amount of current that will flow through a circuit is directlyproportional to the size of the emf applied to it. The circuit diagramsymbols for a cell and a battery are shown in Figs. 1.5(a) and (b)respectively. Note that the positively charged plate (the long line)usually does not have a plus sign written alongside it. Neither doesthe negative plate normally have a minus sign written by it. Thesesigns have been included here merely to indicate (for the first time) thesymbol used for each plate. (a)Fig. 1.5(b)Resistance (R) Although the amount of electrical current that willflow through a circuit is directly proportional to the applied emf, theother property of the circuit (or material) that determines the resultingcurrent is the opposition offered to the flow. This opposition is knownas the electrical resistance, which is measured in ohms ( Ω ). Thusconductors, which have many ‘ free ’ electrons available for currentcarrying, have a low value of resistance. On the other hand, sinceinsulators have very few ‘ free ’ charge carriers then insulators have avery high resistance. Pure semiconductors tend to behave more likeinsulators in this respect. However, in practice, semiconductors tendto be used in an impure form, where the added impurities improvethe conductivity of the material. An electrical device that is designedto have a specified value of resistance is called a resistor. The circuitdiagram symbol for a resistor is shown in Fig. 1.6 .

Fundamentals 11Fig. 1.6Potential Difference (p.d.) Whenever current flows through a resistorthere will be a p.d. developed across it. The p.d. is measured in volts,and is quite literally the difference in voltage levels between two pointsin a circuit. Although both p.d. and emf are measured in volts they arenot the same quantity. Essentially, emf (being the driving force) causescurrent to flow; whilst a p.d. is the result of current flowing through aresistor. Thus emf is a cause and p.d. is an effect. It is a general rulethat the symbol for a quantity is different to the symbol used for theunit in which it is measured. One of the few exceptions to this ruleis that the quantity symbol for p.d. happens to be the same as its unitsymbol, namely V. In order to explain the difference between emf andp.d. we shall consider another analogy.Figure 1.7 represents a simple hydraulic system consisting of apump, the connecting pipework and two restrictors in the pipe.The latter will have the effect of limiting the rate at which the waterflows around the circuit. Also included is a tap that can be used tointerrupt the flow completely. Figure 1.8 shows the equivalentelectrical circuit, comprising a battery, the connecting conductors(cables or leads) and two resistors. The latter will limit theamount of current flow. Also included is a switch that can beused to ‘ break ’ the circuit and so prevent any current flow. As faras each of the two systems is concerned we are going to make someassumptions.p 1 p 2restrictorswaterflowtappumpPFig. 1.7For the water system we will assume that the connecting pipework hasno slowing down effect on the flow, and so will not cause any pressure

12 Fundamental Electrical and Electronic PrinciplesV 1 V 2R 1R 2ISWEFig. 1.8drop. Provided that the pipework is relatively short then this is areasonable assumption. The similar assumption in the electrical circuitis that the connecting wires have such a low resistance that they willcause no p.d. If anything, this is probably a more legitimate assumptionto make. Considering the water system, the pump will provide the totalsystem pressure (P) that circulates the water through it. Using someform of pressure measuring device it would be possible to measure thispressure together with the pressure drops (p l and p 2 ) that would occuracross the two restrictors. Having noted these pressure readings itwould be found that the total system pressure is equal to the sum of thetwo pressure drops. Using a similar technique for the electrical circuit,it would be found that the sum of the two p.d.s ( V 1 and V 2 ) is equal tothe total applied emf E volts. These relationships may be expressed inmathematical form as:andP p p pascal1 2E V 1 V2 volt (1.2)When the potential at some point in a circuit is quoted as having aparticular value (say 10 V) then this implies that it is 10 V abovesome reference level or datum. Compare this with altitudes. If amountain is said to be 5000 m high it does not necessarily meanthat it rises 5000 m from its base to its peak. The figure of 5000 mrefers to the height of its peak above mean sea level. Thus, meansea level is the reference point or datum from which altitudes aremeasured. In the case of electrical potentials the datum is taken tobe the potential of the Earth which is 0 V. Similarly, 10 V means10 V below or less than 0 V.Conventional current and electron flow You will notice in Fig. 1.8that the arrows used to show the direction of current flow indicate thatthis is from the positive plate of the battery, through the circuit, returningto the negative battery plate. This is called conventional current flow.However, since electrons are negatively charged particles, then thesemust be moving in the opposite direction. The latter is called electron

Fundamentals 13flow. Now, this poses the problem of which one to use. It so happensthat before science was sufficiently advanced to have knowledge of theelectron, it was assumed that the positive plate represented the ‘ high ’potential and the negative the ‘ low ’ potential. So the convention wasadopted that the current flowed around the circuit from the high potentialto the low potential. This compares with water which can naturally onlyflow from a high level to a lower level. Thus the concept of conventionalcurrent flow was adopted. All the subsequent ‘rules ’ and conventionswere based on this direction of current flow. On the discovery ofthe nature of the electron, it was decided to retain the concept ofconventional current flow. Had this not been the case then all the otherrules and conventions would have needed to be changed! Hence, trueelectron flow is used only when it is necessary to explain certain effects(as in semiconductor devices such as diodes and transistors). Wheneverwe are considering basic electrical circuits and devices we shall useconventional current flow i.e. current flowing around the circuit from thepositive terminal of the source of emf to the negative terminal.Ohm ’ s Law This states that the p.d. developed between the two endsof a resistor is directly proportional to the value of current flowingthrough it, provided that all other factors (e.g. temperature) remainconstant. Writing this in mathematical form we have:Vα IHowever, this expression is of limited use since we need an equation.This can only be achieved by introducing a constant of proportionality;in this case the resistance value of the resistor.Thus V IR volt(1.3)Vor I amp(1.4)RVand R ohm(1.5)IWorked Example 1.7Q A current of 5.5 mA flows through a 33 k Ω resistor. Calculate the p.d. thus developed across it.AI 5.5 10 3 A;R 33 10 3 ΩV I R volt55 . 103310V 181.5 V Ans3 3

14 Fundamental Electrical and Electronic PrinciplesWorked Example 1.8QIf a p.d. of 24 V exists across a 15 Ω resistor then what must be the current flowing through it?AV 2 4 V; R 15 ΩVI Ramp2415I 1. 6 A AnsInternal Resistance (r) So far we have considered that the emf E voltsof a source is available at its terminals when supplying current to a circuit.If this were so then we would have an ideal source of emf. Unfortunatelythis is not the case in practice. This is due to the internal resistance of thesource. As an example consider a typical 12 V car battery. This consists ofa number of oppositely charged plates, appropriately interconnected to theterminals, immersed in anelectrolyte . The plates themselves, the internalconnections and the electrolyte all combine to produce a small but finiteresistance, and it is this that forms the battery internal resistance.An electrolyte is the chemical ‘ cocktail ’ in which the plates are immersed. In the case ofa car battery, this is an acid/water mixture.In this context, finite simply means measurable.Figure 1.9 shows such a battery with its terminals on open circuit (noexternal circuit connected). Since the circuit is incomplete no current canflow. Thus there will be no p.d. developed across the battery ’ s internalresistance r . Since the term p.d. quite literally means a difference inpotential between the two ends of r , then the terminal A must be at apotential of 12 V, and terminal B must be at a potential of 0 V. Hence, underthese conditions, the full emf 12 V is available at the battery terminals.Figure 1.10 shows an external circuit, in the form of a 2 Ω resistor,connected across the terminals. Since we now have a complete circuitAr0.1 ΩE12 VBFig. 1.9

Fundamentals 15then current I will flow as shown. The value of this current will be5.71 A (the method of calculating this current will be dealt withearly in the next chapter). This current will cause a p.d. across r andalso a p.d. across R . These calculations and the consequences for thecomplete circuit now follow:p.d. across r Ir volt (Ohm’s law applied)571 . 01. 0.571Vp.d. across RIR volt571 . 2 11.42 VNote: 0.571 11.42 11.991 V but this figure should be 12 V. Thevery small difference is simply due to ‘ rounding ’ the figures obtainedfrom the calculator.IAr0.1 ΩE12 VVR2 ΩFig. 1.10BThe p.d. across R is the battery terminal p.d. V. Thus it may be seenthat when a source is supplying current, the terminal p.d. will alwaysbe less than its emf. To emphasise this point let us assume that theexternal resistor is changed to one of 1.5 Ω resistance. The current nowdrawn from the battery will be 7.5 A. Hence:p.d. across r 75 . 01 . 075. Vand p.d. across R 75 . 15 . 1125. VNote that 11.25 0.75 12 V (rounding error not involved). Hence,the battery terminal p.d. has fallen still further as the current drawn hasincreased. This example brings out the following points.1 Assuming that the battery ’ s charge is maintained, then its emfremains constant. But its terminal p.d. varies as the current drawn isvaried, such thatV E Ir volt (1.6)2 Rather than having to write the words ‘ p.d. across R ’ it is moreconvenient to write this as V AB , which translated, means thepotential difference between points A and B.

16 Fundamental Electrical and Electronic Principles3 In future, if no mention is made of the internal resistance of asource, then for calculation purposes you may assume that it is zero,i.e. an ideal source.Worked Example 1.9Q A battery of emf 6 V has an internal resistance of 0.15 Ω . Calculate its terminal p.d. when delivering acurrent of (a) 0.5 A, (b) 2 A, and (c) 10 A.Worked Example 1.10QA(a)E 6 V; r 0.15 ΩV E Ir volt 6 (0.5 0.15) 6 0.075so, V 5.925 V Ans(b) V 6 (2 0.15) 6 0.3so, V 5.7 V Ans(c) V 6 (10 0.15) 6 1.5so, V 4.5 V AnsNote: This example verifies that the terminal p.d. of a source of emf decreases asthe load on it (the current drawn from it) is increased.A battery of emf 12 V supplies a circuit with a current of 5 A. If, under these conditions, the terminalp.d. is 11.5 V, determine (a) the battery internal resistance, (b) the resistance of the external circuit.AE 12 V; I 5 A; V 11. 5 VAs with the vast majority of electrical problems, a simple sketch of the circuitdiagram will help you to visualise the problem. For the above problem thecircuit diagram would be as shown in Fig. 1.11 .(a) EVIrvoltEV IrvoltE Vso, r ohmI12 11.55hence, r 0. 1 Ω AnsIr5 AE12 VV 11.5 VR (External)Fig. 1.11

Fundamentals 17(b)VR ohmI11. 55so, R 23Ω . AnsEnergy (W) This is the property of a system that enables it to dowork. Whenever work is done energy is transferred from that system toanother one. The most common form into which energy is transformedis heat. Thus one of the effects of an electric current is to produce heat(e.g. an electric kettle). J. P. Joule carried out an investigation into thiseffect. He reached the conclusion that the amount of heat so producedwas proportional to the value of the square of the current flowing andthe time for which it flowed. Once more a constant of proportionality isrequired, and again it is the resistance of the circuit that is used. Thusthe heat produced (or energy dissipated) is given by the equationW I 2 Rt joule (1.7)and applying Ohm ’ s law as shown in equations (1.3) to (1.5)W V2tjoule (1.8)RorW VIt joule (1.9)Worked Example 1.11QA current of 200 mA flows through a resistance of 750 Ω for a time of 5 minutes. Calculate (a) the p.d.developed, and (b) the energy dissipated.AI 200 mA 0.2A; t 5 60 300 s; R 750 Ω(a) V I R volt0.2750V 150V Ans(b) W Rt jouleI 20.2750300W 9000 J or 9kJAns

18 Fundamental Electrical and Electronic PrinciplesNote: It would have been possible to use either equation (1.8) or (1.9)to calculate W. However, this would have involved using the calculatedvalue for V. If this value had been miscalculated, then the answers toboth parts of the question would have been incorrect. So, wheneverpossible, make use of data that are given in the question in preferenceto values that you have calculated. Please also note that the time hasbeen converted to its basic unit, the second.Power (P) This is the rate at which work is done, or at which energyis dissipated. The unit in which power is measured is the watt (W).Warning: Do not confuse this unit symbol with the quantity symbol forenergy. In general terms we can say that power is energy divided by time.Wi.e. P watttThus, by dividing each of equations (1.7), (1.8) and (1.9), in turn, by t,the following equations for power result:PP I 2 R watt (1.10)V 2 watt (1.11)Ror P VI watt(1.12)Worked Example 1.12QA resistor of 680 Ω , when connected in a circuit, dissipates a power of 85 mW. Calculate (a) the p.d.developed across it, and (b) the current flowing through it.AR 680 Ω ; P 85 10 3 W(a)so, VV2P R2 PRwattand V PR volt 85103680 57.8so, V 76 . V Ans(b) PI2RwattPso, I2Rand I PRamp85106803so, I 11. 18 mA Ans1.25104

Fundamentals 19Note: Since P VI watt, the calculations may be checked as followsP 76 . 11.1810 3so, P 84.97 mW, which when rounded up to one decimal place gives85.0 mW — the value given in the question.Worked Example 1.13QA current of 1.4 A when flowing through a circuit for 15 minutes dissipates 200 KJ of energy. Calculate(a) the p.d., (b) power dissipated, and (c) the resistance of the circuit.AI 1.4 A; t 15 60 s; W 2 10 5 J(a) W VItjouleWso V voltIt21051.41560V 158. 7 V Ans(b) PVI watt158. 71.4P 222. 2 W Ans(c)VR ohmI158.71.4R 1134. Ω AnsThe Commercial Unit of Energy (kWh) Although the joule is theSI unit of energy, it is too small a unit for some practical uses, e.g.where large amounts of power are used over long periods of time.The electricity meter in your home actually measures the energy consumption.So, if a 3 kW heater was in use for 12 hours the amount ofenergy used would be 129.6 MJ. In order to record this the meter wouldrequire at least ten dials to indicate this very large number. In addition towhich, many of them would have to rotate at an impossible rate. Hencethe commercial unit of energy is the kilowatt-hour (kWh). Kilowatthoursare the ‘ units ’ that appear on electricity bills. The number of unitsconsumed can be calculated by multiplying the power (in kW) by thetime interval (in hours). So, for the heater mentioned above, the numberof ‘ units ’ consumed would be written as 36 kWh. It should be apparentfrom this that to record this particular figure, fewer dials are required,and their speed of rotation is perfectly acceptable.

20 Fundamental Electrical and Electronic PrinciplesWorked Example 1.14QCalculate the cost of operating a 12.5 kW machine continuously for a period of 8.5 h if the cost per unitis 7.902 p.AW 125 . 85. kWhso W 106.25 kWhand cost 106. 257.902 £ 840 . AnsNote: When calculating energy in kWh the power must be expressed inkW, and the time in hours respectively, rather than in their basic unitsof watts and seconds respectively.Worked Example 1.15QAn electricity bill totalled £78.75, which included a standing charge of £15.00. The number of unitscharged for was 750. Calculate (a) the charge per unit, and (b) the total bill if the charge/unit had been9p, and the standing charge remained unchanged.ATotal bill £78.75; standing charge £15.00; units used 750 750 kWh(a) Cost of the energy (units) usedtotalstanding charge £78. 75£ 1500 . £6375 .£ 63.75Cost/unit £0.085750so, cost/unit 85 . p Ans(b) If the cost/unit is raised to 9p, thencost of energy used £0.09 750 £67.50total bill cost of units used standing charge £67.50 £15.00so, total bill £82.50 AnsAlternating and Direct Quantities The sources of emf and resultingcurrent flow so far considered are called d.c. quantities. This isbecause a battery or cell once connected to a circuit is capable ofdriving current around the circuit in one direction only. If it isrequired to reverse the current it is necessary to reverse the batteryconnections. The term d.c., strictly speaking, means ‘ direct current ’ .However, it is also used to describe unidirectional voltages. Thusa d.c. voltage refers to a unidirectional voltage that may only bereversed as stated above.However, the other commonly used form of electrical supply is thatobtained from the electrical mains. This is the supply that is generated

Fundamentals 21and distributed by the power companies. This is an alternating or a.c.supply in which the current flows alternately in opposite directionsaround a circuit. Again, the term strictly means ‘ alternating current ’ ,but the emfs and p.d.s associated with this system are referred to as a.c.voltages. Thus, an a.c. generator (or alternator) produces an alternatingvoltage. Most a.c. supplies provide a sinusoidal waveform (a sinewaveshape). Both d.c. and a.c. waveforms are illustrated in Fig. 1.12 . Thetreatment of a.c. quantities and circuits is dealt with in Chapters 6, andneed not concern you any further at this stage.I(A)steady d.c.a.c.0t(s)varying d.c.Fig. 1.12Factors affecting Resistance The resistance of a sample of materialdepends upon four factors(i)(ii)(iii)(iv)its lengthits cross-sectional area (csa)the actual material usedits temperatureSimple experiments can show that the resistance is directlyproportional to the length and inversely proportional to the csa.Combining these two statements we can write:R α where length (in metres) and A csa (in square metres)AThe constant of proportionality in this case concerns the third factorlisted above, and is known as the resistivity of the material. This isdefined as the resistance that exists between the opposite faces of a 1 mcube of that material, measured at a defined temperature. The symbolfor resistivity is ρ . The unit of measurement is the ohm-metre ( Ω m).Thus the equation for resistance using the above factor isR ρ ohm (1.13)A

22 Fundamental Electrical and Electronic PrinciplesWorked Example 1.16QA coil of copper wire 200 m long and of csa 0.8 mm2has a resistivity of 0.02 μ Ω m at normal workingtemperature. Calculate the resistance of the coil.Al 200 m; ρ 2 10 8 Ω m; A 8 10 7 m 2ρ R ohmA2 108 2008107R 5ΩAnsWorked Example 1.17Q A wire-wound resistor is made from a 250 metre length of copper wire having a circular cross-sectionof diameter 0.5 mm. Given that the wire has a resistivity of 0.018 μΩ m, calculate its resistance value.Al 250 m; d 5 10 4 m ; ρ 1.8 10 8 Ω mρ πd2R ohm, where cross-sectional area, A metreA4πhence, A ( 5 104)21.9635107m241.8hence, R 108 2501.9635107so, R 22. 92 Ω Ans2The resistance of a material also depends on its temperature andhas a property known as its temperature coefficient of resistance.The resistance of all pure metals increases with increase oftemperature. The resistance of carbon, insulators, semiconductorsand electrolytes decreases with increase of temperature. For thesereasons, conductors (metals) are said to have a positive temperaturecoefficient of resistance. Insulators etc. are said to have a negativetemperature coefficient of resistance. Apart from this there isanother major difference. Over a moderate range of temperature, thechange of resistance for conductors is relatively small and is a veryclose approximation to a straight line. Semiconductors on the otherhand tend to have very much larger changes of resistance over thesame range of temperatures, and follow an exponential law. Thesedifferences are illustrated in Fig. 1.13 .Temperature coefficient of resistance is defined as the ratio of thechange of resistance per degree change of temperature, to the resistanceat some specified temperature. The quantity symbol is α and the unit of

semiconductorFundamentals 23R (Ω)conductor0 θ (°C)measurement is per degree, e.g. /°C. The reference temperature usuallyquoted is 0°C, and the resistance at this temperature is referred to asR 0 . Thus the resistance at some other temperature u 1 °C can be obtainedfrom:R RFig. 1.13( 1αθ ) ohm (1.14)1 0 1In general, if a material having a resistance R 0 at 0°C has a resistanceR 1 at θ 1 C and R 2 at θ 2 C, and if α is the temperature coefficient at0°C, thenR R ( 1αθ) and R R ( 1αθ)1 0 1 2 0 2so R R121 αθ1αθ12(1.15)Worked Example 1.18Q The field coil of an electric motor has a resistance of 250 Ω at 15°C. Calculate the resistance if themotor attains a temperature of 45°C when running. Assume that α 0.00428/°C referred to 0°C.AUsing equation (1.15):R 1 250 Ω ; u 1 15°C; u 2 45°C; α 4.28 10 3250 1 4 28 035 ( . 1 1 )R 1( 428 . 10345)2250 0.8923R2R 280. 2Ω Ans2

24 Fundamental Electrical and Electronic PrinciplesWorked Example 1.19QA coil of wire has a resistance value of 350 Ω when its temperature is 0°C. Given that the temperaturecoefficient of resistance of the wire is 4.26 10 3 /°C referred to 0°C, calculate its resistance at atemperature of 60°C.AR 0 350 Ω ; α 4.26 10 3 /°C; u 1 60°CR1 R0( 1αθ 1)ohm; where R1is the resistance at 60C 350 { 1( 426 . 10360)}350 { 10. 22556}3501.2556so, R1 439.6Ω AnsWorked Example 1.20Q A carbon resistor has a resistance value of 120 Ω at a room temperature of 16°C. When it is connectedas part of a circuit, with current flowing through it, its temperature rises to 32°C. If the temperaturecoefficient of resistance of carbon is 0.000 48/°C referred to 0°C, calculate its resistance under theseoperating conditions.Au 1 16°C; u 2 32°C; R 1 12 0 Ω ; α 0.000 48/°CRR1 αθ1αθ1 12 2120 1 0 000 48 16 ( . )R 1( 0. 000 4832)21201.0078R2120R21.0078so, R 119.1 Ω Ans2Use of meters The measurement of electrical quantities is anessential part of engineering, so you need to be proficient in the useof the various types of measuring instrument. In this chapter we willconsider only the use of the basic current and voltage measuringinstruments, namely the ammeter and voltmeter respectively.An ammeter is a current measuring instrument. It has to be connectedinto the circuit in such a way that the current to be measured is forced toflow through it. If you need to measure the current flowing in a sectionof a circuit that is already connected together, you will need to ‘ break ’the circuit at the appropriate point and connect the ammeter in the‘ break ’ . If you are connecting a circuit (as you will frequently have todo when carrying out practical assignments), then insert the ammeter asthe circuit connections are being made. Most ammeters will have theirterminals colour coded: red for the positive and black for the negative.

Fundamentals 25PLEASE NOTE that these polarities refer to conventional current flow,so the current should enter the meter at the red terminal and leave viathe black terminal. The ammeter circuit symbol is shown in Fig. 1.14 .AFig. 1.14As you would expect, a voltmeter is used for measuring voltages; inparticular, p.d.s. Since a p.d. is a voltage between two points in a circuit,then this meter is NOT connected into the circuit in the same way as anammeter. In this sense it is a simpler instrument to use, since it need onlybe connected across (between the two ends of) the component whosep.d. is to be measured. The terminals will usually be colour coded in thesame way as an ammeter, so the red terminal should be connected to themore positive end of the component, i.e. follow the same principle aswith the ammeter. The voltmeter symbol is shown in Fig. 1.15 .VFig. 1.15It is most probable that you will have to make use of meters that arecapable of combining the functions of an ammeter, a voltmeter and anohmmeter. These instruments are known as multimeters. One of the mostcommon examples is the AVO. This meter is an example of the typeknown as analogue instruments, whereby the ‘ readings ’ are indicatedby the position of a pointer along a graduated scale. The other type ofmultimeter is of the digital type (often referred to as a DMM). In this case,the ‘ readings ’ are in the form of a numerical display, using either lightemitting diodes or a liquid crystal, as on calculator displays. Although thedigital instruments are easier to read, it does not necessarily mean that theygive more accurate results. The choice of the type of meter to use involvesmany considerations. At this stage it is better to rely on advice from yourteacher as to which ones to use for a particular measurement.All multimeters have switches, either rotary or pushbutton, that areused to select between a.c. or d.c. measurements. There is also afacility for selecting a number of current and voltage ranges. To gaina proper understanding of the use of these meters you really need tohave the instrument in front of you. This is a practical exercise thatyour teacher will carry out with you. I will conclude this section byoutlining some important general points that you should observe whencarrying out practical measurements.All measuring instruments are quite fragile, not only mechanically(please handle them carefully) but even more so electrically. If aninstrument becomes damaged it is very inconvenient, but more

26 Fundamental Electrical and Electronic Principlesimportantly, it is expensive to repair and/or replace. So whenever youuse them please observe the following rules:1 Do not switch on (or connect) the power supply to a circuit until yourconnections have been checked by the teacher or laboratory technician.2 Starting with all meters switched to the OFF position, select the highestpossible range, and then carefully select lower ranges until a suitabledeflection (analogue instrument) or figure is displayed (DMM).3 When taking a series of readings try to select a range that willaccommodate the whole series. This is not always possible. However,if the range(s) are changed and the results are used to plot a graph, thena sudden unexpected gap or ‘ jump ’ in the plotted curve may well occur.4 When finished, turn off and disconnect all power supplies, and turnall meters to their OFF position.1.7 CommunicationIt is most important that an engineer is a good communicator. He orshe must be capable of transmitting information orally, by the writtenword and by means of sketches and drawings. He or she must also beable to receive and translate information in all of these forms. Most ofthese skills can be perfected only with guidance and practice. Thus,an engineering student should at every opportunity strive to improvethese skills. The art of good communication is a specialised area, andthis book does not pretend to be authoritative on the subject. However,there are a number of points, given below, regarding the presentation ofwritten work, that may assist you.The assignment questions at the end of each chapter are intended tofulfil three main functions. To reinforce your knowledge of the subjectmatter by repeated application of the underlying principles. To giveyou the opportunity to develop a logical and methodical approach tothe solution of problems. To use these same skills in the presentation oftechnical information. Therefore when you complete each assignment,treat it as a vehicle for demonstrating your understanding of thesubject. This means that your method and presentation of the solution,are more important than always obtaining the ‘ correct ’ numericalanswer. To help you to achieve this use the following procedures:1 Read the question carefully from beginning to end in order toensure that you understand fully what is required.2 Extract the numerical data from the question and list this at the topof the page, using the relevant quantity symbols and units. Thisis particularly important when values are given for a number ofquantities. In this case, if you try to extract the data in the midst ofcalculations, it is all too easy to pick out the wrong figure amongstall the words. At the same time, convert all values into their basic

Fundamentals 27units. Another advantage of using this technique is that the resultinglist, with the quantity symbols, is likely to jog your memory as tothe appropriate equation(s) that will be required.3 Whenever appropriate, sketch the relevant circuit diagram, clearlyidentifying all components, currents, p.d.s etc. If the circuit is onein which there are a number of junctions then labelling as shownin Fig. 1.16 makes the presentation of your solution very muchsimpler. For exampleAI1R 2R 1 B 15 ΩC6 ΩI 2 R 3I10 Ωr0.02 ΩEFig. 1.16If the diagram had not been labelled, and you wished to referto the effective resistance between points B and C, then youwould have to write out ‘ the effective resistance of R 2 and R 3 inparallel … ’ . However, with the diagram labelled you need simplywrite ‘R BC … ’ . Similarly, instead of having to write ‘ the currentthrough the 15 Ω resistor … ’, all that is required is ‘ I 1 … ’ .4 Before writing down any figures quote the equation being used,together with the appropriate unit of measurement. This serves toindicate your method of solution. Also note that the units shouldbe written in words. The unit symbols should only be used whenpreceded by a number thus. Thus, V IR volt for equation, and24 V for actual value.5 Avoid the temptation to save space by having numerous ‘ ’ signsalong one line. If the line is particularly short then a maximum oftwo equals signs per line is acceptable.6 Show ALL figures used in the calculations including any subanswersobtained.7 Clearly identify your answer(s) by either underlining or by writing‘ Ans ’ alongside.You will notice that the above procedures have been followed in all ofthe worked examples throughout this book.In addition to written assignments you will be required to undertakeothers. Although these may not entail using exactly the sameprocedures as outlined above, they will still require a logical and

28 Fundamental Electrical and Electronic Principlesmethodical approach to the presentation. Practical assignments willalso be a major feature of your course of study. These will normallyinvolve the use of equipment and measuring instruments in order thatyou may discover certain technical facts or to verify your theories,etc. This form of exercise also needs to be well documented so that,if necessary, another person can repeat your procedure to confirm (orrefute!) your findings. In these cases a more formalised form of writtenreport is required, and the following format is generally acceptable:‘ Title ’Objective:Apparatus:This needs to be a concise and clear statement as towhat it is that you are trying to achieve.This will be a list of all equipment and instrumentsused, quoting serial numbers where appropriate.Diagram(s): The circuit diagram, clearly labelled.Method: This should consist of a series of numbered paragraphsthat describe each step of the procedure carried out.This section of the report (as with the rest) should beimpersonal and in the past tense. Words such as ‘I ’ , ‘we ’ ,‘ they ’ , ‘ us ’ should not be used. Thus, instead of writing aphrase such as ‘ We set the voltage to 0 V and I adjusted itin 0.1 V steps … ’ you should write ‘ The voltage was setto 0 V and then adjusted in 0.1 V steps … ’Results:All measurements and settings should be neatlytabulated. To avoid writing unit symbols alongsidefigures in the body of the table the appropriate unitsneed to be stated at the top of each column. If there ismore than one table of results then each one should beclearly identified. These points are illustrated below:Table 1p.d. ( V) I (mA) R (k Ω )0.0 0.00 —1.0 1.25 0.802.0 1.75 1.14etc.Calculations: This section may not always be required, but if youhave carried out any calculations using the measureddata then these should be shown here.Graphs:In most cases the tabulated results form the basis fora graph or graphs which must be carefully plotted on

Fundamentals 29approved graph paper. Simple lined or squared paperis NOT adequate. Try to use as much of the page aspossible, but at the same time choose sensible scales.For example let the graduations on the graph paperrepresent increments such as 0.1, 0.2, 0.5, 10 000, etc.and not 0.3, 0.4, 0.6, 0.7, 0.8, 0.9 etc. By doing this youwill make it much simpler to plot the graph in the firstplace, and more importantly, very much easier to takereadings from it subsequently.Conclusions: Having gone through the above procedure you need tocomplete the assignment by drawing some conclusionsbased on all the data gathered. Any such conclusionsmust be justified. For example, you may have takenmeasurements of some variable y for correspondingvalues of a second variable x. If your plotted graphhappens to be a straight line that passes through theorigin then your conclusion would be as follows:‘ Since the graph of y versus x is a straight line passingthrough the origin then it can be concluded that y isdirectly proportional to x ’ .Summary of EquationsCharge: Q It coulombOhm ’ s law: V IR voltTerminal p.d.: V E Ir voltEnergy: W VIt I 2 Rt V 2tRjoulePower: P W t VI I 2 R V 2wattRResistance: R ρ A ohmResistance at specified temp.: R 1 R 0 (1αu) ohmor R R121 1 αθ1αθ2

30 Fundamental Electrical and Electronic PrinciplesAssignment Questions1 Convert the following into standard form.(a) 456.3 (b) 902 344 (c) 0.000 285 (d) 8000(e) 0.047 12 (f) 18 0 μ A (g) 38 mV (h) 80 GN(i) 2000 μ F2 Write the following quantities in scientificnotation.(a) 1500 Ω (b) 0.0033 Ω (c) 0.000 025 A(d) 750 V (e) 800 000 V (f) 0.000 000 047 F3 Calculate the charge transferred in 25 minutesby a current of 500 mA.4 A current of 3.6 A transfers a charge of 375 mC.How long would this take?5 Determine the value of charging currentrequired to transfer a charge of 0.55 mC in atime of 600 μs.6 Calculate the p.d. developed across a 750 Ωresistor when the current flowing through it is(a) 3 A, (b) 25 mA.7 An emf of 50 V is applied in turn to thefollowing resistors: (a) 22 Ω , (b) 820 Ω , (c)2.7 M Ω , (d) 330 k Ω . Calculate the current flowin each case.8 The current flowing through a 470 Ωresistance is 4 A. Determine the energydissipated in a time of 2 h. Express youranswer in both joules and in commercial units.9 A small business operates three pieces ofequipment for nine hours continuouslyper day for six days a week. If the threeequipments consume 10 kW, 2.5 kW and 600 Wrespectively, calculate the weekly cost if thecharge per unit is 7.9 pence.10 A charge of 500 μ C is passed through a560 Ω resistor in a time of 1 ms. Under theseconditions determine (a) the current flowing,(b) the p.d. developed, (c) the power dissipated,and (d) the energy consumed in 5 min.11 A battery of emf 50 V and internal resistance0.2 Ω supplies a current of 1.8 A to an externalload. Under these conditions determine (a)the terminal p.d. and (b) the resistance of theexternal load.12 The terminal p.d. of a d.c. source is 22.5 V whensupplying a load current of 10 A. If the emf is24 V calculate (a) the internal resistance and(b) the resistance of the external load.13 For the circuit arrangement specified in Q12above determine the power and energydissipated by the external load resistor in 5minutes.14 A circuit of resistance 4 Ω dissipates a power of16 W. Calculate (a) the current flowing throughit, (b) the p.d. developed across it, and (c) thecharge displaced in a time of 20 minutes.15 In a test the velocity of a body wasmeasured over a period of time, yielding theresults shown in the table below. Plot thecorresponding graph and use it to determinethe acceleration of the body at times t 0,t 5 s and t 9 s. You may assume that thegraph consists of a series of straight lines.v (m/s) 0.0 3.0 6.0 10.0 14.0 15.0 16.0t (s) 0.0 1.5 3.0 4.5 6.0 8.0 10.016 The insulation resistance between aconductor and earth is 30 M Ω . Calculate theleakage current if the supply voltage is 240 V.17 A 3 kW immersion heater is designed tooperate from a 240 V supply. Determine itsresistance and the current drawn from thesupply.18 A 110 V d.c. generator supplies a lighting loadof forty 100 W bulbs, a heating load of 10 kWand other loads which consume a currentof 15 A. Calculate the power output of thegenerator under these conditions.19 Th e field winding of a d.c. motor is connectedto a 110 V supply. At a temperature of 18°C,the current drawn is 0.575 A. After running themachine for some time the current has fallento 0.475 A, the voltage remaining unchanged.Calculate the temperature of the windingunder the new conditions, assuming thatthe temperature coefficient of resistance ofcopper is 0.004 26/°C at 0°C.20 A coil consists of 1500 turns of aluminium wirehaving a cross-sectional area of 0.75 mm2.The mean length per turn is 60 cm and theresistivity of aluminium at the workingtemperature is 0.028 μΩ m. Calculate theresistance of the coil.

Chapter 2D.C. CircuitsLearning OutcomesThis chapter explains how to apply circuit theory to the solution of simple circuits andnetworks by the application of Ohm ’ s law and Kirchhoff’s laws, and the concepts of potentialand current dividers.This means that on completion of this chapter you should be able to:1 Calculate current flows, potential differences, power and energy dissipations for circuitcomponents and simple circuits, by applying Ohm ’ s law.2 Carry out the above calculations for more complex networks using Kirchhoff’s Laws.3 Calculate circuit p.d.s using the potential divider technique, and branch currents using thecurrent divider technique.4 Understand the principles and use of a Wheatstone Bridge.5 Understand the principles and use of a slidewire potentiometer.2.1 Resistors in SeriesResistors cascaded orconnected in seriesorWhen resistors are connected ‘ end-to-end ’ so that the same currentflows through them all they are said to be cascaded or connected inseries . Such a circuit is shown in Fig. 2.1 . Note that, for the sake ofsimplicity, an ideal source of emf has been used (no internal resistance).R 1 R 2 R 3V 1V 2 V 3IEFig. 2.131

32 Fundamental Electrical and Electronic PrinciplesFrom the previous chapter we know that the current flowing throughthe resistors will result in p.d.s being developed across them. We alsoknow that the sum of these p.d.s must equal the value of the appliedemf. ThusV IR volt; V IR volt; and V IR volt1 1 2 2 3 3However, the circuit current I depends ultimately on the applied emf Eand the total resistance R offered by the circuit. HenceE IRvolt.Also, E V V Vvolt1 2 3and substituting for E, V 1 , V 2 and V 3 in this last equationwe haveIR IR1IR2 IR3voltand dividing this last equation by the common factor IR R1R2 R3 ohm (2.1)where R is the total circuit resistance. From this result it may be seenthat when resistors are connected in series the total resistance is foundsimply by adding together the resistor values.Worked Example 2.1QFor the circuit shown in Fig. 2.2 calculate (a) the circuit resistance, (b) the circuit current, (c) the p.d.developed across each resistor, and (d) the power dissipated by the complete circuit.AE 2 4 V; R 1 330 Ω ; R 2 1500 Ω ; R 3 470 ΩR 1 R 2 R 3330 Ω 1.5 kΩ 470 ΩV 1V 2 V 3IE24 VFig. 2.2

D.C. Circuits 33(a)(b)(c)R R 1 R 2 R 3 ohm 330 1500 470R 2300 Ω or 2.3 k Ω AnsI ER amp242300I 10.43 mA AnsV 1 IR 1 volt 10.43 10 3 330V 1 3.44 V AnsV 2 IR 2 volt 10.43 10 3 1500V 2 15.65 volts AnsV 3 IR 3 volt 10.43 10 3 470V 3 4.90 V AnsNote: The sum of the above p.d.s is 23.99 V instead of 24 V due to the roundingerrors in the calculation. It should also be noted that the value quoted for thecurrent was 10.43 mA whereas the calculator answer is 10.4347 mA. This lattervalue was then stored in the calculator memory and used in the calculations forpart (c), thus reducing the rounding errors to an acceptable minimum.(d)P EI watt 24 10.43 10 3P 0.25 W or 250 mW AnsIt should be noted that the power is dissipated by the three resistors in thecircuit. Hence, the circuit power could have been determined by calculating thepower dissipated by each of these and adding these values to give the total. Thisis shown below, and serves, as a check for the last answer.P1I 2 R1watt( 10. 43103)2330P1 35.93mWP 1 1 3 22 ( 043 . 0 ) 1500163.33mWP3( 10. 43103)2470 511. 8mWtotal power: P P1 P2P3wattso P 250.44mw(Note the worsening effect of continuous rounding error)

34 Fundamental Electrical and Electronic PrinciplesWorked Example 2.2QTwo resistors are connected in series across a battery of emf 12 V. If one of the resistors has a valueof 16 Ω and it dissipates a power of 4 W, then calculate (a) the circuit current, and (b) the value of theother resistor.ASince the only two pieces of data that are directly related to each other concernthe 16 Ω resistor and the power that it dissipates, then this information mustform the starting point for the solution of the problem. Using these data we candetermine either the current through or the p.d. across the 16 Ω resistor (and itis not important which of these is calculated first). To illustrate this point bothmethods will be demonstrated. The appropriate circuit diagram, which forms anintegral part of the solution, is shown in Fig. 2.3 .A B 16 Ω CV ABV BCIE12 VFig. 2.3E 12 V; R BC 16 Ω ; P BC 4 W(a)(b)I 2 R BC P BC wattI 2 RBC PBC4 0.2516so I 0.5 A AnsEtotal resistance, R ohm I12 24 Ω05 .R AB R R BC 24 16so R AB 8 Ω AnsAlternatively, the problem can be solved thus:(a)V2BC PBCwattRBCV2BC P BC R BC 4 16 64

D.C. Circuits 35so V BC 8 VI V ampRBCBC 8 16so I 0.5 A Ans(b)V AB E V BC volt 12 8V AB 4 VRABV IAB 405 .so R AB 8 Ω Ans2.2 Resistors in ParallelWhen resistors are joined ‘ side-by-side ’ so that their corresponding endsare connected together they are said to be connected in parallel . Usingthis form of connection means that there will be a number of pathsthrough which the current can flow. Such a circuit consisting of threeresistors is shown in Fig. 2.4 , and the circuit may be analysed as follows:Resistors in Parallelor or orI 1R 1I 2 R 2I 3R 3IEFig. 2.4

36 Fundamental Electrical and Electronic PrinciplesSince all three resistors are connected directly across the batteryterminals then they all have the same voltage developed across them.In other words the voltage is the common factor in this arrangementof resistors. Now, each resistor will allow a certain value of current toflow through it, depending upon its resistance value. ThusI1EEE amp; I2 amp; and I3 ampRRR12The total circuit current I is determined by the applied emf and the totalcircuit resistance R ,Eso I RampAlso, since all three branch currents originate from the battery, then thetotal circuit current must be the sum of the three branch currentsso I I1I2 I3and substituting the above expression for the currents:ERE E E R R R1 2 3and dividing the above equation by the common factor E :1 1 1 1 siemen (2.2)R R1 R2 R3Note: The above equation does NOT give the total resistance of thecircuit, but does give the total circuit conductance (G) which is measuredin Siemens (S). Thus, conductance is the reciprocal of resistance, soto obtain the circuit resistance you must then take the reciprocal ofthe answer obtained from an equation of the form of equation (2.2).3Conductance is a measure of the ‘ willingness ’ of a material or circuit to allow current tofl ow through itThat is 1 Gsiemen; and 1 R ohm (2.3)RGHowever, when only two resistors are in parallel the combinedresistance may be obtained directly by using the following equation:R R1 R2ohm (2.4)R R1 2

D.C. Circuits 37In this context, the wordidentical means having thesame value of resistanceIf there are ‘x ’ identical resistors in parallel the total resistance issimply R/x ohms.Worked Example 2.3QConsidering the circuit of Fig. 2.5 , calculated (a) the total resistance of the circuit, (b) the three branchcurrent, and (c) the current drawn from the battery.I 1 R 1330 ΩI 2 R 21.5 kΩIR 3 3I470 ΩE24 VFig. 2.5A(a)E 2 4 V; R 1 330 Ω ; R 2 1500 Ω ; R 3 470 Ω1 1 1 1 siemenR R R R12 31 1 1 330 1500 470 0.00303 0.000667 0.00213 0.005825 Sso R 171.68 Ω Ans (reciprocal of 0.005825)(b) I 1 E R1amp 24330I 1 72.73 mA AnsI 2 E R2amp 241500I 2 16 mA Ans

38 Fundamental Electrical and Electronic PrinciplesI 3 E R3 24470ampI 3 51.06 mA Ans(c)I I 1 I 2 I 3 amp 72.73 16 51.06 mAso I 139.8 mA AnsAlternatively, the circuit current could have been determined by using thevalues for E and R as followsI E R amp 24171.68I 139.8 mA AnsCompare this example with worked example 2.1 (the same valuesfor the resistors and the emf have been used). From this it should beobvious that when resistors are connected in parallel the total resistanceof the circuit is reduced. This results in a corresponding increase ofcurrent drawn from the source. This is simply because the parallelarrangement provides more paths for current flow.Worked Example 2.4Q Two resistors, one of 6 Ω and the other of 3 Ω resistance, are connected in parallel across a source ofemf of 12 V. Determine (a) the effective resistance of the combination, (b) the current drawn from thesource, and (c) the current through each resistor.AThe corresponding circuit diagram, suitably labelled is shown in Fig. 2.6.II 1I 2ER 16 ΩR 23 Ω12 VFig. 2.6

D.C. Circuits 39E 12 V; R 1 6 Ω ; R 2 3 Ω(a)RR 1 2R ohmR R126 36 3so R 2 Ω Ans189(b)I E R amp 12 2so I 6 A Ans(c) I 1 E R1amp 12 6I 1 2 A AnsI 2 E R2 12 3I 2 4 A AnsWorked Example 2.5QA 10 Ω resistor, a 20 Ω resistor and a 30 Ω resistor are connected (a) in series, and then(b) in parallel with each other. Calculate the total resistance for each of the twoconnections.AR 1 10 Ω ; R 2 2 0 Ω ; R 3 3 0 Ω(a)(b)R R 1 R 2 R 3 ohm 10 20 30so, R 6 0 Ω Ans1 1 1 1 siemenR R R R12 31 1 1 0. 10. 050.03310 20 30so, R 1 0.183 5.46 Ω Ans

40 Fundamental Electrical and Electronic PrinciplesAlternatively,1 1 1 1R 1 6 3 20 20 30 60 1160 Sso, R 6011 5.46 Ω Ans2.3 Potential DividerWhen resistors are connected in series the p.d. developed across eachresistor will be in direct proportion to its resistance value. This is auseful fact to bear in mind, since it means it is possible to calculate thep.d.s without first having to determine the circuit current. Consider tworesistors connected across a 50 V supply as shown in Fig. 2.7 . In orderto demonstrate the potential divider effect we will in this case firstlycalculate circuit current and hence the two p.d.s by applying Ohm’s law:RR1R2ohmR 7525 100ΩEI ampR50I 05. A100V1 IR1volt05 . 75V1 37.5 V AnsV2 IR2volt05 . 25V 12.5 V Ans2IR 1 75 Ω V 1E 50 VR 225 ΩV 2Fig. 2.7

D.C. Circuits 41Applying the potential divider technique, the two p.d.s may be obtained byusing the fact that the p.d. across a resistor is given by the ratio of itsresistance value to the total resistance of the circuit, expressed as a proportionof the applied voltage. Although this sounds complicated it is very simple toput into practice. Expressed in the form of an equation it meansand2.4 Current DividerVV12R1 E volt (2.5)R R1 2R2 E volt (2.6)R R1 2and using the above equations the p.d.s can more simply be calculatedas follows:75V1 50 37.5 V Ans10025and V2 50 12.5 V Ans100This technique is not restricted to only two resistors in series, but maybe applied to any number. For example, if there were three resistors inseries, then the p.d. across each may be found fromR1V1 ER1 R2 R3R2V2 ER1 R2 R3R3and V3 E voltR R R1 2 3It has been shown that when resistors are connected in parallel the totalcircuit current divides between the alternative paths available. So far wehave determined the branch currents by calculating the common p.d.across a parallel branch and dividing this by the respective resistancevalues. However, these currents can be found directly, without the need tocalculate the branch p.d., by using the current divider theory. Considertwo resistors connected in parallel across a source of emf 48 V as shown inFig. 2.8 . Using the p.d. method we can calculate the two currents as follows:II1EE and I2 ampR1R248481224 4A and I 2A1 2

42 Fundamental Electrical and Electronic PrinciplesII 1I 2E48 VR 112 ΩR 224 ΩFig. 2.8It is now worth noting the values of the resistors and the correspondingcurrents. It is clear that R 1 is half the value of R 2 . So, from the calculationwe obtain the quite logical result that I 1 is twice the value of I 2 . That is,a ratio of 2:1 applies in each case. Thus, the smaller resistor carries thegreater proportion of the total current. By stating the ratio as 2:1 we cansay that the current is split into three equal ‘ parts ’. Two ‘ parts ’ are flowingthrough one resistor and the remaining ‘ part ’ through the other resistor.2Thus3 I flows through R 11and3 I flows through R 2Since I 6 A then2I1 64A31I2 62A3In general we can say thatIandI12R2 I(2.7)R R1 2R1 I(2.8)R R1 2Note: This is NOT the same ratio as for the potential divider. If youcompare (2.5) with (2.7) you will find that the numerator in (2.5) is R 1whereas in (2.7) the numerator is R 2 . There is a similar ‘ cross-over ’when (2.6) and (2.8) are compared.Again, the current divider theory is not limited to only two resistors inparallel. Any number can be accommodated. However, with three or

D.C. Circuits 43more parallel resistors the current division method can be cumbersometo use, and it is much easier for mistakes to be made. For this reason itis recommended that where more than two resistors exist in parallel the‘ p.d. method ’ is used. This will be illustrated in the next section, butfor completeness the application to three resistors is shown below.Consider the arrangement shown in Fig. 2.9 :1 1 1 1 1 1 1 4 3 2R R 1 R 2 R 3 3 4 6 12and examining the numerator, we have 4 3 2 9 ‘ parts ’.I 1 R 13 ΩI 2 R 24 ΩI 3R 3I18 A6 ΩFig. 2.9Thus, the current ratios will be 4/9, 3/9 and 2/9 respectively for thethree resistors.432So, I1 18 8A; I2 18 6A; I3 18 4 A9992.5 Series/Parallel CombinationsMost practical circuits consist of resistors which are interconnected inboth series and parallel forms. The simplest method of solving such acircuit is to reduce the parallel branches to their equivalent resistancevalues and hence reduce the circuit to a simple series arrangement.This is best illustrated by means of a worked example.Worked Example 2.6Q For the circuit shown in Fig. 2.10 , calculate (a) the current drawn from the supply, (b) the currentthrough the 6 Ω resistor, and (c) the power dissipated by the 5.6 Ω resistor.AThe first step in the solution is to sketch and label the circuit diagram, clearlyshowing all currents flowing and identifying each part of the circuit as shown in

44 Fundamental Electrical and Electronic Principles6 Ω5.6 Ω4 ΩE64 VFig. 2.10Fig. 2.11 . Also note that since there is no mention of internal resistance it may beassumed that the source of emf is ideal.I 16 ΩA5.6 ΩBR 1CII 24 ΩR 2E64 VFig. 2.11(a)To determine the current I drawn from the battery we need to know thetotal resistance R AC of the circuit.R BC 6 464(usingproductsumfor two resistors in parallel) 24 10so R BC 2.4 ΩThe original circuit may now be redrawn as in Fig 2.12 .A 5.6 Ω B 2.4 Ω CIV ABV BCE64 VFig. 2.12

D.C. Circuits 45R AC R AB R BC ohm (resistors in series) 5.6 2.4so R AC 8 ΩI ER ACso I 8 A Ansamp 648(b) To find the current I 1 through the 6 Ω resistor we may use either of twomethods. Both of these are now demonstrated.p.d. method:V BC IR BC volt ( Fig. 2.12 ) 8 2.4so, V BC 19.2 VI 1 VBC amp ( Fig. 2.11 )R1 192 .6so, I 1 3.2 A AnsThis answer may be checked as follows:I 1 V BCampR219.2448. Aand since I I 1 I 2 3.2 4.8 8 Awhich agrees with the value found in (a).current division method:Considering Fig. 2.11 , the current I splits into the components I 1 and I 2according to the ratio of the resistor values. However, you must bear in mindthat the larger resistor carries the smaller proportion of the total current.I1 R2 I ampR R4 86 4so, I 1 3.2 A Ans12(c)P AB I 2 R AB watt2 8 5.6so, P AB 358.4 W AnsAlternatively, P AB V AB I wattwhere V AB E V BC volt 64 19.2 44.8 VP AB 44.8 8so, P AB 358.4 W Ans

46 Fundamental Electrical and Electronic PrinciplesWorked Example 2.7QFor the circuit of Fig. 2.13 calculate (a) the current drawn from the source, (b) the p.d. across eachresistor, (c) the current through each resistor, and (d) the power dissipated by the 5 Ω resistor.R 43 Ω4 ΩR 2R 35 ΩR 56 ΩR 66 Ω8 ΩAE18 VFig. 2.13The first step in the solution is to label the diagram clearly with letters at thejunctions and identifying p.d.s and branch currents. This shown in Fig. 2.14 .R 4R 33 ΩI4 Ω 3I R 5 5I 4I 1R 1A R 2B 5 ΩC 6 ΩI 2 6 ΩI 6R 68 ΩDIV AB V BC V CDE18 VFig. 2.14(a)RAB RR 1 2R R1R BC 5 Ω24 6ohm4 6 24 . Ω1 1 1 1 1 1 1 RCDR4 R5 R6 3 6 88 4 32415S24R CD 24 1. 6 Ω15R R AB R BC R CD ohmR 2.4 5 1.6 9 ΩEI R amp 189I 2 A Ans

D.C. Circuits 47(b) The circuit has been reduced to its series equivalent as shown in Fig. 2.15 .Using this equivalent circuit it is now a simple matter to calculate the p.d.across each section of the circuit.V AB IR AB volt 2 2.4V AB 4.8 V Ans(this p.d. is common to both R 1 and R 2 )V BC IR BC volt 2 5V BC 10 V AnsV CD IR CD volt 2 1.6V CD = 3.2 V Ans(this p.d. is common to R 4 , R 5 and R 6 )A2.4 Ω 5 Ω 1.6 ΩB CDIV AB V BC V CDE18 VFig. 2.15(c)I 1 V ABR1 48 .4I 1 1.2 A AnsorI 1 I 2 V AB 48 . or I 2 R26I 2 0.8 A AnsI 3 I 2 A AnsI 4 V CD 32 . orR43R26R1 RI 2 102I 1 1.2 A AnsR1R R 4I0 21 2 1I 2 0.8 A Ans1 1 1 1 R R R RCD4 5 61 1 1I 4 1.067 A Ans 3 6 8I 5 V CDR5I 5 0.533 A Ans 132 . 8 4 36248so I 4 215I 6 V CD 32 . R68I 4 1.067 A AnsI 6 0.4 A Ans4I 5 215I 5 0.533 A AnsI 6 315 2I 6 0.4 A Ans524

48 Fundamental Electrical and Electronic PrinciplesNotice that the p.d. method is an easier and less cumbersomeone thancurrent division when more than two resistors are connected in parallel.(d)P 3 I 32R3 watt or V BC I 3 wattorVBC2R3wattand using the first of these alternative equation:P 3 2 2 5P 3 2 0 W AnsIt is left to the reader to confirm that the other two power equations aboveyield the same answer.2.6 Kirchhoff ’s Current LawWe have already put this law into practice, though without stating itexplicitly. The law states that the algebraic sum of the currents at anyjunction of a circuit is zero. Another, and perhaps simpler, way ofstating this is to say that the sum of the currents arriving at a junctionis equal to the sum of the currents leaving that junction. Thus we haveapplied the law with parallel circuits, where the assumption has beenmade that the sum of the branch currents equals the current drawn fromthe source. Expressing the law in the form of an equation we have:where the symbol Σ means ‘ the sum of ’ .I 0 (2.9)Figure 2.16 illustrates a junction within a circuit with a number of currentsarriving and leaving the junction. Applying Kirchhoff ’ s current law yields:I1I2 I3 I4 I5 0where ‘ ’ signs have been used to denote currents arriving and ‘ ’signs for currents leaving the junction. This equation can be transposedto comply with the alternative statement for the law, thus:I1I3 I4 I2 I5I 1I 4I 2I 3I 5Fig. 2.16

D.C. Circuits 49Worked Example 2.8QFor the network shown in Fig. 2.17 calculate the values of the marked currents.I 540 A AFI 230 A10 A80 AI 4B I 1 C DI 3E25 AFig. 2.17AJunction A: I 2 40 10 50 A AnsJunction C: I1I2 80I 50 801so I1 30 A AnsJunction D: I 3 80 30 110 A AnsJunction E: I25I4 3Iso I4411025 85 A AnsJunction F:III5 45I5so I5308530308555A AnsNote: The minus sign in the last answer tells us that the current I 5 is actuallyflowing away from the junction rather than towards it as shown.2.7 Kirchhoff ’ s Voltage LawThis law also has already been used — in the explanation of p.d. and inthe series and series/parallel circuits. This law states that in any closednetwork the algebraic sum of the emfs is equal to the algebraic sum ofthe p.d.s taken in order about the network. Once again, the law soundsvery complicated, but it is really only common sense, and is simpleto apply. So far, it has been applied only to very simple circuits, suchas resistors connected in series across a source of emf. In this casewe have said that the sum of the p.d.s is equal to the applied emf (e.g.V 1 V 2 E ). However, these simple circuits have had only one source

50 Fundamental Electrical and Electronic Principlesof emf, and could be solved using simple Ohm ’ s law techniques.When more than one source of emf is involved, or the network is morecomplex, then a network analysis method must be used. Kirchhoff ’ s isone of these methods.Expressing the law in mathematical form:E IR(2.10)A generalised circuit requiring the application of Kirchhoff ’ s laws isshown in Fig. 2.18 . Note the following:1 The circuit has been labelled with letters so that it is easy to referto a particular loop and the direction around the loop that is beingconsidered. Thus, if the left-hand loop is considered, and you wishto trace a path around it in a clockwise direction, this would bereferred to as ABEFA. If a counterclockwise path was required, itwould be referred to as FEBAF or AFEBA.IA1IB2(I 1 I 2 )R 1R 3E 1CR 2E 2F E DFig. 2.182 Current directions have been assumed and marked on the diagram.As was found in the previous worked example (2.8), it may wellturn out that one or more of these currents actually flows in theopposite direction to that marked. This result would be indicated bya negative value obtained from the calculation. However, to ensureconsistency, make the initial assumption that all sources of emf aredischarging current into the circuit; i.e. current leaves the positiveterminal of each battery and enters at its negative terminal. Thecurrent law is also applied at this stage, which is why the currentflowing through R 3 is marked as ( I 1 I 2 ) and not as I 3 . This is animportant point since the solution involves the use of simultaneousequations, and the fewer the number of ‘ unknowns ’ the simpler thesolution. Thus marking the third-branch current in this way means

D.C. Circuits 51that there are only two ‘ unknowns ’ to find, namely I 1 and I 2 . Thevalue for the third branch current, I 3 , is then simply found by usingthe values obtained for I 1 and I 2 .3 If a negative value is obtained for a current then the minus signMUST be retained in any subsequent calculations. However, whenyou quote the answer for such a current, make a note to the effectthat it is flowing in the opposite direction to that marked, e.g. fromC to D.4 When tracing the path around a loop, concentrate solely on thatloop and ignore the remainder of the circuit. Also note that if youare following the marked direction of current then the resultingp.d.(s) are assigned positive values. If the direction of ‘ travel ’ isopposite to the current arrow then the p.d. is assigned a negativevalue.Let us now apply these techniques to the circuit of Fig. 2.18 .Consider first the left-hand loop in a clockwise direction. Tracingaround the loop it can be seen that there is only one source of emfwithin it (namely E 1 ). Thus the sum of the emfs is simply E 1 volt.Also, within the loop there are only two resistors ( R 1 and R 2 ) whichwill result in two p.d.s, I 1 R 1 and ( I 1 I 2 )R 3 volt. The resulting loopequation will therefore be:ABEFA: E I R ( I I ) R[1]1 1 1 1 2 3Now taking the right-hand loop in a counterclockwise direction it canbe seen that again there is only one source of emf and two resistors.This results in the following loop equation:CBEDC: E I R ( I I ) R[2]2 2 2 1 2 3Finally, let us consider the loop around the edges of the diagram in aclockwise direction. This follows the ‘ normal ’ direction for E 1 but isopposite to that for E 2 , so the sum of the emfs is E 1 E 2 volt. The loopequation is thereforeABCDEFA: E1E2 I1R1 I2R2[3]Since there are only two unknowns then only two simultaneousequations are required, and three have been written. However it is auseful practice to do this as the ‘ extra ’ equation may contain moreconvenient numerical values for the coefficients of the ‘ unknown ’currents.The complete technique for the applications of Kirchhoff ’ s lawsbecomes clearer by the consideration of a worked example containingnumerical values.

52 Fundamental Electrical and Electronic PrinciplesWorked Example 2.9QFor the circuit of Fig. 2.19 determine the value and direction of the current in each branch, and the p.d.across the 10 Ω resistor.3 Ω2 ΩR 1 R 2R 3 10 ΩE 1 E 210 V 4 VFig. 2.19AThe circuit is first labelled and current flows identified and marked by applyingthe current law. This is shown in Fig. 2.20 .I 1I 2(I 1 I 2 )A B C3 Ω2 Ω10 Ω10 V 4 VF E DFig. 2.20ABEFA :1043I2I1so 63I2I……………[ 1]122ABCDEFA :103I 10( I I)2 1 23I 10I 10I2 1 2so 1013I10I……………[ 2]12

D.C. Circuits 53BCDEB :42I 10( I I)2 1 22I 10I 10I2 1 2so 410I12I……………[ 3]12Inspection of equations [1] and [2] shows that if equation [1] is multiplied by 5then the coefficient of I 2 will be the same in both equations. Thus, if the two arenow added then the term containing I 2 will be eliminated, and hence a valuecan be obtained for I 1 .30 15I110I2……………[]1 51013I10I……………[ 2]1240 28I1so I 1 4028 1.429 A AnsSubstituting this value for I 1 into equation [3] yields;414.2912I12I414.29220292 1 .so I0.857 A (charge) Ans12( I I) . 4290. 857 0.572 A Ans1 2 1VCD( I I) R2 2CD0.57210so V CD 572 . V AnsvoltWorked Example 2.10QFor the circuit shown in Fig 2.21 , use Kirchhoff ’ s Laws to calculate (a) the current flowing in eachbranch of the circuit, and (b) the p.d. across the 5 Ω resistor.1.5 Ω 2 Ω5 Ω6 V 4.5 VFig. 2.21

54 Fundamental Electrical and Electronic PrinciplesAFirstly the circuit is sketched and labelled and currents identified usingKirchhoff ’ s current law. This is shown in Fig. 2.22 .A I 1 B I 2 CR 11.5 Ω(I 1 I 2 )R 22 ΩE 1 6 VR 3E 2 4.5 V5 ΩF E DFig. 2.22(a) We can now consider three loops in the circuit and write down thecorresponding equations using Kirchhoff ’ s voltage law:ABEFA:E1I1R1( I1I2)R3volt61. 5I15( I1I ) 1.5I15I15Iso, 6 65 . I1 5I2 ……………[ 1]2 2CBEDC:ABCDEFA:E2 I2R2( I1I2)R3volt4. 52I 5( I1I ) 2I 5I15Iso, 45 . 5I1 7I ……………[ 22 ]2 2 2 2E1E2 I1R1I2R2volt64. 51.5I12I2so, 1. 51. 5I2I……………[3]1Now, any pair of these three equations may be used to solve the problem,using the technique of simultaneous equations. We shall use equations [1]and [3] to eliminate the unknown current I 2 , and hence obtain a value forcurrent I 1 . To do this we can multiply [1] by 2 and [3] by 5, and then add thetwo modified equations together, thus:1213I10I……………..[ 1]2175 . 75 . I 10I……………[ 3]512195 . 205 . I1195.hence, I1 095. 1 A Ans20.522

D.C. Circuits 55Substituting this value for I 1 into equation [3] gives:1. 5( 1. 50. 951)2I21. 51.4272I2hence, 2I21. 4271. 50.0732and I 0.0366 A Ans2Note: The minus sign in the answer for I 2 indicates that this current isactually flowing in the opposite direction to that marked in Fig. 2.22 . Thismeans that battery E 1 is both supplying current to the 5 Ω resistor andcharging battery E 2 .Current through 5 Ω resistor I1 I2 amp 0. 951( 0. 0366)so current through 5 Ω resistor 0. 9510. 0366 0.914A Ans(b)To obtain the p.d. across the 5 Ω resistor we can either subtract the p.d.(voltage drop) across R 1 from the emf E 1 or add the p.d. across R 2 to emfE 2 , because E 2 is being charged. A third alternative is to multiply R 3 by thecurrent flowing through it. All three methods will be shown here, and,provided that the same answer is obtained each time, the correctness ofthe answers obtained in part (a) will be confirmed.so, VVBE E1I 1R1 volt 6( 0. 9511. 5)61.4265BE 4. 574 V AnsOR:so, VVBE E2I 2R2 volt 4. 5( 0. 03662)4. 50.0732BE 4. 573 V AnsOR:VBE( I1 I2) R3 volt 09 . 145so, V 457 . V AnsBEThe very small differences between these three answers is due simplyto rounding errors, and so the answers to part (a) are verified ascorrect.2.8 The Wheatstone Bridge NetworkThis is a network of interconnected resistors or other components,depending on the application. Although the circuit contains only onesource of emf, it requires the application of a network theorem suchas the Kirchhoff ’ s method for its solution. A typical network, suitablylabelled and with current flows identified is shown in Fig. 2.23 .

56 Fundamental Electrical and Electronic PrinciplesB(I 1 I 3 )IAI 1I 2R 1 R 3I 3CR 5RR 42(I 2 I 3 )DE 1Fig. 2.23Notice that although there are five resistors, the current law has beenapplied so as to minimise the number of ‘ unknown ’ currents to three.Thus only three simultaneous equations will be required for thesolution, though there are seven possible loops to choose from. Theseseven loops are:ABCDA; ADCA; ABDCA; ADBCA; ABDA; BCDB; andABCDAIf you trace around these loops you will find that the last three do notinclude the source of emf, so for each of these loops the sum of theemfs will be ZERO! Up to a point it doesn ’ t matter which three loopsare chosen provided that at least one of them includes the source.If you chose to use only the last three ‘ zero emf ’ loops you wouldsucceed only in proving that zero equals zero!The present level of study does not require you to solve simultaneousequations containing three unknowns. It is nevertheless good practicein the use of Kirchhoff’s laws, and the seven equations for the aboveloops are listed below. In order for you to gain this practice it issuggested that you attempt this exercise before reading further, andcompare your results with those shown below.ABCA : E 1 I 1 R 1 ( I 1 I 3 ) R 3ADCA : E 1 I 2 R 2 ( I 2 I 3 ) R 4ABDCA : E 1 I 1 R 1 I 3 R 5 ( I 2 I 3 ) R 4ADBCA : E 1 I 2 R 2 I 3 R 5 ( I 1 I 3 ) R 3ABDA : 0 I 1 R 1 I 3 R 5 I 2 R 2BCDB : 0 ( I 1 I 3 ) R 3 ( I 2 I 3 ) R 4 I 3 R 5ABCDA : 0 I 1 R 1 ( I 1 I 3 ) R 3 ( I 2 I 3 ) R 4 I 2 R 2

D.C. Circuits 57As a check that the current law has been correctly applied, considerjunctions B and C:current arriving at B Itotal current leaving I1I2so I I1 I2current arriving at C ( I1I3) ( I2 I3) I1I3 I2 I3 I1I2 IHence, current leaving battery current returning to battery.Worked Example 2.11Q For the bridge network shown in Fig. 2.24 calculate the current through each resistor, and the currentdrawn from the supply.B(I 1 I 3 )6 Ω 4 ΩI 1I 3A5 ΩCI 2I3 Ω1 Ω(I 2 I 3 )D10 VFig. 2.24AThe circuit is first labelled and the currents identified using the current law asshown in Fig. 2.24 .ABDA :06I15I33I206I1 3I25I3……………[ 1]BDCB :05I31( I2I3) 4( I1I3)5I3I2I34I14I304I I 10I……………[ 2]12 3

58 Fundamental Electrical and Electronic PrinciplesADCA :103I21( I2I3)3I2I2I3104II……………[ 3]2 3Multiplying equation [1] by 2, equation [2] by 3 and then adding them0 12I16I210I3……………[] 1 2012I13I230I3……………[ 2]30 3I40I 3 ……………[ 4]2Multiplying equation [3] by 3, equation [4] by 4 and then adding them30 12I2 3I3……………[ 3]3012I160I……………[ 4]42 330 163I330I3 0.184A Ans163Substituting for I 3 in equation [3]104I20.1844I2 9.81698 . 16I2 2. 454 A Ans4Substituting for I 3 and I 2 in equation [2]0 4I12. 454 1.844I1 4.2944.294I1 1. 074 A Ans4I I1I21. 0742.454I 3. 529 A AnsSince all of the answers obtained are positive values then the currents willflow in the directions marked on the circuit diagram.Worked Example 2.12Q If the circuit of Fig. 2.24 is now amended by simply changing the value of R DC from 1 Ω to 2 Ω , calculatethe current flowing through the 5 Ω resistor in the central limb.AThe amended circuit diagram is shown in Fig. 2.25 .ABDA :06I15I33I206I 3I 5I……………[ ]1 2 31

D.C. Circuits 59B(I 1 I 3 )6 Ω 4 ΩI 1I 3A5 ΩCI 23 Ω2 Ω(I 2 I 3 )D10 VFig. 2.25BDCB :0 5I32( I2I3) 4( I1I3) 5I32I22I34I14I304I1 2I211 I3……………[ 2]Multiplying equation [1] by 2, equation [2] by 3 and adding them0 12I16I210I3……………[] 1 2012I 6I 33I……………[ 2]32 3043Iso I 3 0 A Ans13At first sight this would seem to be a very odd result. Here we have aresistor in the middle of a circuit with current being drawn from thesource, yet no current flows through this particular resistor! Now, in anycircuit, current will flow between two points only if there is a differenceof potential between the two points. So we must conclude that thepotentials at junctions B and D must be the same. Since junction A isa common point for both the 6 Ω and 3 Ω resistors, then the p.d. acrossthe 6 Ω must be the same as that across the 3 Ω resistor. Similarly, sincepoint C is common to the 4 Ω and 2 Ω resistors, then the p.d. acrosseach of these must also be equal. This may be verified as follows.Since I 3 is zero then the 5 Ω resistor plays no part in the circuit. In thiscase we can ignore its presence and re-draw the circuit as in Fig. 2.26 .Thus the circuit is reduced to a simple series/parallel arrangement thatcan be analysed using simple Ohm ’ s law techniques.RABC R1R3ohm 64so RABC 10 ΩEI1RABCamp10so I1 1A10V IR 1 1 volt 166VAB

60 Fundamental Electrical and Electronic PrinciplesB6 Ω 4 ΩAI 1R 1I 2I 1I 2R 3R 2 R 43 Ω2 ΩCD10 VFig. 2.26Similarly, Rand VADCI2AD325Ω10 2A5236VBalance condition refersto that condition when zerocurrent fl ows through thecentral arm of the bridgecircuit, due to a particularcombination of resistorvalues in the four ‘ outer ’arms of the bridgeThus V AB V AD 6 V, so the potentials at B and D are equal. Inthis last example, the values of R l , R 2 , R 3 and R 4 are such to producewhat is known as the balance condition for the bridge. Being able toproduce this condition is what makes the bridge circuit such a usefulone for many applications in measurement systems. The value ofresistance in the central limb has no effect on the balance conditions.This is because, at balance, zero current flows through it. In addition,the value of the emf also has no effect on the balance conditions, butwill of course affect the values for I 1 and I 2 . Consider the general caseof a bridge circuit as shown in Fig. 2.27 , where the values of resistorsR 1 to R 4 are adjusted so that I 3 is zero.B(I 1 I 3 )R 1 R 3CI 1I 3AR 5I 2R 2R 4D(I I 3 )EFig. 2.27

D.C. Circuits 61V I R and V I RAB1 1 AD 2 2but under the balance condition V AB V ADso I 1 R 1 I 2 R 2 ……………[1]Similarly, V BC V DCso ( I 1 I 3 ) R 3 ( I 2 I 3 ) R 4but, I 3 0, so current through R 3 I 1and current through R 4 I 2 , thereforeI 1 R 3 I 2 R 4 ……………[2]Dividing equation [1] by equation [2]:soIR1 1IR1 3RR13IRIRRR2 22 424This last equation may be verified by considering the values used in theprevious example where R 1 6 Ω , R 2 3 Ω , R 3 4 Ω and R 4 2 Ω .i.e. 6 342So for balance, the ratio of the two resistors on the left-hand side of thebridge equals the ratio of the two on the right-hand side.However, a better way to express the balance condition in terms of theresistor values is as follows. If the product of two diagonally oppositeresistors equals the product of the other pair of diagonally oppositeresistors, then the bridge is balanced, and zero current flows throughthe central limbi.e. R R RR(2.11)1 4 2 3and transposing equation (2.11) to make R 4 the subject we haveR4R2R R 3(2.12)Thus if resistors R 1 , R 2 and R 3 can be set to known values, and adjusteduntil a sensitive current measuring device inserted in the central limbindicates zero current, then we have the basis for a sensitive resistancemeasuring device.1

62 Fundamental Electrical and Electronic PrinciplesWorked Example 2.13QA Wheatstone Bridge type circuit is shown in Fig. 2.28 . Determine (a) the p.d. between terminals B andD, and (b) the value to which R 4 must be adjusted in order to reduce the current through R 3 to zero(balance the bridge).BR 120 ΩR 45 ΩAR 38 ΩCR 2R510 Ω 2 ΩDE10 VFig. 2.28AThe circuit is sketched and currents marked, applying Kirchhoff ’ s current law, asshown in Fig. 2.29 .Kirchhoff ’ s voltage law is now applied to any three loops. Note that as in thiscase there are three unknowns (I 1 , I 2 , and I 3 ) then we must have at least threeequations in order to solve the problem.ABDA:020I18I310I2so, 020I 10I 8I……………….........[ 1]12 3BDCB:08I32( I2I3) 5( I1I3)8I32I22I35I15I3so, 05I2I15I ………………........[ 2]123ADCA:1010I22( I2I3)10I22I22I3so, 1012I2I………………………...... . [ 3]2 3Using equations [1] and [2] to eliminate I 1 we have:020I110I28I3……………[] 1020I 8I 60I…………[ 2]412 3and adding, 0 2I 68I ………………...[ 4]2 3

D.C. Circuits 63BI 1(I 1 I 3 )R 1R 3A20 ΩI 38 ΩR 45 ΩCI 2R 5R 22 Ω10 Ω(I 2 I 3 )DE10 VFig. 2.29and now using equations [3] and [4] we can eliminate I 2 as follows:1012I22I3……………[ 3]0 12I408I…………[ 4]610410I2 3310and I3 0.0244 A410VBD I3R3volt0.02448so, VBD 0.195V Ans(b)For balance conditionsRRRso, R RR2 4 1 544RR 1R25ohm20 210 4 Ω Ans2.9 The Wheatstone Bridge InstrumentThis is an instrument used for the accurate measurement of resistanceover a wide range of resistance values. It comprises three arms, theresistances of which can be adjusted to known values. A fourth armcontains the ‘ unknown ’ resistance, and a central limb contains a sensitivemicroammeter (a galvanometer or ‘ galvo ’ ). The general arrangement isshown in Fig. 2.30 . Comparing this circuit with that of Fig. 2.27 and usingequation (2.12), the value of the resistance to be measured ( R x ) is given byRxRmRR v ohmd

64 Fundamental Electrical and Electronic Principles10010001000s100s10R vR dR m1R s10sunitsG110R x10010002 VFig. 2.30R m and R d are known collectively as the ratio arms, where R m is themultiplier and R d is the divider arm. Both of these arms are variablein decade steps (i.e. 1, 10, 100, 1000). This does not mean thatthese figures represent actual resistance values, but they indicate theappropriate ratio between these two arms. Thus, if R d is set to 10 whilstR m is set to 1000, then the resistance value selected by the variable armR v is ‘ multiplied ’ by the ratio 1000/10 100.Worked Example 2.14Q Two resistors were measured using a Wheatstone Bridge, and the following results wereobtained.(a) R m 1000; R d 1; R v 3502 Ω(b) R m 1; R d 1000; R v 296 ΩFor each case determine the value of the resistance being measured.A(a) R x 1 000 3502 3.502 M Ω Ans11(b) R x 1000 296 0.296 Ω Ans

D.C. Circuits 65From the above example it may be appreciated that due to the ratioarms, the Wheatstone Bridge is capable of measuring a very widerange of resistance values. The instrument is also very accuratebecause it is what is known as a null method of measurement.This term is used because no settings on the three arms are usedto determine the value of R x until the galvo (G) in the central limbindicates zero (null reading). Since the galvo is a very sensitivemicroammeter it is capable of indicating fractions of a microamp.Hence, the slightest imbalance of the bridge can be detected. Also,since the bridge is adjusted until the galvo indicates zero, then thiscondition can be obtained with maximum accuracy. The reason forthis accuracy is that before any measurements are made (no currentthrough the galvo) it is a simple matter to ensure that the galvo pointerindicates zero. Thus, only the sensitivity of the galvo is utilised, andits accuracy over the remainder of its scale is unimportant. Included inthe central limb are a resistor and a switch. These are used to limit thegalvo current to a value that will not cause damage to the galvo whenthe bridge is well off balance. When the ratio arms and the variablearm have been adjusted to give only a small deflection of the galvopointer, the switch is then closed to bypass the swamp resistor R s . Thiswill revert the galvo to its maximum sensitivity for the final balancingusing R v . The bridge supply is normally provided by a 2 V cell asshown.Do not confuse accuracy with sensitivity. For an instrument to be accurate it must alsobe sensitive. However, a sensitive instrument is not necessarily accurate. Sensitivity is theability to react to small changes of the quantity being measured. Accuracy is to do withthe closeness of the indicated value to the true value2.10 The Slidewire PotentiometerThis instrument is used for the accurate measurement of smallvoltages. Like the Wheatstone Bridge, it is a null method ofmeasurement since it also utilises the fact that no current can flowbetween points of equal potential. In its simplest form it comprises ametre length of wire held between two brass or copper blocks on a baseboard, with a graduated metre scale beneath the wire. Connected to oneend of the wire is a contact, the other end of which can be placed at anypoint along the wire. A 2 V cell causes current to flow along the wire.This arrangement, including a voltmeter, is shown in Fig. 2.31 . Thewire between the blocks A and B must be of uniform cross-section andresistivity throughout its length, so that each millimetre of its lengthhas the same resistance as the next. Thus it may be considered as anumber of equal resistors connected in series between points A and B.In other words it is a continuous potential divider.

66 Fundamental Electrical and Electronic PrinciplesE2 VACDBVFig. 2.31Let us now conduct an imaginary experiment. If the movable contactis placed at point A then both terminals of the voltmeter will be at thesame potential, and it will indicate zero volts. If the contact is nowmoved to point B then the voltmeter will indicate 2 V. Consider nowthe contact placed at point C which is midway between A and B. Inthis case it is exactly halfway along our ‘ potential divider ’ , so it willindicate 1V. Finally, placing the contact at a point D (say 70 cm fromA), the voltmeter will indicate 1.4 V. These results can be summarisedby the statement that there is a uniform potential gradient along thewire. Therefore, the p.d. ‘ tapped off’ by the moving contact, is in directproportion to the distance travelled along the wire from point A. Sincethe source has an emf of 2 V and the wire is of 1 metre length, then thepotential gradient must be 2 V/m. In general we can say thatVACACAB E volt (2.13)where AC distance travelled along wireAB total length of the wireand E the source voltageUtilising these facts the simple circuit can be modified to become ameasuring instrument, as shown in Fig. 2.32 . In this case the voltmeter2 VABE sE x12GFig. 2.32

D.C. Circuits 67has been replaced by a galvo. The movable contact can be connectedeither to the cell to be measured or the standard cell, via a switch.Using this system the procedure would be as follows:1 The switch is moved to position ‘ 1 ’ and the slider moved along thewire until the galvo indicates zero current. The position of the slideron the scale beneath the wire is then noted. This distance from Arepresents the emf E s of the standard cell.2 With the switch in position ‘ 2 ’ , the above procedure is repeated,whereby distance along the scale represents the emf E x of the cell tobe measured.3 The value of E x may now be calculated fromADEx EsACwhere AC represents the scale reading obtained for the standard celland AD the scale reading for the unknown cell.It should be noted that this instrument will measure the true emf of thecell since the readings are taken when the galvo carries zero current(i.e. no current is being drawn from the cell under test), hence therewill be no p.d. due to its internal resistance.Worked Example 2.15Q A slidewire potentiometer when used to measure the emfs of two cells provided balance conditionsat scale settings of (a) 600 mm and (b) 745 mm. If the standard cell has an emf of 1.0186 V and a scalereading of 509.3 mm then determine the values for the two cell emfs.ALet E s , 1 and 2 represent the scale readings for the standard cell and cells 1 and2 respectively. Hence: s 509.3 mm 1 600 mm; 2 745 mm; E s 1.0186 VEE11l1 Esvoltl s600 1.0186509.31. 2 V AnsEE22l2 Esvoltl s745 1.0186509.31.49 V AnsIt is obviously inconvenient to have an instrument that needs to be onemetre in length and requires the measurements of lengths along a scale.

68 Fundamental Electrical and Electronic PrinciplesIn the commercial version of the instrument the long wire is replacedby a series of precision resistors plus a small section of wire with amovable contact. The standard cell and galvo would also be built-infeatures. Also, to avoid the necessity for separate calculations, therewould be provision for standardising the potentiometer. This meansthat the emf values can be read directly from dials on the front of theinstrument.Summary of EquationsResistors in series: R R 1 R 2 R 3 … ohm1 1 1 1Resistors in parallel: … siemenR R1 R2 R3RRand for ONLY two resistors in parallel, R 1 2 ⎛ohmproduct ⎞R 1 R2⎝⎜sum ⎠⎟R1Potential divider: V1 E voltR1 R2R2Current divider: I1 I ampR R1 2Kirchhoff ’ s laws: ΣI 0 (sum of the currents at a junction 0)ΣE ΣIR (sum of the emfs sum of the p.d.s, in order)Wheatstone Bridge: Balance condition when current through centre limb 0 or R 1 R 4 R 2 R 3(multiply diagonally across bridge circuit)ACSlidewire potentiometer: VAC EAB volt

D.C. Circuits 69Assignment Questions1 Two 560 Ω resistors are placed in series acrossa 400 V supply. Calculate the current drawn.2 When four identical hotplates on a cookerare all in use, the current drawn from a 240 Vsupply is 33 A. Calculate (a) the resistanceof each hotplate, (b) the current drawnwhen only three plates are switched on. Thehotplates are connected in parallel.3 Calculate the total current when six 120 Ωtorch bulbs are connected in parallel across a9 V supply.4 Two 20 Ω resistors are connected in paralleland this group is connected in series with a 4 Ωresistor. What is the total resistance of the circuit?5 A 12 Ω resistor is connected in parallel with a15 Ω resistor and the combination is connectedin series with a 9 Ω resistor. If this circuit issupplied at 12 V, calculate (a) the total resistance,(b) the current through the 9 Ω resistor and(c) the current through the 12 Ω resistor.6 For the circuit shown in Fig. 2.33 calculatethe values for (a) the current through eachresistor, (b) the p.d. across each resistor and(c) the power dissipated by the 20 Ω resistor.7 Determine the p.d. between terminals E and Fof the circuit in Fig. 2.34 .20 V8 Ω15 Ω20 Ω16 ΩAB50 VFig. 2.338 Ω16 ΩFig. 2.348 Ω10 Ω16 ΩCD8 Ω8 ΩEF8 For the circuit of Fig. 2.35 calculate (a) the p.d.across the 8 Ω resistor, (b) the current throughthe 10 Ω resistor and (c) the current throughthe 12 Ω resistor.8 Ω10 Ω24 VFig. 2.3512 Ω18 Ω9 Three resistors of 5 Ω , 6 Ω and 7 Ω respectivelyare connected in parallel. This combinationis connected in series with another parallelcombination of 3 Ω and 4 Ω . If the completecircuit is supplied from a 20 V source, calculate(a) the total resistance, (b) the total current,(c) the p.d. across the 3 Ω resistor and (d) thecurrent through the 4 Ω resistor.10 Two resistors of 18 Ω and 12 Ω are connected inparallel and this combination is connected inseries with an unknown resistor R x . Determinethe value of R x if the complete circuit draws acurrent of 0.6 A from a 12 V supply.11 Three loads, of 24 A, 8 A, and 12 A are suppliedfrom a 200 V source. If a motor of resistance2.4 Ω is also connected across the supply,calculate (a) the total resistance and (b) thetotal current drawn from the supply.12 Two resistors of 15 Ω and 5 Ω connected inseries with a resistor R x and the combination issupplied from a 240 volt source. If the p.d. acrossthe 5 Ω resistor is 20 V calculate the value of R x .13 A 200 V, 0.5 A lamp is to be connected inseries with a resistor across a 240 V supply.Determine the resistor value required for thelamp to operate at its correct voltage.14 A 12 Ω and a 6 Ω resistor are connected inparallel across the terminals of a battery of emf6 V and internal resistance 0.6 Ω . Sketch thecircuit diagram and calculate (a) the currentdrawn from the battery, (b) the terminal p.d.and (c) the current through the 6 Ω resistor.

70 Fundamental Electrical and Electronic PrinciplesAssignment Questions15 An electric cooker element consists of twoparts, each having a resistance of 18 Ω , whichcan be connected (a) in series, (b) in parallel, or(c) using one part only. Calculate the currentdrawn from a 240 V supply for each connection.16 A cell of emf 2 V has an internal resistance0.1 Ω . Calculate the terminal p.d. when (a)there is no load connected and (b) a 2.9 Ωresistor is connected across the terminals.Explain why these two answers are different.17 A battery has a terminal voltage of 1.8 V whensupplying a current of 9 A. This voltage rises to2.02 V when the load is removed. Calculate theinternal resistance.18 Four resistors of values 10 Ω , 20 Ω , 40 Ω , and 40 Ωare connected in parallel across the terminals ofa generator having an emf of 48 V and internalresistance 0.5 Ω . Sketch the circuit diagramand calculate (a) the current drawn from thegenerator, (b) the p.d. across each resistor and(c) the current flowing through each resistor.19 Calculate the p.d. across the 3 Ω resistorshown in Fig. 2.36 given that V AB is 11 V.200 V5 Ω9 Ω12 ΩFig. 2.382 Ω4 Ω22 A circuit consists of a 15 Ω and a 30 Ω resistorconnected in parallel across a battery ofinternal resistance 2 Ω . If 60 W is dissipated bythe 15 Ω resistor, calculate (a) the current inthe 30 Ω resistor, (b) the terminal p.d. and emfof the battery, (c) the total energy dissipatedin the external circuit in one minute and (d)the quantity of electricity through the batteryin one minute.23 Use Kirchhoff’s laws to determine the threebranch currents and the p.d. across the 5 Ωresistor in the network of Fig. 2.39 .6 ΩA4 Ω 1 Ω4 Ω2 Ω5 Ω4 Ω2 Ω3 Ω20 V 10 VBFig. 2.3620 Calculate the p.d. V AB in Fig. 2.37 .10 Ω 5 ΩFig. 2.3924 Determine the value and direction of currentin each branch of the network of Fig. 2.40 , andthe power dissipated by the 4 Ω load resistor.15 Ω8 V6.5 V100 V2 ΩA4 Ω15 Ω2 Ω 1 ΩBFig. 2.3721 For the network shown in Fig. 2.38 , calculate (a)the total circuit resistance, (b) the supply current,(c) the p.d. across the 12 Ω resistor, (d) the totalpower dissipated in the whole circuit and (e) thepower dissipated by the 12 Ω resistor.Fig. 2.4025 Two batteries A and B are connected in parallel(positive to positive) with each other and thiscombination is connected in parallel with abattery C; this is in series with a 25 Ω resistor,the negative terminal of C being connected to

D.C. Circuits 71Assignment Questionsthe positive terminals of A and B. Battery A hasan emf of 108 V and internal resistance 3 Ω , andthe corresponding values for B are 120 V and2 Ω . Battery C has an emf of 30 V and negligibleinternal resistance. Sketch the circuit andcalculate (a) the value and direction of currentin each battery and (b) the terminal p.d. of A.26 For the circuit of Fig. 2.41 determine (a) thecurrent supplied by each battery, (b) thecurrent through the 15 Ω resistor and (c) thep.d. across the 10 Ω resistor.4 Ω2 Ωdo not solve) the loop equations for loopsABDA, ABCDA, ADCA, and CBDC, (b) to whatvalue must the 2 Ω resistor be changed toensure zero current through the 8 Ω resistor?(c) Under this condition, calculate the currentsthrough and p.d.s across the other fourresistors.29 Three resistances were measured usinga commercial Wheatstone Bridge, yieldingthe following results for the settings onthe multiplying, dividing and variablearms. Determine the resistance value ineach case.20 V10 Ω 15 Ω10 VR d 1000 10 100R m 10 100 100R v ( Ω) 349.8 1685 22.5Fig. 2.4127 For the network of Fig. 2.42 , calculate thevalue and direction of all the branch currentsand the p.d. across the 80 Ω load resistor.50 Ω80 V80 Ω100 Ω100 V30 The slidewire potentiometer instrumentshown in Fig. 2.44 when used to measurethe emf of cell E x yielded the followingresults:(a) galvo current was zero when connected tothe standard cell and the movable contactwas 552 mm from A;(b) galvo current was zero when connectedto E s and the movable contact was647 mm from A.Calculate the value of E x , given E s 1.0183 V.Fig. 2.4228 Figure 2.43 shows a Wheatstone Bridgenetwork, (a) For this network, write down (butBIt was found initially that E x was connected theopposite way round and a balance could notbe obtained. Explain this result.2 V10 Ω 5 ΩA8 ΩCAB6 Ω2 ΩE sDGE x2 VFig. 2.43Fig. 2.44

72 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentsAssignment 1Apparatus:Method:Assignment 2Apparatus:Method:Note: Component values and specific items of equipment when quoted here areonly suggestions. Those used in practice will of course depend upon availabilitywithin a given institution.To investigate Ohm ’ s law and Kirchhoff ’ s laws as applied to series and parallelcircuits.Three resistors of different values1 variable d.c. power supply unit (psu)1 ammeter1 voltmeter (DMM)1 Connect the three resistors in series across the terminals of the psu with theammeter connected in the same circuit. Adjust the current (as measuredwith the ammeter) to a suitable value. Measure the applied voltage and thep.d. across each resistor. Note these values and compare the p.d.s to thetheoretical (calculated) values.2 Reconnect your circuit so that the resistors are now connected in parallelacross the psu. Adjust the psu to a suitable voltage and measure, in turn,the current drawn from the psu and the three resistor currents. Note thesevalues and compare to the theoretical values.3 Write an assignment report and in your conclusions justify whetherthe assignment confirms Ohm ’ s law and Kirchhoff ’ s laws, allowing forexperimental error and resistor tolerances.To investigate the application of Kirchhoff ’ s laws to a network containing morethan one source of emf.2 variable d.c. psu3 different value resistors1 ammeter1 voltmeter (DMM)1 Connect the circuit as shown in Fig. 2.45 . Set psu 1 to 2 V and psu 2 to 4 V.Measure, in turn, the current in each limb of the circuit, and the p.d. acrosseach resistor. For each of the three possible loops in the circuit comparethe sum of the p.d.s measured with the sum of the emfs. Carry out a similarexercise regarding the three currents.R 1 R 2psu 1R 3psu 2Fig. 2.45

D.C. Circuits 732 Reverse the polarity of psu 2 and repeat the above.3 Write the assignment report and in your conclusions justify whether or notKirchhoff ’ s laws have been verified for the network.Assignment 3Apparatus:Method:Assignment 4Apparatus:Method:Assignment 5To investigate potential and current dividers.2 decade resistance boxes1 ammeter1 voltmeter (DMM)1 d.c. psu1 Connect the resistance boxes in series across the psu. Adjust one of them( R 1 ) to 3 k Ω and the other ( R 2 ) to 7 k Ω . Set the psu to 10 V and measure thep.d. across each resistor. Compare the measured values with those predictedby the voltage divider theory.2 Reset both R 1 and R 2 to two or more different values and repeat the aboveprocedure.3 Reconnect the two resistance boxes in parallel across the psu and adjust thecurrent drawn from the psu to 10 mA. Measure the current flowing througheach resistance and compare to those values predicted by the currentdivision theory.4 Repeat the procedure of 3 above for two more settings of R 1 and R 2 , but letone of these settings be such that R 1 R 2 .To make resistance measurements using a Wheatstone Bridge.1 commercial form of Wheatstone Bridge3 decade resistance boxes10 Ω , 6.8 k Ω , and 470 k Ω resistors1 centre-zero galvo1 d. c. psu1 Using the decade boxes, galvo and psu (set to 2 V) connect your ownWheatstone Bridge circuit and measure the three resistor values.2 Use the commercial bridge to re-measure the resistors and compare theresults obtained from both methods.Use a slidewire potentiometer to measure the emf of a number of primary cells(nominal emf no more than 1. 5 V ) .


Chapter 3Electric Fields andCapacitorsLearning OutcomesThis chapter deals with the laws and properties of electric fields and their application to electriccomponents known as capacitors.On completion of this chapter you should be able to:1 Understand the properties of electric fields and insulating materials.2 Carry out simple calculations involving these properties.3 Carry out simple calculations concerning capacitors, and capacitors connected in series,parallel and series/parallel combinations.4 Describe the construction and electrical properties of the different types of capacitor.5 Understand the concept of energy storage in an electric field, and perform simple relatedcalculations.3.1 Coulomb ’ s LawA force exists between charged bodies. A force of attraction existsbetween opposite charges and a force of repulsion between likepolarity charges. Coulomb ’ s law states that the force is directlyproportional to the product of the charges and is inversely proportionalto the square of the distance between their centres. So for the twobodies shown in Fig. 3.1 , this would be expressed asFQQd2 1 2In order to obtain a value for the force, a constant of proportionalitymust be introduced. In this case it is the permittivity of free space, ε 0 .This concept of permittivity is dealt with later in this chapter, and neednot concern you for the time being. The expression for the force innewtons becomes75

76 Fundamental Electrical and Electronic PrinciplesQ 1 Q 2dFig. 3.1FQQ 1 2e 0 d2newton (3.1)This type of relationship is said to follow an inverse square lawbecause F 1/ d 2 . The consequence of this is that if the distance ofseparation is doubled then the force will be reduced by a factor offour times. If the distance is increased by a factor of four times, theforce will be reduced by a factor of sixteen times, etc. The practicalconsequence is that although the force can never be reduced to zero,it diminishes very rapidly as the distance of separation is increased.This will continue until a point is reached where the force is negligiblerelative to other forces acting within the system.3.2 Electric FieldsYou are probably more familiar with the concepts and effects ofmagnetic and gravitational fields. For example, you have probablyconducted simple experiments using bar magnets and iron filings todiscover the shape of magnetic fields, and you are aware that forcesexist between magnetised bodies. You also experience the effectsof gravitational forces constantly, even though you probably do notconsciously think about them.Both of these fields are simply a means of transmitting the forcesinvolved, from one body to another. However, the fields themselvescannot be detected by the human senses, since you cannot see, touch,hear or smell them. This tends to make it more difficult to understandtheir nature. An electric field behaves in the same way as theseother two examples, except that it is the method by which forces aretransmitted between charged bodies. In all three cases we can representthe appropriate field by means of arrowed lines. These lines are usuallyreferred to as the lines of force.To illustrate these points, consider Fig. 3.2 which shows two oppositelycharged spheres with a small positively charged particle placed onthe surface of Q 1 . Since like charges repel and unlike charges attracteach other, then the small charged particle will experience a force ofrepulsion from Q 1 and one of attraction from Q 2 .

Electric Fields and Capacitors 77Q 1 Q 2F 2F 2 F 2F 2 qFFFFF 1F 1 F 1F 1Fig. 3.2The force of repulsion from Q 1 will be very much stronger than theforce of attraction from Q 2 because of the relative distance involved.The other feature of the forces is that they will act so as to be at rightangles to the charged surfaces. Hence there will be a resultant forceacting on the particle. Assuming that it is free to move, then it will startto move in the direction of this resultant force. For the sake of clarity,the distance moved in this direction is greatly exaggerated in thediagram. However, when the particle moves to the new position, forceF 1 will have decreased and F 2 will have increased. In addition, thedirection of action of each force will have changed. Thus the directionof action of the resultant will have changed, but its magnitude will haveremained constant. The particle will now respond to the new resultantforce F. This is a continuous process and the particle will trace out acurved path until it reaches the surface of Q 2 . If this ‘ experiment ’ wascarried out for a number of starting points at the surface of Q 1 , the pathstaken by the particle would be as shown in Fig. 3.3 .Fig. 3.3The following points should be noted:1 The lines shown represent the possible paths taken by the positivelycharged particle in response to the force acting on it. Thus they arecalled the lines of electric force. They may also be referred to as thelines of electric flux, ψ .2 The total electric flux makes up the whole electric field existingbetween and around the two charged bodies.3 The lines themselves are imaginary and the field is threedimensional. The whole of the space surrounding the charged

78 Fundamental Electrical and Electronic Principlesbodies is occupied by the electric flux, so there are no ‘ gaps ’ inwhich a charged particle would not be affected.4 The lines of force (flux) radiate outwards from the surface of apositive charge and terminate at the surface of a negative charge.5 The lines always leave (or terminate) at right angles to a chargedsurface.6 Although the lines drawn on a diagram do not actually exist assuch, they are a very convenient way to represent the existenceof the electric field. They therefore aid the understanding of itsproperties and effects.7 Since force is a vector quantity any line representing it must bearrowed. The convention used here is that the arrows point from thepositive to the negative charge.It is evident from Fig. 3.3 that the spacing between the lines of fluxvaries depending upon which part of the field you consider. This meansthat the field shown is non-uniform. A uniform electric field may beobtained between two parallel charged plates as shown in Fig. 3.4 .QQFig. 3.4Note that the electric field will exist in all of the space surrounding thetwo plates, but the uniform section exists only in the space betweenthem. Some non-uniformity is shown by the curved lines at the edges(fringing effect). At this stage we are concerned only with the uniformfield between the plates. If a positively charged particle was placedbetween the plates it would experience a force that would cause itto move from the positive to the negative plate. The value of forceacting on the particle depends upon what is known as the electric fieldstrength.3.3 Electric Field Strength (E)This is defined as the force per unit charge exerted on a test chargeplaced inside the electric field. (An outdated name for this property is‘ electric force ’ ).

Electric Fields and Capacitors 79Hence, field strength forcechargeSo, E F qnewton/coulomb(3.2)or, F E q newton(3.3)where q is the charge on the particle , and not the plates.3.4 Electric Flux ( ψ ) and Flux Density ( D )In the SI system one ‘ line ’ of flux is assumed to radiate from thesurface of a positive charge of one coulomb and terminate at thesurface of a negative charge of one coulomb. Hence the electric fluxhas the same numerical value as the charge that produces it. Thereforethe coulomb is used as the unit of electric flux. In addition, the Greekletter psi is usually replaced by the symbol for charge, namely Q .The electric flux density D is defined as the amount of flux per squaremetre of the electric field. This area is measured at right angles to thelines of force. This gives the following equationD ψAor, D Q coulomb/metre2 (3.4)AWorked Example 3.1QTwo parallel plates of dimensions 30 mm by 20 mm are oppositely charged to a value of 50 mC.Calculate the density of the electric field existing between them.AQ 50 10 3 C; A 30 20 10 6 m 2QD coulomb/metreA50106001036so D 83. 3 C/m2Ans2

80 Fundamental Electrical and Electronic PrinciplesWorked Example 3.2QTwo parallel metal plates, each having a csa of 400 mm 2 , are charged from a constant current sourceof 50 A for a time of 3 seconds. Calculate (a) the charge on the plates, and (b) the density of theelectric field between them.AA 400 10 6 m 2 ; I 50 10 6 A; t 3 s(a) Q It coulomb 50 10 6 3so, Q 150 μ C Ans(b)QD coulomb/metreA50104001066so, D 0. 125C/m2Ans23.5 The Charging Process and Potential GradientWe have already met the concept of a potential gradient whenconsidering a uniform conductor (wire) carrying a current. This conceptformed the basis of the slidewire potentiometer discussed in Chapter 2.However, we are now dealing with static charges that have been inducedon to plates (the branch of science known as electrostatics). Current flowis only applicable during the charging process. The material between theplates is some form of insulator (a dielectric) which could be vacuum,air, rubber, glass, mica, pvc, etc. So under ideal conditions there will beno current flow from one plate to the other via the dielectric. None-thelessthere will be a potential gradient throughout the dielectric.Consider a pair of parallel plates (initially uncharged) that can beconnected to a battery via a switch, as shown in Fig. 3.5 . Note that thenumber of electrons and protons shown for each plate are in no wayelectron flowA E BFig. 3.5

Electric Fields and Capacitors 81representative of the actual numbers involved. They are shown to aidthe explanation of the charging process that will take place when theswitch is closed. On closing the switch, some electrons from plate Awill be attracted to the positive terminal of the battery. In this case, sinceplate A has lost electrons it will acquire a positive charge. This resultsin an electric field radiating out from plate A. The effect of this field isto induce a negative charge on the top surface of plate B, by attractingelectrons in the plate towards this surface. Consequently, the lowersurface of plate B must have a positive charge. This in turn will attractelectrons from the negative terminal of the battery. Thus for everyelectron that is removed from plate A one is transferred to plate B. Thetwo plates will therefore become equally but oppositely charged.This charging process will not carry on indefinitely (in fact it will lastfor only a very short space of time). This is because as the charge onthe plates increases so too does the voltage developed between them.Thus the charging process continues only until the p.d. between theplates, V is equal to the emf, E of the battery. The charging current atthis time will become zero because plates A and B are positive andnegative respectively. Thus, this circuit is equivalent to two batteries ofequal emf connected in parallel as shown in Fig. 3.6 . In this case eachbattery would be trying to drive an equal value of current around thecircuit, but in opposite directions. Hence the two batteries ‘ balance out ’each other, and no current will flow.IIEEFig. 3.6With suitable instrumentation it would be possible to measure the p.d.between plate B and any point in the dielectric. If this was done, thena graph of the voltage versus distance from B would look like that inFig. 3.7 . The slope of this graph is uniform and has units of volts/metrei.e. the units of potential gradientso potential gradient V dvolt/metre(3.5)Now, energy VIt joule, and I Q t amp

82 Fundamental Electrical and Electronic Principlesp.d. (volt)V0 d distance(m)Fig. 3.7therefore energy VQtt VQ joule, and transposing thisenergyV joule/coulombQJi.e. 1volt 1 ……………[ 1]Cbut the joule is the unit used for work done, and work is force distance, i.e. newton metre,so 1 J 1 Nm……………[2]Substituting [2] into [1]:1 volt 1 Nm/C or V Nm C ……………[3]Dividing both sides of [3] by distance of separation d:Vd Nm NCm CReferring back to equation (3.2), we know that electric field strengthE is measured in N/C. So potential gradient and electric field strengthmust be one and the same thing. Now, electric field strength is definedin terms of the ratio of the force exerted on a charge to the value ofthe charge. This is actually an extremely difficult thing to measure.However, it is a very simple matter to measure the p.d. and distancebetween the charged plates. Hence, for practical purposes, electric fieldstrength is always quoted in the units volt/metrei.e. E V dvolt/metre(3.6)Notice that the symbol E has been used for electric field strength. Thisis in order to avoid confusion with the symbol E used for emf.

Electric Fields and Capacitors 83Worked Example 3.3QTwo parallel plates separated by a dielectric of thickness 3 mm acquire a charge of 35 mC whenconnected to a 150 V source. If the effective csa of the field between the plates is 144 mm 2 , calculate (a)the electric field strength and (b) the flux density.Ad 3 10 3 m ; Q 35 10 3 C; V 150 V; A 144 10 6 m 2(a)(b)VE dvolt/metre 1503 10so E50 kV/m AnssoQD coulom /metreAD 243. 1 C/m2Ansb26351014410363.6 Capacitance ( C )We have seen that in order for one plate to be at a different potential tothe other one there is a need for a charge. This requirement is knownas the capacity of the system. For a given system the ratio of the chargerequired to achieve a given p.d. is a constant for that system. This iscalled the capacitance (C) of the systemQi.e. C farad(3.7)Vor Q VC coulomb(3.8)From equation (3.7) it may be seen that the unit for capacitance is thefarad (F). This is defined as the capacitance of a system that requires acharge of one coulomb in order to raise its potential by one volt.The farad is a very large unit, so in practice it is more usual toexpress capacitance values in microfarads ( μ F), nanofarads (nF), orpicofarads (pF).Worked Example 3.4Q Two parallel plates, separated by an air space of 4 mm, receive a charge of 0.2 mC when connected toa 125 V source. Calculate (a) the electric field strength between the plates, (b) the csa of the field2between the plates if the flux density is 15 C/m , and (c) the capacitance of the plates.Ad 4 10 3 m ; Q 2 10 4 C; V 125 V; D 15 C/m2

84 Fundamental Electrical and Electronic Principles(a)(b)VE dvolt/metre 1254 10so, E31.25 kV/m AnsQD AQso, A metre22 10D15thus, A 13310coulomb/metre234.6 2 2m or 133. mm Ans(c) QCV coulombQ 2so, C farad 10V125thus, C 1. 6 μFAns43.7 CapacitorsA capacitor is an electrical component that is designed to have aspecified value of capacitance. In its simplest form it consists of twoparallel plates separated by a dielectric; i.e. exactly the system we havebeen dealing with so far.In order to be able to design a capacitor we need to know what dimensionsare required for the plates, the thickness of the dielectric (the distance ofseparation d ), and the other properties of the dielectric material chosen.Let us consider first the properties associated with the dielectric.3.8 Permittivity of Free Space ( ε o )When an electric field exists in a vacuum then the ratio of the electricflux density to the electric field strength is a constant, known as thepermittivity of free space.The value for ε 8.854 10 12oF/m.Since a vacuum is a well defined condition, the permittivity of freespace is chosen as the reference or datum value from which thepermittivity of all other dielectrics are measured. This is a similarprinciple to using Earth potential as the datum for measuring voltages.3.9 Relative Permittivity (ε r )The capacitance of two plates will be increased if, instead of a vacuumbetween the plates, some other dielectric is used. This difference incapacitance for different dielectrics is accounted for by the relative

Electric Fields and Capacitors 85permittivity of each dielectric. Thus relative permittivity is defined asthe ratio of the capacitance with that dielectric to the capacitance witha vacuum dielectrici.e. ε r C C21(3.9)where C 1 is with a vacuum and C 2 is with the other dielectric.Note: Dry air has the same effect as a vacuum so the relativepermittivity for air dielectrics 1.3.10 Absolute Permittivity (ε )For a given system the ratio of the electric flux density to the electricfield strength is a constant, known as the absolute permittivity of thedielectric being used.so ε D Efarad/metre(3.10)but we have just seen that a dielectric (other than air) is more effectivethan a vacuum by a factor of ε r times, so the absolute permittivity isgiven by:ε ε o ε r farad/metre (3.11)3.11 Calculating Capacitor ValuesFrom the equation (3.10):Dε Ebut DQ/ A and E V/dQdso ε VAalso, QV / capacitance Ctherefore ε C d Aand transposing this for C we haveεAεεAC o rfarad (3.12)d d

86 Fundamental Electrical and Electronic PrinciplesWorked Example 3.5QA capacitor is made from two parallel plates of dimensions 3 cm by 2 cm, separated by a sheet of mica0.5 mm thick and of relative permittivity 5.8. Calculate (a) the capacitance and (b) the electric fieldstrength if the capacitor is charged to a p.d. of 200 V.AA 3 2 10 4 m 2 ; d 5 10 4 m; ε r 5.8; V 200 V(a)εε AC o r faradd8. 854 1015.86105104so C 61.62 pF Ans2 4(b)VE d volt/metre2005 104so E400 kV/m AnsWorked Example 3.6Q A capacitor of value 0.224 nF is to be made from two plates each 75 mm by 75 mm, using a waxedpaper dielectric of relative permittivity 2.5. Determine the thickness of paper required.AC 0.224 10 9 F; A 75 75 10 6 m 2 ; ε r 2.5εε o rAC faraddεε o rAso d C8. 8541012.5757510metre 0.224109so d 56 . 104m056. mm Ans2 6Worked Example 3.7QA capacitor of value 47 nF is made from two plates having an effective csa of 4 cm2and separated bya ceramic dielectric 0.1 mm thick. Calculate the relative permittivity.AC 4.7 10 8 F; A 4 10 4 m 2 ; d 1 10 4 mεε o rAC faraddCd 47 . 108104so εr ε A o 8.854 10 4 10s ε 1327 Anso r12 4

Electric Fields and Capacitors 87Worked Example 3.8QA p.d. of 180 V creates an electric field in a dielectric of relative permittivity 3.5, thickness 3 mm and ofeffective csa 4.2 cm2. Calculate the flux and flux density thus produced.AV 180 V; d 3 10 3 m ; ε r 3.5; A 4.2 10 4 m 2There are two possible methods of solving this problem; either determine thecapacitance and the use Q VC or determine the electric field strength and useD ε o ε r E . Both solutions will be shown.Alternatively:3.12 Capacitors in Parallelεε AC o r 8. 854 10 13. 54.210farad d3103so C 434 . pF and since QVC coulomb thenQ 1804.341012hence, Q 078 . nC AnsQD coulomb/metre278 . 10A42 . 10so D 1.86μC/m2Ans2 4104V180E volt/metre d3 103so E60kV/m and using D εεo rEcoulomb/metreD 8. 854 10123.560103so D 1.86μC/m2AnsQDA coulomb 1. 861064.2104so Q 078 . nC AnsConsider two capacitors that are identical in every way (same platedimensions, same dielectric material and same distance of separationbetween plates) as shown in Fig. 3.8 . Let them now be moved verticallyuntil the top and bottom edges respectively of their plates make contact.We will now effectively have a single capacitor of twice the csa of oneof the original capacitors, but all other properties will remain unchanged.2Fig. 3.8

88 Fundamental Electrical and Electronic PrinciplesSince C ε A/d farad, then the ‘ new ’ capacitor so formed will havetwice the capacitance of one of the original capacitors. The same effectcould have been achieved if we had simply connected the appropriateplates together by means of a simple electrical connection. In otherwords connect them in parallel with each other. Both of the originalcapacitors have the same capacitance, and this figure is doubled whenthey are connected in parallel. Thus we can draw the conclusionthat with this connection the total capacitance of the combinationis given simply by adding the capacitance values. However, thismight be considered as a special case. Let us verify this conclusionby considering the general case of three different value capacitorsconnected in parallel to a d.c. supply of V volts as in Fig. 3.9 .Q 1C 1Q 2 C 2Q 3 C 3VFig. 3.9Each capacitor will take a charge from the supply according to itscapacitance:Q 1 VC 1 ; Q 2 VC 2 ; Q 3 VC 3 coulombbut the total charge drawn from the supply must be:also, total charge, Q VCwhere C is the total circuit capacitance.Q Q 1 Q 2 Q 3Thus, VC VC 1 VC 2 VC 3 , and dividing through by VC C1C2 C3 farad (3.13)Note: This result is exactly the opposite in form to that for resistorsin parallel.

Electric Fields and Capacitors 89Worked Example 3.9QThree capacitors of value 4.7 F, 3.9 F and 2.2 F are connected in parallel. Calculate the resultingcapacitance of this combination.AIn this case, since all of the capacitor values are in μ F then it is not necessaryto show the 10 6 multiplier in each case, since the answer is best quoted in μ F.However it must be made clear that this is what has been done.CC1C2C3microfaradC 47 . 39 . 22.so C 108. μFAns3.13 Capacitors in SeriesThree parallel plate capacitors are shown connected in series inFig. 3.10 . Each capacitor will receive a charge. However, you maywonder how capacitor C 2 can receive any charging current since it issandwiched between the other two, and of course the charging currentcannot flow through the dielectrics of these. The answer lies in theexplanation of the charging process described in section 3.6 earlier.To assist the explanation, the plates of capacitors C 1 to C 3 have beenlabelled with letters.C 1C 2 C 3A B C D EFEFig. 3.10Plate A will lose electrons to the positive terminal of the supply, andso acquires a positive charge. This creates an electric field in thedielectric of C 1 which will cause plate B to attract electrons from plateC of C 2 . The resulting electric field in C 2 in turn causes plate D toattract electrons from plate E . Finally, plate F attracts electrons fromthe negative terminal of the supply. Thus all three capacitors becomecharged to the same value.Having established that all three capacitors will receive the sameamount of charge, let us now determine the total capacitance of thearrangement. Since the capacitors are of different values then eachwill acquire a different p.d. between its plates. This is illustrated inFig. 3.11 .

90 Fundamental Electrical and Electronic PrinciplesQC 1C 2 C 3Q QV 1V 3V 2VFig. 3.11V1QC V QC V Q ; 2 ; 3 voltC123and V V 1 V 2 V 3 (Kirchhoff ’ s voltage law)also, VQ volt, where C is the total circuit capacitance.CQ Q Q QHence, , and dividing by Q:C C C C1 2 31 1 1 1 (3.14)C C1 C2 C3Note: The above equation does not give the total capacitancedirectly. To obtain the value for C the reciprocal of the answerobtained from equation (3.14) must be found. However, if ONLYTWO capacitors are connected in series the total capacitance maybe obtained directly by using the ‘ product/sum ’ formCCi.e. C 1 2C C1 2farad(3.15)Worked Example 3.10Q A 6 F and a 4 F capacitor are connected in series across a 150 V supply. Calculate (a) the totalcapacitance, (b) the charge on each capacitor and (c) the p.d. developed across each.AFigure 3.12 shows the appropriate circuit diagram.C 1 6 μ F; C 2 4 μ F; V 150 V

Electric Fields and Capacitors91C 1C 26 μF 4μFV 1V 2V150 VFig. 3.12(a)C1C2C C C16 424microfarad6410so C 24 . μFAns2farad(b) QVCcoulomb 1502.410so Q 360μCAns(same charge on both)6Since capacitors in series all receive the same value of charge, then this mustbe the total charge drawn from the supply,QVCThis is equivalent to a series resistor circuit where the current drawn fromthe supply is common to all the resistors(c)QV1Cso V 60 V Ans11Similarly, Vso V36010volt 610QCNote that V1 V2 150 V V222 90 V Ans6636010volt 41066Worked Example 3.11QCapacitors of 3 F, 6 F and 12 F are connected in series across a 400 V supply. Determine the p.d.across each capacitor.AFigure 3.13 shows the relevant circuit diagram.

92 Fundamental Electrical and Electronic PrinciplesC 1 C 2 C 33 μF 6μF 12μFV 1 V 2 V 3V400 VFig. 3.13C 1 3 μ F; C 2 6 μF; C 3 12 μ F; V 400 VSimilarly, V1 1 1 1 1 1 1 C C1C2 C3 3 6 124 1 2 7121212so C 1. 714μF7QVC coulomb4001.71410Q 685.7μCQV1C1685.710volt 3106so V 1 228. 6 V Ans2and V3685.71061066685.71012106661143. V Ans 57. 1V Ans63.14 Series/Parallel CombinationsThe techniques required for the solution of this type of circuit are againbest demonstrated by means of a worked example.Worked Example 3.12Q For the circuit shown in Fig. 3.14 , determine (a) the charge drawn from the supply, (b) the charge onthe 8 F capacitor, (c) the p.d. across the 4 F capacitor, and (d) the p.d. across the 3 F capacitor.

Electric Fields and Capacitors933 μF 6μF2 μF4 μF8 μF200 VFig. 3.14AThe first task is to label the diagram as shown in Fig. 3.15 .2 μF3 μF 6μFCAB4 μFDE8 μFF200 VFig. 3.15(a)CBCDCBD3 63 6 2μF (see Fig. 3.16)246μF (see Fig. 3.17)2 μF2 μFAB4 μFDE8 μFF200 VFig. 3.16

94 Fundamental Electrical and Electronic PrinciplesA2 μF 6μFBDE8 μFF200 VFig. 3.17CAD6 26 2 1. 5μF (see Fig. 3.18)CC C1.58ADso C 95 . μFEFQ VCcoulomb 2009.510hence, Q 1.9 mC Ans6A1.5 μFDE8 μFF200 VFig. 3.18(b) Q VCcoulomb 200810so QEFEFEF1. 6 mC Ans6(c)Total charge Q 1.9 mC and Q EF 1.6 mCso Q AD 1.9 1.6 0.3 mC (see Fig. 3.18 ).and referring to Fig. 3.17 , this will be the charge on both the capacitorsshown, i.e.QABQBD03 . mC.Thus, V BD p.d. across 4 μ F capacitor (see Fig. 3.16 )VBDQCBDBDso V 50 V AnsBD03 . 1061036volt

Electric Fields and Capacitors 95(d) QBCD VBDCBCD(see Figs 3. 16 and 3.15)502106100μCand this will be the charge on both the 3 μ F and 6 μ F capacitors, i.e.QThus VBCBCQQCCDBCBCso V 33. 33 V AnsBC1 10310100μCvolt463.15 Multiplate CapacitorsMost practical capacitors consist of more than one pair of parallelplates, and in these cases they are referred to as multiplate capacitors.The sets of plates are often interleaved as shown in Fig. 3.19 . Theexample illustrated has a total of five plates. It may be seen that thiseffectively forms four identical capacitors, in which the three innerplates are common to the two ‘ inner ’ capacitors. Since all the positiveplates are joined together, and so too are the negative plates, then thisarrangement is equivalent to four identical capacitors connected inparallel, as shown in Fig. 3.20 . The total capacitance of four identicalcapacitors connected in parallel is simply four times the capacitanceof one of them. Thus, this value will be the effective capacitance of thecomplete capacitor.The capacitance between one adjacent pair of plates will beAC1 εε o rfaraddFig. 3.19

96 Fundamental Electrical and Electronic PrinciplesC 1C 2C 3C 4Fig. 3.20so, the total for the complete arrangement C 1 4, but we canexpress 4 as (5 1) so the total capacitance isC1εεA51 o r ( )dfaradIn general therefore, if a capacitor has N plates, the capacitance isgiven by the expression:C1εεAN1 o r ( )dfarad(3.16)Since the above equation applies generally, then it must also apply tocapacitors having just one pair of plates as previously considered. Thisis correct, since if N 2 then ( N 1) 1, and the above equationbecomes identical to equation (3.12) previously used.Worked Example 3.13Q A capacitor is made from 20 interleaved plates each 80 mm by 80 mm separated by mica sheets1.5 mm thick. If the relative permittivity for mica is 6.4, calculate the capacitance.AN 20; A 80 80 10 6 m 2 ; d 1.5 10 3 m; ε r 6.4C1εε AN1 o r ( ) faradd8. 854 10126.4 6400106191.5103soC 46 . nF AnsWorked Example 3.14QA 300 pF capacitor has nine parallel plates, each 40 mm by 30 mm, separated by mica of relativepermittivity 5. Determine the thickness of the mica.

Electric Fields and Capacitors 97AN 9; C 3 10 10 F; A 40 30 10 6 m 2 ; ε r 5soεε o rAN( 1)C dfaradεε o rAN( 1)d metreC8.854 10 15 120031010d 1. 42mmAns2106 8Worked Example 3.15QA parallel plate capacitor consists of 11 circular plates, each of radius 25 mm, with an air gap of 0.5 mmbetween each pair of plates. Calculate the value of the capacitor.AN 11; r 25 10 3 m; d 5 10 4 m; ε r 1(air)Aπrmetre π( 2510)so, A 1.9635103m2εε o rAn( 1)C faradd2 2 3 28. 854 10121.9635103105104thus, C 348 . 1010F or 348pFAns3.16 Energy StoredWhen a capacitor is connected to a voltage source of V volts we haveseen that it will charge up until the p.d. between the plates is also Vvolts. If the capacitor is now disconnected from the supply, the chargeand p.d. between its plates will be retained.Consider such a charged capacitor, as shown in Fig. 3.21 , which nowhas a resistor connected across its terminals. In this case the capacitorwill behave as if it were a source of emf. It will therefore drive currentthrough the resistor. In this way the stored charge will be dissipated asthe excess electrons on its negative plate are returned to the positiveplate. This discharge process will continue until the capacitor becomescompletely discharged (both plates electrically neutral). Note that thedischarge current marked on the diagram indicates conventionalcurrent flow.

98 Fundamental Electrical and Electronic PrinciplesVIRFig. 3.21However, if a discharge current flows then work must be done (energyis being dissipated). The only possible source of this energy in thesecircumstances must be the capacitor itself. Thus the charged capacitormust store energy.If a graph is plotted of capacitor p.d. to the charge it receives, the areaunder the graph represents the energy stored. Assuming a constantcharging current, the graph will be as shown in Fig. 3.22 .p.d. (volt)V0 Q charge(coulomb)Fig. 3.221The area under the graph 2 QVbut QCV coulombso energy stored, W 1 CV22joule(3.17)

Electric Fields and Capacitors 99Worked Example 3.16QA 3 μ F capacitor is charged from a 250 V d.c. supply. Calculate the charge and energy stored. Thecharged capacitor is now removed from the supply and connected across an uncharged 6 μ F capacitor.Calculate the p.d. between the plates and the energy now stored by the combination.AC 1 3 μF; V 1 250 V; C 2 6 μ FQV C 1 1 coulomb 250 3 106so Q 075 . mC Ans1 1W C1V21 joule 31025022so W 93. 75 mJ Ans6 2When the two capacitors are connected in parallel the 3 μ F will share its chargewith 6 μ F capacitor. Thus the total charge in the system will remain unchanged,but the total capacitance will now be different:Total capacitance, CC1 C2 farad36so C 9μFQ 75 .V volt 104C 9106so V 83. 33 V Ans1total energy stored, W CV21joul e 91083.3322so W 31.25mJAns6 2Note: The above example illustrates the law of conservation of charge,since the charge placed on the first capacitor is simply redistributedbetween the two capacitors when connected in parallel. The totalcharge therefore remains the same. However, the p.d. now existingbetween the plates has fallen, and so too has the total energy stored.But there is also a law of conservation of energy, so what has happenedto the ‘ lost ’ energy? Well, in order for the 3 μ F capacitor to share itscharge with the 6 μ F capacitor a charging current had to flow from oneto the other. Thus this ‘ lost ’ energy was used in the charging process.Worked Example 3.17QConsider the circuit of Fig. 3.23 , where initially all three capacitors are fully discharged, with theswitch in position ‘ 1 ’ .(a) If the switch is now moved to position ‘ 2 ’ , calculate the charge and energy stored by C 1 .(b) O nce C 1 is fully charged, the switch is returned to position ‘ 1 ’ . Calculate the p.d. now existingacross C 1 and the amount of energy used in charging C 2 and C 3 from C 1 .

100 Fundamental Electrical and Electronic PrinciplesC 2 C 3‘1’6.8 μFC 14.7 μF‘2’10 μFV200 VFig. 3.23A(a) Q VCcoulomb 20010101 1so, Q 2mCAns11W1C1V2so, W 02 . J Ans126joule 0.5102005 2(b)C 2 and C 3 in series is equivalent to C4C2C3 microfaradC C2 368 . 47 .so, C 4 278 . μF68 . 47 .and total capacitance of the whole circuit,CC1C4microfarad102.78hence, C 1278. μFNow, the charge received by the circuit remains constant, although the totalcapacitance has increased.Q2103Thus, V volt C 1278 . 10and V 156. 5 V Ans61Total energy remaining in the circuit, W CV22joule05. 1278 . 101565.and W 0.156J6 2the energy used up must be the difference between the energy first storedby C 1 and the final energy stored in the system, henceEnergy used 0.2 0.156 4 4 mJ Ans

Electric Fields and Capacitors 1013.17 Dielectric Strength and Working VoltageThere is a maximum potential gradient that any insulating material canwithstand before dielectric breakdown occurs.There are of course some applications where dielectric breakdownis deliberately produced e.g. a sparking plug in a car engine, whichproduces an arc between its electrodes when subjected to a p.d. ofseveral kilovolts. This then ignites the air/fuel mixture. However, it isobviously not a condition that is desirable in a capacitor, since it resultsin its destruction.Capacitors normally have marked on them a maximum workingvoltage. When in use you must ensure that the voltage applied betweenits terminals does not exceed this value, otherwise dielectric breakdownwill occur.Dielectric breakdown is the effect produced in an insulating material when the voltageapplied across it is more than it can withstand. The result is that the material is forced toconduct. However when this happens, the sudden surge of current through it will cause itto burn, melt, vaporise or be permanently damaged in some other wayAnother way of referring to this maximum working voltage is to quotethe dielectric strength. This is the maximum voltage gradient that thedielectric can withstand, quoted in kV/m or in V/mm.Worked Example 3.18QA capacitor is designed to be operated from a 400 V supply, and uses a dielectric which (allowing for afactor of safety), has a dielectric strength of 0.5 MV/m. Calculate the minimum thickness of dielectricrequired.AV 400 V; E 0.5 10 6 V/mVE dvolt/metreV400so d metre E05 . 10d 08 . mm Ans6Worked Example 3.19QA 270pF capacitor is to be made from two metallic foil sheets, each of length 20 cm and width 3 cm,separated by a sheet of Teflon having a relative permittivity of 2.1. Determine (a) the thickness ofTeflon sheet required, and (b) the maximum possible working voltage for the capacitor if the Teflonhas a dielectric strength of 350 kV/m.

102 Fundamental Electrical and Electronic PrinciplesAC 270 10 12 F; A 20 3 10 4 m 2 ; ε r 2.1; E 350 10 3 V / m(a)εε o rAC dfaradεε o rAso, d Cmetre8. 8541012.160102701012thus, d 04 . 13mmAns2 4(b)Dielectric strength is the same thing as electric field strength, expressed involt/metre, soVE dvolt/metreand V Edvolt 35010 0.41310so, V 144. 6 V Ans3 3Note: This figure is the voltage at which the dielectric will start to break down,so, for practical purposes, the maximum working voltage would be specifiedat a lower value. For example, if a factor of safety of 20% was required, then themaximum working voltage in this case would be specified as 115 V .3.18 Capacitor TypesThe main difference between capacitor types is in the dielectric used.There are a number of factors that will influence the choice of capacitortype for a given application. Amongst these are the capacitance value,the working voltage, the tolerance, the stability, the leakage resistance,the size and the price.Tolerance is the deviation from the nominal value. This is normally expressed as apercentage. Thus a capacitor of nominal value 2 μ F and a tolerance of 10%, shouldhave an actual value of between 1.8 and 2.2 μ FSince C εA/d , then any or all of these factors can be varied tosuit particular requirements. Thus, if a large value of capacitanceis required, a large csa and/or a small distance of separation will benecessary, together with a dielectric of high relative permittivity.However, if the area is to be large, then this can result in a devicethat is unacceptably large. Additionally, the dielectric cannot be madetoo thin lest its dielectric strength is exceeded. The various capacitortypes overcome these problems in a number of ways.

Electric Fields and Capacitors 103Paper This is the simplest form of capacitor. It utilises two stripsof aluminium foil separated by sheets of waxed paper. The wholeassembly is rolled up into the form of a cylinder (like a Swiss roll).Metal end caps make the electrical connections to the foils, and thewhole assembly is then encapsulated in a case. By rolling up the foiland paper a comparatively large csa can be produced with reasonablycompact dimensions. This type is illustrated in Fig. 3.24 .FoilPaperFoilPaperFig. 3.24Air Air dielectric capacitors are the most common form of variablecapacitor. The construction is shown in Fig. 3.25 . One set of plates isfixed, and the other set can be rotated to provide either more or lessoverlap between the two. This causes variation of the effective csa andhence variation of capacitance. This is the type of device connected tothe station tuning control of a radio.Fig. 3.25‘ Plastic ’ With these capacitors the dielectric can be of polyester,polystyrene, polycarbonate or polypropylene. Each material hasslightly different electrical characteristics which can be used toadvantage, depending upon the proposed application. The constructiontakes much the same form as that for paper capacitors. Examples ofthese types are shown in Fig. 3.26 , and their different characteristicsare listed in Table 3.1 .

104 Fundamental Electrical and Electronic PrinciplesPolystyrene Polycarbonate Tubular polyester Rectangular polyesterFig. 3.26Table 3.1 Capacitor characteristicsType Capacitance Tolerance (%) OthercharacteristicsPaper 1 nF–40 μF 2 Cheap. Poor stabilityAir 5 pF –1 nF 1 Variable. Good stabilityPolycarbonate 100 pF –10 μF 10 Low loss. HightemperaturePolyester 1 nF –2 μF 20 Cheap. Low frequencyPolypropylene 100 pF –10 nF 5 Low loss. HighfrequencyPolystyrene 10 pF–10 nF 2 Low loss. HighfrequencyMixed 1 nF–1 μF 20 General purposeSilvered mica 10 pF–10 nF 1 High stability. Low lossElectrolytic (aluminium) 1–100 000 μF 20 to 80 High loss. Highleakage. d.c. circuitsonlyElectrolytic (tantalum) 0.1 150 μF 20 As for aluminium aboveCeramic 2 pF 100 nF 10 Low temperaturecoeffi cient. HighfrequencySilvered mica These are the most accurate and reliable of the capacitortypes, having a low tolerance figure. These features are usually reflectedin their cost. They consist of a disc or hollow cylinder of ceramicmaterial which is coated with a silver compound. Electrical connectionsare affixed to the silver coatings and the whole assembly is placed into acasing or (more usually) the assembly is encased in a waxy substance.Mixed dielectric This dielectric consists of paper impregnated withpolyester which separates two aluminium foil sheets as in the papercapacitor. This type makes a good general-purpose capacitor, and anexample is shown in Fig. 3.27 .Fig. 3.27

Electric Fields and Capacitors 105Electrolytic This is the form of construction used for the largestvalue capacitors. However, they also have the disadvantages ofreduced working voltage, high leakage current, and the requirementto be polarised. Their terminals are marked and , and thesepolarities must be observed when the device is connected into a circuit.Capacitance values up to 100 000 μ F are possible.A polarised capacitor is one in which the dielectric is formed by passing a d.c. currentthrough it. The polarity of the d.c. supply used for this purpose must be subsequentlyobserved in any circuit in which the capacitor is then used. Thus, they should be usedonly in d.c. circuitsThe dielectric consists of either an aluminium oxide or tantalum oxidefilm that is just a few micrometres thick. It is this fact that allows suchhigh capacitance values, but at the same time reduces the possiblemaximum working voltage. Tantalum capacitors are usually verymuch smaller than the aluminium types. They therefore cannot obtainthe very high values of capacitance possible with the aluminiumtype. The latter consist of two sheets of aluminium separated bypaper impregnated with an electrolyte. These are then rolled up like asimple paper capacitor. This assembly is then placed in a hermeticallysealed aluminium cannister. The oxide layer is formed by passinga charging current through the device, and it is the polarity of thischarging process that determines the resulting terminal polarity thatmust be subsequently observed. If the opposite polarity is applied tothe capacitor the oxide layer is destroyed. Examples of electrolyticcapacitors are shown in Fig. 3.28 .Double-endedSingle-endedSolid tantalumFig. 3.28Summary of EquationsForce between charges: FQQ 1 2 ε d2onewtonElectric field strength: E Fqnewton/coulomb Vdvolt/metre

106Fundamental Electrical and Electronic PrinciplesElectric flux density: D Q Aεε o r E coulomb/metre2Permittivity: ε ε o ε rDEfarad/metreQ εε o rAN( 1)Capacitance: C farad faradVdCapacitors in parallel: C C 1 C 2 C 3 … farad1 1 1 1Capacitors in series: … faradC C C C1 2 3CCFor ONLY two in series: C 1 2C C1 2faradEnergy stored: W 0.5 QV 0.5CV 2 joule⎛⎝⎜1productsum⎞⎠⎟

Electric Fields and Capacitors 107Assignment Questions1 Two parallel plates 25 cm by 35 cm receive acharge of 0.2 μ C from a 250 V supply. Calculate(a) the electric flux and (b) the electric fluxdensity.2 Th e flux density between two plates separatedby a dielectric of relative permittivity 8 is1. 2 μ C/m2. Determine the potential gradientbetween them.3 Calculate the electrical field strength betweena pair of plates spaced 10 mm apart when ap.d. of 0.5 kV exists between them.4 Two plates have a charge of 30 μ C. If theeffective area of the plates is 5 cm 2 , calculatethe flux density.5 A capacitor has a dielectric 0.4 mm thick andoperates at 50 V. Determine the electric fieldstrength.6 A 10 0 μ F capacitor has a p.d. of 400 V across it.Calculate the charge that it has received.7 A 4 7 μ F capacitor stores a charge of 7.8 mCwhen connected to a d.c. supply. Calculate thesupply voltage.8 Determine the p.d. between the plates ofa 470 nF capacitor if it stores a charge of0.141 mC.9 Calculate the capacitance of a pair of plateshaving a p.d. of 600 V when charged to 0.3 μ C.10 The capacitance of a pair of plates is 40 pFwhen the dielectric between them is air. If asheet of glass is placed between the plates(so that it completely fills the space betweenthem), calculate the capacitance of the newarrangement if the relative permittivity of theglass is 6.11 A dielectric 2.5 mm thick has a p.d. of 440 Vdeveloped across it. If the resulting flux densityis 4.7 μ C/m 2 determine the relative permittivityof the dielectric.12 State the factors that affect the capacitance ofa parallel plate capacitor, and explain how thevariation of each of these factors affects thecapacitance.Calculate the value of a two plate capacitorwith a mica dielectric of relative permittivity 5and thickness 0.2 mm. The effective area of theplates is 250 cm2 .13 A capacitor consists of two plates, each ofeffective area 500 cm2, spaced 1mm apart inair. If the capacitor is connected to a 400 Vsupply, determine (a) the capacitance,(b) the charge stored and (c) the potentialgradient.14 A paper dielectric capacitor has two plates,each of effective csa 0.2 m2. If the capacitanceis 50 nF calculate the thickness of the paper,given that its relative permittivity is 2.5.15 A two plate capacitor has a value of 47 nF. If theplate area was doubled and the thickness ofthe dielectric was halved, what then would bethe capacitance?16 A parallel plate capacitor has 20 plates, each50 mm by 35 mm, separated by a dielectric0.4 mm thick. If the capacitance is 1000 pFdetermine the relative permittivity of thedielectric.17 Calculate the number of plates used in a0.5 nF capacitor if each plate is 40 mm square,separated by dielectric of relative permittivity6 and thickness 0.102 mm.18 A capacitor is to be designed to have acapacitance of 4.7 pF and to operate with a p.d.of 120 V across its terminals. The dielectric is tobe teflon ( ε r 2.1) which, after allowing fora safety factor has a dielectric strength of25 kV/m. Calculate (a) the thickness of teflonrequired and (b) the area of a plate.19 Capacitors of 4 μ F and 10 μ F are connected(a) in parallel and (b) in series. Calculate theequivalent capacitance in each case.20 Determine the equivalent capacitance whenthe following capacitors are connected (a) inseries and (b) in parallel(i) 3 μF, 4 μ F and 10 μ F(ii) 0.02 μ F, 0.05 μ F and 0.22 μ F(iii) 20 pF and 470 pF(iv) 0.01 μ F and 220 pF.21 Determine the value of capacitor which whenconnected in series with a 2 nF capacitorproduces a total capacitance of 1.6 nF.22 Th re e 15 μ F capacitors are connected in seriesacross a 600 V supply. Calculate (a) the totalcapacitance, (b) the p.d. across each and (c) thecharge on each.23 Three capacitors, of 6 μF, 8 μ F and 10 μ Frespectively are connected in parallel across a60 V supply. Calculate (a) the total capacitance,

108 Fundamental Electrical and Electronic PrinciplesAssignment Questions(b) the charge stored in the 8 μ F capacitor and(c) the total charge taken from the supply.24 For the circuit of Fig. 3.29 calculate (a) thep.d. across each capacitor and (b) the chargestored in the 3 nF.4nF3nF6nFC 1 C 220 μF 30μFC 3600 VFig. 3.3220 VAFig. 3.2925 Calculate the values of C 2 and C 3 shown inFig. 3.30 .26 μF 26μF 14 μF8 μF C 2C 3BFig. 3.3340 V 70 V 90 V200 VFig. 3.3026 Calculate the p.d. across, and charge stored in,each of the capacitors shown in Fig. 3.31 .29 A 50 pF capacitor is made up of two platesseparated by a dielectric 2 mm thick and ofrelative permittivity 1.4. Calculate the effectiveplate area.30 For the circuit shown in Fig. 3.34 the totalcapacitance is 16 pF. Calculate (a) the valueof the unmarked capacitor, (b) the charge onthe 10 pF capacitor and (c) the p.d. across the40 pF capacitor.6 μF10 μF16 μF40 pFC50 pF400 VFig. 3.3127 A capacitor circuit is shown in Fig. 3.32 . Withthe switch in the open position, calculate thep.d.s across capacitors C 1 and C 2 . When theswitch is closed, the p.d. across C 2 becomes400 V. Calculate the value of C 3 .28 In the circuit of Fig. 3.33 the variable capacitoris set to 60 μ F. Determine the p.d. acrossthis capacitor if the supply voltage betweenterminals AB is 500 V.100 VFig. 3.3410 pF31 A 2 0 μ F capacitor is charged to a p.d. of 250 V.Calculate the energy stored.32 The energy stored by a 400 pF capacitor is 8 μJ.Calculate the p.d. between its plates.33 Determine the capacitance of a capacitor thatstores 4 mJ of energy when charged to a p.d.of 40 V.

Electric Fields and Capacitors 109Assignment Questions34 When a capacitor is connected across a 200 Vsupply it takes a charge of 8 μ C. Calculate (a) itscapacitance, (b) the energy stored and (c) theelectric field strength if the plates are 0.5 mmapart.35 A 4 μ F capacitor is charged to a p.d. of 400 Vand then connected across an uncharged 2 μ Fcapacitor. Calculate (a) the original charge andenergy stored in the 4 μ F, (b) the p.d. across,and energy stored in, the parallel combination.36 Two capacitors, of 4 μ F and 6 μ F are connectedin series across a 250 V supply. (a) Calculate thecharge and p.d. across each. (b) The capacitorsare now disconnected from the supply andreconnected in parallel with each other, withterminals of similar polarity being joinedtogether. Calculate the p.d. and chargefor each.37 A ceramic capacitor is to be made so that it hasa capacitance of 100 pF and is to be operatedfrom a 750 V supply. Allowing for a safetyfactor, the dielectric has a strength of 500 kV/m.Determine (a) the thickness of the ceramic, (b)the plate area if the relative permittivity of theceramic is 3.2, (c) the charge and energy storedwhen the capacitor is connected to its ratedsupply voltage, and (d) the flux density underthese conditions.38 A large electrolytic capacitor of value 100 μ Fhas an effective plate area of 0.942 m2. If thealuminium oxide film dielectric has a relativepermittivity of 6, calculate its thickness.

110 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentAssignment 1Apparatus:Method:To determine the total capacitance of capacitors, when connected in series, andin parallel.Various capacitors, of known values1 capacitance meter, or capacitance bridge1 Using either the meter or bridge, measure the actual value of each capacitor.2 Connect different combinations of capacitors in parallel, and measure thetotal capacitance of each combination.3 Repeat the above procedure, for various series combinations.4 Calculate the total capacitance for each combination, and compare thesevalues to those previously measured.5 Account for any difference between the actual and nominal values, for theindividual capacitors.

Chapter 4Magnetic Fields andCircuitsLearning OutcomesThis chapter introduces the concepts and laws associated with magnetic fields and theirapplication to magnetic circuits and materials.On completion of this chapter you should be able to:1 Describe the forces of attraction and repulsion between magnetised bodies.2 Understand the various magnetic properties and quantities, and use them to solve simpleseries magnetic circuit problems.3 Appreciate the effect of magnetic hysteresis, and the properties of different types ofmagnetic material.4.1 Magnetic MaterialsAll materials may be broadly classified as being in one of two groups.They may be magnetic or non-magnetic, depending upon the degreeto which they exhibit magnetic effects. The vast majority of materialsfall into the latter group, which may be further classified into diamagneticand paramagnetic materials. The magnetic properties of thesematerials are very slight, and extremely difficult even to detect. Thus,for practical purposes, we can say that they are totally non-magnetic.The magnetic materials (based on iron, cobalt and ferrites) are theferromagnetic materials, all of which exhibit very strong magneticeffects. It is with these materials that we will be principally concerned.4.2 Magnetic FieldsMagnetic fields are produced by permanent magnets and by electriccurrent flowing through a conductor. Like the electric field, a magneticfield may be considered as being the medium by which forces are111

112 Fundamental Electrical and Electronic Principlestransmitted. In this case, the forces between magnetised materials.A magnetic field is also represented by lines of force or magnetic flux.These are attributed with certain characteristics, listed below:1 They always form complete closed loops. Unlike lines of electricflux, which radiate from and terminate at the charged surfaces, linesof magnetic flux also exist all the way through the magnet.2 They behave as if they are elastic. That is, when distorted they try toreturn to their natural shape and spacing.3 In the space surrounding a magnet, the lines of force radiate fromthe north (N) pole to the south (S) pole.4 They never intersect (cross).5 Like poles repel and unlike poles attract each other.Characteristics (1) and (3) are illustrated in Fig. 4.1 which shows themagnetic field pattern produced by a bar magnet.Characteristics (2) and (4) are used to explain characteristic (5), asillustrated in Figs. 4.2 and 4.3 .In the case of the arrangement of Fig. 4.2 , since the lines behave as ifthey are elastic, then those lines linking the two magnets try to shortenthemselves. This tends to bring the two magnets together.NSNSFig. 4.1Fig. 4.2NNFig. 4.3The force of repulsion shown in Fig. 4.3 is a result of the unnaturalcompression of the lines between the two magnets. Once more, actingas if they are elastic, these lines will expand to their normal shape. Thiswill tend to push the magnets apart.Permanent magnets have the advantage that no electrical supply isrequired to produce the magnetic field. However, they also have several

Magnetic Fields and Circuits 113disadvantages. They are relatively bulky. The strength of the fieldcannot be varied. Over a period of time they tend to lose some of theirmagnetism (especially if subjected to physical shock or vibration).For many practical applications these disadvantages are unacceptable.Therefore a more convenient method of producing a magnetic field isrequired.In addition to the heating effect, an electric current also produces amagnetic field. The strength of this field is directly proportional to thevalue of the current. Thus a magnetic field produced in this way maybe turned on and off, reversed, and varied in strength very simply. Amagnetic field is a vector quantity, as indicated by the arrows in theprevious diagrams. The field pattern produced by a current flowingthrough a straight conductor is illustrated in Figs. 4.4(a) and (b). Notethat conventional current flow is considered. The convention adoptedto represent conventional current flowing away from the observeris a cross, and current towards the observer is marked by a dot. Thedirection of the arrows on the flux lines can easily be determined byconsidering the X as the head of a cross-head screw. In order to drivethe screw away from you, the screw would be rotated clockwise . Onthe other hand, if you were to observe the point of the screw comingout towards you, it would be rotating anticlockwise . This convention iscalled the screw rule, and assumes a normal right-hand thread.I(a)XI(b)Fig. 4.4It should be noted that the magnetic flux actually extends the wholelength of the conductor, in the same way that the insulation on acable covers the whole length. In addition, the flux pattern extendsoutwards in concentric circles to infinity. However, as with electricand gravitational fields, the force associated with the field follows aninverse square law. It therefore diminishes very rapidly with distance.The flux pattern produced by a straight conductor can be adapted toprovide a field pattern like a bar magnet. This is achieved by windingthe conductor in the form of a coil. This arrangement is known as asolenoid. The principle is illustrated in Figs. 4.5(a) and (b) , whichshow a cross-section of a solenoid. Figure 4.5(a) shows the fluxpatterns produced by two adjacent turns of the coil. However, sincelines of flux will not intersect, the flux distorts to form complete loopsaround the whole coil as shown in Fig. 4.5(b) .

114 Fundamental Electrical and Electronic PrinciplesN S(a)Fig. 4.5(b)4.3 The Magnetic CircuitA magnetic circuit is all of the space occupied by the magnetic flux.Figure 4.6 shows an iron-cored solenoid, supplied with direct current,and the resulting flux pattern. This is what is known as a compositemagnetic circuit, since the flux exists both in the iron core and in thesurrounding air space. In addition, it can be seen that the spacing ofthe lines within the iron core is uniform, whereas it varies in the airspace. Thus there is a uniform magnetic field in the core and a nonuniformfield in the rest of the magnetic circuit. In order to make thedesign and analysis of a magnetic circuit easier, it is more convenientNSIFig. 4.6if a uniform field can be produced. This may be achieved by the use ofa completely enclosed magnetic circuit. One form of such a circuit isan iron toroid, that has a current carrying coil wound round it. A toroidis a ‘ doughnut ’ shape having either a circular or a rectangular crosssection.Such an arrangement is shown in Fig. 4.7 , and from this it can

Magnetic Fields and Circuits 115INFig. 4.7be seen that only the toroid itself forms the magnetic circuit. Providedthat it has a uniform cross-section then the field contained within it willbe uniform.4.4 Magnetic Flux and Flux DensityThe magnetic flux is what causes the observable magnetic effects suchas attraction, repulsion etc. The unit of magnetic flux is the weber(Wb). This was the name of a German scientist so it is pronounced as‘ vayber ’ .The number of webers of flux per square metre of cross-section of thefield is defined as the magnetic flux density ( B ), which is measuredin tesla (T). This sometimes causes some confusion at first, since thelogical unit would appear to be weber/metre 2 . Indeed, this is the wayin which it is calculated: the value of flux must be divided by theappropriate area. Tesla was the name of another scientist, whose nameis thus commemorated. On reflection, it should not be particularlyconfusing, since the logical unit for electrical current would becoulomb/second; but it seems quite natural to use the term ampere.The quantity symbols for magnetic flux and flux density are Φ and Brespectively. Hence, flux density is given by the equation:B Φ tesla (4.1)ANote: references have been made to iron as a core material and as thematerial used for toroids etc. This does not necessarily mean that pureiron is used. It could be mild steel, cast iron, silicon iron, or ferrite etc.The term ‘ iron circuit ’ , when used in this context, is merely a simpleway in which to refer to that part of the circuit that consists of amagnetic material. It is used when some parts of the circuit may beformed from non-magnetic materials.

116 Fundamental Electrical and Electronic PrinciplesWorked Example 4.1QThe pole face of a magnet is 3 cm by 2 cm and it produces a flux of 30 μ Wb. Calculate the flux densityat the pole face.AA 3 2 10 4 m 2 ; Φ 30 10 6 WbΦB teslaA30 1066104so B 50 mT AnsWorked Example 4.2Q A magnetic field of density 0.6 T has an effective csa of 45 10 6 m 2 . Determine the flux.AB 0.6 T; A 45 10 6 m 2ΦSince B tesla, then Φ BA weberAso Φ 06 . 4510627μWbAns4.5 Magnetomotive Force (mmf)In an electric circuit, any current that flows is due to the existence ofan emf. Similarly, in a magnetic circuit, the magnetic flux is due to theexistence of an mmf. The concept of an mmf for permanent magnets isa difficult one. Fortunately it is simple when we consider the flux beingproduced by current flowing through a coil. This is the case for mostpractical magnetic circuits.In section 4.2 we saw that each turn of the coil made a contribution tothe total flux produced, so the flux must be directly proportional to thenumber of turns on the coil. The flux is also directly proportional to thevalue of current passed through the coil.Putting these two facts together we can say that the mmf is the productof the current and the number of turns. The quantity symbol for mmf isF (the same as for mechanical force). The number of turns is justa number and therefore dimensionless. The SI unit for mmf is thereforesimply ampere. However, this tends to cause considerable confusion tostudents new to the subject. For this reason, throughout this book , theunit will be quoted as ampere turns (At).Thus mmf, F NI ampere turn(4.2)

Magnetic Fields and Circuits 117Worked Example 4.3QA 1500 turn coil is uniformly wound around an iron toroid of uniform csa 5 cm 2 . Calculate the mmfand flux density produced if the resulting flux is 0.2 mWb when the coil current is 0.75 A.AN 1500; A 5 10 4 m 2 ; Φ 0.2 10 3 Wb; I 0.75 AF NI ampere turn15000.75so F 1125At AnsΦ 02 . 10B tesla A 510so B 04 . T Ans34Worked Example 4.4Q Calculate the excitation current required in a 600 turn coil in order to produce an mmfof 1500 At.AN 600; F 1500 AtFsince FNIampere turn, then I ampereN1500so I 25. A Ans6004.6 Magnetic Field StrengthThis is the magnetic equivalent to electric field strength inelectrostatics. It was found that electric field strength is the sameas potential gradient, and is measured in volt/metre. Now, the voltis the unit of emf, and we have just seen that mmf and emf arecomparable quantities, i.e. mmf can be considered as the magneticcircuit equivalent of electric potential. Hence magnetic field strengthis defined as the mmf per metre length of the magnetic circuit.The quantity symbol for magnetic field strength is H , the unit ofmeasurement being ampere turn/metre.F NIH ampere turn/metre (4.3)l lwhere is the mean or average length of the magnetic circuit.Thus, if the circuit consists of a circular toroid, then the meanlength is the mean circumference. This point is illustrated inFigs. 4.8(a) and (b) .

118 Fundamental Electrical and Electronic Principlesrd(a)(b)Fig. 4.8Worked Example 4.5QA current of 400 mA is passed through a 550 turn coil wound on a toroid of mean diameter 8 cm.Calculate the magnetic field strength.AI 0.4 A; N 550; d 8 10 2 m πdmetre π81020.251mNI5500.4H ampere turn/metre 0.251so H 875. 35 At/m Ans4.7 Permeability of Free Space ( μ 0 )We have seen in electrostatics that the permittivity of the dielectric is ameasure of the ‘ willingness ’ of the dielectric to allow an electric fieldto exist in it. In magnetic circuits the corresponding quantity is thepermeability of the material.If the magnetic field exists in a vacuum, then the ratio of the fluxdensity to the magnetic field strength is a constant, called thepermeability of free space.μ 0 B henry/metreH(4.4)Dcompare this to ε0 farad/metreEThe value for μ 4π10 7henry/metre0μ 0 is used as the reference or datum level from which the permeabilitiesof all other materials are measured.

Magnetic Fields and Circuits 1194.8 Relative Permeability ( μ r )Consider an air-cored solenoid with a fixed value of current flowing it.The mmf will produce a certain flux density in this air core. If an ironcore was now inserted, it would be found that the flux density would bevery much increased. To account for these different results for differentcore materials, a quantity known as the relative permeability is used.This is defined as the ratio of the flux density produced in the iron, tothat produced in the air, for the same applied mmf.i.e. μr B B21(4.5)where B 2 is the flux density produced in the iron and B 1 is the fluxdensity produced in the air.Compare this to the equation ε r C 2used in electrostatics.C1As with ε r , μ r has no units, since it is simply a ratio.Note: For air or any other NON-MAGNETIC material, μ r 1.In other words, all non-magnetic materials have the same magneticproperties as a vacuum.4.9 Absolute Permeability (μ )The absolute permeability of a material is the ratio of the flux densityto magnetic field strength, for a given mmf.Thus, μ B henry/metreH(4.6)but since μ 0 is the reference value, then μ μ 0 μ r .compare this to the equation ε ε 0 ε rBTherefore, μμ 0 r so, Bμμ0 r H tesla (4.7)H2This equation compares directly with D ε 0 ε r E coulomb/m .Worked Example 4.6QA solenoid with a core of csa of 15 cm 2 and relative permeability 65, produces a flux of 200 μ Wb. Ifthe core material is changed to one of relative permeability 800 what will be the new flux and fluxdensity?AA 15 10 4 m 2 ; μ r1 65; Φ 1 2 10 4 Wb; μ r2 800

120 Fundamental Electrical and Electronic PrinciplesB1so B1Φ12 1 0teslaA 1510 0.133T44Now, the original core is 65 times more effective than air. The second core is800 times more effective than air. Therefore, we can say that the second corewill produce a greater flux density. The ratio of the two flux densities will be800/65 12.31:1. Thus the second core will result in a flux density 12.31 timesgreater than produced by the first coreThus Bso Band Φ123 . 1 B 123 . 10.1332 12so Φ1. 641T Ans B Aweber1.64115102 22 2.462 mWb Ans4Worked Example 4.7QA toroid of mean radius 40 mm, effective csa 3 cm2, and relative permeability 150, is wound with a900 turn coil that carries a current of 1.5 A. Calculate (a) the mmf, (b) the magnetic field strength and(c) the flux and flux density.Ar 0.04 m; A 3 10 4 m 2 ; μ r 150; N 900; I 1.5 A(a) F NI ampere turn900 1.5so F 1350 At Ans(b)FH ampere turn/metre, where 2πr metre13502π 0.04so, H 5371. 5 At/m Ans(c) BμμH 0 r tesla 4π10 71505371.5so B 1. 0125T AnsΦ BA weber1.0125310so Φ 303.75 μWbAns4Worked Example 4.8Q A steel toroid of the dimensions shown in Fig. 4.9 is wound with a 500 turn coil of wire. What valueof current needs to be passed through this coil in order to produce a flux of 250 μ Wb in the toroid, ifunder these conditions the relative permeability of the toroid is 300.

Magnetic Fields and Circuits 121A 4.5 cm 2r30 cmAr 3 10 2 m ; A 4.5 10 4 m 2 ; N 500; Φ 250 10 6 Wb; μ r 300Effective length of the toroid, l 2 π r metre 2 π 3 10 2 m 0.18 8 mΦ 25010B teslaA 45 . 10so, B 0.556 TNow, Bμμ H tesla0 r64Band, H ampere turns/weberμμ0 rFig. 4.90.5564π 107 300so, H 1 474 At/mFHampere turns14740.188thus, F 277 AtF 277and, I ampN 500so, I 055 . A AnsWorked Example 4.9QA coil is made by winding a single layer of 0.5 mm diameter wire onto a cylindrical wooden dowel,which is 5 cm long and of csa 7 cm2. When a current of 0.2 A is passed through the coil, calculate(a) the mmf produced, (b) the flux density, and (c) the flux produced.A(a) I 0.2 A; l 5 10 2 m ; A 7 10 4 m 2 ; d 0.5 10 3 ; μ r 1 (wood)Since F NI ampere turn, we first need to calculate the number of turns of wireon the coil. Consider Fig. 4.10 which represents the coil wound onto the dowel.From Fig. 4.10 it may be seen that the number of turns may be obtained bydividing the length of the dowel by the diameter (thickness) of the wire. 50Thus, N d 05 .so, N 100Thus, F 1000.2 20At Ans

122 Fundamental Electrical and Electronic Principlesd 0.5 mm5 cmFig. 4.10(b)ΦB tesla, or Bμμ0 r H teslaAbut since we do not yet know the value for the flux, but can calculate the valuefor H , then the second equation needs to be used.F20H ampere turn/metre 5 102so, H 400 At/mand, B 4π 1071400 5.02610so, B 503 μTAns4(c)Φ BA weber 50310710thus, Φ 0. 352 μWbAns6 44.10 Magnetisation (B/H) CurveA magnetisation curve is a graph of the flux density produced ina magnetic circuit as the magnetic field strength is varied. SinceH NI / , then for a given magnetic circuit, the field strength may bevaried by varying the current through the coil.If the magnetic circuit consists entirely of air, or any other non-magneticmaterial then, the resulting graph will be a straight line passing throughthe origin. The reason for this is that since μ r 1 for all non-magneticmaterials, then the ratio B /H remains constant. Unfortunately, therelative permeability of magnetic materials does not remain constant forall values of applied field strength, which results in a curved graph.This non-linearity is due to an effect known as magnetic saturation.The complete explanation of this effect is beyond the scope of thisbook, but a much simplified version of this afforded by Ewing ’smolecular theory. This states that each molecule in a magnetic materialmay be considered as a minute magnet in its own right. When thematerial is unmagnetised, these molecular magnets are orientated in acompletely random fashion. Thus, the material has no overall magneticpolarisation. This is similar to a conductor in which the free electronsare drifting in a random manner. Thus, when no emf is applied, nocurrent flows. This random orientation of the molecular magnets isillustrated in Fig. 4.11 where the arrows represent the north poles.

Magnetic Fields and Circuits 123However, as the coil magnetisation current is slowly increased, sothe molecular magnets start to rotate towards a particular orientation.This results in a certain degree of polarisation of the material, asshown in Fig. 4.12 . As the coil current continues to be increased, sothe molecular magnets continue to become more aligned. Eventually,the coil current will be sufficient to produce complete alignment. Thismeans that the flux will have reached its maximum possible value.Further increase of the current will produce no further increase offlux. The material is then said to have reached magnetic saturation, asillustrated in Fig. 4.13 .un-magnetisedpartially magnetisedFig. 4.11Fig. 4.12saturationFig. 4.13Typical magnetisation curves for air and a magnetic material are shownin Fig. 4.14. Note that the flux density produced for a given value of His very much greater in the magnetic material. The slope of the graph isB / H μ 0 μ r , and this slope varies. Since μ 0 is a constant, then the valueof μ r for the magnetic material must vary as the slope of the graphvaries.B(T)magnetic materialair0H (At/m)Fig. 4.14

124 Fundamental Electrical and Electronic PrinciplesThe variation of μ r with variation of H may be obtained from theB / H curve, and the resulting μ r / H graph is shown in Fig. 4.15 . Themagnetisation curves for a range of magnetic materials is given inFig. 4.16 .μrOFig. 4.15H (At/m)1.61.4B (T)MildsteelStalloy1.2Radiometal1.00.8CastironMumetal0.60.4Ferrite0.20 1000 2000 3000 4000 5000 6000 7000 8000Fig. 4.16H (At/m)

Magnetic Fields and Circuits 125For a practical magnetic circuit, a single value for μ r cannot bespecified unless it is quoted for a specified value of B or H . Thus B / Hdata must be available. These may be presented either in the form of agraph as in Fig. 4.14 , or in the form of tabulated data, from which therelevant section of the B / H curve may be plotted.Worked Example 4.10Q An iron toroid having a mean radius of 0.1 m and csa of π cm 2 is wound with a 1000 turn coil. The coilcurrent results in a flux of 0.1775 mWb in the toroid. Using the following data, determine (a) the coilcurrent and (b) the relative permeability of the toroid under these conditions.AH(At/m) 80 85 90 95 100B(T) 0.50 0.55 0.58 0.59 0.6The first step in the solution of the problem is to plot the section of B /H graphfrom the given data.Note: This must be plotted as accurately as possible on graph paper. The valuesused in this example have been obtained from such a graph.Φ 0.177510B teslaA π 104so B 0.565 T3and from the plotted graph, when B 0.565 T, H 88 At/m.(a)Now, the length of the toroid, 2 π r metre 0.27 π mNIand H ampere turn/metrelHl 88so I amp 0. 2πN 1000hence I 553. mA Ans(b) Bμμ H tesla0 rB 0.565so μr μ H 4π 107 88hence μ 5109Ansr0Worked Example 4.11Q A cast iron toroid of mean length 15 cm is wound with a 2500 turn coil, through which a magnetisingcurrent of 0.3 A is passed. Calculate the resulting flux density and relative permeability of the toroidunder these conditions.ASince B / H data are necessary for the solution, but none have been quoted, thenthe B / H curve for cast iron shown in Fig. 4.16 will be used. 0.15 m; N 2500; I 0.3 A

126 Fundamental Electrical and Electronic PrinciplesNI25000.3H ampere turn/metre l 0.15so H 5000 At/mand from the graph for cast iron in Fig. 4.16 , the corresponding flux density isB 075 . T AnsBut, Bμμ 0 rHteslaB075 .so μr μ H 4 070 π 1 5000μ 119. 4Ansr4.11 Composite Series Magnetic CircuitsMost practical magnetic circuits consist of more than one materialin series. This may be deliberate, as in the case of an electric motoror generator, where there have to be air gaps between the stationaryand rotating parts. Sometimes an air gap may not be required, but themethod of construction results in small but unavoidable gaps.In other circumstances it may be a requirement that two or moredifferent magnetic materials form a single magnetic circuit.Let us consider the case where an air gap is deliberately introducedinto a magnetic circuit. For example, making a sawcut through a toroid,at right angles to the flux path.Worked Example 4.12Q A mild steel toroid of mean length 18.75 cm and csa 0.8 cm 2 is wound with a 750 turn coil. (a) Calculatethe coil current required to produce a flux of 112 μ Wb in the toroid. (b) If a 0.5 mm sawcut is now madeacross the toroid, calculate the coil current required to maintain the flux at its original value.AI 0.1875 m; A 8 10 5 m 2 ; N 750; Φ = 112 10 6 Wb; gap 0.5 10 3 m(a)Φ 11210B tesla A 81065S o B 1.4 T, and from the graph for mild steel in Fig. 4.16 , the correspondingvalue for H is 2000 At/mFso FFeFe Hl ampere turn20000.1875 375 AtFe is the chemical symbol for iron. In Worked Example 4.12 the mmfrequired to produce the flux in the ‘ iron ’ part of the circuit has been referredto as F Fe . This will distinguish it from the mmf required for the air gap whichis shown as F gap

Magnetic Fields and Circuits 127FI Fe 375amp N 750so I 05 . A Ans(b)When the air gap is introduced into the steel the effective length ofthe steel circuit changes by only 0.27%. This is a negligible amount, sothe values obtained in part (a) above for H and F Fe remain unchanged.However, the introduction of the air gap will produce a considerablereduction of the circuit flux. Thus we need to calculate the extra mmf, andhence current, required to restore the flux to its original value.Since the relative permeability for air is a constant ( 1), then a B / H graphis not required. The csa of the gap is the same as that for the steel, andthe same flux exists in it. Thus, the flux density in the gap must also be thesame as that calculated in part (a) above. Hence the value of H required tomaintain this flux density in the gap can be calculated from:so Hand Halso, since Fthen FBμ Hgapgap0 gapB1.4 ampere turn/metre μ4π 10701.1110 Hgap gap gapgapl6At/m111 . 1060.51 03557AtTotal circuit mmf, F F F375557Fegapso, F 932AtF 932I N 750so I 1. 243 A AnsWorked Example 4.13Q A magnetic circuit consists of two stalloy sections A and B as shown in Fig. 4.17 . The mean lengthand csa for A are 25 cm and 11.5 cm 2 , whilst the corresponding values for B are 15 cm and 12 cm 2respectively. A 1000 turn coil wound on section A produces a circuit flux of 1.5 mWb. Calculatethe coil current required.ABFig. 4.17

128 Fundamental Electrical and Electronic PrinciplesA A 0.25 m; A A 11.5 10 4 m 2 ; B 0.15 mA B 12 10 4 m 2 ; Φ 1.5 10 3 Wb; N 1000BAΦΦ tesla and BB teslaAAA1.51031.5 1011.51041210so B 1. 3T and B 1.25TABBFrom the B / H curve for stalloy in Fig. 4.16 , the corresponding H values are:HA1470 At/m and HB845At/mFA HAI A ampere turn/metre, and FB HBIBampere turn/metre14700. 25 8450.15F 367. 5AtF 126. 75AtAtotal circuit mmf, FF FAso, F 494.25AtF 494.25and I N 1000so I 0.494 A AnsB34BFrom the last two examples it should now be apparent that in a seriesmagnetic circuit the only quantity that is common to both (all) sectionsis the magnetic flux Φ . This common flux is produced by the currentflowing through the coil, i.e. the total circuit mmf F . Also, if thelengths, csa and/or the materials are different for the sections, thentheir flux densities and H values must be different. For these reasons itis not legitimate to add together the individual H values. It is correct,however, to add together the individual mmfs to obtain the total circuitmmf F . This technique is equivalent to adding together the p.d.s acrossresistors connected in series in an electrical circuit. The sum of thesep.d.s then gives the value of emf required to maintain a certain currentthrough the circuit.For example, if a current of 4 A is to be maintained through tworesistors of 10 Ω and 20 Ω connected in series, then the p.d.s would be40 V and 80 V respectively. Thus, the emf required would be 120 V.4.12 Reluctance (S)Comparisons have already been made between the electric circuit andthe magnetic circuit. We have compared mmf to emf; current to flux ;and potential gradient to magnetic field strength. A further comparisonmay be made, as follows.The resistance of an electric circuit limits the current that can flow fora given applied emf. Similarly, in a magnetic circuit, the flux producedby a given mmf is limited by the reluctance of the circuit. Thus,

Magnetic Fields and Circuits 129the reluctance of a magnetic circuit is the opposition it offers to theexistence of a magnetic flux within it.Current is a movement of electrons around an electric circuit. A magnetic flux merelyexists in a magnetic circuit; it does not involve a fl ow of particles. However, both currentand fl ux are the direct result of some form of applied forceemfIn an electric circuit, current resistancemmfso in a magnetic circuit, flux reluctanceThus, Φ FSNIS weberFSo reluctance, S ΦNIamp turn/weberΦ(4.8)but NI HHso S Φalso, Φ BAHso S BAH 1 1and since B μ μ μ0 rthen S amp turn/metreμμA0 r(4.9)Let us continue the comparison between series electrical circuitsand series magnetic circuits. We know that the total resistance in theelectrical circuit is obtained simply by adding together the resistorvalues. The same technique may be used in magnetic circuits, such thatthe total reluctance of a series magnetic circuit, S is given byS S S S… (4.10)1 2 3Assume that the physical dimensions of the sections, and the relativepermeabilities (for the given operating conditions) of each sectionare known. In this case, equations (4.9), (4.10) and (4.8) enable analternative form of solution.Worked Example 4.142Q An iron ring of csa 8 cm and mean diameter 24 cm, contains an air gap 3 mm wide. It is requiredto produce a flux of 1.2 mWb in the air gap. Calculate the mmf required, given that the relativepermeability of the iron is 1200 under these operating conditions.

130 Fundamental Electrical and Electronic PrinciplesAA Fe A gap 8 10 4 m 2 ; Φ 1.2 10 3 ; μ r 1200; gap 3 10 3 m; Fe 0.24 π mFor the iron circuit: SFor the air gap: SFelFe ampere turn/Wbμμ A0 r024 . π4π101200810so S 625 . 105At/Wbso SFegapgapl gapμμ A0 r7 431034π1018107 42. 984 10 6 At/WbTotal circuit reluctance, SSFeSgap625 . 1052984 . 106so S 36 . 110 6 AtFSince Φ weberSthen FΦ S ampere turn (compare to V IRvolt)so F 4331 At AnsThe above example illustrates the quite dramatic increase of circuitreluctance produced by even a very small air gap. In this example, theair gap length is only 0.4% of the total circuit length. Yet its reluctanceis almost five times greater than that of the iron section. For thisreason, the design of a magnetic circuit should be such as to try tominimise any unavoidable air gaps.Worked Example 4.15Q A magnetic circuit consists of three sections, the data for which is given below. Calculate (a) the circuitreluctance and (b) the current required in a 500 turn coil, wound onto section 1, to produce a flux of 2 mWb.Section length (cm) csa (cm 2 ) μ r1 85 10 6002 65 15 9503 0.1 12.5 1A(a)so Sl1S1 ampere turn/weberμμ A10 r1085 .4 10 π 600 107 31.12710 6 At/Wb

Magnetic Fields and Circuits 131S2so SS2so Sl2μμ A0 r2065 .4 10 π 950 15 107 4363 . 105At/Wb33l3μμ A0 r31034 10 π 1 12.5 107 4637 . 105At/WbTotal reluctance, S S S S12 36 5 51. 2710 3. 6310 6.3710so S 2.1310 6 At/Wb Ans(b) F Φ S ampere turn 2 10 3 2.13 10 6so F 4254 AtF 4254and I amp N 500so I 85 . 1 A AnsMagnetic flux, like most other things in nature tends to take the easiestpath available. For flux this means the lowest reluctance path. This isillustrated in Fig. 4.18 . The reluctance of the soft iron bar is very muchless than the surrounding air. For this reason, the flux will opt to distortfrom its normal pattern, and make use of this lower reluctance path.copperSNsoftironplasticFig. 4.184.13 Comparison of Electrical, Magnetic and ElectrostaticQuantitiesAlthough a number of comparisons have already been made in the text,it is useful to list these comparisons on one page. This helps to put thethree systems into context, and Table 4.1 serves this purpose.

132 Fundamental Electrical and Electronic PrinciplesTable 4.1 Comparison of QuantitiesElectrical Magnetic ElectrostaticQuantity Symbol Unit Quantity Symbol Unit Quantity Symbol Unitemf E V mmf F At emf E Vcurrent I A fl ux Φ Wb fl ux Q Cresistance R Ω reluctance S At/Wb resistance R Ωresistivity ρ Ωm permeability μ H/m permittivity ε F/mpotentialgradientcurrentdensity— V/m fi eldstrengthH At/m fi eldstrengthJ A/m 2 fl ux density B T fl ux density D C/m 2Also listed below are some comparable equationsE IR volt; F ΦS ampere turnJ I A amp/metre2 ; B Φ A tesla; D Q A coulomb/metre2R R 1 R 2 … ohm; S S 1 S 2 … ampere turn/weberEV/mpotential gradient E d volt/metre; H F lampere turn/metre;E V d volt/metreB μ 0 μ r H tesla; D ε 0 ε r E coulomb/metre 2μ rBB21D2; ε r orD1CC21μ r 1 for all non-magnetic materals; ε r 1 air onlyμ 0 4π 10 7 henry/metre; ε 0 8.854 10 12 farad/metreNote: Although the concept of current density has not been coveredpreviously, it may be seen from Table 4.1 that it is simply the value ofcurrent flowing through a conductor divided by the csa of the conductor.4.14 Magnetic HysteresisHysteresis comes from a Greek word meaning ‘ to lag behind ’ . It isfound that when the magnetic field strength in a magnetic material isvaried, the resulting flux density displays a lagging effect.Consider such a specimen of magnetic material that initially iscompletely unmagnetised. If no current flows through the magnetisingcoil then both H and B will initially be zero. The value of H isnow increased by increasing the coil current in discrete steps. Thecorresponding flux density is then noted at each step. If these values

Magnetic Fields and Circuits 133B(T)ACHD0GH (At/m)FEBFig. 4.19are plotted on a graph until magnetic saturation is achieved, the dottedcurve (the initial magnetisation curve) shown in Fig. 4.19 results.Let the current now be reduced (in steps) to zero, and thecorresponding values for B again noted and plotted. This would resultin the section of graph from A to C . This shows that when the currentis zero once more (so H 0), the flux density has not reduced tozero. The flux density remaining is called the remanent flux density(OC). This property of a magnetic material, to retain some flux afterthe magnetising current is removed, is known as the remanance orretentivity of the material.Let the current now be reversed, and increased in the oppositedirection. This will have the effect of opposing the residual flux.Hence, the latter will be reduced until at some value of H it reacheszero (point D on the graph). The amount of reverse magnetic fieldstrength required to reduce the residual flux to zero is known as thecoercive force. This property of a material is called its coercivity.If we now continue to increase the current in this reverse direction, thematerial will once more reach saturation (at point E). In this case it willbe of the opposite polarity to that achieved at point A on the graph.Once again, the current may be reduced to zero, reversed, and thenincreased in the original direction. This will take the graph from pointE back to A, passing through points F and G on the way. Note thatresidual flux density shown as OC has the same value, but oppositepolarity, to that shown as OF. Similarly, coercive force OD OG.In taking the specimen through the loop ACDEFGA we have taken itthrough one complete magnetisation cycle. The loop is referred to as thehysteresis loop. The degree to which a material is magnetised dependsupon the extent to which the ‘ molecular magnets ’ have been aligned.Thus, in taking the specimen through a magnetisation cycle, energymust be expended. This energy is proportional to the area enclosed bythe loop, and the rate (frequency) at which the cycle is repeated.

134 Fundamental Electrical and Electronic PrinciplesMagnetic materials may be subdivided into what are known as ‘ hard ’and ‘ soft ’ magnetic materials. A hard magnetic material is one whichpossesses a large remanance and coercivity. It is therefore one whichretains most of its magnetism, when the magnetising current isremoved. It is also difficult to demagnetise. These are the materialsused to form permanent magnets, and they will have a very ‘ fat ’ loopas illustrated in Fig. 4.20(a) .A soft magnetic material, such as soft iron and mild steel, retains verylittle of the induced magnetism. It will therefore have a relatively ‘ thin ’hysteresis loop, as shown in Fig. 4.20(b) . The soft magnetic materialsare the ones used most often for engineering applications. Examplesare the magnetic circuits for rotating electric machines (motors andgenerators), relays, and the cores for inductors and transformers.B(T)B(T)H (At/m)H (At/m)(a)(b)4.15 Parallel Magnetic CircuitsFig. 4.20When a magnetic circuit is subjected to continuous cycling throughthe loop a considerable amount of energy is dissipated. This energyappears as heat in the material. Since this is normally an undesirableeffect, the energy thus dissipated is called the hysteresis loss. Thus, thethinner the loop, the less wasted energy. This is why ‘ soft ’ magneticmaterials are used for the applications listed above.The BTEC syllabus at this level does not require the treatment ofparallel magnetic circuits, but since many practical circuits take thisform, a brief coverage now follows.We have seen that the magnetic circuit may be treated in much thesame manner as its electrical circuit equivalent. The same is true forparallel circuits in the two systems.Two equivalent circuits are shown in Fig. 4.21 , and from this thefollowing points emerge:1 In the electrical circuit, the current supplied by the source of emfsplits between the two outer branches according to the resistances

Magnetic Fields and Circuits 135Φ 1Φ 2IR 1I 1 I 2R 2EN(a)(b)ΦFig. 4.21offered. In the magnetic circuit, the flux produced by the mmf splitsbetween the outer limbs according to the reluctances offered.2 If the two resistors in the outer branches are identical, the currentsplits equally. Similarly, if the reluctances of the outer limbs are thesame then the flux splits equally between them.However, a note of caution. In the electric circuit it has been assumed thatthe source of emf is ideal (no internal resistance) and that the connectingwires have no resistance. The latter assumption cannot be applied to themagnetic circuit. All three limbs will have a value of reluctance that mustbe taken into account when calculating the total circuit reluctance.Summary of EquationsΦMagnetic flux density: B μμ o r H teslaAMagnetomotive force (mmf): F NI Φ S ampere turnFMagnetic field strength: H NIampere turn/metreBPermeability: μ μ o μ r henry/metreH NI FReluctance: S ampere turn/weberμμAΦ ΦorSeries magnetic circuit: S S S S… ampere turn/weber1 2 3

136 Fundamental Electrical and Electronic PrinciplesAssignment Questions1 The pole faces of a magnet are 4 cm 3 cmand produce a flux of 0.5 mWb. Calculate theflux density.2 A flux density of 1.8 T exists in an air gap ofeffective csa 11 c m 2 . Calculate the value ofthe flux.3 I f a fl ux of 5 mWb has a density of 1.25 T,determine the csa of the field.4 A magnetising coil of 850 turns carriesa current of 25 mA. Determine theresulting mmf.5 It is required to produce an mmf of 1200 Atfrom a 1500 turn coil. What will be therequired current?6 A current of 2.5 A when flowing through acoil produces an mmf of 675 At. Calculate thenumber of turns on the coil.7 A toroid has an mmf of 845 At applied toit. If the mean length of the toroid is 15 cm,determine the resulting magnetic fieldstrength.8 A magnetic field strength of 2500 At/m existsin a magnetic circuit of mean length 45 mm.Calculate the value of the applied mmf.9 Calculate the current required in a 500 turncoil to produce an electric field strengthof 4000 At/m in an iron circuit of meanlength 25 cm.10 A 400 turn coil is wound onto an iron toroid ofmean length 18 cm and uniform csa 4.5 cm 2 .If a coil current of 2.25 A results in a flux of0.5 mWb, determine (a) the mmf, (b) the fluxdensity, (c) the magnetic field strength.11 An air-cored coil contains a flux density of25 mT. When an iron core is inserted the fluxdensity is increased to 1.6 T. Calculate therelative permeability of the iron under theseconditions.12 A magnetic circuit of mean diameter 12 cmhas an applied mmf of 275 At. If the resultingflux density is 0.8 T, calculate the relativepermeability of the circuit under theseconditions.13 A toroid of mean radius 35 mm, effective csa4 c m 2 and relative permeability 200, is woundwith a 1000 turn coil that carries a current of1.2 A. Calculate (a) the mmf, (b) the magneticfield strength, (c) the flux density and (d) theflux in the toroid.14 A magnetic circuit of square cross-section1.5 1.5 cm and mean length 20 cm is woundwith a 500 turn coil. Given the B / H databelow, determine (a) the coil current requiredto produce a flux of 258.8 μ Wb and (b) therelative permeability of the circuit under theseconditions.B(T) 0.9 1.1 1.2 1.3H(At/m) 250 450 600 82515 For the circuit of Question 14 above, a 1.5 mmsawcut is made through it. Calculate thecurrent now required to maintain the flux atits original value.16 A cast steel toroid has the following B / Hdata. Complete the data table for thecorresponding values of μ r and hence plot theμ r / H graph, and (a) from your graph determinethe values of magnetic field strength at whichthe relative permeability of the steel is 520,and (b) the value of relative permeabilitywhen H 1200 At/m.B(T) 0.15 0.35 0.74 1.05 1.25 1.39H(At/m) 250 500 1000 1500 2000 2500μ r17 A magnetic circuit made of radiometal issubjected to a magnetic field strength of5000 At/m. Using the data given in Fig. 4.16 ,determine the relative permeability under thiscondition.18 A magnetic circuit consists of two sectionsas shown in Fig. 4.22 . Section 1 is made ofmild steel and is wound with a 100 turn coil.Section 2 is made from cast iron. Calculatethe coil current required to produce a fluxof 0.72 mWb in the circuit. Use the B/H datagiven in Fig. 4.16 .

Magnetic Fields and Circuits 137Assignment QuestionsSection 120 cm4 cmSection 220 A closed magnetic circuit made from siliconsteel consists of two sections, connected inseries. One is of effective length 42 mm andcsa 85 mm2, and the other of length 17 mmand csa 65 mm2. A 50 turn coil is wound onto the second section and carries a current of0.4 A. Determine the flux density in the 17 mmlength section if the relative permeability ofthe silicon iron under this condition is 3000.8 cm4 cm 3 cmFig. 4.223 cm21 A magnetic circuit of csa 0.45 cm2consistsof one part 4 cm long and μ r of 1200; and asecond part 3 cm long and μ r of 750. A 100turn coil is wound onto the first part and acurrent of 1.5 A is passed through it. Calculatethe flux produced in the circuit.19 A circular toroid of mean diameter 25 cm andcsa 4 cm 2 has a 1.5 mm air gap in it. The toroidis wound with a 1200 turn coil and carries aflux of 0.48 mWb. If, under these conditions,the relative permeability of the toroid is 800,calculate the coil current required.

138 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentsNote: These assignments are qualitative in nature.Assignment 1Apparatus:Method:To compare the effectiveness of different magnetic core materials.1 coil of wire of known number of turns1 d.c. psu1 ammeter1 set of laboratory weights1 set of different ferromagnetic cores, suitable for the coil used.1 Connect the circuit as shown in Fig. 4.23 .2 Adjust the coil current carefully until the magnetic core just holds thesmallest weight in place. Note the value of current and weight.3 Using larger weights, in turn, increase the coil current until each weight isjust held by the core. Record all values of weight and corresponding current.4 Repeat the above procedure for the other core materials.5 Tabulate all results. Calculate and tabulate the force of attraction and mmf ineach case.6 Write an assignment report, commenting on your findings, and comparingthe relative effectiveness of the different core materials.coreAId.c.p.s.u.NweightFig. 4.23Assignment 2Apparatus:To plot a magnetisation curve, and initial section of a hysteresis loop, for amagnetic circuit.1 magnetic circuit of known length, and containing a coil(s) of known numberof turns1 variable d.c. psu1 Hall Effect probe1 ammeter1 DVM

Magnetic Fields and Circuits 139Method:1 Ensure that the core is completely demagnetised before starting.2 Zero the Hall probe, monitoring its output with the DVM.3 Connect the circuit as in Fig. 4.24 .4 Increase the coil current in 0.1 A steps, up to 2 A. Record the DVM reading ateach step.Note: If you ‘ overshoot ’ a desired current setting. DO NOT then reduce thecurrent back to that setting. Record the value actually set, together with thecorresponding DVM reading.5 Reduce the current from 2 A to zero, in 0.1 A steps. Once more, if youovershoot a desired current setting, DO NOT attempt to correct it.6 Reverse the connections to the psu, and increase the reversed current insmall steps until the DVM indicates zero.Note: The Hall effect probe output (as measured by the DVM) represents theflux density in the core. The magnetic field strength, H , may be calculatedfrom NI / .AprobeHallProbeCircuitVd.c.p.s.u.magneticcircuitFig. 4.247 Plot a graph of DVM readings ( B ) versus H .8 Submit a full assignment report.


Chapter 5ElectromagnetismLearning OutcomesThis chapter concerns the principles and laws governing electromagnetic induction and theconcepts of self and mutual inductance.On completion of this chapter you should be able to use these principles to:1 Understand the basic operating principles of motors and generators.2 Carry out simple calculations involving the generation of voltage, and the production offorce and torque.3 Appreciate the significance of eddy current loss.4 Determine the value of inductors, and apply the concepts of self and mutual inductance tothe operating principles of transformers.5 Calculate the energy stored in a magnetic field.6 Explain the principle of the moving coil meter, and carry out simple calculations for theinstrument.7 Describe the operation of a wattmeter and simple ohmmeter.5.1 Faraday ’s Law of Electromagnetic InductionIt is mainly due to the pioneering work of Michael Faraday, in thenineteenth century, that the modern technological world exists aswe know it. Without the development of the generation of electricalpower, such advances would have been impossible. Thus, although theconcepts involved with electromagnetic induction are very simple, theyhave far-reaching influence. Faraday ’ s law is best considered in twointerrelated parts:1 The value of emf induced in a circuit or conductor is directlyproportional to the rate of change of magnetic flux linking with it.2 The polarity of such an emf, induced by an increasing flux, isopposite to that induced by a decreasing flux.141

142 Fundamental Electrical and Electronic PrinciplesThe key to electromagnetic induction is contained in part one of thelaw quoted above. Here, the words ‘ rate of change ’ are used. If there isno change in flux, or the way in which this flux links with a conductor,then no emf will be induced. The proof of the law can be very simplydemonstrated. Consider a coil of wire, a permanent bar magnet and agalvanometer as illustrated in Figs. 5.1 and 5.2 .Fig. 5.1Fig. 5.2Consider the magnet being moved so that it enters the centre ofthe coil. When this is done it will be seen that the pointer of thegalvo deflects in one direction. This deflection of the pointer is onlymomentary, since it only occurs whilst the magnet is moving. Thegalvo is of course a current measuring device. However, any currentflowing through it must be due to a voltage between its terminals. Sincethere is no other source of emf in the circuit, then it must be concludedthat an emf has been induced or created in the coil itself. The resultingcurrent indicated by the galvo depends on the value of this emf. It willalso be observed that when the magnet is stationary (either inside oroutside the coil) the galvo does not deflect. Hence, emf is induced intothe coil only when the magnet is in motion.

Electromagnetism 143When the magnet is withdrawn from the coil, the galvo will again beseen to deflect momentarily. This time, the deflection will be in theopposite direction. Provided that the magnet is removed at the samerate as it was inserted, then the magnitudes of the deflections will bethe same. The polarities of the induced emfs will be opposite to eachother, since the current flow is reversed. Thus far, we have confirmationthat an emf is induced in the coil when a magnetic flux is movingrelative to it. We also have confirmation of part two of the law.In order to deduce the relationship between the value of inducedemf and the rate of change of flux, the magnet needs to be moved atdifferent speeds into and out of the coil. When this is done, and theresulting magnitudes of the galvo deflection noted, it will be found thatthe faster the movement, the greater the induced emf.This simple experiment can be further extended in three ways. If themagnet is replaced by a more powerful one, it will be found that forthe same speed of movement, the corresponding emf will be greater.Similarly, if the coil is replaced with one having more turns, then for agiven magnet and speed of movement, the value of the emf will again befound to be greater. Finally, if the magnet is held stationary within thecoil, and the coil is then moved away, it will be found that anemf is once more induced in the coil. In this last case, it will also befound the emf has the same polarity as obtained when the magnet wasfirst inserted into the stationary coil. This last effect illustrates the pointthat it is the relative movement between the coil and the flux that inducesthe emf.The experimental procedure described above is purely qualitative.However, if it was refined and performed under controlled conditions,then it would yield the following results:The magnitude of the induced emf is directly proportional to the valueof magnetic flux, the rate at which this flux links with the coil, and thenumber of turns on the coil. Expressed as an equation we have:Ne d Φvoltdt(5.1)Notes:1 The symbol for the induced emf is shown as a lower-case letter e.This is because it is only present for the short interval of timeduring which there is relative movement taking place, and so hasonly a momentary value.2 The term d Φ /d t is simply a mathematical means of stating ‘ therate of change of flux with time ’ . The combination NΦ/ d t is oftenreferred to as the ‘ rate of change of flux linkages ’ .3 The minus sign is a reminder that Lenz ’ s law applies. This law isdescribed in the next section.

144 Fundamental Electrical and Electronic Principles4 Equation (5.1) forms the basis for the definition of the unit ofmagnetic flux, the weber, thus:The weber is that magnetic flux which, linking a circuit of one turn,induces in it an emf of one volt when the flux is reduced to zero at auniform rate in one second.In other words, 1 volt 1 weber/second or 1 weber 1 volt second.5.2 Lenz ’ s LawThis law states that the polarity of an induced emf is always such that itopposes the change which produced it. This is similar to the statementin mechanics, that for every force there is an opposite reaction.5.3 Fleming ’s Righthand RuleThis is a convenient means of determining the polarity of an induced emfin a conductor. Also, provided that the conductor forms part of a completecircuit, it will indicate the direction of the resulting current flow.The first finger, the second finger and the thumb of the right hand areheld out mutually at right angles to each other (like the three edges ofa cube as shown in Fig. 5.3 ). The F irst finger indicates the direction ofthe F lux, the thu M b the direction of M otion of the conductor relativeto the flux, and the s EC ond finger indicates the polarity of the inducedE mf, and direction of Current flow. This process is illustrated inFig. 5.4 , which shows the cross-section of a conductor beingthuMbfirst fingerseCond fingerFig. 5.3vxFig. 5.4

Electromagnetism 145moved vertically upwards at a constant velocity through the magneticfield.Note: The thumb indicates the direction of motion of the conductorrelative to the flux. Thus, the same result would be obtained from thearrangement of Fig. 5.4 if the conductor was kept stationary and themagnetic field was moved down.Worked Example 5.1Q The flux linking a 100 turn coil changes from 5 mWb to 15 mWb in a time of 2 ms. Calculate the averageemf induced in the coil; see Fig. 5.5 .Φ(mWb)1510dΦ5dt0 1Fig. 5.52t(ms)AN 100; d Φ (15 5) 10 3 Wb; d t 2 10 3 sNe dΦ 100 ( 15 volt 5) 10dt2103 100 10 102103so e 500 V Ans33Note that if the flux was reduced from 15 mWb to 5 mWb, then theterm shown in brackets above would be 10. The resulting emf wouldbe 500 V. When quoting equation (5.1), the minus sign should alwaysbe included. However, since it is often the magnitude of the inducedemf that is more important, it is normal practice to ignore the minussign in the subsequent calculation. One of the major exceptions to thispractice arises when considering the principles of operation of thetransformer.

146 Fundamental Electrical and Electronic PrinciplesWorked Example 5.2QA 250 turn coil is linked by a magnetic flux that varies as follows: an increase from zero to 20 mWb ina time of 0.05 s; constant at this value for 0.02 s; followed by a decrease to 4 mWb in a time of 0.01 s.Assuming that these changes are uniform, draw a sketch graph (i.e. not to an accurate scale) of thevariation of the flux and the corresponding emf induced in the coil, showing all principal values.AFirstly, the values of induced emf must be calculated for those periods when theflux changes.d Φ 1 (20 0) 10 3 Wb; d t 1 0.05 sd Φ 2 (4 20) 10 3 Wb; d t 2 0.01 sN 1e1 dΦvoltdt1andeN dΦdtvolt1 250 20 03 250 ( 16)10005 .00 . 1so e 100 V and e 400 V1The resulting sketch graph is shown in Fig. 5.6 .22223Φ(mWb)204050 70 80t(ms)e (V)4000t(ms)100Fig. 5.6

Electromagnetism 147Worked Example 5.3Q A coil when linked by a flux which changes at the rate of 0.1 Wb/s, has induced in it an emf of 100 V.Determine the number of turns on the coil.Ae100 V; d Φ 0. 1 Wb/sdtNe dΦd t volteso N d /dtturns 100Φ0.1N 1000AnsNote that the minus sign has been ignored in the calculation. A negativevalue for number of turns makes no sense.5.4 EMF Induced in a Single Straight ConductorConsider a conductor moving at a constant velocity v metre per secondat right angles to a magnetic field having the dimensions shown inFig. 5.7 . The direction of the induced emf may be obtained usingFleming ’ s righthand rule, and is shown in the diagram. Equation (5.1)is applicable, and in this case, the value for N is 1.dVFig. 5.7dΦThus, e volt, and since Φ is constantdtΦthen e volttbut Φ BA weberBAso e t

148 Fundamental Electrical and Electronic Principlesalso, the csa of the field, A d metre2B dso e and sincetdt velocity v, thene B v volt (5.2)The above equation is only true for the case when the conductor ismoving at right angles to the magnetic field. If the conductor movesthrough the field at some angle less than 90°, then the ‘ cutting ’action between the conductor and the flux is reduced. This results ina consequent reduction in the induced emf. Thus, if the conductoris moved horizontally through the field, the ‘ cutting ’ action is zero,and so no emf is induced. To be more precise, we can say that onlythe component of the velocity at 90° to the fl ux is responsible for theinduced emf. In general therefore, the induced emf is given by:eB v sin θ volt(5.3)where v sin θ is the component of velocity at 90° to the field, asillustrated in Fig. 5.8 .θv cosθv sinθvFig. 5.8This equation is simply confirmed by considering the previous twoextremes; i.e. when conductor moves parallel to the flux, θ 0°; sinθ 0; so e 0. When it moves at right angles to the flux, θ 90°;sin θ 1; so we are back to equation (5.2).Note: is known as the effective length of the conductor, since it isthat portion of the conductor that actually links with the flux. The totallength of the conductor may be considerably greater than this, butthose portions that may extend beyond the field at either end will nothave any emf induced.

Electromagnetism 149Worked Example 5.4QA conductor is moved at a velocity of 5 m/s at an angle of 60° to a uniform magnetic field of 1.6 mWb.The field is produced by a pair of pole pieces, the faces of which measure 10 cm by 4 cm. If theconductor length is parallel to the longer side of the field, calculate the emf induced; see Fig. 5.9 .10 cmd60°4 cm5 m/sAFig. 5.9v 5 m/s; u 60°; Φ 1.6 10 3 Wb; l 0.1 m; d 0.04 m (see Fig. 5.9 )Φ 1.6B 103tesla 04 . TA 0. 10.04e Bvsin θ volt 04 . 0.15si n 60so e 0.173V AnsWorked Example 5.5QA conductor of effective length 15 cm, when moved at a velocity of 8 m/s at an angle of 55° to auniform magnetic field, generates an emf of 2.5 V. Determine the density of the field.Al 0.15 m ; v 8 m/s; u 55°; e 2.5 VeeBv sin θ volt, so Bteslavsin θ25 .hence, B 0.158sin55B 2. 543 T AnsWorked Example 5.6QThe axle of a lorry is 2.2 m long, and the vertical component of the Earth ’ s magnetic field density,through which the lorry is travelling, is 38 T. If the speed of the lorry is 80 km/h, then calculate theemf induced in the axle.

150 Fundamental Electrical and Electronic PrinciplesA80l 2.2 m; v 103m/s ; B 38 10 6 T; sin u 1, since u 90°6060eB v sin θ volt38 10 2.2 80 103600so, e 1. 86 mV Ans6 3This section, covering the induction or generation of an emf in aconductor moving through a magnetic field, forms the basis of thegenerator principle. However, most electrical generators are rotatingmachines, and we have so far considered only linear motion of theconductor.Consider the conductor now formed into the shape of a rectangularloop, mounted on to an axle. This arrangement is then rotated betweenthe poles of a permanent magnet. We now have the basis of a simplegenerator as illustrated in Fig. 5.10 .(a)(b)Fig. 5.10The two sides of the loop that are parallel to the pole faces will eachhave an effective length metre. At any instant of time, these sidesare passing through the field in opposite directions. Applying therighthand rule at the instant shown in Fig. 5.10 , the directions ofthe induced emfs will be as marked, i.e. of opposite polarities.However, if we trace the path around the loop, it will be seen thatboth emfs are causing current to flow in the same direction aroundthe loop. This is equivalent to two cells connected in series asshown in Fig. 5.11 .The situation shown in Fig. 5.10 applies only to one instant in onerevolution of the loop (it is equivalent to a ‘ snapshot ’ at that instant).

Electromagnetism 151IeIeRFig. 5.11If we were to plot a graph of the total emf generated in the loop,for one complete revolution, it would be found to be one cycle of asinewave, i.e. an alternating voltage. This result should not come asany surprise though, since the equation for the emf generated in eachside of the loop is e Blv sinθ volt. This very simple arrangementtherefore is the basis of a simple form of a.c. generator or alternator.Exactly the same principles apply to a d.c. generator, but the way inwhich the inherent a.c. voltage is converted into d.c. automatically bythe machine is dealt with in detail in Further Electrical and ElectronicPrinciples .Note: the ‘ ends ’ of the loop attached to the axle do not have emfinduced in them, since they do not ‘ cut ’ the flux. Additionally, currentcan only flow around the loop provided that it forms part of a closedcircuit.5.5 Force on a Current-Carrying ConductorFigure 5.12(a) shows the field patterns produced by two pole pieces, andthe current flowing through the conductor. Since the lines of flux obeythe rule that they will not intersect, then the flux pattern from the poleswill be distorted as illustrated in Fig. 5.12(b) . Also, since the lines of fluxtend to act as if elastic, then they will try to straighten themselves. Thisresults in a force being exerted on the conductor, in the direction shown.(a)(b)FFig. 5.12The direction of this force may be more simply obtained by applyingFleming ’ s lefthand rule. This rule is similar to the righthand rule. Themajor difference is of course that the fingers and thumb of the left hand

152 Fundamental Electrical and Electronic Principlesare now used. In this case, the F irst finger indicates the direction of themain F lux (from the poles). The se C ond finger indicates the directionof C urrent flow. The thu M b shows the direction of the resulting forceand hence consequent M otion. This is shown in Fig. 5.13 .thuMbFirstfingerseCond fingerFig. 5.13Simple experiments can be used to confirm that the force exerted on theconductor is directly proportional to the flux density produced by the polepieces, the value of current flowing through the conductor, and the lengthof conductor lying inside the field. This yields the following equation:Force, F BI newton(5.4)The determination of the effective length of the conductor is exactlythe same as that for the generator principle previously considered.So any conductor extending beyond the main field does not contributeto the force exerted.Equation (5.4) also only applies to the condition when the conductoris perpendicular to the main flux. If it lies at some angle less than 90°,then the force exerted on it will be reduced. Thus, in general, the forceexerted is given byF BI sin θ newton(5.5)Worked Example 5.7Q A conductor of effective length 22 cm lies at right angles to a magnetic field of density 0.35 T. Calculatethe force exerted on the conductor when carrying a current of 3 A.Al 0.22 m; B 0.35 T; I 3 A; θ 90°F BI sin θ newton035 . 3022. 1so, F 023 . 1 N Ans

Electromagnetism 153Worked Example 5.8QA pair of pole pieces 5 cm by 3 cm produce a flux of 2.5 mWb. A conductor is placed in this fieldwith its length parallel to the longer dimension of the poles. When a current is passed throughthe conductor, a force of 1.25 N is exerted on it. Determine the value of the current.If the conductor was placed at 45° to the field, what then would be the force exerted?AΦ 2.5 10 3 Wb; l 0.05 m; d 0.03 m; F 1.25 Ncsa of the field, A 005 . 003 . 1.5103 m2Φ 25 . 103flux density, B tesla 1.667 TA 1.5103and since θ90, then sin θ1FFBIsin θ newton, so I ampB1.25therefore I 15 A Ans1. 6670.05F BIsinθ newton, where θ45so F 1. 667150. 050.707hence, F 0. 884 N AnsThe principle of a force exerted on a current carrying conductoras described above forms the basis of operation of a linear motor.However, since most electric motors are rotating machines, the abovesystem must be modified.5.6 The Motor PrincipleOnce more, consider the conductor formed into the shape of arectangular loop, placed between two poles, and current passed throughit. A cross-sectional view of this arrangement, together with the fluxpatterns produced is shown in Fig. 5.14 .The flux patterns for the two sides of the loop will be in oppositedirections because of the direction of current flow through it. The resultis that the main flux from the poles is twisted as shown in Fig. 5.15 .This produces forces on the two sides of the loop in oppositedirections. Thus there will be a turning moment exerted on the loop, in acounterclockwise direction. The distance from the axle (the pivotal point)is r metre, so the torque exerted on each side of the loop is given byT Fr newton metrebut F BIsin θ newton, and sin θ1

154 Fundamental Electrical and Electronic PrinciplesrFFFig. 5.14Fig. 5.15so torque on each side BIr. Since the torque on each side is exertinga counterclockwise turning effect then the total torque exerted on theloop will beT 2BI r newton metre (5.6)Worked Example 5.9Q A rectangular single-turn loop coil 1.5 cm by 0.6 cm is mounted between two poles, which produce aflux density of 1.2 T, such that the longer sides of the coil are parallel to the pole faces. Determine thetorque exerted on the coil when a current of 10 mA is passed through it.Al 0.015 m; B 1. 2 T; I 10 2 Aradius of rotation, r 0. 006/ 2 0.003 mT 2BIrnewton metre 21. 21020. 0150.003so T 1.08 mNmAnsFrom the above example it may be seen that a single-turn loopproduces a very small amount of torque. It is acknowledged that thedimensions of the coil specified, and the current flowing through it, arealso small. However, even if the coil dimensions were increased by afactor of ten times, and the current increased by a factor of a thousandtimes (to 10 A), the torque would still be only a very modest 0.108 Nm.The practical solution to this problem is to use a multi-turn coil, asillustrated in Fig. 5.16 . If the coil now has N turns, then each side has

Electromagnetism 155rFig. 5.16an effective length of N . The resulting torque will be increased bythe same factor. So, for a multi-turn coil the torque is given byT 2NBI r newton metreThe term 2r in the above expression is equal to the area ‘enclosed ’ by thecoil dimensions, so this is the effective csa A , of the field affecting the coil.Thus, 2r A metre2, and the above equation may be writtenT BANI newton metre (5.7)The principle of using a multi-turn current-carrying coil in a magneticfield is therefore used for rotary electric motors. However, the sameprinciples apply to the operation of analogue instruments known asmoving coil meters. The classic example of such an instrument is theAVO meter, mentioned earlier, in Chapter 1.Worked Example 5.10QThe coil of a moving coil meter consists of 80 turns of wire wound on a former of length 2 cm and radius1.2 cm. When a current of 45 A is passed through the coil the movement comes to rest when the springsexert a restoring torque of 1.4 Nm. Calculate the flux density produced by the pole pieces.AN 80; l 0.02 m; r 0.012 m ; I 45 10 6 ; T 1.4 10 6 NmThe meter movement comes to rest when the deflecting torque exerted on thecoil is balanced by the restoring torque of the springs.T BANInewton metreTso, B teslaANI1.4106002 . 200 . 12804510 so, B 08 . 1 T Ans6

156 Fundamental Electrical and Electronic Principles5.7 Force between Parallel ConductorsWhen two parallel conductors are both carrying current their magneticfields will interact to produce a force of attraction or repulsion betweenthem. This is illustrated in Fig. 5.17 .attractionrepulsionFig. 5.17In order to determine the value of such a force, consider first a singleconductor carrying a current of I ampere. The magnetic field producedat some distance d from its centre is shown in Fig. 5.18 .dIn general, HFig. 5.18NI ampere turn/metrebut in this case, N 1 (one conductor) and 2π d metre (thecircumference of the dotted circle), soHNI 2πd

Electromagnetism157Now, flux density B μ 0 μ r H tesla, and as the field exists in air, thenμ r 1. Thus, the flux density at distance d from the centre is given byIB μo πd tesla……………[ 1]2Consider now two conductors Y and Z carrying currents I 1 and I 2respectively, at a distance of d metres between their centres as inFig. 5.19 .YI 1 I 2ZdFig. 5.19Using equation [1] we can say that the flux density acting on Z due tocurrent I 1 flowing in Y is:B1 I μ0 π dtesla12and the force exerted on Z B 1 I 2 newton, or B 1 I 2 newton per metrelength of Z.Hence, force/metre length acting on Zμ0II1 2 newton2πd4π 107 II 1 22πdso force/metre length acting on Z2 107 II 1 2newton (5.8)dNow, the current I 2 flowing in Z also produces a magnetic field whichwill exert a force on Y. Using the same reasoning as above, it can beshown that:force/metre length acting on Y2 10 7II 1 2newtondso if I 1 I 2 1 A, and d l m, thenforce exerted on each conductor 2 10 7newton

158 Fundamental Electrical and Electronic PrinciplesThis value of force forms the basis for the definition of the ampere,namely: that current, which when maintained in each of two infinitelylong parallel conductors situated in vacuo , and separated one metrebetween centres, produces a force of 2 10 7newton per metre lengthon each conductor.Worked Example 5.11QTwo long parallel conductors are spaced 35 mm between centres. Calculate the force exerted betweenthem when the currents carried are 50 A and 40 A respectively.Ad 0.035 m; I 1 50 A; I 2 40 A2F 107II2 071 21 50 40newtond0.035so F 11. 4 mN AnsWorked Example 5.12QCalculate the flux density at a distance of 2 m from the centre of a conductor carrying a current of1000 A. If the centre of a second conductor, carrying 300 A, was placed at this same distance, whatwould be the force exerted?Ad 2 m ; I 1 1000 A; I 2 300 AμB 0I14π10 7 1000teslad2so, B 0. 628 mT Ans2F 107II 1 22newton 107 300 1000d2so F 30 mN Ans5.8 The Moving Coil MeterMost analogue (pointer-on-scale) instruments rely on three factors fortheir operation: a defl ecting torque; a restoring torque; and a dampingtorque.Deflecting Torque Essentially, a moving coil meter is a currentmeasuring device. The current to be measured is passed through amulti-turn coil suspended between the poles of a permanent magnet.The coil is made from fine copper wire which is wound on to a lightaluminium former . Thus the motor effect, described in section 5.6, isutilised. Since the wire is of small diameter, and the aluminium former

Electromagnetism 159A formeris also very light, then this assembly has very little inertia. This isan essential requirement, to ensure that the instrument is sufficientlysensitive. This means that it can respond to very small deflectingtorques. To illustrate this point, refer back to Worked Example 5.9, inwhich the torque exerted on a small coil with a current of 10 mA wasfound to be only 1.08 μ Nm. In order to improve the sensitivity further,the friction of the coil pivots must be minimised. This is achieved bythe use of jewelled bearings as shown in Fig. 5.20 .coil onformerpointerjewelledbearingFig. 5.20When a current is passed through the coil it will rotate under theinfluence of the deflecting torque. However, if the ‘ starting ’ position ofthe coil is as shown in Fig. 5.21 , then it will finally settle in the verticalposition, regardless of the actual value of the current. The reason isthat the effective perpendicular distance from the pivot point (the termr sin θ in the expression F 2 NBIr sin θ ) decreases to zero at thisposition. Another way to explain this is to consider the forces acting oneach side of the coil. These will always act at right angles to the mainflux. Thus, when the coil reaches the vertical position, these forces areno longer producing a turning effect. This means that the instrument iscapable of indicating only the presence of a current, but not the actual

160 Fundamental Electrical and Electronic PrinciplesFFFFFig. 5.21value. Hence, the deflecting torque (which is dependent upon the valueof the coil current) needs to be counter-balanced by another torque.Restoring Torque This is the counter-balancing torque mentionedabove. It is provided by two contrawound spiral springs, one endof each being connected to the top and bottom respectively, of thespindle that carries the coil former. The greater the current passedthrough the coil, the greater will be the deflecting torque. The restoringtorque provided by the springs will increase in direct proportion to thedeflection applied to them. The pointer, carried by the coil spindle willtherefore move to a point at which the deflecting and restoring torquesbalance each other. The springs also serve as the means of passing thecurrent to and from the coil. This avoids the problem of the coil havingto drag around a pair of trailing leads.The word contrawound means ‘ wound in opposite directions ’ . The reason for windingthem in this way is to prevent the pointer position from being affected by temperaturechanges. If the temperature increases, then both springs tend to expand, by the sameamount. Since one spring is acting to push the pointer spindle in one direction, and theother one in the opposite direction, then the effects cancel outWe now have a system in which the deflection obtained depends uponthe value of the coil current. There are still other problems to overcomethough. One of these is that the deflecting torque, due to a given valueof current, will vary with the coil position. This would have the effectof non-linear deflections for linear increments of current. If we couldensure that the coil always lies at right angles to the field, regardlessof its rotary position, then this problem would be resolved. The way inwhich this is achieved is by the inclusion of a soft iron cylinder insidethe coil former. This cylinder does not touch the former, but causes aradial flux pattern in the air gap in which the coil rotates. This patternresults from the fact that the lines of flux will take the path(s) of leastreluctance, and so cross the gaps between the pole faces and ironcylinder by the shortest possible path, i.e. at 90° to the surfaces. Thiseffect is illustrated in Fig. 5.22 .

Electromagnetism 161Fig. 5.22Damping Torque When current is passed through the coil, thedeflecting torque accelerates the pointer away from the zero position.Now, although the coil and pointer assembly is very light, it will stillhave sufficient inertia to ‘ overshoot ’ its final position on the graduatedscale. It is also likely to under- and overshoot several times beforesettling. To prevent this from happening, the movement needs to beslowed down, or damped. This effect is achieved automatically by thegeneration of eddy currents in the aluminium coil former as it rotatesin the magnetic field. The full description of eddy currents is dealtwith later in this chapter. However, being induced currents means theyare subject to Lenz ’ s law. They will therefore flow in the coil formerin such a direction as to oppose the change that produced them; thatis the rapid deflection of the coil. If the dimensions of the formerare correctly chosen, then the result will be either one very smallovershoot, or the overshoots are just prevented from occurring.In the latter case the instrument is said to be critically damped, or‘ dead beat ’.The main advantages of the moving coil instrument are:1 Good sensitivity: this is due to the low inertia of the coil and pointerassembly. Typically, a current of 50 μA through the coil is sufficientto move the pointer to the extreme end of the scale (full-scaledeflection or fsd).2 Linear scale: from equation (5.7) we know that T BANI .For a given instrument, B,A , and N are fixed values, so T I.Thus the deflecting torque is directly proportional to the coilcurrent.The main disadvantage is the fact that the basic meter movement so fardescribed can be used only for d.c. measurements. If a.c. was appliedto the coil, the pointer would try to deflect in opposite directionsalternately. Thus, if only a moderate frequency such as 50 Hz isapplied, the pointer cannot respond quickly enough. In this case onlya very small vibration of the pointer, about the zero position, might beobserved.The complete arrangement for a moving coil meter is illustrated inFig. 5.23 .

162 Fundamental Electrical and Electronic PrinciplesFig. 5.235.9 Shunts and MultipliersThe basic instrument so far described is capable of measuring onlysmall d.c. currents (up to 50 μ A typically). This obviously severelylimits its usefulness. The figure of 50 μ A quoted is the value of coilfull-scale deflection current ( I fsd ) for an AVO meter. As you willnow be aware, this meter is in fact capable of measuring up to 10 A,by using the current range switch. This is achieved by means of whatare known as shunts. Similarly, this type of meter is also capableof measuring a range of voltages. This is achieved by means ofmultipliers. Both shunts and multipliers are incorporated into theinstrument when it is assembled.5.10 ShuntsA shunt is a device connected in parallel with the meter ’ s movingcoil, in order to extend the current reading range of the instrument.It consists of a very low resistance element, often made from a smallstrip of copper or aluminium. Being connected in parallel with thecoil, it forms an alternative path for current flow. It can thereforedivert excessive current from the coil itself. In this instance, ‘ excessivecurrent ’ means current in excess of that required to produce full-scaledeflection of the pointer. The latter is known as the full-scale deflectioncurrent, which is normally abbreviated to I fsd .The application for a shunt is best illustrated by a worked example.

Electromagnetism 163Worked Example 5.13QA moving coil meter has a coil resistance of 40 Ω , and requires a current of 0.5 mA to produce f.s.d.Determine the value of shunt required to extend the current reading range to 3 A.AA sketch of the appropriate circuit should be made, and such a diagram isshown in Fig. 5.24 .II fsdR c3 A 0.5 mA 40V cI sR sFig. 5.24R C 40 Ω ; I tsd 5 10 4 A; I 3 AVIR volt 510440C fsd Cso V 002 . VCI IIamp 3( 5104)sfsdso I s 2.995 ARsVItherefore, R 667 . m Ω AnssCs002 .ohm2.99955.11 MultipliersFrom the previous worked example, it may be seen that the p.d.across the coil ( V c ) when I fsd flows through it is only 20 mV. This istherefore the maximum voltage that may be applied to the coil. It alsorepresents the maximum voltage measurement that can be made.In order to extend the voltage reading range, a multiplier must beemployed. This is a high value resistance, connected in series with themeter coil. Thus, when a voltage in excess of V c is applied to the meterterminals, the multiplier will limit the current to I fsd .Worked Example 5.14Q Considering the same meter movement specified in Worked Example 5.13, determine the multiplierrequired to extend the voltage reading range to 10 V.AR C 40 Ω ; I fsd 5 10 4 A; V 10 V

164 Fundamental Electrical and Electronic PrinciplesFrom worked Example 5.13, we know that the p.d. across the coil with I fsdflowing is V C 0.02 V. The circuit diagram is shown in Fig. 5.25 .0.5 mAR m R c40I fsdV mV c0.02 VV10 VFig. 5.25V 10Total resistance, R volt I5 10therefore, Rfsdso R 20 kΩbut RR Rohmmc4so R RR20 00040mmc1 996 . kΩ AnsThis problem may also be solved by the following method:VmVVcvolt 100.02so Vm 998 . VVm998 .Rm ohm I5 104fsdtherefore, R m 19. 96kΩ AnsWorked Example 5.15QThe coil of a moving coil multimeter has a resistance of 1.5 kΩ , and it requires a current of 75 Athrough it in order to produce full-scale deflection. Calculate (a) the value of shunt required to enablethe meter to indicate current up to the value of 5 A, and (b) the value of multiplier required to enableit to indicate voltage up to 10 V.AI fsdR c75 μAV cI sR sI5 AFig. 5.26

Electromagnetism 165Coil resistance, R C 1500 Ω ; I fsd 75 10 6 A; (a) I 5 A; (b) V 10 V(a) VC IfsdRCvolt 751061500so, V 0.1125VCI IIamp 5(7510 6 )Sfsdand, IS 4.999 925 AVCRS ohmIS0.11254.999 925and, RS 22.5 m Ω AnsNote: Since the value of R S will be very small, all the digits obtained for I S areretained in the calculation in order to minimise any rounding error.(b)From part (a), V C 0.1125 V, and the circuit is now equivalent to thatshown in Fig. 5.27 .R mR cV mV cI fsd75 μAV10 VThus, V VV100.1125so, VRmmm 9.8875 VVImfsdC9.8875ohm7510Fig. 5.27so, R m 131. 83 k Ω AnsAlternatively, total resistance, R R m R C ohmV10and also, R ohm I75 10fsdso, R 133.33 kΩand, R RR(133. 331. 5)10mthus, 131.83 k Ω AnsR mC663When the voltage range switch on an AVO is rotated, it simply connectsthe appropriate value of multiplier in series with the moving coil.Similarly, when the current range switch is operated, it connects theappropriate value shunt in parallel with the coil. In order to protectthe instrument from an electrical overload, you must always start themeasurement from the highest available range. Progressively lower rangesettings can then be selected, until a suitable deflection is obtained.

166 Fundamental Electrical and Electronic PrinciplesNote: The range setting marked on any instrument indicates the value ofapplied voltage or current that will cause full-scale deflection on that range.5.12 Figure of Merit and Loading EffectWhen a moving coil meter is used as a voltmeter, the total resistancebetween its terminals will depend upon the value of multiplierconnected. This will of course vary with the range selected. This totalresistance is called the input resistance of the instrument ( R in ). Anideal voltmeter would draw zero current from the circuit to which it isconnected. Thus, for a practical instrument, the higher the value of R in ,the closer it approaches the ideal.Since R in changes as the voltage ranges are changed, some other meansof indicating the ‘ quality ’ of the instrument is required. This is thefigure of merit, which is quoted in ohms/volt.R in VI fsdohm; whereV is the voltage range selectedSo,1IfsdRin ohms/volt;Vand1I fsdis the figure of meritThus, the reciprocal of the full-scale deflection current gives the figureof merit for a moving coil instrument, when used as a voltmeter.The lowest current range on an AVO is 50 μ A, which happens to be itsI fsd . Hence, its figure of merit1 2050 106k Ω/VSince R in figure of merit V, then on the 0.3 V range, R in 20 000 0.3 6 k Ω . The figures for other voltage ranges will therefore be:1 V range, R 20 k Ω; 100 V range, R 2M Ω; etcinIt may therefore be seen that the higher the voltage range selected,the closer R in approaches the ideal of infinity. In order to illustrate thepractical significance of this, let us consider another example.inWorked Example 5.16Q A simple potential divider circuit is shown in Fig. 5.28 . The p.d. V 2 is to be measured by an AVO, usingthe 10 V range. Calculate the p.d. indicated by this meter, and the percentage error in this reading.

Electromagnetism 16712 V30 kR 1V70 kR 2V 20VFig. 5.28AFirstly, we can calculate the p.d. developed across R 2 by applying the potentialdivider theory, thusVso, the true value of V22R2R R V 7 volt 12 01 2 1 84 . VHowever, when the voltmeter is connected across R 2 , the circuit is effectivelymodified to that shown in Fig. 5.29 . (Since it is an AVO, switched to the 10 Vrange, then the value for R in is 200 k Ω .)12 VA30 k R 1VB70 kR 2 V 2R in200 kCOVFig. 5.29Rso RBCBCRR 2 in 70 200ohmkΩR R2702in 51.85 kΩand using the potential divider technique:51.85V 2 2 7 65185 30 1. . V;

168 Fundamental Electrical and Electronic Principlesthis is the p.d. that will be indicated by the voltmeter.so,V 2 76 .V AnsThe percentage error in the reading is defined as:indicated valuetrue valueerror 100%true value7. 68.4so, error 1 00% 9. 52%Ans84 .Hence, it can be seen that the meter does not indicate the true p.d.across R 2 , since it indicates a lower value. This is known as the loadingeffect. The reason for this effect is that, in order to operate, the meterhas to draw current from the circuit. The original circuit conditionshave therefore been altered, as shown in Fig. 5.29 .The loading effect does not depend entirely on the value of R in .Thevalue of R in , relative to the resistance of the component whose p.d.is being measured, is equally important. It is left to the reader toverify that, if resistors R 1 and R 2 in Fig. 5.28 were 300 Ω and 700 Ωrespectively, the loading error would be only 0.004%.Digital multimeters have an input resistance of 10 to 20 M Ω . Thisfigure remains sensibly constant regardless of the voltage rangeselected. The loading effect is therefore much less than for an AVO,especially when using the lower voltage ranges. Also, as the indicatedvalues are easier to read, they tend to be used more often than movingcoil meters. However, there are other factors that need to be consideredwhen selecting a meter for a given measurement. These are covered inFurther Electrical and Electronic Principles.Worked Example 5.17QTwo resistors, of 10 k Ω and 47 k Ω respectively, are connected in series across a 10 V d.c. supply. The p.d.developed across each resistor is measured with a moving coil multimeter having a figure of merit of2 0 k Ω /V, and switched to its 10 V range. For each measurement calculate (a) the p.d. indicated by themeter, and (b) the loading error involved in each case.AWith a figure of merit of 20 k Ω /V the meter internal resistance, R in will beR in 20 k Ω 10 200 k ΩWithout the meter connected, the circuit will be as shown in Fig. 5.30 , and theactual p.d. across each resistor will be V 1 and V 2 respectively.R 1 10 k V 1V10 VR 247 kV 2Fig. 5.30

Electromagnetism 169R110 1V1 V 0voltR R1047so, V 1.754 VV121R247 V volt 10R R1047and, V 8.246 V2122(a)With the meter connected across R 1 , the meter will indicate V AB volt asshown in Fig. 5.31 .AV10 VR 1R 210 kB47 kV ABR in200 kFig. 5.31Rso, RVand, VABABABABRinR120010 ohmkΩRin R1210 9.524 kΩRAB9.524 10 Vvolt R R9.524 47AB21.69 V AnsIn this case the voltmeter reading has been rounded up because it would not bepractical to read the scale to two decimal places. Indeed, the indication wouldprobably be read as 1. 7 V.With the meter now connected across R 2 , the circuit is now modified to that asshown in Fig. 5.32 .V 10 VR 110 kBR 247 kV BCR in200 kCRso, Rand,VBCBCBCV BCFig. 5.32RR 2 in 47 200ohmkΩR R2472in 38.057 kΩRBCR RBC1 792 . V Ans38.05710Vvolt 48.057

170 Fundamental Electrical and Electronic Principles(b) error indicated valuetrue valuetrue value 100%so, error for V 1 measurement 1 . 691 . 754 100%1.754hence, error 3.65% Ans7. 928.246error for V 2 measurement 100%8.246and, error 3.95% AnsNote: Although the two error figures are fairly close to each other invalue, the percentage error in the second measurement is about 10%greater than that for the first. This illustrates the fact that the higher themeter internal resistance, compared with the resistance across whichit is placed, the smaller the loading effect, and hence higher accuracyis obtained. Since the internal resistance of DVMs is usually in theorder of megohms, in many cases they are used in preference to thetraditional moving coil instrument.5.13 The OhmmeterAs the name implies, this instrument is used for the measurement ofresistance. This feature is normally included in multimeters. In the caseof the AVO, the moving coil movement is also utilised for this purpose.The theory is based simply on Ohm ’ s law. If a known emf is appliedto a resistor, the resulting current flow is inversely proportional tothe resistance value. The basic arrangement is shown in Fig. 5.33 .The battery is incorporated into the instrument, as is the variableresistance, R.metermovementERFig. 5.33To use the instrument in this mode, the terminals are first ‘ shorted ’together, using the instrument leads. The resistor R is then adjusted sothat the pointer indicates zero on the ohms scale. Note that the zeroposition on this scale is at the righthand extreme of the scale. Having

Electromagnetism 171‘ zeroed ’ the meter, the ‘ short ’ is removed. Finally, the resistance to bemeasured is connected between the terminals. The resistance value willthen be indicated by the position of the pointer on the ohms scale.It is not recommended that resistance values be measured using thistechnique! The scale is extremely cramped over the higher resistancerange, and is extremely non-linear. This makes accurate measurementof resistance virtually impossible. However, the use of this facility forcontinuity checking is useful.If you wish to measure a resistance value fairly accurately, then use thisfacility on a digital multimeter. The same basic principle, of applying aknown emf and measuring the resulting current, may still be employed.The internal electronic circuitry then converts this into a display ofresistance value. By definition, the scale cannot be cramped, and iseasy to read. If a resistance has to be measured with a high degree ofaccuracy, then a Wheatstone Bridge must be used.5.14 WattmeterThis instrument is used to measure electrical power. In its traditionalform it is called a dynamometer wattmeter. It is also available in apurely electronic form. The dynamometer type utilises the motorprinciple. The meter consists of two sets of coils. One set is fixed, andis made in two identical parts. This is the current coil, and is madefrom heavy gauge copper wire. The resistance of this coil is thereforelow. The voltage coil is wound from fine gauge wire, and therefore hasa relatively high resistance. The voltage coil is mounted on a circularformer, situated between the two parts of the current coil. The basicarrangement is illustrated in Fig. 5.34 .voltagecoilcurrentcoilFig. 5.34

172 Fundamental Electrical and Electronic PrinciplesThe current coil is connected to a circuit so as to allow the circuitcurrent to flow through it. The voltage coil is connected in parallel withthe load, in the same way as a voltmeter. Both coils produce magneticfields, which interact to produce a deflecting torque on the voltage coil.Restoring torque is provided by contrawound spiral springs, as in themoving coil meter. The interacting fields are proportional to the circuitcurrent and voltage respectively. The deflecting torque is thereforeproportional to the product of these two quantities; i.e. VI which is thecircuit power.The voltage coil may be connected either on the supply side or theload side of the current coil. The choice of connection depends onother factors concerning the load. This aspect is dealt with in FurtherElectrical and Electronic Principles. The manner of connecting thewattmeter into a circuit is shown in Fig. 5.35 .currentcoilvoltagecoilsupplyloadFig. 5.35For d.c. circuits, a wattmeter is not strictly necessary. Since powerP VI watt, then V and I may be measured separately by a multimeter.The power can then be calculated simply by multiplying together thesetwo meter readings. However, in an a.c. circuit, this simple techniquedoes not yield the correct value for the true power. Thus a wattmeter isrequired for the measurement of power in a.c. circuits.5.15 Eddy CurrentsConsider an iron-cored solenoid, as shown in cross-section in Fig. 5.36 .Let the coil be connected to a source of emf via a switch. When theswitch is closed, the coil current will increase rapidly to some steadyvalue. This steady value will depend upon the resistance of the coil.The coil current will, in turn, produce a magnetic field. Thus, this fluxpattern will increase from zero to some steady value. This changingflux therefore expands outwards from the centre of the iron core. Thismovement of the flux pattern is shown by the arrowed lines pointingoutwards from the core.Since there is a changing flux linking with the core, then an emf willbe induced in the core. As the core is a conductor of electricity, then

Electromagnetism 173Fig. 5.36the induced emf will cause a current to be circulated around it. This isknown as an eddy current, since it traces out a circular path similar tothe pattern created by an eddy of water. The direction of the inducedemf and eddy current will be as shown in Fig. 5.36 . This has beendetermined by applying Fleming ’ s righthand rule. Please note, that toapply this rule, we need to consider the movement of the conductorrelative to the flux. Thus, the effective movements of the left and righthalves of the core are opposite to the arrows showing the expansion ofthe flux pattern.As the eddy current circulates in the core, it will produce a heatingeffect. This is normally an undesirable effect. The energy thusdissipated is therefore referred to as the eddy current loss. If thesolenoid forms part of a d.c. circuit, this loss is negligible. This isbecause the eddy current will flow only momentarily—when the circuitis first connected, and again when it is disconnected. However, if ana.c. supply is connected to the coil, the eddy current will be flowingcontinuously, in alternate directions. Under these conditions, the core isalso being taken through repeated magnetisation cycles. This will resultin a hysteresis loss also.In order to minimise the eddy current loss, the resistance of the coreneeds to be increased. On the other hand, the low reluctance needs to beretained. It would therefore be pointless to use an insulator for the corematerial, since we might just as well use an air core! The technique usedfor devices such as transformers, used at mains frequency, is to makethe core from laminations of iron. This means that the core is made upof thin sheets (laminations) of steel, each lamination being insulatedfrom the next. This is illustrated in Fig. 5.37 . Each lamination, beingthin, will have a relatively high resistance. Each lamination will have aneddy current, the circulation of which is confined to that lamination. If

174 Fundamental Electrical and Electronic PrinciplesFig. 5.37the values of these individual eddy currents are added together, it will befound to be less than that for the solid core.The hysteresis loss is proportional to the frequency f of the a.c. supply.The eddy current loss is proportional to f 2 . Thus, at higher frequencies(e.g. radio frequencies), the eddy current loss is predominant. Underthese conditions, the use of laminations is not adequate, and the eddycurrent loss can be unacceptably high. For this type of application, irondust cores or ferrite cores are used. With this type of material, the eddycurrents are confined to individual ‘ grains ’ , so the eddy current loss isconsiderably reduced.5.16 Self and Mutual InductanceThe effects of self and mutual inductance can be demonstrated byanother simple experiment. Consider two coils, as shown in Fig. 5.38 .Coil 1 can be connected to a battery via a switch. Coil 2 is placedclose to coil 1, but is not electrically connected to it. Coil 2 has a galvoconnected to its terminals.coil 1 coil 2EFig. 5.38When the switch is closed, the current in coil 1 will rapidly increasefrom zero to some steady value. Hence, the flux produced bycoil 1 will also increase from zero to a steady value. This changingflux links with the turns of coil 2, and therefore induces an emf into it.This will be indicated by a momentary deflection of the galvo pointer.

Electromagnetism 175Similarly, when the switch is subsequently opened, the flux producedby coil 1 will collapse to zero. The galvo will again indicate that amomentary emf is induced in coil 2, but of the opposite polarity tothe first case. Thus, an emf has been induced into coil 2, by achanging current (and flux) in coil 1. This is known as a mutuallyinduced emf.If the changing flux can link with coil 2, then it must also link withthe turns of coil 1. Thus, there must also be a momentary emfinduced in this coil. This is known as a self-induced emf. Anyinduced emf obeys Lenz ’ s law. This self-induced emf must thereforebe of the opposite polarity to the battery emf. For this reason, it isalso referred to as a back emf. Unfortunately, it is extremely difficultto demonstrate the existence of this back emf. If a voltmeter wasconnected across coil 1, it would merely indicate the terminal voltageof the battery.5.17 Self-InductanceSelf-inductance is that property of a circuit or component whichcauses a self-induced emf to be produced, when the current through itchanges. The unit of self-inductance is the henry, which is defined asfollows:A circuit has a self-inductance of one henry (1 H) if an emf of one voltis induced in it, when the circuit current changes at the rate of oneampere per second (1 A/s).The quantity symbol for self-inductance is L . From the abovedefinition, we can state the following equationeL d/d i thenryLdior, self-induced emf, e volt (5.9)dtNotes:1 The minus sign again indicates that Lenz ’ s law applies.2 The emf symbol is e, because it is only a momentary emf.3 The current symbol is i, because it is the change of current that isimportant.4 The term di/dt is the rate of change of current.

176 Fundamental Electrical and Electronic PrinciplesWorked Example 5.18QA coil has a self-inductance of 0.25 H. Calculate the value of emf induced, if the current through itchanges from 100 mA to 350 mA, in a time of 25 ms.AL 0.25 H; d I (350 100) 10 3 A; d t 25 10 3 sL ie d 0.25 250 10volt dt25 103so e 25 . V Ans3Worked Example 5.19QCalculate the inductance of a circuit in which an emf of 30 V is induced, when the circuit currentchanges at the rate of 200 A/s.Aie 30 V;d 200 A/SdtL ie dvoltdteso, L d/d i thenry 30200therefore, L 0.15H AnsWorked Example 5.20Q A circuit of self-inductance 50 mH has an emf of 8 V induced into it. Calculate the rate of change of thecircuit current that induced this emf.AL 50 10 3 H; e 8 VL ie dvoltdtdie8So, amp/s dtL50 10dihence 160A/s Ansdt35.18 Self-Inductance and Flux LinkagesConsider a coil of N turns, carrying a current of I amp. Let us assumethat this current produces a flux of Φ weber. If the current nowchanges at a uniform rate of d i /dt ampere per second, it will causea corresponding change of flux of d f /dt weber per second. Let usalso assume that the coil has a self-inductance of L henry.

Electromagnetism 177The self-induced emf may be determined from equation (5.9):L ie ddtvolt……………[ 1]However, the induced emf is basically due to the rate of change of fluxlinkages. Thus, the emf may also be calculated by using equation (5.1),namely:Ne d φvolt……………[ 2]dtSince both equations [1] and [2] represent the same induced emf, then[1] must be equal to [2]. ThusLdiNdφ (the minus signs cancel out)dtdtNdso, L φ henry(5.10)diA coil which is designed to have a specific value of self-inductanceis known as an inductor. An inductor is the third of the main passiveelectrical components. The other two are the resistor and the capacitor.A passive component is one which (a) requires an external source of emf in order toserve a useful function, and (b) does not provide any amplifi cation of current or voltageNow, a resistor will have a specific value of resistance, regardless ofwhether it is in a circuit or not. Similarly, an inductor will have somevalue of self-inductance, even when the current through it is constant.In other words, an inductor does not have to have an emf induced init, in order to possess the property of self-inductance. For this reason,equation (5.10) may be slightly modified as follows.If the current I through an N turn coil produces a flux of Φ weber, thenits self-inductance is given by the equationLN φ henry (5.11)IIn other words, although no change of current and flux is specified,the coil will still have some value of inductance. Strictly speaking,equation (5.11) applies only to an inductor with a non-magneticcore. The reason is that, in this case, the flux produced is directlyproportional to the coil current. However, it is a very closeapproximation to the true value of inductance for an iron-coredinductor which contains an air gap in it.

178 Fundamental Electrical and Electronic PrinciplesWorked Example 5.21QA coil of 150 turns carries a current of 10 A. This current produces a magnetic flux of 0.01 Wb. Calculate(a) the inductance of the coil, and (b) the emf induced when the current is uniformly reversed in a timeof 0.1 s.AN 150; I 10 A; Φ 0.01 Wb ; d t 0.1 s(a)(b)NΦ1500.10L henry I10so, L 1. 5 H AnsSince current is reversed then it will change from 10 A to 10 A, i.e.a change of 10 ( 10). So, d I 20 A.L ie d 1.520volt dt0.1therefore, e 300 V AnsWorked Example 5.22QA current of 8 A, when flowing through a 3000 turn coil, produces a flux of 4 mWb. If the current isreduced to 2 A in a time of 100 ms, calculate the emf thus induced in the coil. Assume that the flux isdirectly proportional to the current.AI 1 8 A; N 3000; Φ 1 4 10 3 Wb; I 2 2 A; d t 0.1 sThis problem may be solved in either of two ways. Both methods will bedemonstrated.L iNe dvolt, where L dtIhenry30004 103so, L 1. 5 H; di826A81.5 6therefore, e 90 V Ans0.1Φ 11Alternatively, Φ I so Φ 1 I 1 and Φ 2 I 2therefore,ΦΦ22hence, Φ2II21and Φ2Φ1II2 4 03 1110 3Wb;8and dφ( 41)103 Ndφ3000310e volt dt0.1so, e 90 V Ans1Wb23

Electromagnetism 1795.19 Factors Affecting InductanceConsider a coil of N turns wound on to a non-magnetic core, of2uniform csa A metre and mean length metre. The coil carries acurrent of I amp, which produces a flux of Φ weber. From equation(5.11), we know that the inductance will beNΦL henry, but Φ BA weberINBAtherefore, L henry……………[ 1]IAlso, magnetic field strength, HNI l; so IH lNand substituting this expression for I into equation [1]NBA BANL Hl/N Hl2……………[ 2]Now, equation [2] contains the term B H , which equals μ oμ r0 rN Atherefore, L μμ 2henryl(5.12)We also know thatlreluctanceμμ 0 r A , SNhence L 2 henry(5.13)SNotes:εεAN1 Equation (5.12) compares with C o r ( 1)farad for acapacitor.d2 If the number of turns is doubled, then the inductance isquadrupled, i.e. L N 2 .3 The terms A and in equation (5.12) refer to the dimensions ofthe core, and NOT the coil.Worked Example 5.232Q A 600 turn coil is wound on to a non-magnetic core of effective length 45 mm and csa 4 cm .(a) Calculate the inductance, (b) The number of turns is increased to 900. Calculate the inductancevalue now produced. (c) The core of the 900 turn coil is now replaced by an iron core having a relativepermeability of 75, and of the same dimensions as the original. Calculate the inductancein this case.

180 Fundamental Electrical and Electronic PrinciplesAN 1 600; 45 10 3 m; A 4 10 4 m 2 ; μ r1 1N 2 900; μ r2 1; N 3 900; μ r3 75(a)μμ N20 r1 1AL1henrylso, L 402 . mH Ans14π1 1 1 0 600 4 0451037 2 4(b) Since L1 N12 , and L N , then2 2 2LLNN2 2 2 21 1LN2 L23 21 4.0210900so, 2 N260021therefore, L 9. 045 mH Ans2(c) Since μ r3 75 μ r2 , and there are no other changes, then L 3 75 L 2therefore, L 3 0.678 H Ans5.20 Mutual InductanceWhen a changing current in one circuit induces an emf in another separatecircuit, then the two circuits are said to possess mutual inductance. Theunit of mutual inductance is the henry, and is defined as follows.Two circuits have a mutual inductance of one henry, if the emf inducedin one circuit is one volt, when the current in the other is changing atthe rate of one ampere per second.The quantity symbol for mutual inductance is M, and expressing theabove definition as an equation we haveinduced emf in coil 2M rate of change of current in coil 1 e 2d i /dthenryand transposing this equation for emf e 21e2M i d 1volt (5.14)dt

Electromagnetism 181This emf may also be expressed in terms of the flux linking coil 2. Ifall of the flux from coil 1 links with coil 2, then we have what is called100% flux linkage. In practice, it is more usual for only a proportionof the flux from coil 1 to link with coil 2. Thus the flux linkage isusually less than 100%. This is indicated by a factor, known as thecoupling factor, k . 100% coupling is indicated by k 1. If there is noflux linkage with coil 2, then k will have a value of zero. So if zeroemf is induced in coil 2, the mutual inductance will also be zero. Thus,the possible values for the coupling factor k , lie between zero and 1.Expressed mathematically, this is written as0k1Consider two coils possessing mutual inductance, and with a couplingfactor 1. Let a change of current d i 1 /d t amp/s in coil 1 produce achange of flux d φ 1 /d t weber/s. The proportion of this flux changelinking coil 2 will be d φ 2 /d t weber/s. If the number of turns on coil 2 isN 2 , thene2N d φdt2 2volt(5.15)However, equations (5.14) and (5.15) both refer to the same inducedemf. Therefore we can equate the two expressionsMdidtand transposing for M , we haveN dφdt1 2 2MN dφdi 2 21henry(5.16)As with self-inductance for a single coil, mutual inductance is aproperty of a pair of coils. They therefore retain this property,regardless of whether or not an emf is induced. Hence, equation (5.16)may be modified toMN Φ henry (5.17)I 2 21

182 Fundamental Electrical and Electronic PrinciplesWorked Example 5.24Q Two coils, A and B, have 2000 turns and 1500 turns respectively. A current of 0.5 A, flowing in A,produces a flux of 60 Wb. The flux linking with B is 83% of this value. Determine (a) the selfinductanceof coil A, and (b) the mutual inductance of the two coils.AN A 2000; N B 1500; I A 0.5 A; Φ A 60 10 6 Wb(a)LAANAΦIso, L 024 . H AnsAA20006010henry 05 .6(b) M N BΦIso, M 0. 149H AnsAB15000.836010henry 05 .65.21 Relationship between Self- and Mutual-InductanceConsider two coils of N 1 and N 2 turns respectively, wound on to acommon non-magnetic core. If the reluctance of the core is S ampereturns/weber, and the coupling coefficient is unity, thentherefore,N1 NL12 and L2SS2 2N1 NLL2 21 2 2……………[ 1]S2MN 2Φ henry, and multiplying byIM1NN 1 2NI1 1The above expression contains the termΦNI1 11SΦNN11so, M NN S1 2N NM2 1 2 2therefore, 2……………[ 2]S2and comparing equations [1] and [2]M2 L L1 2therefore, M L 1 L 2 henry(5.18)

Electromagnetism 183the above equation is correct only provided that there is 100% couplingbetween the coils; i.e. k 1. If k 1, then the general form of theequation, shown below, applies.M k L1L2henry (5.19)Worked Example 5.25Q A 400 turn coil is wound onto a cast steel toroid having an effective length of 25 cm and csa 4.5 cm2.If the steel has a relative permeability of 180 under the operating conditions, calculate the selfinductanceof the coil.AN 400; l 0.25 m ; A 4.5 10 4 m2; μ r 180μμ N AL 0 r 4π10 henry 180 400 4.5 10025 .so, L 65 mH Ans2 7 2 4Worked Example 5.26QConsidering Example 5.25, a second coil of 650 turns is wound over the first, and the current throughcoil 1 is changed from 2 A to 0.5 A in a time of 3 ms. If 95% of the flux thus produced links with coil 2,then calculate (a) the self-inductance of coil 2, (b) the value of mutual inductance, (c) the self-inducedemf in coil 1, and (d) the mutually induced emf in coil 2.AL 1 65 10 3 H; d I 1 2 0.5 1.5 A; d t 3 10 3 s; k 0.95(a)From the equation for inductance we know that L N 2 , and since all otherfactors for the two coils are the same, thenLLso, Land, LNN2 2 2 21 1N2 2 L1 henryN22216502654002mH172mH AnsIt is left to the student to confirm this answer by using the equationN AL μμ 2o r(b) M k L1 L2 henry 095 . 65172mHthus, M 100mH Ans

184 Fundamental Electrical and Electronic Principles(c)edi1 1L dtvolt 65 03. 531031 1so, e 32. 5 V Ans1(d)and, ede M i 1 0052 dtvolt 1 1.31032 50 V Ans5.22 Energy StoredAs with an electric field, a magnetic field also stores energy. When thecurrent through an inductive circuit is interrupted, by opening a switch,this energy is released. This is the reason why a spark or arc occursbetween the contacts of the switch, when it is opened.Consider an inductor connected in a circuit, in which the currentincreases uniformly, to some steady value I amp. This current changeis illustrated in Fig. 5.39 . The magnitude of the emf induced by thischange of current is given bye LItvoltcurrentI0tFig. 5.39timeThe average power input to the coil during this time is:average power eaverage currentFrom the graph, it may be seen that the average current over this timeis I/2 amp.Therefore,1 LII LI2average power eI 2 2t2twatt

Electromagnetism 185but, energy stored average power timepower LI 2t2tjoulethus, energy stored, W 1 LI2joule2(5.20)Equation 5.20 applies to a single inductor. When two coils possessmutual inductance, and are connected in series, both will store energy.In this situation, the total energy stored is given by the equation1 1W L I L I MI I2 21 1 2 2 2 2 1 2joule (5.21)Worked Example 5.27Q Calculate the energy stored in a 50 mH inductor when it is carrying a current of 0.75 A.AL 50 10 3 H; I 0.75 A1W L 250 10 0.75I joule22therefore, W 14. 1mJAns3 2Worked Example 5.28Q Two inductors, of inductance 25 mH and 40 mH respectively, are wound on a common ferromagneticcore, and are connected in series with each other. The coupling coefficient, k, between them is 0.8.When the current flowing through the two coils is 0.25 A, calculate (a) the energy stored in each,(b) the total energy stored when the coils are connected (i) in series aiding, and (ii) in series opposition.AL 1 25 10 3 H; L 2 40 10 3 H; I 1 I 2 0.25 A; k 0.81(a) W1 L1I 21 joule 0. 5251030.2522so, W1 078 . mJ Ans1W2 L2I 22 joule 0. 5401030.2522so, W 1.25 mJ Ans2(b)The general equation for the energy stored by two inductors with fluxlinkage between them is:W 1 L 111 I2L2I2 2 MI1I2joule2 2

186 Fundamental Electrical and Electronic PrinciplesWhen the coils are connected in series such that the two fluxes produced actin the same direction, the total flux is increased and the coils are said to beconnected in series aiding. In this case the total energy stored in the systemwill be increased, so the last term in the above equation is added, i.e. the signapplies. If, however, the connections to one of the coils are reversed, then thetwo fluxes will oppose each other, the total flux will be reduced, and the coilsare said to be in series opposition. In this case the sign is used. These twoconnections are shown in Fig. 5.40 .L 1L 2IΦ 1Φ 2aidingL 1L 2Φ 1 Φ 2IIopposingFig. 5.40IM k L1 L2 henry 0.8 2540mHso, M 25.3 mHThe values for 1 L211 I and 1 L22 2 I 22 have been calculated in part (a) andMI1I225. 31030. 2521. 58 mJ(i)For series aiding:W 078 . 1. 251.58mJso, W 36 . mJ Ans(ii)For series opposition:W 078 . 1. 251.58mJso, W 045 . mJ Ans5.23 The Transformer PrincipleA transformer is an a.c. machine, which utilises the mutual inductancebetween two coils, or windings. The two windings are wound on to

Electromagnetism 187a common iron core, but are not electrically connected to each other. Thepurpose of the iron core is to reduce the reluctance of the magnetic circuit.This ensures that the flux linkage between the coils is almost 100%.a.c. means alternating current, i.e. one which fl ows alternately, fi rst in one direction, thenin the opposite direction. It is normally a sinewaveSince it is an a.c. machine, an alternating flux is produced in the core.The core is therefore laminated, to minimise the eddy current loss.Indeed, the transformer is probably the most efficient of all machines.Efficiencies of 98% to 99% are typical. This high efficiency is duemainly to the fact that there are no moving parts.The general arrangement is shown in Fig. 5.41 . One winding, called theprimary, is connected to an a.c. supply. The other winding, the secondary,is connected to a load . The primary will draw an alternating current I 1from the supply. The flux, Φ , produced by this winding, will thereforeA load is any device or circuit connected to some source of emf. Thus, a load will drawcurrent from the source. The term load is also loosely used to refer to the actual currentdrawn from a sourcealso be alternating; i.e. it will be continuously changing. Assuming100% flux linkage, then this flux is the only common factor linking thetwo windings. Thus, a mutually induced emf, E 2 , will be developedacross the secondary. Also, there will be a back emf, E l , induced acrossΦI 1 I 2V 1 E 1 E 2 R L V 2N 1 N 2Fig. 5.41

188 Fundamental Electrical and Electronic Principlesthe primary. If the secondary is connected to a load, then it will cause thesecondary current I 2 to flow. This results in a secondary terminal voltage,V 2 . Figure 5.42 shows the circuit symbol for a transformer.N 1 : N 2V 2V 1Fig. 5.425.24 Transformer Voltage and Current RatiosLet us consider an ideal transformer. This means that the resistanceof the windings is negligible, and there are no core losses due tohysteresis and eddy currents. Also, let the secondary be connected to apurely resistive load, as shown in Fig. 5.43 .I 1N 1 :N 2 I 2V 2V1 E 1 E 2R LFig. 5.43Under these conditions, the primary back emf, E l , will be of the samemagnitude as the primary applied voltage, V 1 . The secondary terminalvoltage, V 2 , will be of the same magnitude as the secondary inducedemf, E 2 . Finally, the output power will be the same as the input power.The two emfs are given byE1N1 d Φvolt, and Edt2N2 dΦvolt so,dt

Electromagnetism 189dΦdtand d ΦdtENEN1122……………[]1……………[ 2]Since both equations [1] and [2] refer to the same rate of change of fluxin the core, then [1] [2]:hence,ENEE1112ENNN2212and since E 1 V 1 , and E 2 V 2 , thenVV12N1 (5.22)N2From this equation, it may be seen that the voltage ratio is the sameas the turns ratio. This is perfectly logical, since the same flux linksboth windings, and each induced emf is directly proportional to itsrespective number of turns. This is the main purpose of the transformer.It can therefore be used to ‘ step up ’ or ‘ step down ’ a.c. voltages,depending upon the turns ratio selected.Worked Example 5.29Q A transformer is to be used to provide a 60 V output from a 240 V a.c. supply. Calculate (a) the turnsratio required, and (b) the number of primary turns, if the secondary is wound with 500 turns.AV 2 6 0 V; V 1 240 V; N 2 500(a)VV1 12 2N14so, turns ratio, or 4 : 1 AnsN 12N N24060(b)therefore, NN14500 11 2000 AnsSince the load is purely resistive, then the output power, P 2 , is given byPand the input power, P V I2 2 21 V I1 1wattwatt

190 Fundamental Electrical and Electronic PrinciplesAlso since the transformer has been considered to be 100% efficient(no losses), thenPtherefore, VIII P2 1 VI2 2 1 212VV21butVV21NN21hence, I I12N2 (5.23)N1i.e. The current ratio is the inverse of the turns ratio.This result is also logical. For example, if the voltage was ‘stepped up ’by the ratio N 2 /N 1 , then the current must be ‘ stepped down ’ by the sameratio. If this was not the case, then we would get more power out thanwas put in! Although this result would be very welcome, it is a physicalimpossibility. It would require the machine to be more than 100% efficient.Worked Example 5.30Q A 15 Ω resistive load is connected to the secondary of a transformer. The terminal p.d. at the secondaryis 240 V. If the primary is connected to a 600 V a.c. supply, calculate (a) the transformer turns ratio, (b)the current and power drawn by the load, and (c) the current drawn from the supply. Assume an idealtransformer.AR L 15 Ω ; V 2 240 V; V 1 600 VThe appropriate circuit diagram is shown in Fig. 5.43 .(a)(b)N1 V1600 N2 V2240so, turns ratio, N / N 25 . : Ans1 2 1V2240I2 ohmRL15so, I216A AnsP2 V2I2 watt 24016therefore, P 2 384 . k W Ans(c) P1 P2 384. kWand, P1V1I1 wattP13840therefore, I1 amp V 600hence, I 1 64 . A Ans1

Electromagnetism 191Alternatively, using the inverse of the turns ratio:N2 16I1I2 N 1 25 .so, I 64 . A Ans1Summary of EquationsSelf-induced emf: eN d φd tvoltEmf in a straight conductor: e Bυ sin φ voltForce on a current carrying conductor: F BI sin φ voltMotor principle: T BANI newton metre2 IIForce between current carrying conductors: F 10 71 2newtondVoltmeter figure of merit:1I fsdohm/voltdSelf-inductance: Self-induced emf, e L i dtEnergy stored: W 0.5 LI 2 joulevoltdφNΦL N henryd i IL NSμμN A o rhenry2 2dMutual inductance: Mutually induced emf, e M i 12 voltdtd2N ΦM N2 di I12 21henryM k L1L2 henryEnergy stored: W 0.5L I 0.5 L I MI I joule1 1 2 2 2 2 1 2Transformer: Voltage ratio, V VCurrent ratio, I I2121NNNN2122

192 Fundamental Electrical and Electronic PrinciplesAssignment Questions1 The flux linking a 600 turn coil changes uniformlyfrom 100 mWb to 50 mWb in a time of 85 ms.Calculate the average emf induced in the coil.rotation2 An average emf of 350 V is induced in a 1000turn coil when the flux linking it changes by200 μ Wb. Determine the time taken for theflux to change.SN3 A fl ux of 1.5 mWb, linking with a 250 turncoil, is uniformly reversed in a time of 0.02 s.Calculate the value of the emf so induced.4 A coil of 2000 turns is linked by a magneticflux of 400 μ Wb. Determine the emf induced inthe coil when (a) this flux is reversed in 0.05 s,and (b) the flux is reduced to zero in 0.15 s.5 When a magnetic flux linking a coil changes,an emf is induced in the coil. Explain thefactors that determine (a) the magnitude ofthe emf, and (b) the direction of the emf.6 State Lenz ’ s law, and hence explain the term‘ back emf ’ .7 A coil of 15 000 turns is required to producean emf of 15 kV. Determine the rate of changeof flux that must link with the coil in order toprovide this emf.8 A straight conductor, 8 cm long, is movedwith a constant velocity at right angles toa magnetic field. If the emf induced in theconductor is 40 mV, and its velocity is 10 m/s,calculate the flux density of the field.9 A conductor of effective length 0.25 m ismoved at a constant velocity of 5 m/s, througha magnetic field of density 0.4 T. Calculate theemf induced when the direction of movementrelative to the magnetic field is (a) 90°, (b) 60°,and (c) 45°.10 Figure 5.44 represents two of the armatureconductors of a d.c. generator, rotating in aclockwise direction. Copy this diagram andhence:(a) Indicate the direction of the field pattern ofthe magnetic poles.An armature is the rotating part of a d.c. machine.If the machine is used as a generator, it containsthe coils into which the emf is induced. In the caseof a motor, it contains the coils through whichcurrent must be passed, to produce the torqueFig. 5.44(b) Indicate the direction of induced emf ineach side of the coil.(c) If this arrangement was to be used as amotor, with the direction of rotation asshown, indicate the direction of currentflow required through the coil.11 A conductor of effective length 0.5 m is placedat right angles to a magnetic field of density0.45 T. Calculate the force exerted on theconductor if it carries a current of 5 A.12 A conductor of effective length 1.2 m is placedinside a magnetic field of density 250 mT.Determine the value of current flowingthrough the conductor, if a force of 0.75 N isexerted on the conductor.13 A conductor, when placed at right angles to amagnetic field of density 700 mT, experiencesa force of 20 mN, when carrying a current of200 mA. Calculate the effective length of theconductor.14 A conductor, 0.4 m long, lies between twopole pieces, with its length parallel to the polefaces. Determine the force exerted on theconductor, if it carries a current of 30 A, andthe flux density is 0.25 T.15 The coil of a moving coil meter is woundwith 75 turns, on a former of effective length2.5 cm, and diameter 2 cm. The former rotatesat right angles to the field, which has a fluxdensity of 0.5 T. Determine the deflectingtorque when the coil current is 50 μ A.16 A moving coil meter has a coil of 60 turnswound on to a former of effective length22.5 mm and diameter 15 mm. If the fluxdensity in the air gap is 0.2 T, and the coilcurrent is 0.1 mA. Calculate (a) the force actingon each side of the coil, and (b) the restoringtorque exerted by the springs for the resultingdeflection of the coil.

Electromagnetism 193Assignment Questions17 Two long parallel conductors, are spaced12 cm between centres. If they carry 100 Aand 75 A respectively, calculate the force permetre length acting on them. If the currentsare flowing in opposite directions, will this bea force of attraction or repulsion? Justify youranswer by means of a sketch of the magneticfield pattern produced.18 The magnetic flux density at a distance of 1.4 mfrom the centre of a current carrying conductoris 0.25 mT. Determine the value of the current.19 A moving coil meter has a coil resistance of2 5 Ω , and requires a current of 0.25 mA toproduce full-scale deflection. Determine thevalues of shunts required to extend its currentreading range to (a) 10 mA, and (b) 1 A. Sketchthe relevant circuit diagram.20 For the meter movement described in question19 above, show how it may be adapted to serveas a voltmeter, with voltage ranges of 3 V and10 V. Calculate the values for, and name, anyadditional components required to achieve this.Sketch the relevant circuit diagram.21 Explain what is meant by the term ‘ loading effect ’ .22 A voltmeter, having a figure of merit of 15 k Ω /volt, has voltage ranges of 0.1 V, 1 V, 3 V and10 V. If the resistance of the moving coil is 30 Ω ,determine the multiplier values required foreach range. Sketch a circuit diagram, showinghow the four ranges could be selected.23 Figure 5.45 shows a circuit in which the p.d.across resistor R 2 is to be measured. Thevoltmeter available for this measurement hasa figure of merit of 20 k Ω /V, and has voltageranges of 1 V, 10 V and 100 V. Determine thepercentage error in the voltmeter reading,when used to measure this p.d.90 V0 VFig. 5.45R 1R 25 k5 k24 Calculate the self-inductance of a 700 turncoil, if a current of 5 A flowing through itproduces a flux of 8 mWb.25 A coil of 500 turns has an inductance of 2.5 H.What value of current must flow through it inorder to produce a flux of 20 mWb?26 When a current of 2.5 A flows through a0.5 H inductor, the flux produced is 80 μWb.Determine the number of turns.27 A 1000 turn coil has a flux of 20 mWb linking itwhen carrying a current of 4 A. Calculate thecoil inductance, and the emf induced when thecurrent is reduced to zero in a time of 25 ms.28 A coil has 300 turns and an inductance of5 mH. How many turns would be required toproduce an inductance of 0.8 mH, if the samecore material were used?29 If an emf of 4.5 V is induced in a coil having aninductance of 200 mH, calculate the rate ofchange of current.30 An iron ring having a mean diameter of 300 mmand cross-sectional area of 500 mm 2 is woundwith a 150 turn coil. Calculate the inductance, ifthe relative permeability of the ring is 50.31 An iron ring of mean length 50 cm and csa0.8 cm 2 is wound with a coil of 350 turns. Acurrent of 0.5 A through the coil produces aflux density of 0.6 T in the ring. Calculate (a)the relative permeability of the ring, (b) theinductance of the coil, and (c) the value of theinduced emf if the current decays to 20% ofits original value in 0.01 s, when the current isswitched off.32 When the current in a coil changes from 2 A to12 A in a time of 150 ms, the emf induced intoan adjacent coil is 8 V. Calculate the mutualinductance between the two coils.33 The mutual inductance between two coilsis 0.15 H. Determine the emf induced in onecoil when the current in the other decreasesuniformly from 5 A to 3 A, in a time of 10 ms.34 A coil of 5000 turns is wound on to a nonmagnetictoroid of csa 100 cm , and mean2circumference 0.5 m. A second coil of 1000turns is wound over the first coil. If a currentof 10 A flows through the first coil, determine(a) the self-inductance of the first coil, (b)the mutual inductance, assuming a couplingfactor of 0.45, and (c) the average emf inducedin the second coil if interruption of the currentcauses the flux to decay to zero in 0.05 s.

194 Fundamental Electrical and Electronic PrinciplesAssignment Questions35 Two air-cored coils, A and B, are wound with100 and 500 turns respectively. A current of 5 Ain A produces a flux of 15 μ Wb. Calculate (a)the self-inductance of coil A, (b) the mutualinductance, if 75% of the flux links with B, and(c) the emf induced in each of the coils, whenthe current in A is reversed in a time of 10 ms.36 Two coils, of self-inductance 50 mH and 85 mHrespectively, are placed parallel to each other.If the coupling coefficient is 0.9, calculate theirmutual inductance.37 The mutual inductance between two coilsis 250 mH. If the current in one coil changesfrom 14 A to 5 A in 15 ms, calculate (a) the emfinduced in the other coil, and (b) the change offlux linked with this coil if it is wound with 400turns.38 The mutual inductance between the twowindings of a car ignition coil is 5 H. Calculatethe average emf induced in the high tensionwinding, when a current of 2.5 A, in the lowtension winding, is reduced to zero in 1 ms. Youmay assume 100% flux linkage between thetwo windings.39 Sketch the circuit symbol for a transformer,and explain its principle of operation. Why isthe core made from laminations? Is the corematerial a ‘ hard ’ or a ‘ soft ’ magnetic material?Give the reason for this.40 A transformer with a turns ratio of 20:1 has240 V applied to its primary. Calculate thesecondary voltage.41 A 4 :1 voltage ‘ step-down ’ transformer isconnected to a 110 V a.c. supply. If the currentdrawn from this supply is 100 mA, calculate thesecondary voltage, current and power.42 A transformer has 450 primary turns and 80secondary turns. It is connected to a 240 Va.c. supply. Calculate (a) the secondaryvoltage, and (b) the primary current when thetransformer is supplying a 20 A load.43 A coil of self-inductance 0.04 H has a resistanceof 15 Ω . Calculate the energy stored when it isconnected to a 24 V d.c. supply.44 The energy stored in the magnetic field of aninductor is 68 mJ, when it carries a current of1.5 A. Calculate the value of self-inductance.45 What value of current must flow through a 20 Hinductor, if the energy stored in its magneticfield, under this condition, is 60 J?

Electromagnetism 195Suggested Practical AssignmentsNote: The majority of these assignments are only qualitative in nature.Assignment 1Apparatus:Method:To investigate Faraday ’ s laws of electromagnetic induction.Several coils, having different numbers of turns2 permanent bar magnets1 galvanometer1 Carry out the procedures outlined in section 5.1 at the beginning of thischapter.2 Write an assignment report, explaining the procedures carried out, andstating the conclusions that you could draw from the observed results.Assignment 2Apparatus:Method:Assignment 3Apparatus:Method:Force on a current carrying conductor.1 current balance1 variable d.c. psu1 ammeter1 Assemble the current balance apparatus.2 Adjust the balance weight to obtain the balanced condition, prior toconnecting the psu.3 With maximum length of conductor, and all the magnets in place, varythe conductor current in steps. For each current setting, re-balance theapparatus, and note the setting of the balance weight.4 Repeat the balancing procedure with a constant current, and maximummagnets, but varying the effective length of the conductor.5 Repeat once more, this time varying the number of magnets. The currentmust be maintained constant, as must the conductor length.6 Tabulate all results obtained, and plot the three resulting graphs.7 Write an assignment report. This should include a description of theprocedures carried out, and conclusions drawn, regarding the relationshipsbetween the force produced and I, , and B.To investigate the loading effect of a moving coil meter.1 10 k Ω rotary potentiometer, complete with circular scale1 d.c. psu1 Heavy Duty AVO1 DVM (digital voltmeter)1 Connect the circuit as shown in Fig. 5.46 , with the psu set to 10 V d. c.2 Start with the potentiometer moving contact at the ‘ zero ’ end. Measure thep.d. indicated, in turn, by both the DVM and the AVO, for every 30° rotationof the moving contact.Note: Do NOT connect both meters at the same time; connect them IN TURN.

196 Fundamental Electrical and Electronic PrinciplesSuggested Practical Assignments10 Vp.s.u0VFig. 5.463 Tabulate both voltmeter readings. Plot graphs (on the same axes) for thevoltage readings versus angular displacement.4 Determine the percentage loading error of the AVO, for displacements of 0°,180°, and 270°. Write an assignment report, and include comment regardingthe variation of loading error found.Assignment 4Apparatus:Method:Assignment 5Apparatus:Method:To demonstrate mutual inductance and coupling coefficient.Several coils, having different numbers of turns.Ferromagnet core1 galvo1 d.c. psu1 Place the two coils as close together as possible. Connect the galvo to onecoil, and connect the other coil to the psu via a switch.2 Close the switch, and note the deflection obtained on the galvo.3 Repeat this procedure for increasing distances of separation, and fordifferent coils.4 Mount two of the coils on a common magnetic core, and repeat the procedure.5 Write an assignment report, explaining the results observed.To determine the relationship between turns ratio and voltage ratio for a simpletransformer.Either 1 single-phase transformer with tappings on both windings;or Several different coils with a ferromagnetic core.Either a low voltage a.c. supply;or 1 a.c. signal generator.1 DVM (a.c. voltage ranges)1 Connect the primary to the a.c. source.2 Measure both primary and secondary voltages, and note the correspondingnumber of turns on each winding.3 Vary the number of turns on each winding, and note the correspondingvalues of the primary and secondary voltages.4 Tabulate all results. Write a brief report, explaining your findings.

Chapter 6Alternating QuantitiesLearning OutcomesThis chapter deals with the concepts, terms and definitions associated with alternatingquantities. The term alternating quantities refers to any quantity (current, voltage, flux, etc.),whose polarity is reversed alternately with time. For convenience, they are commonly referredto as a.c. quantities. Although an a.c. can have any waveshape, the most common waveformis a sinewave. For this reason, unless specified otherwise, you may assume that sinusoidalwaveforms are implied.On completion of this chapter you should be able to:1 Explain the method of producing an a.c. waveform.2 Defi ne all of the terms relevant to a.c. waveforms.3 Obtain values for an a.c., both from graphical information and when expressed inmathematical form.4 Understand and use the concept of phase angle.5 Use both graphical and phasor techniques to determine the sum of alternating quantities.6.1 Production of an Alternating WaveformFrom electromagnetic induction theory, we know that the average emfinduced in a conductor, moving through a magnetic field, is given bye B v sin θ volt ⋯⋯⋯⋯⋯[ 1]Where B is the flux density of the field (in tesla) is the effective length of conductor (in metre)v is the velocity of the conductor (in metre/s)u is the angle at which the conductor ‘ cuts ’ the lines ofmagnetic flux (in degrees or radians)i.e. v sin u is the component of velocity at right angles to the flux.197

198 Fundamental Electrical and Electronic PrinciplesConsider a single-turn coil, rotated between a pair of poles, asillustrated in Figs. 6.1(a) and (b) . Figure (a) shows the generalarrangement. Figure (b) shows a cross-section at one instant in time,such that the coil is moving at an angle of u° to the flux.WZXNS(a)Y(b)Fig. 6.1Considering Fig. 6.1(b) , each side of the coil will have the same valueof emf induced, as given by equation [1] above. The polarities of theseemfs will be as shown (Fleming ’ s righthand rule). Although these emfsare of opposite polarities, they both tend to cause current to flow in thesame direction around the coil. Thus, the total emf generated is given by:e2 B v sin θ volt ⋯⋯⋯⋯⋯[2]Still considering Fig. 6.1(b) , at the instant the coil is in the plane W Y,angle u 0°. Thus the emf induced is zero. At the instant that it is inthe plane X Z, u 90°. Thus, the emf is at its maximum possiblevalue, given by:e2 B v volt ⋯⋯⋯⋯⋯[3]Let us consider just one side of the coil, starting at position W. After90° rotation (to position X), the emf will have increased from zero to itsmaximum value. During the next 90° of rotation (to position Y), the emffalls back to zero. During the next 180° rotation (from Y to Z to W), theemf will again increase to its maximum, and reduce once more to zero.However, during this half revolution, the polarity of the emf is reversed.If the instantaneous emf induced in the coil is plotted, for one completerevolution, the sinewave shown in Fig. 6.2 will be produced. Forconvenience, it has been assumed that the maximum value of the coilemf is 1 V, and that the plot starts with the coil in position W.

Alternating Quantities 199e(V)1.0E ppE mFig. 6.20.51 cycle030 60 90 120 150 210 240 270 300 3301 cyclerotation(deg)0.51.0When the coil passes through one complete revolution, the waveformreturns to its original starting point. The waveform is then said to havecompleted one cycle. Note that one cycle is the interval between any twocorresponding points on the waveform. The number of cycles generatedper second is called the frequency, f, of the waveform. The unit forfrequency is the hertz (Hz). Thus, one cycle per second is equal to 1 Hz.For the simple two-pole arrangement considered, one cycle of emfis generated in one revolution. The frequency of the waveform istherefore the same as the speed of rotation, measured in revolution persecond (rev/s). This yields the following equationf np hertz (6.1)where p the number of pole pairsTherefore, if the coil is rotated at 50 rev/s, the frequency will bef50 1 Hz ( one pair of poles) 50HzThe time taken for the waveform to complete one cycle is called theperiodic time, T . Thus, if 50 cycles are generated in one second, thenone cycle must be generated in 1/50 of a second. The relationshipbetween frequency and period is therefore

200 Fundamental Electrical and Electronic PrinciplesT 1 second, or f 1 hertzfT(6.2)The maximum value of the emf in one cycle is shown by the peaks ofthe waveform. This value is called, either the maximum or peak value,or the amplitude of the waveform. The quantity symbol used may beeither Ê , or E m .The voltage measured between the positive and negative peaks is calledthe peak-to-peak value. This has the quantity symbol E pkpk , or E pp .6.2 Angular Velocity and FrequencyIn SI units, angles are measured in radians , rather than degrees.Similarly, angular velocity is measured in radian per second, ratherthan revolutions per second. The quantity symbol for angular velocityis ω (lower case Greek omega).A radian is the angle subtended at the centre of a circle, by an arc on the circumference,which has length equal to the radius of the circle. Since the circumference 2 πr, thenthere must be 2 π such arcs in the circumference. Hence there are 2 π radians in onecomplete circle; i.e. 2 π rad 360°If the coil is rotating at n rev/s, then it is rotating at 360 n degrees/second. Since there are 2π radians in 360°, then the coil must be rotatingat 2 π n radian per second.Thus, angular velocity, ω 2 πn rad/sbut for a 2-pole system, n frequency, f hertz,therefore, ω 2πf rad/s(6.3)and,f ω 2πhertz(6.4)If the coil is rotating at ω rad/s, then in a time of t seconds, it willrotate through an angle of ω t radian. Hence the waveform diagram maybe plotted to a base of degrees, radians, or time. In the latter case, thetime interval for one cycle is, of course, the periodic time, T . These areshown in Fig. 6.3 .6.3 Standard Expression for an Alternating QuantityAll the information regarding an a.c. can be presented in the formof a graph. The information referred to here is the amplitude,

Alternating Quantities 201emf (V)0T/4/2T/2 3 T/4 T t (s) 3 /2 2 t (rad)Fig. 6.3frequency, period, and value at any instant. The last is normallycalled the instantaneous value. However, presenting the informationin this way is not always very convenient. Firstly, the graph has to beplotted accurately, on graph paper. This in itself is a time-consumingprocedure. In addition, obtaining precise information from the graphis difficult. The degree of accuracy depends on the suitability ofthe scales chosen, and the individual ’ s interpretation. For example,if several people are asked to obtain a particular value from thegraph, their answers are likely to differ slightly from one another. Toovercome these difficulties, the a.c. needs to be expressed in a moreconvenient form. This results in an equation, sometimes referred to asthe algebraic form of the a.c. More correctly, it should be called thetrigonometric form. Since many students are put off by these terms, weshall refer to it simply as the standard expression for a waveform.The emf for an N -turn coil is:e2NBvsin θ volt, where θ is in degreesor, e2NBvsin( ωt) volt, where ωtis in radiansand the emf is at its maximum value when sin( ω t ), or sin u is equalto 1. Therefore, E m 2 NBv volt, so the expression becomes:e sin θ voltE mor, eE sin( ωt) voltm(6.5)(6.6)or, e Emsin( 2π ft) volt(6.7)All three of the above equations are the so-called standard expressionsfor this a.c. voltage. Equations (6.6), and (6.7) in particular, arethose most commonly used. Using these, all the relevant informationconcerning the waveform is contained in a neat mathematicalexpression. There is also no chance of ambiguity.

202 Fundamental Electrical and Electronic PrinciplesWorked Example 6.1QAn alternating voltage is represented by the expression v 35 sin(314.2 t ) volt. Determine, (a) themaximum value, (b) the frequency, (c) the period of the waveform, and (d) the value 3.5 ms after itpasses through zero, going positive.A(a)v 35 sin(314.2t) voltand comparing this to the standard,vVmsin( 2πft) volt we can see that:V 35 V Ansm(b)Again, comparing the two expressions:2πf 314.2314.2so, f 50Hz Ans2π(c)1 1T secondf 50so, T 20 ms Ans(d) When t 35 . ms; then:v 35 sin( 2π 503.510 35 sin( 1.099)*350.891therefore, v 311. 9 V Ans3) volt*The term inside the brackets is an angle in RADIAN. You must thereforeremember to switch your calculator into the RADIAN MODE.So far, we have dealt only with an alternating voltage. However, allof the terms and definitions covered are equally applicable to anyalternating quantity. Thus, exactly the same techniques apply to a.c.currents, fluxes, etc. The same applies to mechanical alternatingquantities involving oscillations, vibrations, etc.Worked Example 6.2Q For a current, i 75 sin(200 π t ) milliamp, determine (a) the frequency, and (b) the time taken for it toreach 35 mA, for the first time, after passing through zero.A(a) i75 sin( 200πt) milliampiImsin( 2πft) ampso, 2πf 200π200πand f 100Hz Ans2π

Alternating Quantities 203(b) 3575 sin( 200πt) milliamp35sin( 200πt ) 0.466775therefore, 200πt sin10.4667* 0.4855 rad0.4855so, t 0. 773 ms Ans200π* Remember , use RADIAN mode on your calculator.6.4 Average ValueFigure 6.4 shows one cycle of a sinusoidal current.0t (s)Fig. 6.4From this it is apparent that the area under the curve in the positive halfis exactly the same as that for the negative half. Thus, the average valueover one complete cycle must be zero. For this reason, the averagevalue is taken to be the average over one half cycle. This average maybe obtained in a number of ways. These include, the mid-ordinate rule,the trapezoidal rule, Simpson ’ s rule, and integral calculus. The simplestof these is the mid-ordinate rule, and this will be used here to illustrateaverage value; see Fig. 6.5 .I mI avt (s)i ni 1 i 3Fig. 6.5A number of equally spaced intervals are selected, along the time axis ofthe graph. At each of these intervals, the instantaneous value is determined.

204 Fundamental Electrical and Electronic PrinciplesThis results in values for a number of ordinates, i 1 , i 2 , …, i n , where n is thenumber of ordinates chosen. The larger the number of ordinates chosen,the more accurate will be the final average value obtained. The average issimply found by adding together all the ordinate values, and then dividingthis figure by the number of ordinates chosen, thusi1 i i iIav 2 3 ⋯ nnThe average value will of course depend upon the shape of thewaveform, and for a sinewave only it is2Iav Im 0. 637 Im(6.8)πWorked Example 6.3Q A sinusoidal alternating voltage has an average value of 3.5 V and a period of 6.67 ms. Write down thestandard (trigonometrical) expression for this voltage.AV av 3.5 V; T 6.67 10 3 sThe standard expression is of the form v V m sin (2 π ft) voltVav 0.637 VmvoltVav35 .so, Vm volt 0.637 0.637Vm 55 . V11f hertzHzT 667 . 103and, f 150Hzv55 . sin ( 2π150t) voltso, v 5. 5 sin ( 300πt) volt AnsWorked Example 6.4QFor the waveform specified in Example 6.3 above, after the waveform passes through zero, goingpositive, determine its instantaneous value (a) 0.5 ms later, (b) 4.5 ms later, and (c) the time taken forthe voltage to reach 3 V for the first time.A(a) t 0.5 10 3 s; (b) t 4.5 10 3 s; (c) v 3 V(a) vV m sin ( 300π 0.510 55 . sin 047 . 1255 . 0454.thus, v 25 . V Ans3) volt

Alternating Quantities 205(b) v 55 . sin ( 300π 45 . 10 3) volt 55 . sin 424 . 155 . ( 089. 1)and,v 49 . V AnsNote: Remember that the expression inside the brackets is an angle in RADIAN.(c) 35.5 sin ( 300πt) volt3so, sin ( 300πt) 0.545555 .300πt sin10. 54550.5769 rad0.5769t 6.1210300π4and, t 06 . 12ms AnsA sketch graph illustrating these answers is shown in Fig. 6.6 .v (V)5.5V m3.5V av2.54.56.670 t (ms)0.54.9Fig. 6.66.5 r.m.s. ValueThe r.m.s. value of an alternating current is equivalent to that value ofdirect current, which when passed through an identical circuit, willdissipate exactly the same amount of power. The r.m.s. value of an a.c.thus provides a means of making a comparison between a.c. and d.c.systems.The term r.m.s. is an abbreviation of the square Root of the MeansSquared. The technique for finding the r.m.s. value may be basedon the same ways as were used to find the average value. However,the r.m.s. value applies to the complete cycle of the waveform. Forsimplicity, we will again consider the use of the mid-ordinate ruletechnique.

206 Fundamental Electrical and Electronic PrinciplesConsidering Fig. 6.5 , the ordinates would be selected and measuredin the same way as before. The value of each ordinate is then squared.The resulting values are then summed, and the average found. Finally,the square root of this average (or mean) value is determined. This isillustrated below:Irmsi i i in1 2 2 2 3 2 n21and, for a sinewave only, Irms Im 0. 707Im(6.9)2Other waveforms will have a different ratio between r.m.s. and peak values.Note: The r.m.s. value of an a.c. is the value normally used and quoted.For example, if reference is made to a 240 V a.c. supply, then 240 V isthe r.m.s. value. In general therefore, if an unqualified value for an a.c.is given, then the assumption is made that this is the r.m.s. value. Sincer.m.s. values are those commonly used, the subscript letters r.m.s. arenot normally included. I rms has been used above, simply for emphasis.The following convention is used:i, v, e, represent instantaneous valuesI av , V av , E av , represent average valuesI m , V m , E m , represent maximum or peak values, or amplitudeI, V, E, represent r.m.s. values6.6 Peak FactorThis is defined as the ratio of the peak or maximum value, to the r.m.s.value, of a waveform. Thus, for a sinewave onlymaximum valuepeak factor r.m.s. valueVm 2 or 1.4140.707 VmWorked Example 6.5Q Calculate the amplitude of the household 240 V supply.ASince this supply is sinusoidal, then the peak factor will be2 , soVso, Vmm 2 Vvolt 2 240 339. 4 V Ans

Alternating Quantities 207Worked Example 6.6QA non-sinusoidal waveform has a peak factor of 2.5, and an r.m.s. value of 240 V. It is proposed to usea capacitor in a circuit connected to this supply. Determine the minimum safe working voltage ratingrequired for the capacitor.Apeak factor 2.5; V 240VVm25 . Vvolt 25 . 240so, V 600 VmThus the absolute minimum working voltage must be 600 V AnsIn practice, a capacitor having a higher working voltage would beselected. This would then allow a factor of safety.6.7 Form FactorAs the name implies, this factor gives an indication of the form orshape of the waveform. It is defined as the ratio of the r.m.s. value tothe average value.Thus, for a sinewave ,r.m.s. valueform factor 0 . 707average value 0.637so, form factor 111 .For a rectangular waveform (a squarewave), form factor 1,since the r.m.s. value, the peak value, and the average value are allthe same.Worked Example 6.7Q A rectangular coil, measuring 25 cm by 20 cm, has 80 turns. The coil is rotated, about an axisparallel with its longer sides, in a magnetic field of density 75 mT. If the speed of rotation is3000 rev/min, calculate, from first principles, (a) the amplitude, r.m.s. and average values ofthe emf, (b) the frequency and period of the generated waveform, (c) the instantaneous value,2 ms after it is zero.A 0. 25 m; d0. 2 m; N80; B0.075 T3000n rev/s; t210 3s60(a)e 2 NBlv sin (2 π ft ) voltNow, we know the rotational speed n , but the above equation requires thetangential velocity, v , in metre per second. This may be found as follows.

208 Fundamental Electrical and Electronic PrinciplesrotationvdFig. 6.7Consider Fig. 6.7 ,which shows the path travelled by the coil sides.The circumference of rotation π d metre 0.2 π metre.The coil sides travel this distance in one revolution.The rotational speed n 3000/60 50 rev/s.Hence the coil sides have a velocity, v 50 0.2 π m/s.Therefore, e 2 80 0.075 0.25 50 0.27 π sin (2 π ft ) voltand emf is a maximum value when sin (2 π ft ) 1so, EEmm2800. 0750. 25500.2π 94.25 V AnsAssuming a sinusoidal waveform:E0. 707 Em0. 70794.25so, E 66.64 V AnsE 0. 637 E 0.6379425.avmso, E av 60.04 V AnsAssuming a 2-pole field system, then f n therefore f 50 H z Ans1 1(b) T sf 50so T 20 ms Ans(c) eEmsin( 2πft) volt94.25 sin ( 2π502103)94. 250.5878so, e 55. 4 V Ans6.8 RectifiersA rectifier is a circuit which converts a.c. to d.c. The essentialcomponent of any rectifier circuit is a diode. This is a semiconductordevice, which allows current to flow through it in one direction only.

Alternating Quantities 209It is the electronic equivalent of a mechanical valve, for example thevalve in a car tyre. This device allows air to be pumped into the tyre,but prevents the air from escaping.Fig. 6.8The circuit symbol for a diode is shown in Fig. 6.8 . The ‘ arrow head ’part of the symbol is known as the anode. This indicates the directionin which conventional current can flow through it. The ‘ plate ’ partof the symbol is the cathode, and indicates that conventional currentis prevented from entering at this terminal. Thus, provided that theanode is more positive than the cathode, the diode will conduct. Thisis known as the forward bias condition. If the cathode is more positivethan the anode, the diode is in its blocking mode, and does not conduct.This is known as reverse bias.Note: The potentials at anode and cathode do not have to be positiveand negative. Provided that the anode is more positive than the cathode,the diode will conduct. So if the anode potential is (say) 10 V, and thecathode potential is 8 V, then the diode will conduct. Similarly, ifthese potentials are reversed, the diode will not conduct.6.9 Half-wave RectifierThis is the simplest form of rectifier circuit. It consists of a singlediode, placed between an a.c. supply and the load, for which d.c. isrequired. The arrangement is shown in Fig. 6.9 , where the resistor Rrepresents the load.linputRoutputinputRoutputFig. 6.9Fig. 6.10Let us assume that, in the first half cycle of the applied voltage, theinstantaneous polarities at the input terminals are as shown in Fig. 6.10 .Under this condition, the diode is forward biased. A half sinewave ofcurrent will therefore flow through the load resistor, in the directionshown.In the next half cycle of the input waveform, the instantaneouspolarities will be reversed. The diode is therefore reverse biased, andno current will flow. This is illustrated in Fig. 6.11 .

210 Fundamental Electrical and Electronic PrinciplesinputRoutputFig. 6.11The graphs of the applied a.c. voltage, and the corresponding loadcurrent, are shown in Fig. 6.12 . The load p.d. will be of exactly thesame waveshape as the load current. Both of these quantities are unidirectional,and so by definition, are d.c. quantities. The ‘ quality ’ of thed.c. so produced is very poor, since it exists only in pulses of current.The average value of this current is determined over the time period,0 to t 2 . The average value from 0 to time t 1 will be 0.637 I m . From t 1to t 2 it will be zero. The average value of the d.c. will therefore be,I av 0.318 I m .input0t (s)I mI av0t (s)t 1 t 2Fig. 6.126.10 Full-wave Bridge RectifierBoth the ‘ quality ’ and average value of the d.c. need to be improved.This may be achieved by utilising the other half cycle of the a.c.supply. The circuit consists of four diodes, connected in a ‘ bridge ’configuration, as shown in Fig. 6.13 .We will again assume the instantaneous polarities for the first halfcycle as shown. In this case, diodes D 1 and D 4 will be forward biased.Diodes D 2 and D 3 will be reverse biased. Thus, D 1 and D 4 allowcurrent to flow, as shown. In the next half cycle, the polarities arereversed. Hence, D 2 and D 3 will conduct, whilst D 1 and D 4 are reversebiased. Current will therefore flow as shown in Fig. 6.14 . Notice that

Alternating Quantities 211IinputD1D2D3RD4Fig. 6.13inputID1D2RD3D4Fig. 6.14the current through the load resistor is in the same direction, for thewhole cycle of the a.c. supply.The relevant waveforms are shown in Fig. 6.15 . It should be apparentthat the average value of the a.c. will now be twice that in the previouscircuit. That is I av 0.637 I m .It is this type of rectifier circuit that is incorporated in multimeters.This enables the measurement of a.c. voltages and currents.input0t (s)I mI av0t (s)Fig. 6.15

212 Fundamental Electrical and Electronic Principles6.11 Rectifier Moving Coil MeterThe circuit arrangement for such a meter is illustrated in Fig. 6.16 . Thesymbol, with the letter M in it, represents the moving coil movement.The current through the coil will therefore be a series of half-sinewavepulses, as in Fig. 6.15 .a.c. to bemeasuredMFig. 6.16Due to the inertia of the meter movement, the coil will respond tothe average value of this current. Thus, the pointer would indicate theaverage value of the a.c. waveform being measured. However, thenormal requirement is for the meter to indicate the r.m.s. value.Thus, some form of ‘ correction factor ’ is required.The majority of a.c. quantities to be measured are sinewaves. Forthis reason, the meter is calibrated to indicate the r.m.s. values ofsinewaves. Now, the ratio between r.m.s. and average values is theform factor. For the sinewave, the form factor has a value of 1.11. Thevalues chosen for shunts and multipliers, used on the a.c. ranges, aretherefore modified by this factor. Thus, although the pointer positioncorresponds to the average value, the scale indication will be ther.m.s. value.Worked Example 6.8QA moving coil meter has a figure of merit of 10 k Ω / V. The coil has a resistance of 50 Ω . Calculate thevalue of multiplier required for (a) the 10 V d.c. range, (b) the 10 V a.c. range.AR c 5 0 Ω ; figure of merit 10 kΩ /V; V 10 V(a)I fsd 1/10 000 10 0 μ Atotal meter resistance, R VI fsd10ohm1 0 4so, R100k Ω; and since R R R, then:Rm100 00050 99. 95 k Ω Ansmc

Alternating Quantities 213(b)I fsd 10 0 μ A I avtherefore, r.m.s. value,I 1.11100111μAtherefore,10so, R 11110R m6 90 09050 90. 04 k Ω Ans 90.09 kΩThe meter is therefore calibrated for the measurement of sinewaves.If any other waveform is measured, the meter reading will be in error,because the waveform will have a different form factor. Provided thatthe form factor is known, then the true r.m.s. value can be calculated.Worked Example 6.9QA moving coil meter is used to measure both a squarewave and a triangular wave voltage. The meterreading is 5 V in each case. Calculate the true r.m.s. values.A(a)(b)The form factor for a squarewave is 1. The meter, however, has beencalibrated for a form factor of 1.11. Thus the indicated reading will be toohigh by a factor of 1.11:11Therefore the true value 5111 . 4. 504 V AnsIf the form factor for the triangular wave is 1.15, then the meter reading willbe too low, by a factor of 1.15:1.11.11 . 5Therefore the true value 5111 . 5. 18V AnsA further complication arises when the meter is used to measure smalla.c. voltages. When conducting, each diode will have a small p.d.developed across it. This forward voltage drop will be in the order of0.6 to 0.7 V. At any instant, two diodes are conducting. Thus, therewill be a total forward voltage drop of 1.2 to 1.4 V across the diodes.This voltage drop is ‘ lost ’ as far as the meter coil is concerned. Thiseffect also must be taken into account when determining the values ofmultipliers, for the lower a.c. voltage ranges.6.12 Phase and Phase AngleConsider two a.c. voltages, of the same frequency, as shown inFig. 6.17 . Both voltages pass through the zero on the horizontal axis atthe same time. They also reach their positive and negative peaks at the

214 Fundamental Electrical and Electronic Principlessame time. Thus, the two voltages are exactly synchronised, and aresaid to be in phase with each other.V 1V 20/2 3 /22t(rad)Fig. 6.17Figure 6.18 shows the same two voltages, but in this case let v 2 reachits maximum value π /2 radian (90°) after v 1 . It is necessary to considerone of the waveforms as the reference waveform. It is normal practiceto consider the waveform that passes through zero, going positive, atthe beginning of the cycle, as the reference waveform. So, for the twowaveforms shown, v 1 is taken as the reference. In this case, v 2 is saidto lag v 1 by π /2 radian or 90°. The standard expressions for the twovoltages would therefore be written as follows:v1 Vm1 sin( 2πft), or Vm1sin θ voltv V sin( 2πftπ / 2), or V m2 (sin θ 90) volt2 m2V 1V 20t(rad)/2Fig. 6.18The minus signs, in the brackets of the above expressions, indicate thatv 2 lags the reference by the angle quoted. This angle is known as thephase angle , or phase difference , between the two waveforms.In general, the standard expression for an a.c. voltage is:vVmsin( 2πftφ) volt,or vV sin( ωtφ) voltm(6.10)

Alternating Quantities 215Although it would be usual to take v 1 as the reference in the aboveexample, it is not mandatory. Thus, if for some good reason v 2 waschosen as the reference, the expressions would be written as:v2 Vm2 sin( ωt), or Vm2sin θ voltv V sin( ωtπ / 2), or V sin( θ 90) volt1 m1 m1Note: when the relevant phase angle, f , is quoted in the standardexpression, do not mix degrees with radians. Thus, if the initial angulardata is in radian ( ω t or 2 π ft ), then f must also be expressed in radian.Similarly, if the angular data is initially in degrees ( u ), the f must alsobe quoted in degrees.Worked Example 6.10Q Three alternating currents are specified below. Determine the frequency, and for each current,determine its phase angle, and amplitude.Ai 1 5 sin (80 π t π /6) ampi 2 3 sin 80 π t ampi 3 6 sin (80 π t π /4) ampAll three waveforms have the same value of ω , namely 80 π rad/s. Thus all threehave the same frequency:ω2πf80πrad/s80πtherefore, f 40Hz Ans2πSince zero phase angle is quoted for i 2 , then this is the reference waveform, ofamplitude 3 A AnsIIm1 5 A, and leads i2by π/ 6 rad ( 30 ) Ans 6 A, and lags i 2 by π/ 4 rad ( 45)Ansm3The majority of people can appreciate the relative magnitudes ofangles, when they are expressed in degrees. Angles expressed inradians are more difficult to appreciate. Some of the principal anglesencountered are listed below. This should help you to gain a better‘ feel ’ for radian measure.degrees radians radians degrees360 2π 6.28 0.1 5.73270 3π /2 4.71 0.2 11.46180 π 3.14 0.3 17.19120 2π /3 2.09 0.4 22.9290 π /2 1.57 0.5 28.6560 π /3 1.05 1.0 57.3045 π/40.79 1.5 85.9430 π /6 0.52 2.0 114.60

216 Fundamental Electrical and Electronic Principles6.13 Phasor RepresentationA phasor is a rotating vector. Apart from the fact that a phasor rotatesat a constant velocity, it has exactly the same properties as any othervector. Thus its length corresponds to the magnitude of a quantity. Ithas one end arrowed, to show the direction of action of the quantity.Consider two such rotating vectors, v 1 and v 2 , rotating at the sameangular velocity, ω rad/s. Let them rotate in a counterclockwisedirection, with v 2 lagging behind v 1 by π /6 radian (30°). This situationis illustrated in Fig. 6.19 . (rad/s)/6V V 1 1V 2V 20/2 3 /22(rad)/6Fig. 6.19The instantaneous vertical height of each vector is then plotted for onecomplete revolution. The result will be the two sinewaves shown.Notice that the angular difference between v 1 and v 2 is also maintainedthroughout the waveform diagram. Also note that the peaks of the twowaveforms correspond to the magnitudes, or amplitudes, of the twovectors. In this case, these two waveforms could equally well representeither two a.c. voltages, or currents. If this were the case, then the twoa.c. quantities would be of the same frequency. This is because thevalue of ω is the same for both. The angular difference, of π /6 radian,would then be described as the phase difference between them.We can therefore, represent an alternating quantity by means of aphasor. The length of the phasor represents the amplitude. Its angle,with respect to some reference axis, will represent its phase angle.Considering the two waveforms in Fig. 6.19 , the plot has been startedwith V 1 in the horizontal position (vertical component of V 1 0). Thishorizontal axis is therefore taken as being the reference axis. Thus, ifthese waveforms represent two voltages, v 1 and v 2 , then the standardexpressions would be:v1 Vm1sin( ωt) voltand v V sin( ωtπ/ ) volt2 m2 6The inconvenience of representing a.c. quantities in graphical form waspointed out earlier, in section 6.3. This section introduced the concept

Alternating Quantities 217of using a standard mathematical expression for an a.c. However, avisual representation is also desirable. We now have a much simplermeans of providing a visual representation. It is called a phasordiagram. Thus the two voltages we have been considering above maybe represented as in Fig. 6.20 . (rad/s)/6V 1mV 2mFig. 6.20Notice that v 1 has been chosen as the reference phasor. This is becausethe standard expression for this voltage has a phase angle of zero (thereis no f term in the bracket). Also, since the phasors are rotatingcounterclockwise, and v 2 is lagging v 1 by π /6 radian, then v 2 is shownat this angle below the reference axis.Notes1 Any a.c. quantity can be represented by a phasor, provided that it isa sinewave.2 Any number of a.c. voltages and/or currents may be shown on thesame phasor diagram, provided that they are all of the same frequency.3 Figure 6.20 shows a counterclockwise arrow, with ω rad/s. This hasbeen shown here to emphasise the point that phasors must rotate inthis direction only. It is normal practice to omit this from the diagram.4 When dealing with a.c. circuits, r.m.s. values are used almostexclusively. In this case, it is normal to draw the phasors to lengthsthat correspond to r.m.s. values.Worked Example 6.11Q Four currents are as shown below. Draw to scale the corresponding phasor diagram.i1 25 . sin( ωt π/ 4) amp; i2 4sin( ωt π/ 3) amp;i 6 sin ωtamp; i cos ωtamp3A4 3Before the diagram is drawn, we need to select a reference waveform (if oneexists). The currents i 1 and i 2 do not meet this criterion, since they both have anassociated phase angle.This leaves the other two currents. Neither of these has a phase angle shown.However, i 3 is a sinewave, whilst i 4 is a cosine waveform. Now, a cosine waveleads a sinewave by 90°, or π /2 radian.Therefore, i 4 may also be expressed as i 4 3 sin ( ωt π /2) amp. Thus i 3 is chosenas the reference waveform, and will therefore be drawn along the horizontal axis.

218 Fundamental Electrical and Electronic PrinciplesThe resulting phasor diagram is shown in Fig. 6.21 .3A2.5A/4/36Ascale 1 cm: 1A4AFig. 6.21Worked Example 6.12QThe phasor diagram representing four alternating currents is shown in Fig. 6.22 , where the length ofeach phasor represents the amplitude of that waveform. Write down the standard expression foreach waveform.Î 1 7A7050Î 2 6Î 4 4AÎ 3 5AFig. 6.22AIIIm1 1m2 2m370π7 A; φ 70 1.22 rad1806 A; φ 00rad50 5 A; φ350 π0.873 rad180904 A; φ 90 π1. 5 7 1 rad180I m 4 4hence,i 1 7 sin ( ω t 1.22) amp Ansi 2 6 sin ω t amp Ansi 3 5 sin ( ω t 0.873) amp Ansi 4 4 sin ( ω t 1.571) amp Ans

Alternating Quantities 2196.14 Addition of Alternating QuantitiesConsider two alternating currents, i 1 I m1 sin ω t amp and i 2 I m2 sin(ω t π /4) amp, that are to be added together. There are three methodsof doing this, as listed below.(a) Plotting them on graph paper. Their ordinates are then addedtogether, and the resultant waveform plotted. This is illustratedin Fig. 6.23 . The amplitude, I m , and the phase angle, f, of theresultant current are then measured from the two axes.I mii 1i 20t(rad)Fig. 6.23Thus, iI sin( ωtφ) ampmNote: Although i i 1 i 2 , the AMPLITUDE of the resultant is NOTI ml I m2 amp. This would only be the case if i 1 and i 2 were in phasewith each other.(b) Drawing a scaled phasor diagram, as illustrated in Fig. 6.24 . Theresultant is found by completing the parallelogram of vectors. Theamplitude and phase angle are then measured on the diagram.(c) Resolving the two currents, into horizontal and vertical components,and applying Pythagoras ’ theorem. This method involves using asketch of the phasor diagram, followed by a purely mathematicalprocess. This phasor diagram, including the identification of thehorizontal and vertical components, is shown in Fig. 6.25 ./4I m1I m2 cos /4/4I m1I m2I mScale: 1 cm x amp.I m2 sin /4I m2Fig. 6.24Fig. 6.25

220 Fundamental Electrical and Electronic PrinciplesHorizontal Component ( H.C .):Vertical Component ( V.C .):H.C. Im1 cos 0Im2cos π/4so, H.C. I ( 0.707I) ampm1 m2V.C. Im1 sin 0Im2sin π/4so, V.C. 0 ( 0.707I) ampThe triangle of H.C., V.C., and the resultant current, is shown inFig. 6.26 . From this, we can apply Pythagoras ’ theorem to determinethe amplitude and phase angle, thus:m2H.C.V.C.I mFig. 6.26I m H.C.2V.C. 2ampV.C.and tan so tan V.C.φ , φ1H.C.H.C.The final answer, regardless of the method used, would then beexpressed in the form i I m sin ( ω t f ) amp.Let us now compare the three methods, for speed, convenience, andaccuracy.The graphical technique is very time-consuming (even for the additionof only two quantities). The accuracy also leaves much to be desired;in particular, determining the exact point for the maximum value ofthe resultant. The determination of the precise phase angle is also verydifficult. This method is therefore not recommended.A phasor diagram, drawn to scale, can be the quickest method of solution.However, it does require considerable care, in order to ensure a reasonabledegree of accuracy. Even so, the precision with which the length—and(even more so) the angle—can be measured, leaves a lot to be desired.This is particularly true when three or more phasors are involved. Thismethod is therefore recommended only for a rapid estimate of the answer.The use of the resolution of phasors is, with practice, a rapid technique,and yields a high degree of accuracy. Unless specified otherwise it isthe technique you should use. Although, at first acquaintance, it mayseem to be rather a complicated method, this is not the case. Witha little practice, the technique will be found to be relatively simple andquick. Two worked examples now follow.

Alternating Quantities 221Worked Example 6.13QDetermine the phasor sum of the two voltages specified below.v 1 25 sin (314 t π /3), and v 2 15 sin (314 t π /6) voltAFigure 6.27 shows the sketch of the phasor diagram.25 V/3/6Fig. 6.2715 VNote: Always sketch a phasor diagram.H.C. 25 cos π/ 315 cos ( π/6)( 250. 5) ( 150.866)125 . 129. 9so, H.C. 25.49 VV.C. 25 sin π/ 315 sin ( π/6)( 250. 866) ( 15( 0. 5))21. 657.5so, V.C. 14.15VFigure 6.28 shows the phasor diagram for H.C., V.C. and V m .V.C.V mH.C.Fig. 6.28Vso, Vmm H.C. V.C. 25. 49 14.15 29.15V2 2 2 2V.C. 14.15tan φ 0.555*H.C. 25.49so, φ tan 10. 5550.507 radtherefore, v 29. 15 sin( 314t 0.507) volt Ans*radian mode required

222 Fundamental Electrical and Electronic PrinciplesWorked Example 6.14Q Calculate the phasor sum of the three currents listed below.i1 6 sin ωtampi 8 sin( ωtπ/ 2) ampi23 4 sin( ωtπ/ 6) ampAThe relevant phasor diagrams are shown in Figs. 6.29 and 6.30.4A/66AH.C.8AV.C.I mFig. 6.29Fig. 6.30H.C. 6 cos 08 cos ( π/ 2)4 cos π/6( 61) (8 0) ( 40.866) 6 346 .so, H.C. 946 . AV.C. 6 sin 08 sin ( π/ 2)4 sin π/6 ( 60) ( 8[ 1] ) ( 40.5)82so, V.C. 6AI H.C.2V.C.2m 9.46 ( 6)2 2so, I m 11.2 A6 V.C.φ tan1tan 1 tan 1 0.6342H.C. 946 .so, φ 0.565 radtherefore, i11. 2 sin( ωt0.565) amp AnsWorked Example 6.15Q Three alternating voltages and one current are as specified in the expressions below.v 1 10 sin(628t π /6) voltv 2 8 sin(628t π /3) voltv 3 12 sin(628t π /4) volti 6 sin(628t) amp(a)(b)For each voltage determine the frequency, phase angle and amplitude.Determine the phasor sum of the three voltages.

Alternating Quantities 223A(a)All four waveforms have the same value of ω 628 rad/s, so they are all ofthe same frequency, henceω2πf628628so, f Hz2πand, f 100Hz Ansfor v , φ π/ 6 rad or 30; and V 0 V Ans1 1 m 1for v , φ π/ 3 rad or 60; and V 8V Ans2 2for v , φ π / 4 rad or 45; and V m 12 V Ans3 3m(b)Firstly the phasor diagram (Fig. 6.31 ) is sketched, very roughly to scale. Inorder to do this a reference waveform needs to be selected, and since thecurrent has a zero phase angle, this is chosen as the reference. However, if thecurrent waveform had not been specified, the horizontal axis would still betaken as the reference from which all phase angles are measured. Since v 2 andv 3 have positive phase angles, and phasors rotate anticlockwise, then thesetwo phasors will appear above the reference axis. The voltage v 1 , having anegative phase angle will appear below the reference axis. Also shown on thephasor diagram are the horizontal and vertical components of each voltage.8 V12 V/3/6/4I m (ref)10 VFig. 6.31H.C. 12 cos π/ 48 cos π/ 310 cos π/6( 12 0. 707) ( 8 0. 5) (100. 866)8. 4848.66so, H.C. 21.44 VV.C. 12 sin π/ 48 sin π/ 310sin π/ 6( 120. 707) ( 80. 866) ( 100.5)8. 486.9285and V.C. 10. 4 12VThe phasor diagram for H.C. and V.C., and the resultant phasor sum is Fig. 6.32 .

224 Fundamental Electrical and Electronic PrinciplesV.C.V mH.C.I m (ref)6.15 The Cathode Ray OscilloscopeFig. 6.32Vm H.C.2V.C.2 21. 44210.4122so, Vm 23.83 Vtan1V.C.φ tan1104 . 12 tan10.4856H.C. 21.44and, φ 0.452 radHence, the phasor sum, v 23.83 sin (628 t 0.452) volt AnsThe name of this instrument is more often abbreviated to the oscilloscope,the ‘ scope, or CRO. It is a very versatile instrument, that may be used tomeasure both a.c. and d.c. voltages. For d.c. measurements, a voltmeteris usually more convenient to use. The principal advantages of theoscilloscope when used to measure a.c. quantities are:1 A visual indication of the waveform is produced.2 The frequency, period and phase angle of the waveform(s) can bedetermined.3 It can be used to measure very high frequency waveforms.4 Any waveshape can be displayed, and measured with equal accuracy.5 The input resistance (impedance) is of the same order as a digitalvoltmeter. It therefore applies minimal loading effect to a circuit towhich it is connected.6 Some oscilloscopes can display two or more waveformssimultaneously.Cathode Ray Tube The basic arrangement of a crt is shown inFig. 6.33 . The main components are contained within an evacuatedelectrongunanodesXdeflectionY deflectiongraphitecoatingFig. 6.33

Alternating Quantities 225glass tube. These components are: the electron gun; a focusing system;a beam deflection system; and a screen. Each of these will be verybriefly described.Electron gun assembly This component produces a beam ofelectrons. This beam can then be accelerated, down the axis of thetube, by a series of high potential anodes.Focusing system The beam consists entirely of electrons. Sincethey are all negatively charged, then they will tend to repel each other.The beam will therefore tend to spread out, and this would result in avery fuzzy display. The focusing system consists of a series of highpotential anodes. These also provide the acceleration for the electronbeam. Each successive anode along the tube, towards the screen, is ata higher potential than the previous one. The electric fields, betweenthese anodes, will be of the same shape as a double convex optical lens.This is referred to as an electron lens, and causes the beam to convergeto a small spot by the time it reaches the screen.Deflection system Two sets of parallel plates are situated after thelast anode. One set is mounted in the horizontal, and the other set in thevertical plane. These are the X-plates and the Y-plates. When a p.d. isdeveloped between a pair of plates, an electric field will exist betweenthem. This electric field will cause the electrons in the beam to bedeflected, towards the more positive of the two plates. Thus, the beam canbe made to deflect in both planes. This effect is illustrated in Fig. 6.34 .Fig. 6.34The screen The inner surface of the screen is coated with a phosphor.Wherever the electron beam strikes the phosphor, it will glow verybriefly. This is because the kinetic energy of the bombarding electronsis converted into ‘ light ’ energy. On the inside of the ‘ bell ’ shape of thetube is a graphite coating. This provides a conducting path to return theelectrons to the internal power supply, and hence complete the circuitback to the electron gun.In order for a waveform to be displayed on the screen, the beam mustbe swept at a constant speed across the screen. At the same time, thebeam has to be deflected up and down. You can demonstrate this for

226 Fundamental Electrical and Electronic Principlesyourself, as follows. Take a pencil and a sheet of paper. At the lefthandedge of the paper, move the pencil up and down, at as constanta rate as possible. Maintaining this up-and-down rhythm, now movethe pencil across the page, again at as constant a rate as possible. Thepattern traced on to the paper should now resemble a sinewave.The electron beam in the crt is subjected to similar forces, exerted bythe deflecting plates, when displaying a sinusoidal waveform. TheX-plates cause the beam to be swept across the screen at a constantrate. The Y-plates cause the beam to deflect up and down, in sympathywith the voltage being displayed.6.16 Operation of the OscilloscopeIn addition to the crt, the other main components in the oscilloscope enablethe user to adjust the display, by means of controls on the front panel.A simplified block diagram of the oscilloscope is shown in Fig. 6.35 .brightnessYinputExt.trig.PowersupplyY-amp.focusX-ampXinputTriggerpulsegen.Timebasegen.Fig. 6.35The Power Supply This provides the high potentials required for theanodes. It also provides the d.c. supplies for the amplifiers, the electrongun, the timebase and trigger pulse generators.The Focus Control This control allows the potentials applied to theanodes to be varied. This allows the shape of the electron lens to bealtered, and hence achieve a sharp clear trace on the screen.The Brightness Control This varies the potentials applied to theelectron gun. The number of electrons forming the beam are thuscontrolled, which determines the brightness of the display.

Alternating Quantities 227The Timebase Control This controls the speed at which the beam isswept across the screen, from left to right. It does this by setting the‘ sweep ’ time of the Timebase generator.This generator produces a sawtooth voltage, as illustrated in Fig. 6.36 .This waveform is applied to the X-plates. During the sweep time,the beam is steadily deflected across the screen. During the ‘ flyback ’time, the beam is rapidly returned to the left-hand side of the screen,ready for the next sweep. The steepness of the sweep section of thiswaveform determines the speed of the sweep.VsweepflybacktFig. 6.36At the front of the screen is a graticule, marked out in centimetresquares. The timebase control is marked in units of time/cm. Thus, ifthis control is set to (say) 10 ms/cm, then each centimetre graduationacross the graticule represents a time interval of 10 ms. This facilityenables the measurement of the periodic time (and hence frequency),of the displayed waveform.Trigger Control This enables the user to obtain a single stationaryimage of the trace on the screen. If this control is incorrectly set, thenthe trace will scroll continuously across the screen. Alternatively,multiple overlapping traces are displayed, which may also be scrolling.In either case, measurement of the periodic time is impossible. Thiscontrol determines the point in time at which the sweep cycle of thesawtooth waveform commences. If required an external trigger inputmay be used.X-Amplifier This amplifies the sawtooth waveform. This ensures thatthe voltage applied to the X-plates is sufficiently large to deflect thebeam across the full width of the screen. There is also provision for theapplication of an external timebase signal.Y-Amplifier The waveform to be displayed is applied to thisamplifier. Thus, small amplitude signals can be amplified to givea convenient height of the trace. The gain or amplification of thisamplifier is determined by the user. The control on the front panelis marked in units of volt/cm. Thus, if this control was set to (say)100 mV/cm, then each vertical graduation on the screen graticulerepresents a voltage of 100 mV. This enables the measurement of theamplitude of the displayed waveform.

228 Fundamental Electrical and Electronic PrinciplesX and Y Shift Controls These controls enable the trace position onthe screen to be adjusted. This makes the measurement of period andamplitude easier.6.17 Dual Beam OscilloscopesThese instruments are widely used, and are more versatile than the singlebeam type described. They have the advantage that two waveforms canbe displayed simultaneously. This enables waveforms to be compared, interms of their amplitudes, shape, phase angle or frequency.The principles of operation are exactly the same as for the single beaminstrument. They contain two electron gun assemblies, which havecommon brightness and focusing controls. The timebase generatoris also common to both channels. There will be two separateY-amplifiers, each controlling its own set of vertical deflectionplates. The inputs to these two amplifiers are usually marked aschannel 1 and channel 2, or as channels A and B.Worked Example 6.16Q The traces obtained on a double beam oscilloscope are shown in Fig. 6.37 . The graticule is marked in1 cm squares. The channel 1 input is displayed by the upper trace. If settings of the controls for thetwo channels are as follows, determine the amplitude, r.m.s. value, and frequency of each input.Channel 1: timebase of 0.1 ms/cm; Y-amp setting of 5 V/cmChannel 2: timebase of 10 μ s/cm; Y-amp setting of 0.5 V/cmFig. 6.37AChannel 1: peak to peak occupies 3 cm, soVppVpp 15the amplitude 2 2therefore, V 75 . V Ansm3515V

Alternating Quantities 229As the waveform is a sinewave, then r.m.s. value V V m I 275 .therefore, V 53. V Ans21 cycle occupies 4 cm, so T 4 0.1 0.4 ms1 1f Hz T 04 . 10so, f 25 . kHz AnsChannel 2: peak to peak occupies 2 cm, so3Vp p 20. 5 1 V, and Vm0.5 V AnsSince it is a squarewave, then r.m.s. value amplitude,hence V 05 . V Ans2 cycles occur in 3 cm, so 1 cycle occurs in 2/3 cmtherefore, T 0.6667 10 6.667 μ s1 1f Hz T 6.66710so, f 150kHz Ans6Summary of EquationsFrequency generated: f np hertzPeriodic time: T 1 secondfAngular velocity: ω 2π f rad/secondStandard expression for a sinewave: e Emsin( θφ) Emsin( ωtφ) E sin( 2πftφ) voltAverage value for a sinewave: Iave2IπImR.m.s. value for a sinewave: I 0.707I2mm0637. ImmPeak factor for a sinewave:max. valuer.m.s. value 1.414Form factor for a sinewave: r.m.s.valueave value 111 .

230 Fundamental Electrical and Electronic PrinciplesAssignment Questions1 A coil is rotated between a pair of poles.Calculate the frequency of the generatedemf if the rotational speed is (a) 150 rev/s,(b) 900 rev/minute, (c) 200 rad/s.2 An alternator has 8 poles. If the motor windingis rotated at 1500 rev/minute, determine(a) the frequency of the generated emf, and(b) the speed of rotation required to producefrequency of 50 Hz.3 A frequency of 240 Hz is to be generated bya coil, rotating at 1200 rev/min. Calculate thenumber of poles required.4 A sinewave is shown in Fig. 6.38 . Determine itsamplitude, periodic time and frequency.I (mA)5055 A sinusoidal current has a peak-to-peak valueof 15 mA and a frequency of 100 Hz. (a) Plotthis waveform, to a base of time, and (b)write down the standard expression for thewaveform.6 A sinusoidal voltage is generated by an 85turn coil, of dimensions 20 cm by 16 cm. Thecoil is rotated at 3000 rev/min, with its longersides parallel to the faces of a pair of poles. Ifthe flux density produced by the poles is 0.5 T,calculate (a) the amplitude of the generatedemf, (b) the frequency, (c) the r.m.s. andaverage values.7 Write down the standard expression for avoltage, of r.m.s. value 45 V, and frequency1.5 kHz. Hence, calculate the instantaneousvalue, 38 μ s after the waveform passesthrough its zero value.8 For each of the following alternatingquantities, determine (a) the amplitude andr.m.s. value, and (b) the frequency and period.(i) e 250 sin 50πt volt(ii) i 75 sin 628.3t milliamp25Fig. 6.3850 t(s)(iii) f 20 sin 100πt milliweber(iv) v 6.8 sin (9424.8tf) volt.9 For a current of r.m.s. value 5 A, and frequency2 kHz, write down the standard expression.Hence, calculate (a) the instantaneous value15 0 μ s after it passes through zero, and (b) thetime taken for it to reach 4 A, after passingthrough zero for the first time.10 Calculate the peak and average values for a250 V sinusoidal supply.11 A sinusoidal current has an average value of3.8 mA. Calculate its r.m.s. and peak values.12 An alternating voltage has an amplitude of500 V, and an r.m.s. value of 350 V. Calculatethe peak factor.13 A waveform has a form factor of 1.6, and anaverage value of 10 V. Calculate its r.m.s.value.14 A moving coil voltmeter, calibrated forsinewaves, is used to measure a voltagewaveform having a form factor of 1.25.Determine the true r.m.s. value of this voltage,if the meter indicates 25 V. Explain why themeter does not indicate the true value.15 Explain why only sinusoidal waveforms can berepresented by phasors.16 Sketch the phasor diagram for the twowaveforms shown in Fig. 6.39 .I (mA)53017 Sketch the phasor diagram for the twovoltages represented by the followingexpressions:v1 12 sin 314tvolt,v 8 sin( 314tπ/ 3) volt.2Fig. 6.392 t(rad)

Alternating Quantities 231Assignment Questions18 Determine the phasor sum of the two voltagesspecified in Question 17 above.19 Three currents, in an a.c. circuit, meet at ajunction. Calculate the phasor sum, if thecurrents are:iii 0 sin ωtamp5 sin( ωtπ/ 4) amp14 sin( ωtπ/ 3) amp.1 123vV sin( ωtφ)mv 2 sin t, and v 8 sin( ωtπ/ 6) volt.1 1223 The waveform displayed on an oscilloscopeis as shown in Fig. 6.40 . The timebase is setto 100 μs/cm, and the Y-amp is set to 2 V/cm.Determine the amplitude, r.m.s. value, periodictime and frequency of this waveform.20 Determine the phasor sum of the followingvoltages, all of which are sinewaves of thesame frequency:v 1 has an amplitude of 25 volt, and a phaseangle of zero.v 2 has an amplitude of 13.5 volt, and lags v 1by 25°.v 3 has an amplitude of 7.6 volt, and leads v 2by 40°.21 By means of a phasor diagram, drawn to scale,check your answer to Question 19 above.22 Plot, on the same axes, the graphs of thefollowing two voltages. By adding ordinates,determine the sum of these voltages. Expressthe result in the formFig. 6.40

232 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentsThe principal practical exercise relating to this chapter is the usage of theoscilloscope. The actual exercises carried out are left to the discretion of yourteacher. Using an oscilloscope is not difficult, but does require some practice;particularly in obtaining a clear, stationary trace, from which measurements canbe made.

Chapter 7D.C. MachinesLearning OutcomesThis chapter covers the operating principles of d.c. generators and motors, their characteristicsand applications. On completion you should be able to:1 Understand and explain generator/motor duality.2 Appreciate the need for a commutator.3 Identify the different types of d.c. generator, and describe their characteristics. Carry outpractical tests to compare the practical and theoretical characteristics.7.1 Motor/Generator DualityAn electric motor is a rotating machine which converts an electricalinput power into a mechanical power output. A generator converts amechanical power input into an electrical power output. Since oneprocess is the converse of the other, a motor may be made to operate asa generator, and vice versa. This duality of function is not confined tod.c. machines. An alternator can be made to operate as a synchronousa.c. motor, and vice versa.To demonstrate the conversion process involved, let us reconsidertwo simple cases that were met when dealing with electromagneticinduction.Consider a conductor being moved at constant velocity, through amagnetic field of density B tesla, by some externally applied force Fnewton. This situation is illustrated in Fig. 7.1 .Work done in moving the conductor,W Fd newton metre233

234 Fundamental Electrical and Electronic PrinciplesNdFSFig. 7.1Wmechanical power input, P1 twattso,Fd P 1 watttand since d /t is the velocity, v at which the conductor is moved,thenP1 Fv watt.............[ 1]However, when the conductor is moved, an emf will be induced intoit. Provided that the conductor forms part of a closed circuit, thenthe resulting current flow will be as shown in Fig. 7.2 . This inducedcurrent, i, produces its own magnetic field, which reacts with the mainfield, producing a reaction force, F r , in direct opposition to the appliedforce, F .Now, F r Bi newtonNF rSFig. 7.2

D.C. Machines 235Assuming no frictional or other losses, then the applied force has onlyto overcome the reaction force, such that:F Fr Bilnewtonso eqn [ 1] becomes P1 Bilvwatt............ [2]Also, induced emf, eBlvvoltso generated power, P2 ei watttherefore, P Bilvwatt............[ ]2 3Since [3] [2], then the electrical power generated is equal to themechanical power input (assuming no losses). Now consider theconductor returned to its original starting position. Let an externalsource of emf, e volt pass a current of i ampere through the conductor.Provided that the direction of this current is opposite to that shown inFig. 7.2 , then the conductor will experience a force that will propelit across the field. In this case, the same basic arrangement exhibitsthe motor effect, since the electrical input power is converted intomechanical power.Although the above examples involve linear movement of theconductor, exactly the same principles apply to a rotating machine.7.2 The Generation of d.c. VoltageWe have seen in Chapter 6 already that, if a single-loop coil isrotated between a pair of magnetic poles, then an alternating emf isinduced into it. This is the principle of a simple form of alternator.Of course, this a.c. output could be converted to d.c. by employinga rectifier circuit. Indeed, that is exactly what is done with vehicleelectrical systems. However, in order to have a truly d.c. machine,this rectification process needs to be automatically accomplishedwithin the machine itself. This process is achieved by means of acommutator, the principle and action of which will now be described.Consider a simple loop coil the two ends of which are connected to asingle ‘ split ’ slip-ring, as illustrated in Fig. 7.3 . Each half of this slipringis insulated from the other half, and also from the shaft on whichit is mounted. This arrangement forms a simple commutator, where theconnections to the external circuit are via a pair of carbon brushes. Therectifying action is demonstrated in the series of diagrams of Fig. 7.4 .In these diagrams, one side of the coil and its associated commutatorsegment are identified by a thickened line edge. For the sake ofclarity, the physical connection of each end of the coil, to its associatedcommutator segment, is not shown. Figure 7.4(a) shows the instantwhen maximum emf is induced in the coil. The current directionshave been determined by applying Fleming ’ s right-hand rule. At this

236 Fundamental Electrical and Electronic PrinciplesrotationΦbrushcommutatorRFig. 7.3IIII(a)(b)Fig. 7.4(c)instant current will be fed out from the coil, through the external circuitfrom right to left, and back into the other side of the coil. As the coilcontinues to rotate from this position, the value of induced emf andcurrent will decrease. Figure 7.4(b) shows the instant when the brushesshort-circuit the two commutator segments. However, the induced emfis also zero at this instant, so no current flows through the externalcircuit. Further rotation of the coil results in an increasing emf, but ofthe opposite polarity to that induced before. Figure 7.4(c) shows theinstant when the emf has reached its next maximum. Although thegenerated emf is now reversed, the current through the external circuitwill be in the same direction as before. The load current will thereforebe a series of half-sinewave pulses, of the same polarity. Thus thecommutator is providing a d.c. output to the load, whereas the armaturegenerated emf is alternating.A single-turn coil will generate only a very small emf. An increasedamplitude of the emf may be achieved by using a multi-turn coil.

D.C. Machines 237E (V)0t (s)Fig. 7.5The resulting output voltage waveform is shown in Fig. 7.5 . Althoughthis emf is unidirectional, and may have a satisfactory amplitude,it is not a satisfactory d.c. waveform. The problem is that we havea concentrated winding. In a practical machine the armature has anumber of multi-turn coils. These are distributed evenly in slots aroundthe periphery of a laminated steel core. Each multi-turn coil has itsown pair of slots, and the two ends are connected to its own pair ofcommutator segments. Figure 7.6 shows the armature construction(before the coils have been inserted). The riser is the section of theriserslotsegmentsFig. 7.6commutator to which the ends of the coils are soldered. Due to thedistribution of the coils around the armature, their maximum inducedemfs will occur one after the other, i.e. they will be out of phase witheach other. Figure 7.7 illustrates this, but for simplicity, only three coilshave been considered.E (V)0t (s)Fig. 7.7Nevertheless, the effect on the resultant machine output voltage isapparent, and is shown by the thick line along the peaks of the waveform.

238 Fundamental Electrical and Electronic PrinciplesWith a large number of armature coils the ripple on the resultantwaveform will be negligible, and a smooth d.c. output is produced.7.3 Construction of d.c. MachinesThe various parts of a small d.c. machine are shown separately inFig. 7.8 , with the exception that neither the field nor armature windingshave been included. The frame shell (bottom left) contains the polepieces, around which the field winding would be wound. One endframe (top left) would simply contain a bearing for the armature shaft.Fig. 7.8The other end frame (bottom right) contains the brushgear assemblyin addition to the other armature shaft bearing. The armature (topright) construction has already been described. The slots are skewed toprovide a smooth starting and slow-speed torque.7.4 Classification of GeneratorsD.C. generators are classified according to whether the field windingis electrically connected to the armature winding, and if it is, whetherit is connected in parallel with or in series with the armature. The fieldcurrent may also be referred to as the excitation current. If this currentis supplied internally, by the armature, the machine is said to be selfexcited.When the field current is supplied from an external d.c. source,the machine is said to be separately excited. The circuit symbol used

D.C. Machines 239for the field winding of a d.c. machine is simply the same as that usedto represent any other form of winding. The armature is representedby a circle and two ‘ brushes ’ . The armature conductors, as such, arenot shown.7.5 Separately Excited GeneratorThe circuit diagram of a separately excited generator is shown inFig. 7.9 . The rheostat, R 1 , is included so that the field excitationcurrent, I f , can be varied. This diagram also shows the armature beingdriven at constant speed by some primemover. Since the armatureof any generator must be driven, this drive is not normally shown.The load, R L , being supplied by the generator may be connected ordisconnected by switch S 2 . The resistance of the armature circuit isrepresented by R a .S 1 I a I LRR LEVFig. 7.9Consider the generator being driven, with switches S 1 and S 2 bothopen. Despite the fact that there will be zero field current, a small emfwould be measured. This emf is due to the small amount of residualmagnetism retained in the poles. With switch S 1 now closed, the fieldcurrent may be increased in discrete steps, and the correspondingvalues of generated emf noted. A graph of generated emf versus fieldcurrent will be as shown in Fig. 7.10 , and is known as the open-circuitcharacteristic of the machine.emfE 20 I f1 I f2 I fE 1Fig. 7.10

240 Fundamental Electrical and Electronic PrinciplesIt will be seen that the shape of this graph is similar to the magnetizationcurve for a magnetic material. This is to be expected, since theemf will be directly proportional to the pole flux. The ‘ flattening ’ of theemf graph indicates the onset of saturation of the machine ’ s magneticcircuit. When the machine is used in practice, the field current wouldnormally be set to some value within the range indicated by I f1 andI f2 on the graph. This means that the facility exists to vary the emfbetween the limits E 1 and E 2 volts, simply by adjusting rheostat R 1 .Let the emf be set to some value E volt, within the range specifiedabove. If the load is now varied, the corresponding values of terminalvoltage, V and load current I L may be measured. Note that with thismachine the armature current is the same as the load current. The graphof V versus I L is known as the output characteristic of the generator,and is shown in Fig. 7.11 . The terminal p.d. of the machine will be lessthan the generated emf, by the amount of internal voltage drop due toR a , such that:V E I R volt (7.1)aavoltageEVI a R a0I L I aFig. 7.11Ideally, the graph of E versus I L would be a horizontal line. However,an effect known as armature reaction causes this graph to ‘ droop ’ at thehigher values of current. The main advantage of this type of generatoris that there is some scope for increasing the generated emf in orderto offset the internal voltage drop, I a R a , as the load is increased. Thebig disadvantage is the necessity for a separate d.c. supply for the fieldexcitation.7.6 Shunt GeneratorThis is a self-excited machine, where the field winding is connectedin parallel (shunt) with the armature winding. The circuit diagram is

D.C. Machines 241shown in Fig. 7.12 , and from this it may be seen that the armature hasto supply current to both the load and the field, such that:Ia IL Ifamp (7.2)I fI LI aR aSR fEVR LFig. 7.12This self-excitation process can take place only if there is someresidual flux in the poles, and if the resistance of the field circuit isless than some critical value. The open-circuit characteristic isillustrated in Fig. 7.13 .E0I fFig. 7.13The resistance of the field winding, R f , is constant and of a relativelyhigh value compared with R a . Typically, I f will be in the order of 1Ato 10A, and will remain reasonably constant. The shunt machine istherefore considered to be a constant-flux machine. When switchS is closed, the armature current will increase in order to supply thedemanded load current, I L . Thus I a I L , and as the load current isincreased, so the terminal voltage will fall, according to the equation,V E I a R a volt. The output characteristic will therefore followmuch the same shape as that for the separately excited generator andis shown in Fig. 7.14 . This condition applies until the machine isproviding its rated full-load output. If the load should now demandeven more current, i.e. the machine is overloaded, the result is that thegenerator simply stops generating. This effect is shown by the dottedlines in the output characteristics.

242 Fundamental Electrical and Electronic PrinciplesvoltageEVI a R a0 Ifull-load LFig. 7.14The shunt generator is the most commonly used d.c. generator, since itprovides a reasonably constant output voltage over its normal operatingrange. Its other obvious advantage is the fact that it is self-exciting,and therefore requires only some mechanical means of driving thearmature.7.7 Series GeneratorIn this machine the field winding is connected in series with thearmature winding and the load, as shown in Fig. 7.15 . In this case,I L I a I f , so this is a variable-flux machine. Since the field windingmust be capable of carrying the full-load current (which could bein hundreds of amps for a large machine), it is usually made from afew turns of heavy gauge wire or even copper strip. This also has theadvantage of offering a very low resistance. This generator is a selfexcitedmachine, provided that it is connected to a load when started.Note that a shunt generator will self-excite only when disconnectedfrom its load.When the load on a series generator is increased, the flux produced willincrease, in almost direct proportion. The generated emf will thereforeI LI a I f I LR fI fR aEI aVR LFig. 7.15

D.C. Machines 243increase with the demanded load. The increase of flux, and hencevoltage, will continue until the onset of magnetic saturation, as shownin the output characteristic of Fig. 7.16 . The terminal voltage is relatedto the emf by the equation:V EI ( R R) volt (7.3)a a fThe variation of terminal voltage with load is not normally arequirement for a generator, so this form of machine is seldom used.However, the rising voltage characteristic of a series-connectedfield winding is put to good use in the compound machine, which isdescribed in Further Electrical and Electronic Principles .emfEV0 I LI a (R a R f )Fig. 7.16Worked Example 7.1Q The resistance of the field winding of a shunt generator is 200 Ω . When the machine is delivering80 kW the generated emf and terminal voltage are 475 V and 450 V respectively. Calculate (a) thearmature resistance, and (b) the value of generated emf when the output is 50 kW, the terminalvoltage then being 460 V.AR f 200 Ω ; P o 80 10 3 watt; V 450 V; E 475 VThe circuit diagram is shown in Fig. 7.17 . It is always good practice to sketch theappropriate circuit diagram when solving machine problems.I aI LI fR aE475VR f200 ΩV450V80kWP oFig. 7.17

244 Fundamental Electrical and Electronic Principles(a) P VIPowatt; so I ampVo L L80therefore IL 1031 77.8 A450aV 450If amp 225. ARf200I I Iamp 180.05Aa L fI R EVvolt 47545025Va25therefore R a ohm0. 139Ω Ans180.05(b) When Po50103W, V 460V50thus L 103I108.7A460V 460If 23 . ARf200hence, Ia108. 72.3111AEVIaRavolt 460(1110.139)therefore E 475. 4 V AnsNote : Although the load had changed by about 60%, the field current haschanged by only about 2.2%. This justifies the statement that a shunt generatoris considered to be a constant-flux machine.7.8 D.C. MotorsAll of the d.c. generators so far described could be operated as motors,provided that they were connected to an appropriate d.c. supply. Whenthe machine is used as a motor, the armature generated emf is referredto as the back-emf, E b , which is directly proportional to the speed ofrotation. However, the speed is inversely proportional to the field flux Φ .In addition, the torque produced by the machine is proportional to boththe flux and the armature current. Bearing these points in mind, we cansay that:Speed, ω E bΦ(7.4)and torque, T Φ I a(7.5)7.9 Shunt MotorWhen the machine reaches its normal operating temperature, R f willremain constant. Since the field winding is connected directly to a

I fR fR aD.C. Machines 245I LI aVE bFig. 7.18fixed supply voltage, V volt, then I f will be fixed. Thus, the shuntmotor ( Fig. 7.18 ) is a constant-flux machine.As the back-emf will have the same shape graph as that for thegenerator emf, and using (7.4) and (7.5) the graphs of speed and torqueversus current will be as in Fig. 7.19 . Note that when the machineis used as a motor, the supply current is identified as I L . In this case,the subscript ‘ L ’ represents the word ‘ line ’ . Thus I L identifies the linecurrent drawn from the supply, and I a is directly proportional to I L .speed ortorqueTω0 I LFig. 7.197.10 Series MotorShunt motors are used for applications where a reasonably constantspeed is required, between no-load and full-load conditions.Like the series generator, this machine is a variable-flux machine.Despite this, the back-emf of this motor remains almost constant,from light-load to full-load conditions. This fact is best illustrated byconsidering the circuit diagram ( Fig. 7.20 ), with some typical values.E V I ( R R) volt (7.6)b a a f

246 Fundamental Electrical and Electronic PrinciplesR fIL I aFig. 7.20R aVE bLet us assume the following: V 200 V; R a 0.15 Ω ; R f 0.03 Ω ;I a 5 A on light-load; I a 50 A on full-loadLight-load: Eb200 5( 0. 150. 03) 199.1VFull-load: E 200 50( 0. 150.03) 191VbFrom the above figures, it may be seen that although the armaturecurrent has increased tenfold, the back-emf has decreased by only 4%.Hence, E b remains sensibly constant.Since ω E b /Φ , and E b is constant, then:ω 1 ...............[ 1]ΦSimilarly, T ΦI a , and since Φ I a until the onset of magneticsaturation,then T I 2 a ..................................[2]and after saturation, T ................[ 3]Using [1] to [3] above, the speed and torque characteristics shown inFig. 7.21 may be deduced.I aspeed ortorqueTT I aT I 2 aω0 I LFig. 7.21Note: From the speed characteristic it is clear that, on very light loads,the motor speed would be excessive. Theoretically, the no-load speed

D.C. Machines 247would be infinite! For this reason a series motor must NEVER bestarted unless it is connected to a mechanical load sufficient to preventa dangerously high speed. Similarly, a series motor must not be used tooperate belt-driven machinery, lifting cranes etc., due to the possibilityof the load being suddenly disconnected. If a series motor is allowedto run on a very light load, its speed builds up very quickly. Theprobable outcome of this is the distintegration of the machine, with theconsequent dangers to personnel and plant.The series motor has a high starting torque due to the ‘ square-law ’response of the torque characteristic. For this reason, it tends to beused mainly for traction purposes. For example, an electric train enginerequires a very large starting torque in order to overcome the massiveinertia of a stationary train.Summary of EquationsGenerators:Shunt generator: I a I L I f ampV EI R voltSeries generator: I a I L I f ampaaV EI ( R R) volta a fMotors:Shunt motor: E b V I a R a voltSeries motor: E b V I a (R a R f ) voltEbSpeed equation: n ΦEbrev/second; or ω Φrad/secondTorque equation: T ΦI a newton metre

248 Fundamental Electrical and Electronic PrinciplesAssignment Questions1 A shunt generator supplies a current of85 A at a terminal p.d. of 380 V. Calculatethe generated emf if the armature and fieldresistances are 0.4 Ω . and 95 Ω respectively.2 A generator produces an armature current of50 A when generating an emf of 400 V. If theterminal p.d. is 390 V, calculate (a) the valueof the armature resistance, and (b) the powerloss in the armature circuit.3 A d.c. shunt generator supplies a 50 kW loadat a terminal voltage of 250 V. The armatureand field circuit resistances are 0.15 Ω and5 0 Ω respectively. Calculate the generatedemf.

Chapter 8D.C. TransientsLearning OutcomesThis chapter explains the response of capacitor-resistor, and inductor-resistor circuits, whenthey are connected to and disconnected from, a d.c. supply.On completion of this chapter you should be able to:1 Show how the current and capacitor voltage in a series C-R circuit varies with time, whenconnected to/disconnected from a d.c. supply.2 Show how the current through, and p.d. across an inductor in a series L-R circuit varies withtime, when connected to/disconnected from a d.c. supply.3 Defi ne the term time constant for both types of above circuits.8.1 Capacitor-Resistor Series Circuit (Charging)Before dealing with the charging process for a C-R circuit, let usfirstly consider an analogous situation. Imagine that you need to inflatea ‘ flat ’ tyre with a foot pump. Initially it is fairly easy to pump airinto the tyre. However, as the air pressure inside the tyre builds up,it becomes progressively more difficult to force more air in. Also, asthe internal pressure builds up, the rate at which air can be pumpedin decreases. Comparing the two situations, the capacitor (which is tobe charged) is analogous to the tyre; the d.c. supply behaves like thepump; the charging current compares to the air flow rate; and the p.d.developed between the plates of the capacitor has the same effect asthe tyre pressure. From these comparisons we can conclude that as thecapacitor voltage builds up, it reacts against the emf of the supply, soslowing down the charging rate. Thus, the capacitor will charge at anon-uniform rate, and will continue to charge until the p.d. between itsplates is equal to the supply emf. This last point would also apply totyre inflation, when the tyre pressure reaches the maximum pressureavailable from the pump. At this point the air flow into the tyre would249

250 Fundamental Electrical and Electronic Principlescease. Similarly, when the capacitor has been fully charged, thecharging current will cease.Let us now consider the C-R charging circuit in more detail. Such acircuit is shown in Fig. 8.1 . Let us assume that the capacitor is initiallyfully discharged, i.e. the p.d. between its plates ( v C ) is zero, as will bethe charge, q. Note that the lowercase letters v and q are used because,during the charging sequence, they will have continuously changingvalues, as will the p.d. across the resistor (v R ) and the charging current,i. Thus these quantities are said to have transient values.RCiV RV C‘B’‘A’EFig. 8.1At some time t 0, let the switch be moved from position ‘ A ’ toposition ‘ B ’ . At this instant the charging current will start to flow.Since there will be no opposition offered by capacitor p.d. (v C 0),then only the resistor, R, will offer any opposition. Consequently, theinitial charging current ( I 0 ) will have the maximum possible value forthe circuit. This initial charging current is therefore given by:EI0 Ramp (8.1)Since we are dealing with a series d.c. circuit, then the followingequation must apply at all times:thus, at time t 0E v v volt...............[ 1]RCE v R0i.e. the full emf of E volt is developed across the resistor at the instantthe supply is connected to the circuit. Since v R iR, and at time t 0,i I 0 , this confirms equation (8.1) above.Let us now consider the situation when the capacitor has reached itsfully-charged state. In this case, it will have a p.d. of E volt, a charge ofQ coulomb, and the charging current, i 0. If there is no current flowthen the p.d. across the resistor, v R 0, and eqn [1] is:E 0v C

D.C. Transients 251Having confirmed the initial and final values for the transients, we nowneed to consider how they vary, with time, between these limits. It hasalready been stated that the variations will be non-linear (i.e. nota straight line graph). In fact the variations follow an exponential law.Any quantity that varies in an exponential fashion will have a graphlike that shown in Fig. 8.2(a) if it increases with time, and as inFig. 8.2(b) for a decreasing function.XXinitial rate of changeXX 00.632initial rate of change0.3680τ2τ 3τ 4τ 5τ t(s)(a)0τ2τ 3τ 4τ 5τ t(s)(b)Fig. 8.2In Fig. 8.2(a) , X represents the final steady state value of the variable x,and in Fig. 8.2(b) , X 0 represents the initial value of x . In each case thestraight line (tangent to the curve at time t 0) indicates the initial rateof change of x . The time interval shown as τ shown on both graphs isknown as the time constant, which is defined as follows:The time constant is the time that it would take the variable to reach itsfinal steady state if it continued to change at its initial rate.From the above Figures it can be seen that for an increasingexponential function, the variable will reach 63.2% of its final valueafter one time constant, and for a decreasing function it will fall to36.8% of its initial value after τ seconds.Note: Considering any point on the graph, it would take one timeconstant for the variable to reach its final steady value if it continued tochange at the same rate as at that point. Thus an exponential graph maybe considered as being formed from an infinite number of tangents,each of which represents the slope at a particular instant in time. Thisis illustrated in Fig. 8.3 .Also, theoretically, an exponential function can never actually reachits final steady state. However, for practical purposes it is assumedthat the final steady state is achieved after 5 time constants. This isjustifiable since the variable will be within 0.67% of the final valueafter 5 τ seconds. So for Fig. 8.2(a) , after 5 τ seconds, x 0.9973 X.

252 Fundamental Electrical and Electronic PrinciplesXX0τ 2τ 3τ 4τ 5τ t(s)Fig. 8.3Considering the circuit of Fig. 8.1 , assuming that the capacitor is fullydischarged, let the switch be moved to position ‘ B ’ . The capacitor willnow charge via resistor R until the p.d. between its plates, v C E volts.Once fully charged, the circuit current will be zero. The variations ofcapacitor voltage and charge, p.d. across the resistance and chargingcurrent are shown in Figs. 8.4 to 8.7.V c (V)Einitial rate of changeq (C)Qinitial rate of change0.632 E0.632 Q0τ2τ 3τ 4τ 5τ t(s)0τ2τ 3τ 4τ 5τ t(s)Fig. 8.4Fig. 8.5V R (V)i (A)EI 0initial rate of changeinitial rate of change0.386 E0.368 I 00τ2τ 3τ 4τ 5τ t(s)0τ2τ 3τ 4τ 5τ t(s)Fig. 8.6Fig. 8.7

D.C. Transients 253For such a CR circuit the time constant, τ (Greek letter tau), is CRseconds. It may appear strange that the product of capacitance andresistance yields a result having units of time. This may be justified byconsidering a simple dimensional analysis, as follows.Q ItC and R V VIt Vso, CR tsecondsV IHence,VIτ CR seconds(8.2)Worked Example 8.1Q An 8 μ F capacitor is connected in series with a 0.5 M Ω resistor, across a 200 V d.c. supply. Calculate(a) the circuit time constant, (b) the initial charging current, (c) the p.d.s across the capacitor andresistor 4 seconds after the supply is connected. You may assume that the capacitor is initially fullydischarged.AC 8 10 6 F; R 0.5 10 6 Ω ; E 200 V0.5 MΩ 8 μF(a)(b)200 VFig. 8.8τ CR second8100.510so τ 4 s AnsItherefore I6 60 60ERamp 20005 . 10 400 μAAns(c) After τ seconds, v 0. 632 E volt 0.632200vvCCRR126.4 V Ans EvCvolt2001 26.4so v 73. 6 V Ans8.2 Capacitor-Resistor Series Circuit (Discharging)Consider the circuit of Fig. 8.1 , where the switch has been in position‘ B ’ for sufficient time to allow the charging process to be completed.

254 Fundamental Electrical and Electronic PrinciplesThus the charging current will be zero, the p.d. across the resistor willbe zero, the p.d. across the capacitor will be E volt, and it will havestored a charge of Q coulomb.At some time t 0, let the switch be moved back to position ‘ A ’ . Thecapacitor will now be able to discharge through resistor R. The generalequation for the voltages in the circuit will still apply.In other words E v vbut, at the instant the switch is moved to position ‘ A ’ , the source ofemf is removed. Applying this condition to the general equation aboveyields:0 vR vC; where vC E and vR I0Rso 0 IRE0hence I0RCE amp(8.3)RThis means that the initial discharge current has the same value asthe initial charging current, but (as you would expect) it flows in theopposite direction.Since the capacitor is discharging, then its voltage will decay fromE volt to zero; its charge will decay from Q coulomb to zero; and thedischarge current will also decay from I 0 to zero. The circuit timeconstant will be the same as before i.e. τ CR seconds.The graphs for v C and i are shown in Fig. 8.9 .V c (V)EV c0τ5τt(s)iI 0iFig. 8.9

D.C. Transients 255Note: The time constant for the C-R circuit was defined previously interms of the capacitor charging. However, a time constant also appliesto the discharge conditions. It is therefore better to define the timeconstant in a more general manner, as follows:The time constant of a circuit is the time that it would have taken forany transient variable to change, from one steady state to a new steadystate, if it had maintained its rate of change existing at the time of thefirst steady state.Worked Example 8.2QA C-R charge/discharge circuit is shown in Fig. 8.10 . The switch has been in position ‘ 1 ’ for a sufficienttime to allow the capacitor to become fully discharged.(a) If the switch is now moved to position ‘ 2 ’ , calculate the time constant and initial charging current.(b) After the capacitor has completely charged the switch is moved back to position ‘ 2 ’ . Calculate thetime constant and the p.d across R 2 at this time.R 1R 2C220 kΩ0.5 μF‘2’ ‘1’110 kΩE150 VFig. 8.10AC 0.5 μ F ; R 1 220 k Ω ; R 2 110 k Ω ; E 150 V(a)When charging, only resistor R 1 is connected in series with the capacitor,so R 2 may be ignored.τ CR 1 seconds 0.5 10 22010so τ 0. 11 s AnsEI0 ampR150220103I 682 μA Ans06 3(b)When discharging, both R 1 and R 2 are connected in series with thecapacitor, so their combined resistance R R 1 R 2 , will determine thedischarge time constant.τ CR seconds 0.510330 10so, τ 0.16sAns6 3

256 Fundamental Electrical and Electronic PrinciplesAfter one time constant the discharge current will have fallen to 0.368 I 0II0 30E 150 amp R 33010 454.5 μAi 0.36845410i 167.26 μAv R 2 2v R 2 16 iR volt 167.26101101084 . V Ans6 38.3 Inductor-Resistor Series Circuit (Connection to Supply)Consider the circuit of Fig. 8.11. At some time t 0, the switch ismoved from position ‘ A ’ to position ‘ B ’ . The connection to the supply isnow complete, and current will start to flow, increasing towards its finalRLiV RV L‘B’‘A’EFig. 8.11steady value. However, whilst the current is changing it will induce aback-emf across the inductor, of e volt. From electromagnetic inductiontheory we know that this induced emf will have a value given by:eL di dtvoltBeing a simple series circuit, Kirchhoff ’ s voltage law will apply, suchthat the sum of the p.d.s equals the applied emf. Also, since we areconsidering a perfect inductor (the resistor shown may be considered asthe coil ’ s resistance), the p.d. across the inductor will be exactly equalbut opposite in polarity to the induced emf.dTherefore, v e L i L dtvolthence, E vRvLvoltdor, E iRL idvolt...............[ 1]tComparing this equation with that for the C-R circuit, it may be seenthat they are both of the same form. Using the analogy technique, we

D.C. Transients 257can conclude that both systems will respond in a similar manner. In thecase of the L-R circuit, the current will increase from zero to its finalsteady value, following an exponential law.At the instant that the switch is moved from ‘ A ’ to ‘ B ’ ( t 0), thecurrent will have an instantaneous value of zero, but it will have acertain rate of change, d i /d t amp/s. From eqn [1] above, this initial rateof change can be obtained, thus:dE 0L i dtso, initial d i E amp/s(8.4)dtLWhen the current reaches its final steady value, there will be no backemfacross the inductor, and hence no p.d. across it. Thus the onlylimiting factor on the current will then be the resistance of the circuit.The final steady current is therefore given by:IE amp (8.5)RThe time constant of the circuit is obtained by dividing the inductanceby the resistance.Thus τ L secondsR(8.6)The above equation may be confirmed by using a simple form ofdimensional analysis, as follows.LI VtIn general, V ; so L tIVand R IL Vt Itherefore, tsecondsR I VThe time constant of the circuit may be defined in the general termsgiven in the ‘Note ’ , in the previous section, dealing with the C-Rcircuit.The rate of change of current will be at its maximum value at time t 0,so the p.d. across the inductor will be at its maximum value at this time.This p.d. therefore decays exponentially from E volt to zero. The graphsfor i, v R , and v L are shown in Figs. 8.12 to 8.14 respectively.

258 Fundamental Electrical and Electronic Principlesi (A)IinitialdidtV R (V)E0.632 I0.632 E0 2τ 3τ 4τ 5τ t(s)τ0τ2τ 3τ 4τ 5τ t(s)Fig. 8.12Fig. 8.13V L (V)Einitial rate of change0.368 E0τ 2τ 3τ 4τ 5τ t(s)Fig. 8.14Worked Example 8.3QThe field winding of a 110 V, d.c. motor has an inductance of 1.5 H, and a resistance of 220 Ω . Fromthe instant that the machine is connected to a 110 V supply, calculate (a) the initial rate of changeof current, (b) the final steady current, and (c) the time taken for the current to reach its final steadyvalue.AE 110 V; L 1.5 H; R 220 ΩThe circuit diagram is shown in Fig. 8.15 .220 Ω 1.5 HiE110 VFig. 8.15

D.C. Transients 259(a)initial d idtEL110amp/s 1.5so, initial d i 73. 33A/sAnsdtE 110(b) final current, I amp R 220therefore, I 05 . A Ans(c)Lτ R second 1. 5220hence, τ 682 . msSince the system takes approximately 5 τ seconds to reach its new steady state,then the current will reach its final steady value in a time:t 56. 82 ms 34.1 ms Ans8.4 Inductor-Resistor Series Circuit (Disconnection)Figure 8.16 shows such a circuit, connected to a d.c. supply. Assumethat the current has reached its final steady value of I amps. Let theswitch now be returned to position ‘ A ’ (at time t 0). The current willnow decay to zero in an exponential manner. However, the decayingcurrent will induce a back-emf across the coil. This emf must opposethe change of current. Therefore, the decaying current will flow inthe same direction as the original steady current. In other words, theback-emf will try to maintain the original current flow. The graph ofthe decaying current, with respect to time, will therefore be as shownin Fig. 8.17. The time constant of the circuit will, of course, still be L / Rsecond, and the current will decay from a value of I E / R amp. Theinitial rate of decay will also be E / L amp/s.i (A)IRLi0.368 I‘B’‘A’EFig. 8.160τ 2τ 3τ 4τ 5τ t(s)Fig. 8.17

260 Fundamental Electrical and Electronic PrinciplesSummary of EquationsC-R circuit:Time constant, τ CR secondEInitial current, I0 Rampsteady-state conditions after approx. 5 τ secondafter, τ second, v C 0.632E volt; and i 0.368I 0 ampL-R circuit:Time constant, τ L secondRInitial rate of change of current, d i E amp/seconddtLEfinal current flowing, I ampRsteady-state conditions after approx. 5 τ secondafter τ second, V L 0.368 E volt; and i 0.632 I amp

D.C. Transients 261Assignment Questions1 A 4 7 μ F capacitor is connected in series witha 39 k Ω resistor, across a 24 V d.c. supply.Calculate (a) the circuit time constant, (b) thevalues for initial and final charging current,and (c) the time taken for the capacitor tobecome fully charged.2 A 150 mH inductor of resistance 50 Ω isconnected to a 50 V d.c. supply. Determine(a) the initial rate of change of current, (b) thefinal steady current, and (c) the time taken forthe current to change from zero to its finalsteady value.3 An inductor of negligible resistance and ofinductance 0.25 H, is connected in series with a1.5 kΩ resistor, across a 24 V d.c. supply. Calculatethe current flowing after one time constant.4 A 5 H inductor has a resistance R ohm.This inductor is connected in series witha10 Ω resistor, across a 140 V d.c. supply. Ifthe resulting circuit time constant is 0.4 s,determine (a) the value of the coil resistance,and (b) the final steady current.5 D e fine the time constant of a capacitor-resistorseries circuit.Such a circuit comprises a 50 μ F capacitor anda resistor, connected to a 100 V d.c. supply viaa switch. If the circuit time constant is to be5 s, determine (a) the resistor value, (b) theinitial charging current.6 The dielectric of a 20 μ F capacitor has aresistance of 65 M Ω . This capacitor is fullycharged from a 120 V d.c. supply. Calculatethe time taken, after disconnection fromthe supply, for the capacitor to become fullydischarged.

262 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentsAssignment 1Apparatus:Method:To investigate the variation of capacitor voltage and current during charge anddischarge cycles.1 10 μ F capacitor1 10 M Ω resistor1 2-pole switch1 d.c. power supply1 ammeter (microammeter)1 DVM (with highest possible input resistance)1 stopwatchV10 μFA10 MΩ‘1’ ‘2’250 VFig. 8.181 Connect the circuit of Fig. 8.18 , and adjust the psu output to 250 V.2 Simultaneously move the switch to position ‘ 1 ’ and start the stopwatch.3 Record the circuit current and capacitor p.d. at 10 s intervals, for the first 60 s.4 Continue recording the current and voltage readings, at 20 s intervals, for afurther 4 minutes. Reset the stopwatch to zero. Reverse the connections tothe ammeter.5 Move the switch back to position ‘ 2 ’ , and repeat the procedures ofparagraphs (3) and (4) above.6 Plot graphs of current and capacitor p.d., versus time, for both the chargingand discharging cycles.7 Submit a complete assignment report, which should include the following:(i) The comparison of the actual time constant (determined from theplotted graphs) to the theoretical value. Explain any discrepancy found.(ii) Explain why both the charging and discharging currents tend to ‘ leveloff ’ at some small value, rather than continuing to decrease to zero.

Chapter 9Semiconductor Theoryand DevicesLearning OutcomesThis chapter explains the behaviour of semiconductors and the way in which they areemployed in diodes.On completion of this chapter you should be able to:1 Understand the way in which conduction takes place in semiconductor materials.2 Understand how these materials are employed to form devices such as diodes.3 Understand the action of a zener diode and perform basic calculations involving a simpleregulator circuit.9.1 Atomic StructureIn Chapter 1 it was stated that an atom consists of a central nucleuscontaining positively charged protons, and neutrons, the latter beingelectrically neutral, surrounded by negatively charged electronsorbiting in layers or shells. Electrons in the inner orbits or shellshave the least energy and are tightly bound into their orbits due tothe electrostatic force of attraction between them and the nucleus.Electrons in the outermost shell experience a much weaker bindingforce, and are known as valence electrons.In conductors it is these valence electrons that can gain sufficientenergy to break free from their parent atoms. They thus become‘ free ’ electrons which are available to drift through the material underthe influence of an emf and hence are mobile charge carriers whichproduce current flow.The shells are identified by letters of the alphabet, beginning with theletter K for the innermost shell, L for the next and so on. Each shellrepresents a certain energy level, and each shell can contain only up to a263

264 Fundamental Electrical and Electronic Principlescertain maximum number of electrons. This maximum possible numberof electrons contained in a given shell is governed by the relationship2n 2 , where n is the number of the shell. Thus the maximum number ofelectrons in the first four shells will be as shown in Table 9.1 .Table 9.1Shell n Max. No. ofelectronsK 1 2 1 2 2L 2 2 2 2 8M 3 2 3 2 18N 4 2 4 2 32All things in nature tend to stabilise at their lowest possible energylevel, and atoms and electrons are no exception. This results in thelowest energy levels (shells) being filled first until all the electronsbelonging to that atom are accommodated. Another feature of thesystem is that if the outermost shell of an atom is completely full(contains its maximum permitted number of electrons) then the bindingforce on these valence electrons is very strong and the atom is verystable. To illustrate this consider the inert gas neon. The term inert isused because it is very difficult to make it react to external influences.A neon atom has a total of 10 electrons, two of which are in the Kshell and the remaining eight completely fill the L shell. Having a fullvalence shell is the reason why neon, krypton and xenon are inert gases.In contrast, a hydrogen atom has only one electron, so its valence shellis almost empty and it is a highly reactive element. One further pointto bear in mind is that the electrons in the shells (from L onwards)may exist at slightly different energy levels known as subshells. Thesesubshells may also contain only up to a certain maximum number ofelectrons. This is shown, for the L, M and N shells, in Table 9.2 .Table 9.2Shell L M NSubshell 2s 2p 3s 3p 3d 4s 4p 4d 4fMax. No. 2 6 2 6 10 2 6 10 14Total 8 18 329.2 Intrinsic (Pure) SemiconductorsSemiconductors are group 4 elements, which means they have fourvalence electrons. For this reason they are also known as tetravalent

Semiconductor Theory and Devices 265elements. Among this group of elements are carbon (C), silicon (Si),germanium (Ge) and tin (Sn). Of these only silicon and germaniumare used as intrinsic semiconductors, with silicon being the mostcommonly used. Carbon is not normally considered as a semconductorbecause it can exist in many different forms, from diamond to graphite.Similarly, tin is not used because at normal ambient temperatures itacts as a good conductor. The following descriptions of the behaviourof semiconductor materials will be confined to silicon although thegeneral properties and behaviour of germanium are the same. Thearrangement of electrons in the shells and subshells of silicon areshown in Table 9.3 .Table 9.3K L M1s 2s 2p 3s 3p 3d2 2 6 2 2 —From Table 9.3 it may be seen that the four valence electrons arecontained in the M shell, where the 3 s subshell is full but the 3psubshell contains only two electrons. However, from Table 9.2 it canbe seen that a 3p subshell is capable of containing up to a maximum ofsix electrons before it is full, so in the silicon atom there is space for afurther four electrons to be accommodated in this outermost shell.Silicon has an atomic bonding system known as covalent bondingwhereby each of the valence electrons orbits not only its ‘ parent ’ atom,but also orbits its closest neighbouring atom. This effect is illustratedin Fig. 9.1 , where the five large circles represent the nucleus and shellsK and L of five adjacent atoms (identified as A, B, C, D and E) and thesmall circles represent their valence electrons, where the letters a, b, cetc. identify their ‘ parent ’ atoms.Concentrating on the immediate space surrounding atom A, it may beseen that there are actually eight valence electrons orbiting this atom;four of its own plus one from each of its four nearest neighbours. Thisfigure is only two-dimensional and is centred on atom A. However, thesame arrangement would be found if the picture was centred on any givenatom in the crystal lattice. In addition, the actual lattice is of course threedimensional.In this case imagine atom A being located at the centre ofan imaginary cube with the other four neighbouring atoms being at fourof the corners of the cube. Each of these ‘ corner ’ atoms is in turn at thecentre of another imaginary cube, and so on throughout the whole crystallattice. The result is what is known as the diamond crystal lattice.From the above description it may be seen that each silicon atom hasan apparent valency of 8, which is the same as for the inert gases

266 Fundamental Electrical and Electronic PrinciplesEaedaD A BabcaCFig. 9.1such as neon. The covalent bonding system is a very strong one so thevalence electrons are quite tightly bound into it. It is for this reason thatintrinsic silicon is a relatively poor conductor of electrical current, andis called a semiconductor.9.3 Electron-Hole Pair Generation and RecombinationAlthough the covalent bond is strong, it is not perfect. Thus, whena sample of silicon is at normal ambient temperature, a few valenceelectrons will gain sufficient energy to break free from the bond and sobecome free electrons available as mobile charge carriers. Wheneversuch an electron breaks free and drifts away from its parent atom itleaves behind a space in the covalent bond, and this space is referredto as a hole. Thus, whenever a bond is broken an electron-hole pairis generated. This effect is illustrated in Fig. 9.2 , where the shortstraight lines represent electrons and the small circle represents acorresponding hole. The large circles again represent the silicon atomscomplete with their inner shells of electrons.The atom which now has a hole in its valence band is effectively apositive ion because it has lost an electron which would normallyoccupy that space. On the atomic scale, the ion is very massive,is locked into the crystal lattice, and so cannot move. However,electron-hole pair generation will be taking place in a random mannerthroughout the crystal lattice, and a generated free electron will at

Semiconductor Theory and Devices 267SiSiSiSiSiholeSifree electronSiSiFig. 9.2some stage drift into the vicinity of one of these positive ions, and becaptured, i.e. the hole will once more be filled by an electron. Thisprocess is known as recombination, and when it occurs the normalcharge balance of that atom is restored.The hole-pair generation and recombination processes occurcontinuously, and since heat is a form of energy, will increase as thetemperature increases. This results in more mobile charge carriers beingavailable, and accounts for the fact that semiconductors have a negativetemperature coefficient of resistance, i.e. as they get hotter they conductmore easily. It must be borne in mind that although these thermallygenerated mobile charge carriers are being produced, the sample ofmaterial as a whole still remains electrically neutral. In other words, if a‘ head count ’ of all the positive and negative charged particles could bemade, there would still be a balance between positive and negative, i.e.for every free electron there will be a corresponding hole.The concept of the drift of free electrons through the material may bereadily understood, but the concept of hole mobility is more difficult toappreciate. In fact the holes themselves cannot move — they are merelygenerated and filled. However, when a bond breaks down the electronthat drifts away will at some point fill a hole elsewhere in the lattice.Thus the hole that has been filled is replaced elsewhere by the newlygenerated hole, and will appear to have drifted to a new location. Inorder to simplify the description of conduction in a semiconductor, theholes are considered to be mobile positive charge carriers whilst thefree electrons are of course mobile negative charge carriers.9.4 Conduction in Intrinsic SemiconductorsFigure. 9.3 illustrates the effect when a source of emf is connectedacross a sample of pure silicon. The electric field produced by

268 Fundamental Electrical and Electronic PrincipleselectronsIholesI(electrons)the battery will attract free electrons towards the positive plateand the corresponding holes towards the negative plate. Since theexternal circuit is completed by conductors, and holes exist onlyin semiconductors, then how does current actually flow around thecircuit without producing an excess of positive charge (the holes) atthe left-hand end of the silicon? The answer is quite simple. For everyelectron that leaves the right-hand end and travels to the positive plateof the battery, another is released from the negative plate and entersthe silicon at the left-hand end, where a recombination can occur. Thisrecombination will be balanced by fresh electron-hole pair generation.Thus, within the silicon there will be a continuous drift of electronsin one direction with a drift of a corresponding number of holes in theopposite direction. In the external circuit the current flow is of coursedue only to the drift of electrons.9.5 Extrinsic (Impure) Semiconductors9.6 n-type SemiconductorFig. 9.3Although pure silicon and germanium will conduct, as explained inthe previous section, their characteristics are still closer to insulatorsthan to conductors. In order to improve their conduction very smallquantities (in the order of 1 part in 10 8 ) of certain other elements areadded. This process is known as doping. The impurity elements thatare added are either pentavalent (have five valence electrons) or aretrivalent (have three valence electrons) atoms. Depending upon whichtype is used in the doping process determines which one of the twotypes of extrinsic semiconductor is produced.To produce this type of semiconductor, pentavalent impurities areemployed. The most commonly used are arsenic (As), phosphorus (P),and antimony (Sb). When atoms of such an element are added to thesilicon a bonding process takes place such that each impurity atomjoins the covalent bonding system of the silicon. However, since eachimpurity atom has five valence electrons, one of these cannot find aplace in a covalent bond. These ‘ extra ’ electrons then tend to drift

Semiconductor Theory and Devices 269away from their parent atoms and become additional free electrons inthe lattice. Since these impurities donate an extra free electron to thematerial they are also known as donor impurities.As a consequence of each donor atom losing one of its valence electrons,they become positive ions locked into the crystal lattice. Note that freeelectrons introduced by this process do not leave a corresponding hole,although thermally generated electron-hole pairs will still be created inthe silicon. The effect of the doping process is illustrated in Fig. 9.4 .SiSiSi5 th ‘free’electronSiPSiSiSiFig. 9.4Since the extra charge carriers introduced by the impurity atom arenegatively charged electrons, and these will be in addition to theelectron-hole pairs, then there will be more mobile negative chargecarriers than positive, which is why the material is known as n-typesemiconductor. In this case the electrons are the majority chargecarriers and the holes are the minority charge carriers. It should againbe noted that the material as a whole still remains electrically neutralsince for every extra donated free electron there will be a fixed positiveion in the lattice. Thus a sample of n-type semiconductor may berepresented as consisting of a number of fixed positive ions with acorresponding number of free electrons, in addition to the thermallygenerated electron-hole pairs. This is shown in Fig. 9.5 .e–h pairfixedpositiveionfreeelectronFig. 9.5

270 Fundamental Electrical and Electronic PrinciplesholeselectronsIFig. 9.6The circuit action when a battery is connected across the materialis illustrated in Fig. 9.6 . Once more, only electrons flow aroundthe external circuit, whilst within the semiconductor there will bemovement of majority carriers in one direction and minority carriersin the opposite direction.9.7 p-type SemiconductorIn this case a trivalent impurity such as aluminium (Al), gallium(Ga), or indium (In) is introduced. These impurity atoms also jointhe covalent bonding system, but since they have only three valenceelectrons there will be a gap or hole in the bond where an electronwould normally be required. Due to electron-hole pair generation inthe lattice, this hole will soon become filled, and hence the hole willhave effectively drifted off elsewhere in the lattice. Since each impurityatom will have accepted an extra electron into its valence band theyare known as acceptor impurities, and become fixed negative ions. Theresult of the doping process is illustrated in Fig. 9.7 .SiSiSiSiAlSie–hpairSiSiFig. 9.7

Semiconductor Theory and Devices 271We now have the situation whereby there will be more mobile holesthan there are free electrons. Since holes are positive charge carriers,and they will be in the majority, the doped material is called p-typesemiconductor, and it may be considered as consisting of a numberof fixed negative ions and a corresponding number of mobile holes asshown in Fig. 9.8 .e–h pairfixednegativeionmobileholeFig. 9.8The circuit action when a battery is connected across the material isshown in Fig. 9.9 . As the holes approach the left-hand end they arefilled by incoming electrons from the battery. At the same time, freshelectron-hole pairs are generated; the electrons being swept to andout of the right-hand end, and the holes drift to the left-hand end tobe filled. Once more, the current flow in the semiconductor is due tothe movement of holes and electrons in opposite directions, and onlyelectrons in the external circuit. As with the n-type material, p-type isalso electrically neutral.holesIelectrons9.8 The p-n JunctionFig. 9.9When a sample of silicon is doped with both donor and acceptorimpurities so as to form a region of p-type and a second region ofn-type material in the same crystal lattice , the boundary where the tworegions meet is called a p-n junction. This is illustrated in Fig. 9.10 .

272 Fundamental Electrical and Electronic PrinciplespnFig. 9.10Due to their random movement some of the electrons will diffuseacross the junction into the p-type, and similarly some of the holes willdiffuse across into the n-type. This effect is illustrated in Fig. 9.11 , andfrom this figure it may be seen that region x acquires a net negativecharge whilst region y acquires an equal but positive net charge.The region between the dotted lines is only about 1 μ m wide, and thenegative charge on x prevents further diffusion on electrons from the n-type. Similarly the positive charge on y prevents further diffusion of holesfrom the p-type. This redistribution of charge results in a potential barrieracross the junction. In the case of silicon this barrier potential will bein the order of 0.6 to 0.7 V, and for germanium about 0.2 to 0.3 V. Onceagain note that although there has been some redistribution of charge, thesample of material as a whole is still electrically neutral (count up thenumbers of positive and negative charges shown in Fig. 9.11 ).xypnChargeFig. 9.119.9 The p-n Junction DiodeA diode is so called because it has two terminals: the anode, whichis the positive terminal, and the cathode, which is the negative

Semiconductor Theory and Devices 273terminal. In the case of a p-n junction diode the anode is the p-typeand the cathode is the n-type. In Chapter 6 it was stated that a diodewill conduct in one direction but not in the other. This behaviour isexplained as follows.9.10 Forward-biased DiodeFigure 9.12 shows a battery connected across a diode such that thepositive terminal is connected to the anode and the negative terminal tothe cathode.pnFig. 9.12The electric field produced by the battery will cause holes andelectrons to be swept toward the junction, where recombinations willtake place. For each of these an electron from the battery will enter thecathode. This would have the effect of disturbing the charge balancewithin the semicoductor, so to counterbalance this a fresh electron-holepair will be created in the p-type. This newly freed electron will thenbe attracted to the positive plate of the battery, whilst the hole will beswept towards the junction. Thus the circuit is complete, with electronsmoving through the external circuit, and a movement of holes andelectrons in the semiconductor. Hence, when the anode of the diode ismade positive with respect to the cathode it will conduct, and it is saidto be forward biased.9.11 Reverse-biased DiodeConsider what now happens when the battery connections are reversed( Fig. 9.13 ). The electric field of the battery will now sweep all themobile holes into the p-type and all the free electrons into then-type. This leaves a region on either side of the junction which hasbeen depleted of all of its mobile charge carriers. This layer thus acts

274 Fundamental Electrical and Electronic PrinciplespdepletionlayernFig. 9.13as an insulator, and is called the depletion layer. There has been aredistribution of charge within the semiconductor, but since the circuithas an insulating layer in it, current cannot flow. The diode is said to bein its blocking mode.However, there is no such thing as a perfect insulator, and the depletionlayer is no exception. Although all the mobile charge carriers providedby the doping process have been swept to opposite ends of thesemiconductor, there will still be some thermally generated electron-holepairs. If such a pair is generated in the p-type region, the electron will beswept across the junction by the electric field of the battery. Similarly,if the pair is generated in the n-type, the hole will be swept across thejunction. Thus a very small reverse current (in the order of microamps)will flow, and is known as the reverse leakage current. Since this leakgecurrent is the result of thermally generated electron-hole pairs, then asthe temperature is increased so too will the leakage current.9.12 Diode CharacteristicsThe characteristics of a device such as a diode can be best illustratedby means of a graph (or graphs) of the current flow through it versusapplied voltage. Circuits for determining both the forward and reversecharacteristics are shown in Fig. 9.14 .RARARV1VRV1V(a) Forward bias(b) Reverse biasFig. 9.14

Semiconductor Theory and Devices 275Note the change of position of the voltmeter for the two differenttests. In (a) the voltmeter measures only the small p.d. across thediode itself, and not any p.d. across the ammeter. In (b) the ammetermeasures only the leakage current of the diode, and does not includeany current drawn by the voltmeter. Ideally the voltmeter should notdraw any current at all, so it is recommended that a DVM is used ratherthan a moving coil instrument such as an AVOmeter. The procedurein each case is to vary the applied voltage, in steps, by means of RV1and record the corresponding current values. When these resultsare plotted, for both silicon and germanium diodes, the graphs willtypically be as shown in Fig. 9.15 . The very different scales for bothcurrent and voltage for the forward and reverse bias conditions shouldbe noted. Also, the actual values shown for the forward current scaleand the reverse voltage scale can vary considerably from those shown,depending upon the type of diode being tested, i.e. whether it be asmall signal diode or a power rectifying diode. In the case of the latter,the forward current would usually be in amperes rather than milliamps.I A (mA)GeSiForwardV A (V)300200 100050.25 0.5 0.75 1.0V A (V)10SiGe1520I A (μA)Fig. 9.15The sudden increase in reverse current occurs at a reverse voltageknown as the reverse breakdown voltage. The effect occurs because theintensity of the applied electric field causes an increase in electron-holepair generation. These electron-hole pairs are not due to temperature,but the result of electrons being torn from bonds by the electric field.This same field will rapidly accelerate the resulting charge carriers andas they cross the junction they will collide with atoms. These collisionswill free more charge carriers, and the whole process builds up veryrapidly. For this reason the effect is known as avalanche breakdown,and will usually result in the destruction of the diode.When the impurity doping of the semiconductor is heavier than ‘normal ’ ,the depletion layer produced is much thinner. In this case, whenbreakdown occurs, the charge carriers can pass through the depletion

276 Fundamental Electrical and Electronic Principleswith very little chance of collisions taking place. This type of breakdownis known as zener breakdown and such diodes are called zener diodes.9.13 The Zener DiodeThe main feature of the zener diode is its ability to operate in thereverse breakdown mode without sustaining permanent damage. Inaddition, during manufacture, the precise breakdown voltage (zenervoltage) for a given diode can be predetermined. For this reason theyare also known as voltage reference diodes. The major application forthese devices is to limit or stabilise a voltage between two points in acircuit. Diodes are available with zener voltages from 2.6 V to about200 V. The circuit symbol for a zener diode is shown in Fig. 9.16 .orFig. 9.16The forward characteristic for a zener diode will be the same as for anyother p-n junction diode, and also, since the device is always used in itsreverse bias mode, only its reverse characteristic need be considered.Such a characteristic is shown in Fig. 9.17 .V D (V)V ZOI Z (min)actualidealI Z (mA)Fig. 9.17In Fig. 9.17 , V Z represents the zener breakdown voltage, and if it werean ideal device, this p.d. across it would remain constant, regardlessof the value of current, I Z , flowing through it. In practice the graphwill have a fairly steep slope as shown. The inverse of the slope of thegraph is defined as the diode slope resistance, r Z , as followsrZV δ δIDZohm (9.1)

Semiconductor Theory and Devices 277Typical values for r Z range from 0.5 Ω to about 150 Ω . For satisfactoryoperation the current through the diode must be at least equal to I Z(min) .Due to the diode slope resistance, the p.d. across the diode will vary bya small amount from the ideal of V Z volt as the diode current changes.For example, if r Z 1 Ω and V Z 15 V, a change in diode current of30 mA would cause only a 0.02% change in the diode p.d. This figuremay be verified by applying equation (9.1).The value of current that may be allowed to flow through the devicemust be limited so as not to exceed the diode power rating. This powerrating is always quoted by the manufacturer, and zener diodes areavailable with power ratings up to about 75 W.Consider now the application of a zener diode to provide simplevoltage stabilisation to a load. A circuit is shown in Fig. 9.18 .I SI ZV Z V oR SI LVR LFig. 9.18In order for satisfactory operation the supply voltage, V S , needs to beconsiderably greater than the voltage required at the load. The purposeof the series resistor R S is to limit the maximum diode current to asafe value, bearing in mind the diode ’ s power rating. ConsideringFig. 9.18 , the diode current will be at its maximum when the load isdisconnected, because under this condition all of the current fromthe supply will flow through the diode, i.e. I Z I S . When the load isconnected it will draw a current I L , and since I Z I S – I L , then underthis condition the diode current will decrease, since it must divertcurrent to the load. The output voltage, however, will remain virtuallyunchanged. Knowing the diode power rating a suitable value for R Smay be calculated as shown in the following worked example. Thisexample also demonstrates the stabilising action of the circuit.Worked Example 9.1Q A 9.1 V, 500 mW zener diode is used in the circuit of Fig. 9.18 to supply a 2.5 k Ω load. The diode has aslope resistance of 1.5 Ω , and the input supply has a nominal value of 12 V.(a) Calculate a suitable value for the series resistor R s .(b) Calculate the value of diode current when the load resistor is connected to the circuit.(c) If the input supply voltage decreases by 10%, calculate the percentage change in the p.d. acrossthe load.

278 Fundamental Electrical and Electronic PrinciplesAV Z 9.1 V; P Z 0.5 W; r Z 1. 5 Ω ; V 12 V; R L 2500 Ω(a) P V IIso, IZ Z ZZZPZ05 . amp VZ9.1I54.95 mASwatt( I S I Z because for this condition the load is disconnected)VSVVZvolt 129.1VS 29 . VVS29 .RS ohmIS54.95103R 52.78 ΩSA resistor of this precise value would not be readily available, so the nearestpreferred value resistor would be chosen. However, to ensure that thediode power rating cannot be exceeded, the nearest preferred value greaterthan 52.78 Ω would be chosen. Thus a 56 Ω resistor would be chosen.In order to protect the resistor, its own power rating must be taken intoaccount. In this circuit, the maximum power dissipated by R S is:P max I S2 R S watt (54.95 10 3 ) 2 56P max 0 .169 W, so a 0.25 W resistor would be chosen,and the complete answer to part (a) is:R S should be a 56 Ω , 0.25 W resistor Ans(b)(c)With R L 2500 Ω and V o 9.1 VVo9.1IL amp RL25 . 103IL 364 . mAIZ IS ILamp ( 553.64) mAI 51.36 mA AnsZWhen V falls by 10% from its nominal value, thenV 12( 0. 112)121.2hence, V 108. VV VZS 108 . 9.1Iamp RS56 30 . 36 mAI SThe current for the load must still be diverted from the diode, soIZ IS ILamp ( 30. 363. 64) mA 26.72 mAtherefore, δI Z ( 51. 362672 . ) mA 2464. mA, and from equation ( 10.1):δVvolt 24 64 10 3Z δIZrZ . 1.5δV 0.037 VZ

Semiconductor Theory and Devices 279Thus the voltage applied to the load changes by 0.037 V, which expressedas a percentage change is:0.037change 1009.1change 04 . 1% Ans (compared with a 10% change in supply)Worked Example 9.2QA d.c. voltage of 15 V 5% is required to be supplied from a 24 V unstabilised source. This is to beachieved by the simple regulator circuit of Fig. 9.19 .24 VR SI SV SV ZI ZVV o0 VFig. 9.19The available diodes and resistors are listed below.(a) For each diode listed determine the appropriate resistor required and hence determine the totalunit cost for each circuit.(b) In order to satisfy the specified output voltage tolerance of 5%, determine which of the threecircuits will meet the specification at lowest cost.DiodeNo.V Z (V)Sloperesistance ( Ω )Max power(W)Unitcost (£)1 15 30 0.5 0.072 15 15 1.3 0.203 15 2.5 5.0 0.67Resistors are available in the following values and unit costs18, 27, 56, 100, 120, 150, 220, 270, and 330 Ω0.25 W £0.0260.5 W £0.0381.0 W £0.0552.5 W £0.2607.5 W £0.280

280 Fundamental Electrical and Electronic PrinciplesA(a)For all three diodes:V S ( V V Z ) volt 24 15 9 VDiode 1:PZ05 .I Z amp VZ15I Z 33.3 mAVS9RS ohmI Z 33. 310RS 270 ΩV2S81PS watt RS270P 03 . WSand, R S 270 Ω , 0.5 W Ans3total unit cost £(0.07 0.038) £0.045 AnsDiode 2:1.3I Z amp 86.7 mA159RS103.85 Ω , so choose the 120 Ω resistor86.710381PS 0. 675 W, so choose 1.0 W rating120hence, R S 12 0 Ω , 1.0 W Anstotal unit cost £(0.20 0.055) £0.26 AnsDiode 3:5I Z 333. 3 mA159RS 27 Ω333.310381PS 3 W so choose 7.5 W resistor27hence, R S 27 Ω , 7. 5 W Anstotal unit cost £ (0.67 0.28) £0.95 Ans5(b) Allowable δV Z 15 075 . V100Diode 1: δ δI volt 30103 30 0. 9 V, which is unacceptableV ZzrzDiode 2:δV 30 03Z 1 1 50.45 V, which is acceptableDiode 3:δV 30 03Z 1 25 . 0075. V, which is acceptableThus to meet the specification at lowest cost, circuit 2 would be adopted Ans

Semiconductor Theory and Devices 281Assignment Questions1 A simple voltage stabiliser circuit is shown inFig. 9.20 . The zener diode is a 7.5 V, 500 mWdevice and the supply voltage is 12 V.Calculate (a) a suitable value for R S , and (b) thevalue of zener current when R L 470 Ω .VR SR L2 Using the circuit of Fig. 9.20 with a 10 V, 1.3 Wzener diode having a slope resistance of 2.5 Ω ,and a supply voltage of 24 V, calculate (a) asuitable value for R S , (b) the value of zenercurrent when on no-load, and (c) the variationof zener voltage when R L is changed from500 Ω to 200 Ω .Fig. 9.20

282 Fundamental Electrical and Electronic PrinciplesSuggested Practical AssignmentsAssignment 1Assignment 2Apparatus:Method:To obtain the forward and reverse characteristics for silicon and germanium p-njunction diodes. An investigation into the effects of increased temperature onthe reverse leakage current could also be undertaken.To investigate the operation of the zener diode.1 5.6 V, 400 mW zener diode1 9.1 V, 400 mW zener diode1 470 Ω resistor1 variable d.c. power supply unit (psu)1 voltmeter (DVM)1 ammeter1 Connect the circuit of Fig. 9.21 using the 5.6 V diode.2 Vary the input voltage in 1 V steps from 0 V to 15 V, and note thecorresponding values of V Z and I.3 Tabulate your results and plot the reverse characteristic for the diode.470 ΩlpsuV iV o V ZFig. 9.214 Repeat steps 1 to 3 above for the 9.1 V diode.5 From the plotted characteristics, determine the diode slope resistance ineach case.

Appendix APhysical Quantities with SI and other preferred unitsGeneral quantities Symbol UnitsAcceleration, linear a2m/s (metre/second/second)Area A m 2 (square metre)Energy or work W J (joule)Force F N (newton)Length l m (metre)Mass m kg (kilogram)Power P W (watt)Pressure p Pa (pascal)Temperature value θ K or °C (Kelvin or degreeCelsius)Time t s (second)Torque T Nm (newton metre)Velocity, angular ω rad/s (radian/second)Velocity, linear v or u m/s (metre/second)Volume V m 3 (cubic metre)Wavelength λ m metreElectrical quantities Symbol UnitsAdmittance Y Ω (ohm)Charge (quantity) Q C (coulomb)Conductance G S (siemen)Current I A (ampere)Current density J2A/m (ampere/square metre)Electromotive force (emf) E V (volts)Frequency f Hz (hertz)Impedance Z Ω (ohm)Period T s (second)Potential difference (p.d.) V V (volt)Power, active P W (watt)Power, apparent S VA (volt ampere)Power, reactive Q VAr (volt ampere reactive)Reactance X Ω (ohm)Resistance R Ω (ohm)Resistivity ρ Ω m (ohm metre)Time constant τ s (second)283

284 Appendix AElectrostatic quantities Symbol UnitCapacitance C F (farad)Field strength E V/m (volt/metre)Flux ψ C (coulomb)Flux density D2C/m (coulomb/square metre)Permittivity, absolute e F/m (farad/metre)General quantities Symbol UnitPermittivity, relative e r no unitsPermittivity, of free space e 0 F/m (farad/metre)Electromagnetic quantities Symbol UnitField strength H(1)A/m (ampere/metre)Flux Φ Wb (weber)Flux density B T (tesla)Inductance, mutual M H (henry)Inductance, self L H (henry)Magnetomotive force (mmf) F(2)A (ampere)Permeability, absolute μ H/m (henry/metre)Permeability, relative μ r no unitsPermeability, of free space μ 0 H/m (henry/metre)Reluctance S(3)A/Wb (ampere/weber)(1) At/m (ampere turn/metre) in this book(2) (ampere turn) in this book(3) At/Wb (ampere turn/weber) in this book

Answers to Assignment QuestionsChapter 11.1 (a) 4.563 10 2(b) 9.023 10 5(c) 2.85 10 4(d) 8 10 3(e) 4.712 10 2(f) 1.8 10 4 A 2(g) 3.8 10 V(h) 8 l0 10 N(i) 2 10 3 F1.2 (a) 1.5 k Ω(b) 3.3 mΩ(c) 25 μ A(d) 750 V or 0.75 kV(e) 800 kV(f) 47 nF1.3 750 C1.4 0.104 s1.5 0.917A1.6 (a) 2.25 kV(b) 18.75 V1.7 (a) 2.273 A(b) 61 mA(c) 18.52 μ A(d) 0.512 mA1.8 54.14 MJ or 15.04 kWh1.9 £55.881.10 (a) 0.5 A(b) 280 V(c) 140 W(d) 42 kJ1.11 (a) 49.64 V(b) 27.58 Ω1.12 (a) 0.15 Ω(b) 2.25 Ω1.13 225 W; 67.5 kJ1.14 (a) 2 A( b) 8 V(c) 2.4 kC1.15 2 m/s2; 2.67 m/s20.5 m/s 21.16 8 μ A1.17 19.2 Ω ; 12.5 A1.18 15.65 kW1.19 71.72°C1.20 33.6 ΩChapter 22.1 0.375 A2.2 (a) 29.1 Ω(b) 24.75 A2.3 0.45 A2.4 1 4 Ω2.5 (a) 15.67 Ω(b) 0.766 A(c) 0.426 A2.6 (a) 2.371 A; 1.779 A; 1.804 A; 1.443 A; 0.902 A(b) 35.37 V (15 Ω and 20 Ω ); 14.43 V (8 Ω , 10 Ω ,1 6 Ω )(c) 63.26 W2.7 2.5 V2.8 (a) 12.63 V(b) 2.4 A(c) 0.947 A2.9 (a) 3.68 Ω(b) 5.44 A(c) 9.32 V(d) 2.33 A2.10 12.8 Ω2.11 (a) 1.57 Ω(b) 127.33 Ω2.12 4 0 Ω2.13 8 0 Ω2.14 (a) 1.304 A(b) 5.217 V(c) 0.87 A2.15 (a) 6.67 A(b) 26.67 A(c) 13.33 A2.16 (a) 2 V(b) 1.933 V285

286 Answers to Assignment Questions2.17 0.024 Ω2.18 (a) 9.6 A(b) 43.2 V(c) 4.32 A; 2.16 A; 1.08 A2.19 3.094 V2.20 6.82 V2.21 (a) 20 Ω(b) 10 A(c) 60 V(d) 2 kW(e) 300 W2.22 (a) 1 A( b) 3 0 V; 3 6 V(c) 54 kJ(d) 180 C2.23 2.368 A; 0.263 A; 2.105 A; 10.526 V2.24 1 A (discharge); 0.5 A (discharge); 1.5 A; 9 W2.25 (a) 0.183 A (charge); 5.725 A (discharge);5.542 A (discharge)(b) 108.55 V2.26 (a) 2.273 A (discharge); 0.455 A (charge);(b) 1.32 A(c) 11.363 V2.27 0.376 A; 0.388 A; 0.764 A; 61.14 V2.28 ( b) 3 Ω(c) 0.133 A (10 Ω and 5 Ω ); 0.222 A (6 Ω and 3 Ω );1.33 V (10 Ω and 6 Ω ); 0.67 V (5 Ω and 3 Ω )2.29 3.498 Ω ; 16.85 k Ω ; 22.5 Ω2.30 1.194 VChapter 33.1 (a) 0.2 μ C2(b) 2.29 μ C/m3.2 16.94 kV/m3.3 50 kV/m3.4260 mC/m3.5 125 kV/m3.6 4 0 m C3.7 165.96 V3.8 300 V3.9 500 pF or 0.5 nF3.10 240 pF or 0.24 nF3.11 33.12 5.53 nF3.13 (a) 442.7pF(b) 0.177 μ C(c) 400 kV/m3.14 0.089 mm3.15 188 nF3.16 1.363.17 2 53.18 (a) 4.8 mm(b) 1.213 10 3m 23.19 (a) 14 μ F(b) 2.86 μF3.20 (i) 1.463 μF; 1 7 μ F(ii) 0.013 μ F; 0.29 μ F(iii) 19.18 pF; 490 pF(iv) 215 pF; 10.2 nF3.21 8 nF3.22 (a) 5 μ F(b) 200 V(c) 3 mC3.23 (a) 24 μ F(b) 480 μ C(c) 1.44 mC3.24 (a) 13.85 V (4 nF); 6.15 V(b) 18.46 nC3.25 C 2 4.57 μF; C 3 3.56 μ F3.26 200 V; 200 V; 1.2 mC; 2 mC; 3.2 mC3.27 V 1 360 V; V 2 240 V; C 3 40 μ F3.28 200 V3.29 80.67 cm23.30 (a) 48 pF(b) 267 pC(c) 40 V3.31 625 mJ3.32 200 V3.33 5 μ F3.34 (a) 40 nF(b) 0.8 mJ(c) 400 kV/m3.35 (a) 1.6 mC; 0.32 J(b) 266.7 V; 0.213 J3.36 (a) 0.6 mC; 150 V; 100 V(b) 120 V; 0.48 mC; 0.72 mC3.37 (a) 1.5 mm(b) 52.94 cm2

Answers to Assignment Questions 287(c) 75 nC; 28 μ J(d) 14.2 μ C/m23.38 0.5 μmChapter 44.1 0.417 T4.2 1.98 mWb4.3 4 0 c m 24.4 21.25 At4.5 0.8 A4.6 2704.7 5633 At/m4.8 112.5 At4.9 2 A4.10 (a) 900 At(b) 1.11 T(c) 5000 At/m4.11 6 44.12 2784.13 (a) 1200 At(b) 5457 At/m(c) 1.37 T(d) 549 μWb4.14 (a) 0.206 A(b) 17774.15 2.95 A4.16 (a) 360 At/m; 1830 At/m(b) 5824.17 176.74.18 5.14 A4.19 1.975 A4.20 1.54 T4.21 11.57 μWbChapter 55.1 352.9 V5.2 0.571 ms5.3 37.5 V5.4 (a) 32 V(b) 5.33 V5.7 1 Wb/s5.8 0.05 T5.9 (a) 0.5 V(b) 0.433 V(c) 0.354 V5.11 1.125 N5.12 2.5 A5.13 0.143 m5.14 3 N5.15 0.94 μ Nm5.16 (a) 27 μ N(b) 0.405 μN m5.17 12.5 mN5.18 1750 A5.19 (a) 0.641 Ω(b) 6.25 mΩ5.20 11.975 k Ω ; 39.975 k Ω5.22 1.47 k Ω ; 14.97 k Ω ; 44.97 k Ω ; 149.97 k Ω5.23 0.125%5.24 1.12 H5.25 4 A5.26 15 6255.27 5 H; 800 V5.28 1205.29 22.5 A/s5.30 0.75 mH5.31 (a) 1364(b) 33.6 mH(c) 1.344 V5.32 0.12 H5.33 3 0 V5.34 (a) 628.3 mH(b) 56.5 mH(c) 5.09 V5.35 (a) 0.3 mH(b) 1.125 mH(c) 0.3 V; 1.125 V5.36 58.67 mH5.37 (a) 150 V(b) 5.625 mWb5.38 12.5 kV5.40 1 2 V5.41 27.5 V; 400 mA; 11 W5.42 (a) 42.67 V(b) 3.56 A

288 Answers to Assignment Questions5.43 51.2 mJ5.44 60.4 mH5.45 2.45 AChapter 66.1 (a) 150 Hz( b) 1 5 H z(c) 31.83 Hz6.2 (a) 100 Hz(b) 12.5 rev/s6.3 2 46.4 5 mA; 50 μ s; 20 kHz6.5 ( b) i 7.5 sin (200 πt ) milliamp6.6 (a) 427.3 V( b) 5 0 H z(c) 302 V; 272.2 V6.7 υ 63.64 sin (3000 πt ) volt; 22.31 V6.8 (a) 250 V; 176.8 V; 75 mA; 53 mA; 20 mWb;14.14 mWb; 6.8 V; 4.81 V(b) 25 Hz; 100 Hz; 50 Hz; 1.5 kHz6.9 i 7.07 sin (4000 πt ) amp(a) 6.724 A(b) 47.86 μ s6.10 353.6 V; 159.25 V6.11 4.22 mA; 5.97 mA6.12 1.4296.13 1 6 V6.14 22.2 V6.18 υ 17.44 sin (314 t 0.409) volt6.19 i 22.26 sin ( ω t 0.396) amp6.20 υ 43.06 sin ( ω t 0.019) volt6.23 3.2 V; 2.26 V; 400 μ s ; 2.5 kHzChapter 77.1 415.6 V7.2 (a) 0.2 Ω(b) 500 W7.3 280.75 VChapter 88.1 (a) 0.183 s(b) 6.15 mA; 0 A(c) 0.915 s8.2 (a) 333.3 A/s(b) 1 A(c) 15 ms8.3 (a) 10.1 mA8.4 (a) 2.5 Ω8.5 (a) 100 Ω( b) 1 m A8.6 1.8 hChapter 99.1 (a) 67.5 Ω [NPV 68 Ω , 0.5 W](b) 50.7 mA9.2 (a) 108 Ω [NPV 120 Ω , 2 W ](b) 117 mA(c) 12 mV

IndexAbsolute permeability , 119Absolute permittivity , 85Acceptor impurity , 270Alternating quantities , 21 , 192angular velocity of , 200addition of , 218amplitude of , 200average value of , 203expression for , 201 , 214form factor of , 207frequency of , 199maximum value of , 200peak factor of , 206periodic time of , 199phase angle of , 214phasor representation of , 216production of , 197rms value of , 205Alternator , 151Ammeter , 22moving coil , 159rectifier moving coil , 212Ampere , 9Atom , 7acceptor , 270Bohr model , 7donor , 269shell structure , 263Back-emf , 244Battery , 10B / H curve , 133Bridge rectifier , 210Capacitance , 83Capacitors , 84dielectric strength , 101energy stored in , 97in parallel , 87in series , 89in series/parallel , 92multiplate , 95types of , 102Cathode ray oscilloscope (CRO) , 224Cathode ray tube (CRT) , 224Cell , 10Charge (Q) , 8Coercive force , 133Coercivity , 133Commercial unit of energy (kWh) , 19Commutator , 235Coulomb , 8Coulomb’s law , 75Coupling factor , 181Covalent bond , 265Current , 9divider , 41ratio , 190Cycle , 199D.C. , 21D.C. circuits , 31parallel , 35series , 31series/parallel , 43D.C. generators , 235armature , 235commutator , 235construction , 238self excitation , 241separately excited , 239series , 242shunt , 240D.C. transients , 249C-R series circuits , 249L-R series circuits , 256time constant , 251 , 255 , 257Diode , 272characteristics , 274p-n junction , 272zener , 276Eddy currents , 172Electriccharge , 8current , 9fi eld , 76fi eld strength , 78fl ux , 79fl ux density , 79289

290 IndexElectromagnetic induction , 141 , 147Electromotive force (emf) , 10Electron-hole pair , 266Energy , 17dissipated by resistance , 17stored in a capacitor , 97stored in an inductor , 184Extrinsic semiconductor , 268Farad , 83Faraday’s laws , 141Ferrite , 174Ferromagnetic material , 111Fieldelectric , 76magnetic , 111Field strengthelectric , 78magnetic , 111Figure of merit , 166Fleming’s lefthand rule , 152Fleming’s righthand rule , 144Fluxelectric , 79magnetic , 111Flux densityelectric , 79magnetic , 115Force between charged bodies , 75Force between conductors , 156Form factor , 207Frequency , 199Fringing , 78Full-wave rectifier , 210Galvanometer (galvo) , 63Generation of emf , 147 , 150 , 235Half-wave rectifier , 209Henry , 175Hysteresis , 132loop , 133loss , 134Induced emf , 142Inductance , 175mutual , 180self , 175Inductor , 177energy storage , 184Internal resistance , 14Instantaneous value , 198 , 206Intrinsic semiconductor , 264Ionnegative , 7 , 270positive , 7 , 266 , 269Iron circuit , 115Iron dust core , 174Kilowatt-hour (kWh) , 19Kirchhoff’s laws , 48 , 49Laminations , 173Lenz’s law , 144Loading effect on voltmeter , 166Magnetic circuits , 114comparison with other circuits , 132composite , 126parallel , 134series , 126Magneticfi eld , 111fi eld strength (H) , 117fl ux , 115fl ux density , 115hysteresis , 132reluctance , 128saturation , 123Magnetisation curve (B/H) , 122Magnetomotive force (mmf) , 116Meterdamping torque , 161deflecting torque , 159full-scale deflection (fsd) , 161multiplier , 163ohmmeter , 170restoring torque , 160shunt , 162wattmeter , 171Motor principle , 153Motors , 244series , 245shunt , 244Motor/generator duality , 233Moving coil meter , 159n-type semiconductor , 268Ohm , 10Ohmmeter , 170Ohm’s law , 13

Index 291Peak factor , 206Peak value , 200Periodic time , 199Permeability , 118absolute , 119of free space , 118relative , 119Permittivity , 84absolute , 85of free space , 84relative , 84Phase and phase angle , 214Phasor , 216Phasor diagram , 217p-n junction , 271p-n junction diode , 272forward characteristics , 273reverse characteristics , 273Potential difference (pd) , 11Potential divider , 40Potential gradient , 80Potentiometer , 65Power , 18Proton , 7p-type semiconductor , 270Recombination , 266Rectifier , 208bridge , 210full-wave , 210half-wave , 209instrument , 212Relative permeability , 119Relative permittivity , 84Reluctance , 128Remanance , 133Remanent flux density , 133Resistance , 10internal , 14Resistivity , 21Resistors , 10in parallel , 35in series , 31in series/parallel , 43Root of the meanssquared (rms) value , 205Saturation , 123Sawtooth waveform , 227Scientific notation , 2Self-excitation , 244Self-inductance , 175Semiconductors , 23 , 263Separately-excited generator , 239Series generator , 242Series motor , 245Shunt generator , 240Shunt motor , 244Sinusoidal waveform , 21Slidewire potentiometer , 65Solenoid , 113Standard form notation , 2Temperature coefficient of resistance , 22Time constant , 251 , 255 , 257Transient response , 250Transformer , 186current ratio , 190principle , 186turns ratio , 189voltage ratio , 189Units, 1, Appendix AValence shell , 264Volt , 10Voltage drop (pd) , 11Voltmeter , 25analogue , 25digital (DVM or DMM) , 25fi gure of merit , 166loading effect , 166multiplier , 163Watt , 18Wattmeter , 171Weber , 115Wheatstone bridge , 55balance conditions , 60 , 61circuit , 56instrument , 63Zener diode , 276

Fundamental Electrical and Electronic Principles, Third Edition

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