Electrical Circuit Theory and...

Electrical Circuit Theory and Technology

In Memory of Elizabeth

Electrical Circuit Theory and Technology

Fifth edition

John Bird, BSc (Hons), CEng, CSci, CMath, FITE, FIMA, FCollT

Fifth edition published 2014by Routledge2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Routledge711 Third Avenue, New York, NY 10017

Routledge is an imprint of the Taylor & Francis Group, an informa business

2014 John Bird

The right of John Bird to be identified as author of this work has been asserted by him in accordance with sections 77 and 78of the Copyright, Designs and Patents Act 1988.

All rights reserved. No part of this book may be reprinted or reproduced or utilized in any form or by any electronic, mechanical,or other means, now known or hereafter invented, including photocopying and recording, or in any information storage orretrieval system, without permission in writing from the publishers.

Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identificationand explanation without intent to infringe.

First edition published by Newnes 1997Fourth edition published by Routledge 2010

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging in Publication DataBird, J. O.Electrical circuit theory and technology / John Bird. 5th edition.pages cmIncludes index.1. Electric circuits. 2. Electrical engineering. I. Title.TK454.B48 2013621.3192dc232013016404

ISBN: 978-0-415-66286-4 (pbk)ISBN: 978-1-315-88334-2 (ebk)

Typeset in Times byServis Filmsetting Ltd, Stockport, Cheshire

Contents

Preface xi

Part 1 Basic electrical engineeringprinciples 1

1 Units associated with basic electricalquantities 3

1.1 SI units 31.2 Charge 41.3 Force 41.4 Work 41.5 Power 51.6 Electrical potential and e.m.f. 61.7 Resistance and conductance 61.8 Electrical power and energy 61.9 Summary of terms, units and their symbols 7

2 An introduction to electric circuits 82.1 Standard symbols for electrical

components 92.2 Electric current and quantity of

electricity 92.3 Potential difference and resistance 102.4 Basic electrical measuring

instruments 102.5 Linear and non-linear devices 112.6 Ohms law 112.7 Multiples and sub-multiples 112.8 Conductors and insulators 132.9 Electrical power and energy 132.10 Main effects of electric current 162.11 Fuses 162.12 Insulation and the dangers of constant

high current flow 17

3 Resistance variation 183.1 Resistor construction 193.2 Resistance and resistivity 193.3 Temperature coefficient of resistance 213.4 Resistor colour coding and ohmic values 23

4 Batteries and alternative sources of energy 264.1 Introduction to batteries 274.2 Some chemical effects of electricity 274.3 The simple cell 284.4 Corrosion 29

4.5 E.m.f. and internal resistance of a cell 294.6 Primary cells 314.7 Secondary cells 324.8 Cell capacity 334.9 Safe disposal of batteries 354.10 Fuel cells 354.11 Alternative and renewable energy sources 35

Revision Test 1 37

5 Series and parallel networks 385.1 Series circuits 395.2 Potential divider 405.3 Parallel networks 425.4 Current division 445.5 Loading effect 475.6 Potentiometers and rheostats 485.7 Relative and absolute voltages 515.8 Earth potential and short circuits 525.9 Wiring lamps in series and in parallel 52

6 Capacitors and capacitance 546.1 Introduction to capacitors 556.2 Electrostatic field 556.3 Electric field strength 566.4 Capacitance 566.5 Capacitors 566.6 Electric flux density 576.7 Permittivity 576.8 The parallel plate capacitor 596.9 Capacitors connected in parallel and series 606.10 Dielectric strength 646.11 Energy stored 656.12 Practical types of capacitor 656.13 Supercapacitors 676.14 Discharging capacitors 68

7 Magnetic circuits 697.1 Introduction to magnetism and

magnetic circuits 707.2 Magnetic fields 707.3 Magnetic flux and flux density 717.4 Magnetomotive force and magnetic

field strength 717.5 Permeability and BH curves 727.6 Reluctance 73

vi Contents

7.7 Composite series magnetic circuits 757.8 Comparison between electrical and

magnetic quantities 787.9 Hysteresis and hysteresis loss 78

Revision Test 2 80

8 Electromagnetism 818.1 Magnetic field due to an electric current 828.2 Electromagnets 838.3 Force on a current-carrying conductor 858.4 Principle of operation of a simple

d.c. motor 888.5 Principle of operation of a moving-coil

instrument 898.6 Force on a charge 89

9 Electromagnetic induction 919.1 Introduction to electromagnetic induction 929.2 Laws of electromagnetic induction 929.3 Rotation of a loop in a magnetic field 959.4 Inductance 969.5 Inductors 979.6 Energy stored 989.7 Inductance of a coil 999.8 Mutual inductance 100

10 Electrical measuring instruments andmeasurements 103

10.1 Introduction 10410.2 Analogue instruments 10410.3 Moving-iron instrument 10410.4 The moving-coil rectifier instrument 10510.5 Comparison of moving-coil, moving-iron

and moving-coil rectifier instruments 10510.6 Shunts and multipliers 10610.7 Electronic instruments 10710.8 The ohmmeter 10810.9 Multimeters 10810.10 Wattmeters 10810.11 Instrument loading effect 10910.12 The oscilloscope 11110.13 Virtual test and measuring instruments 11610.14 Virtual digital storage oscilloscopes 11610.15 Waveform harmonics 12010.16 Logarithmic ratios 12010.17 Null method of measurement 12310.18 Wheatstone bridge 12310.19 D.c. potentiometer 12310.20 A.c. bridges 12410.21 Measurement errors 125

11 Semiconductor diodes 12811.1 Types of material 12911.2 Semiconductor materials 12911.3 Conduction in semiconductor materials 13111.4 The pn junction 13111.5 Forward and reverse bias 13211.6 Semiconductor diodes 13511.7 Characteristics and maximum ratings 13611.8 Rectification 13611.9 Zener diodes 13611.10 Silicon controlled rectifiers 13811.11 Light emitting diodes 13811.12 Varactor diodes 13911.13 Schottky diodes 139

12 Transistors 14112.1 Transistor classification 14212.2 Bipolar junction transistors (BJTs) 14212.3 Transistor action 14312.4 Leakage current 14412.5 Bias and current flow 14512.6 Transistor operating configurations 14512.7 Bipolar transistor characteristics 14512.8 Transistor parameters 14712.9 Current gain 14812.10 Typical BJT characteristics and maximum

ratings 14912.11 Field effect transistors 15012.12 Field effect transistor characteristics 15012.13 Typical FET characteristics and maximum

ratings 15212.14 Transistor amplifiers 15212.15 Load lines 155

Revision Test 3 159

Main formulae for Part 1 Basic electrical andelectronic principles 161

Part 2 Electrical principles andtechnology 163

13 D.c. circuit theory 16513.1 Introduction 16513.2 Kirchhoffs laws 16613.3 The superposition theorem 16913.4 General d.c. circuit theory 17213.5 Thvenins theorem 17413.6 Constant-current source 17813.7 Nortons theorem 17813.8 Thvenin and Norton equivalent networks 18113.9 Maximum power transfer theorem 184

Contents vii

14 Alternating voltages and currents 18714.1 Introduction 18814.2 The a.c. generator 18814.3 Waveforms 18914.4 A.c. values 19014.5 Electrical safety insulation and fuses 19314.6 The equation of a sinusoidal waveform 19314.7 Combination of waveforms 19614.8 Rectification 19914.9 Smoothing of the rectified output waveform 200

Revision Test 4 202

15 Single-phase series a.c. circuits 20315.1 Purely resistive a.c. circuit 20415.2 Purely inductive a.c. circuit 20415.3 Purely capacitive a.c. circuit 20515.4 RL series a.c. circuit 20615.5 RC series a.c. circuit 20915.6 RLC series a.c. circuit 21115.7 Series resonance 21415.8 Q-factor 21515.9 Bandwidth and selectivity 21715.10 Power in a.c. circuits 21715.11 Power triangle and power factor 219

16 Single-phase parallel a.c. circuits 22116.1 Introduction 22216.2 RL parallel a.c. circuit 22216.3 RC parallel a.c. circuit 22316.4 LC parallel a.c. circuit 22416.5 LRC parallel a.c. circuit 22616.6 Parallel resonance and Q-factor 22916.7 Power factor improvement 233

17 D.c. transients 23817.1 Introduction 23917.2 Charging a capacitor 23917.3 Time constant for a CR circuit 24017.4 Transient curves for a CR circuit 24017.5 Discharging a capacitor 24417.6 Camera flash 24617.7 Current growth in an LR circuit 24617.8 Time constant for an LR circuit 24717.9 Transient curves for an LR circuit 24717.10 Current decay in an LR circuit 24917.11 Switching inductive circuits 25117.12 The effect of time constant on a

rectangular waveform 251

18 Operational amplifiers 25318.1 Introduction to operational amplifiers 25418.2 Some op amp parameters 255

18.3 Op amp inverting amplifier 25618.4 Op amp non-inverting amplifier 25818.5 Op amp voltage-follower 25918.6 Op amp summing amplifier 25918.7 Op amp voltage comparator 26018.8 Op amp integrator 26118.9 Op amp differential amplifier 26218.10 Digital to analogue (D/A) conversion 26418.11 Analogue to digital (A/D) conversion 264

Revision Test 5 266

19 Three-phase systems 26719.1 Introduction 26819.2 Three-phase supply 26819.3 Star connection 26819.4 Delta connection 27119.5 Power in three-phase systems 27319.6 Measurement of power in three-phase

systems 27419.7 Comparison of star and delta connections 27919.8 Advantages of three-phase systems 279

20 Transformers 28020.1 Introduction 28120.2 Transformer principle of operation 28120.3 Transformer no-load phasor diagram 28320.4 E.m.f. equation of a transformer 28520.5 Transformer on-load phasor diagram 28720.6 Transformer construction 28820.7 Equivalent circuit of a transformer 28820.8 Regulation of a transformer 29020.9 Transformer losses and efficiency 29120.10 Resistance matching 29320.11 Auto transformers 29620.12 Isolating transformers 29820.13 Three-phase transformers 29820.14 Current transformers 29920.15 Voltage transformers 300

Revision Test 6 301

21 D.c. machines 30221.1 Introduction 30321.2 The action of a commutator 30321.3 D.c. machine construction 30421.4 Shunt, series and compound windings 30421.5 E.m.f. generated in an armature winding 30521.6 D.c. generators 30621.7 Types of d.c. generator and their

characteristics 307

viii Contents

21.8 D.c. machine losses 31121.9 Efficiency of a d.c. generator 31121.10 D.c. motors 31221.11 Torque of a d.c. machine 31321.12 Types of d.c. motor and their

characteristics 31421.13 The efficiency of a d.c. motor 31821.14 D.c. motor starter 32021.15 Speed control of d.c. motors 32121.16 Motor cooling 323

22 Three-phase induction motors 32422.1 Introduction 32522.2 Production of a rotating magnetic field 32522.3 Synchronous speed 32722.4 Construction of a three-phase induction

motor 32822.5 Principle of operation of a three-phase

induction motor 32822.6 Slip 32922.7 Rotor e.m.f. and frequency 33022.8 Rotor impedance and current 33122.9 Rotor copper loss 33122.10 Induction motor losses and efficiency 33222.11 Torque equation for an induction motor 33322.12 Induction motor torquespeed

characteristics 33522.13 Starting methods for induction motors 33622.14 Advantages of squirrel-cage induction

motors 33722.15 Advantages of wound rotor induction

motor 33822.16 Double cage induction motor 33822.17 Uses of three-phase induction motors 338

Revision Test 7 339

Main formulae for Part 2 Electrical principlesand technology 340

Part 3 Advanced circuit theoryand technology 343

23 Revision of complex numbers 34523.1 Introduction 34523.2 Operations involving Cartesian complex

numbers 34723.3 Complex equations 34923.4 The polar form of a complex number 34923.5 Multiplication and division using complex

numbers in polar form 350

23.6 De Moivres theorem powers and rootsof complex numbers 352

24 Application of complex numbers to seriesa.c. circuits 354

24.1 Introduction 35424.2 Series a.c. circuits 35524.3 Further worked problems on series

a.c. circuits 361

25 Application of complex numbers to parallela.c. networks 366

25.1 Introduction 36625.2 Admittance, conductance and susceptance 36725.3 Parallel a.c. networks 37025.4 Further worked problems on parallel

a.c. networks 374

26 Power in a.c. circuits 37726.1 Introduction 37726.2 Determination of power in a.c. circuits 37826.3 Power triangle and power factor 38026.4 Use of complex numbers for

determination of power 38126.5 Power factor improvement 385

Revision Test 8 390

27 A.c. bridges 39127.1 Introduction 39127.2 Balance conditions for an a.c. bridge 39127.3 Types of a.c. bridge circuit 39327.4 Worked problems on a.c. bridges 397

28 Series resonance and Q-factor 40128.1 Introduction 40228.2 Series resonance 40228.3 Q-factor 40428.4 Voltage magnification 40628.5 Q-factors in series 40828.6 Bandwidth 40928.7 Small deviations from the resonant

frequency 413

29 Parallel resonance and Q-factor 41629.1 Introduction 41629.2 The LRC parallel network 41729.3 Dynamic resistance 41829.4 The LRCR parallel network 41829.5 Q-factor in a parallel network 41929.6 Further worked problems on parallel

resonance and Q-factor 423

Revision Test 9 426

Contents ix

30 Introduction to network analysis 42730.1 Introduction 42730.2 Solution of simultaneous equations using

determinants 42830.3 Network analysis using Kirchhoffs laws 429

31 Mesh-current and nodal analysis 43731.1 Mesh-current analysis 43731.2 Nodal analysis 441

32 The superposition theorem 44832.1 Introduction 44832.2 Using the superposition theorem 44832.3 Further worked problems on the

superposition theorem 453

33 Thvenins and Nortons theorems 45833.1 Introduction 45833.2 Thvenins theorem 45933.3 Further worked problems on Thvenins

theorem 46533.4 Nortons theorem 46933.5 Thvenin and Norton equivalent networks 476

Revision Test 10 481

34 Deltastar and stardelta transformations 48234.1 Introduction 48234.2 Delta and star connections 48234.3 Deltastar transformation 48334.4 Stardelta transformation 491

35 Maximum power transfer theorems andimpedance matching 495

35.1 Maximum power transfer theorems 49635.2 Impedance matching 501


36 Complex waveforms 50536.1 Introduction 50636.2 The general equation for a complex

waveform 50636.3 Harmonic synthesis 50736.4 Fourier series of periodic and non-periodic

functions 51436.5 Even and odd functions and Fourier series

over any range 51936.6 R.m.s. value, mean value and the form

factor of a complex wave 52336.7 Power associated with complex waves 52636.8 Harmonics in single-phase circuits 52836.9 Further worked problems on harmonics

in single-phase circuits 532

36.10 Resonance due to harmonics 53636.11 Sources of harmonics 538

37 A numerical method of harmonic analysis 54237.1 Introduction 54237.2 Harmonic analysis on data given in tabular

or graphical form 54237.3 Complex waveform considerations 546

38 Magnetic materials 54938.1 Revision of terms and units used with

magnetic circuits 55038.2 Magnetic properties of materials 55038.3 Hysteresis and hysteresis loss 55238.4 Eddy current loss 55638.5 Separation of hysteresis and eddy current

losses 55938.6 Non-permanent magnetic materials 56138.7 Permanent magnetic materials 562


39 Dielectrics and dielectric loss 56439.1 Electric fields, capacitance and permittivity 56439.2 Polarization 56539.3 Dielectric strength 56539.4 Thermal effects 56639.5 Mechanical properties 56739.6 Types of practical capacitor 56739.7 Liquid dielectrics and gas insulation 56739.8 Dielectric loss and loss angle 567

40 Field theory 57140.1 Field plotting by curvilinear squares 57240.2 Capacitance between concentric cylinders 57540.3 Capacitance of an isolated twin line 58040.4 Energy stored in an electric field 58340.5 Induced e.m.f. and inductance 58540.6 Inductance of a concentric cylinder (or

coaxial cable) 58540.7 Inductance of an isolated twin line 58840.8 Energy stored in an electromagnetic field 590

41 Attenuators 59341.1 Introduction 59441.2 Characteristic impedance 59441.3 Logarithmic ratios 59641.4 Symmetrical T- and -attenuators 59841.5 Insertion loss 60341.6 Asymmetrical T- and -sections 606

x Contents

41.7 The L-section attenuator 60941.8 Two-port networks in cascade 61141.9 ABCD parameters 61441.10 ABCD parameters for networks 61741.11 Characteristic impedance in terms of

ABCD parameters 623


42 Filter networks 62642.1 Introduction 62642.2 Basic types of filter sections 62742.3 The characteristic impedance and the

attenuation of filter sections 62942.4 Ladder networks 63042.5 Low-pass filter sections 63142.6 High-pass filter sections 63742.7 Propagation coefficient and time delay in

filter sections 64242.8 m-derived filter sections 64842.9 Practical composite filters 653

43 Magnetically coupled circuits 65643.1 Introduction 65643.2 Self-inductance 65643.3 Mutual inductance 65743.4 Coupling coefficient 65843.5 Coils connected in series 65943.6 Coupled circuits 66243.7 Dot rule for coupled circuits 667

44 Transmission lines 67444.1 Introduction 67444.2 Transmission line primary constants 67544.3 Phase delay, wavelength and velocity of

propagation 67644.4 Current and voltage relationships 67744.5 Characteristic impedance and

propagation coefficient in terms of theprimary constants 679

44.6 Distortion on transmission lines 68344.7 Wave reflection and the reflection

coefficient 68544.8 Standing-waves and the standing-wave

ratio 688

45 Transients and Laplace transforms 69345.1 Introduction 69445.2 Response of RC series circuit to a step

input 694

45.3 Response of RL series circuit to a stepinput 696

45.4 LRC series circuit response 69945.5 Introduction to Laplace transforms 70245.6 Inverse Laplace transforms and the

solution of differential equations 70645.7 Laplace transform analysis directly from

the circuit diagram 71245.8 LRC series circuit using Laplace

transforms 72145.9 Initial conditions 724


Main formulae for Part 3 Advanced circuittheory and technology 729

Part 4 General reference 735

Standard electrical quantities their symbolsand units 737

Greek alphabet 740

Common prefixes 741

Resistor colour coding and ohmic values 742

Answers to Practice Exercises 743

Index 763

On the Website

Some practical laboratory experiments

1 Ohms law 22 Seriesparallel d.c. circuit 33 Superposition theorem 44 Thvenins theorem 65 Use of a CRO to measure voltage,

frequency and phase 86 Use of a CRO with a bridge rectifier circuit 97 Measurement of the inductance of a coil 108 Series a.c. circuit and resonance 119 Parallel a.c. circuit and resonance 1310 Charging and discharging a capacitor 15

To download and edit go to:www.routledge.com/cw/bird

http://www.routledge.com/cw/bird

Preface

Electrical Circuit Theory and Technology 5th Edi-tion provides coverage for a wide range of coursesthat contain electrical principles, circuit theory andtechnology in their syllabuses, from introductory todegree level and including Edexcel BTEC Levels 2to 5 National Certificate/Diploma and Higher NationalCertificate/Diploma in Engineering.

The text is set out in four parts as follows:

PART 1, involving Chapters 1 to 12, contains BasicElectrical Engineering Principles which any studentwishing to progress in electrical engineering wouldneed to know. An introduction to units, electricalcircuits, resistance variation, batteries and alternativesources of energy, series and parallel circuits, capacitorsand capacitance, magnetic circuits, electromagnetism,electromagnetic induction, electrical measuring instru-ments and measurements, semiconductor diodes andtransistors are all included in this section.

PART 2, involving Chapters 13 to 22, contains Electri-cal Principles and Technology suitable for NationalCertificate, National Diploma and City and Guildscourses in electrical and electronic engineering. D.c.circuit theory, alternating voltages and currents, single-phase series and parallel circuits, d.c. transients, oper-ational amplifiers, three-phase systems, transformers,d.c. machines and three-phase induction motors are allincluded in this section.

PART 3, involving Chapters 23 to 45, containsAdvanced Circuit Theory and Technology suit-able for Degree, Foundation degree, Higher NationalCertificate/Diploma and City and Guilds courses inelectrical and electronic/telecommunications engineer-ing. The two earlier sections of the book will pro-vide a valuable reference/revision for students at thislevel.

Complex numbers and their application to series andparallel networks, power in a.c. circuits, a.c. bridges,series and parallel resonance and Q-factor, networkanalysis involving Kirchhoffs laws, mesh and nodalanalysis, the superposition theorem, Thvenins andNortons theorems, deltastar and stardelta transforms,maximum power transfer theorems and impedance

matching, complex waveforms, Fourier series, har-monic analysis, magnetic materials, dielectrics anddielectric loss, field theory, attenuators, filter networks,magnetically coupled circuits, transmission line theoryand transients and Laplace transforms are all includedin this section.

PART 4 provides a short General Reference for stan-dard electrical quantities their symbols and units, theGreek alphabet, common prefixes and resistor colourcoding and ohmic values.

At the beginning of each of the 45 chapters a briefexplanation as to why it is important to understand thematerial contained within that chapter, together withlearning objectives, is listed.

At the end of each of the first three parts of the text is ahandy reference of the main formulae used.

There are a number of internet downloads freely avail-able to both students and lecturers/instructors; these arelisted on page xii.

It is not possible to acquire a thorough understandingof electrical principles, circuit theory and technologywithout working through a large number of numericalproblems. It is for this reason that Electrical Cir-cuit Theory and Technology 5th Edition contains some700 detailed worked problems, together with nearly1000 further problems (with answers at the back ofthe book), arranged within 177 Exercises that appearevery few pages throughout the text. Over 1100 linediagrams further enhance the understanding of thetheory.

Fourteen Revision Tests have been included, inter-spersed within the text every few chapters. For example,Revision Test 1 tests understanding of Chapters 1 to4, Revision Test 2 tests understanding of Chapters 5to 7, Revision Test 3 tests understanding of Chapters8 to 12, and so on. These Revision Tests do not haveanswers given since it is envisaged that lecturers/instruc-tors could set the Revision Tests for students to attemptas part of their course structure. Lecturers/instructorsmay obtain a complimentary set of solutions of the Revi-sion Tests in an Instructors Manual available from thepublishers via the internet see below.

xii Preface

Learning by example is at the heart of ElectricalCircuit Theory and Technology 5th Edition.

JOHN BIRDDefence School of Marine Engineering,

HMS Sultan,formerly University of Portsmouthand Highbury College, Portsmouth

John Bird is the former Head of Applied Electronicsin the Faculty of Technology at Highbury College,Portsmouth, UK. More recently, he has combinedfreelance lecturing at the University of Portsmouth withexaminer responsibilities for Advanced Mathematicswith City and Guilds, and examining for the Interna-tional Baccalaureate Organisation. He is the author ofover 120 textbooks on engineering and mathematicswith worldwide sales of around one million copies.He is currently a Senior Training Provider at theDefence School of Marine Engineering in the DefenceCollege of Technical Training at HMS Sultan, Gosport,Hampshire, UK.

Companion Website

The following support material is available fromhttp://www.routledge.com/cw/bird

For Students:

1. Full solutions to all 1000 further questions inthe Practice Exercises.

2. A set of formulae for the first three sections ofthe text.

3. Multiple choice questions/answer sheet foreach of the first 23 chapters.

4. Information on 37 engineers/scientists men-tioned in the text.

For Lecturers/Instructors:

1. Full solutions to all 1000 further questions inthe Practice Exercises.

2. Full solutions and marking schemes for eachof the 14 Revision Tests; also, each test maybe downloaded.

3. Lesson Plans and revision material. Typical30-week lesson plans for Electrical and Elec-tronic Principles, Unit 5, and Further Electri-cal Principles, Unit 67 are included, togetherwith two practice examination question papers(with solutions) for each of the modules.

4. Ten practical Laboratory Experiments areavailable. It may be that tutors will wantto edit these experiments to suit their ownequipment/component availability.

5. A set of formulae for each of the three sectionsof the text.

6. Multiple choice questions/answer sheet foreach of the first 23 chapters.

7. Information on 37 engineers/scientists men-tioned in the text.

8. All 1100 illustrations used in the textmay be downloaded for use in PowerPointpresentations.


Part 1

Basic electrical engineeringprinciples

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Chapter 1

Units associated with basicelectrical quantities

Why it is important to understand: Units associated with basic electrical quantities

The relationship between quantities can be written using words or symbols (letters), but symbols arenormally used because they are much shorter; for example, V is used for voltage, I for current and R forresistance. Some of the units have a convenient size for electronics, but most are either too large or too smallto be used directly so they are used with prefixes. The prefixes make the unit larger or smaller by the valueshown; for example, 25 mA is read as 25 milliamperes and means 25 103A = 25 0.001A = 0.025A.Knowledge of this chapter is essential for future studies and provides the basis of electrical units andprefixes; some simple calculations help understanding.

At the end of this chapter you should be able to:

state the basic SI units recognize derived SI units understand prefixes denoting multiplication and division state the units of charge, force, work and power and perform simple calculations involving these units state the units of electrical potential, e.m.f., resistance, conductance, power and energy and perform simple

calculations involving these units

1.1 SI units

The system of units used in engineering and science isthe Systme Internationale dUnits (international sys-tem of units), usually abbreviated to SI units, and isbased on the metric system. This was introduced in 1960and is now adopted by the majority of countries as theofficial system of measurement.

The basic units in the SI system are listed with theirsymbols in Table 1.1.

Derived SI units use combinations of basic units andthere are many of them. Two examples are:

Velocity metres per second (m/s) Acceleration metres per second squared (m/s2)

SI units may be made larger or smaller by using prefixeswhich denote multiplication or division by a particu-lar amount. The six most common multiples, with theirmeaning, are listed in Table 1.2. For a more completelist of prefixes, see page 741.

Electrical Circuit Theory and Technology. 978-0-415-66286-4, 2014 John Bird. Published by Taylor & Francis. All rights reserved.

Part

1

4 Electrical Circuit Theory and Technology

Table 1.1 Basic SI units

Quantity Unit

Length metre, m

Mass kilogram, kg

Time second, s

Electric current ampere, A

Thermodynamic temperature kelvin, K

Luminous intensity candela, cd

Amount of substance mole, mol

1.2 Charge

The unit of charge is the coulomb (C), whereone coulomb is one ampere second (1 coulomb =6.241018 electrons). The coulomb is defined as thequantity of electricity which flows past a given pointin an electric circuit when a current of one ampereismaintained for one second. Thus,

charge, in coulombs Q=Itwhere I is the current in amperes and t is the time inseconds.

Problem 1. If a current of 5 A flows for 2minutes, find the quantity of electricity transferred.

Quantity of electricity Q = It coulombsI =5 A, t =260=120 s

Hence Q =5120=600 C

Table 1.2

Prefix Name Meaning

M mega multiply by 1 000 000 (i.e. 106)k kilo multiply by 1000 (i.e. 103)m milli divide by 1000 (i.e. 103) micro divide by 1 000 000 (i.e. 106)n nano divide by 1 000 000 000 (i.e. 109)p pico divide by 1 000 000 000 000 (i.e. 1012)

Who were Coulomb, Ampere, Newton and Joule? Go towww.routledge.com/cw/bird

1.3 Force

The unit of force is the newton(N) where one newtonis one kilogram metre per second squared. The newtonis defined as the force which, when applied to a mass ofone kilogram, gives it an acceleration of one metre persecond squared. Thus,

force, in newtons F=mawhere m is the mass in kilograms and a is the accelera-tion in metres per second squared. Gravitational force,or weight, is mg, where g =9.81 m/s2.

Problem 2. A mass of 5000 g is accelerated at2 m/s2 by a force. Determine the force needed.

Force=massacceleration

=5 kg2 m/s2 =10 kg ms2

=10 N

Problem 3. Find the force acting verticallydownwards on a mass of 200 g attached to a wire.

Mass=200 g=0.2 kg and acceleration due to gravity,g =9.81 m/s2Force acting downwards=weight=massacceleration

=0.2 kg9.81 m/s2=1.962 N

1.4 Work

The unit of work or energy is the joule(J), whereone joule is one newton metre. The joule is defined as the


Part

1

Units associated with basic electrical quantities 5

work done or energy transferred when a force of onenewton is exerted through a distance of one metre in thedirection of the force. Thus

work done on a body, in joules W =Fswhere F is the force in newtons and s is the distance inmetres moved by the body in the direction of the force.Energy is the capacity for doing work.

1.5 Power

The unit of power is the watt (W) where one watt isone joule per second. Power is defined as the rate ofdoing work or transferring energy. Thus,

power in watts, P = Wt

where W is the work done or energy transferred in joulesand t is the time in seconds. Thus

energy in joules, W =Pt

Problem 4. A portable machine requires a forceof 200 N to move it. How much work is done if themachine is moved 20 m and what average power isutilized if the movement takes 25 s?

Work done = force distance = 200 N 20 m= 4000 Nm or 4 kJ

Power = work donetime taken

= 4000 J25s

= 160 J/s=160 W

Problem 5. A mass of 1000 kg is raised through aheight of 10 m in 20 s. What is (a) the work doneand (b) the power developed?

(a) Work done=forcedistanceforce=massacceleration

Hence, work done= (1000 kg9.81 m/s2) (10 m)=98 100 Nm=98.1 kNm or 98.1 kJ

(b) Power= work donetime taken

= 98100 J20 s

= 4905 J/s

=4905 W or 4.905 kW

Who was Watt? Go to www.routledge.com/cw/bird

Now try the following Practice Exercise

Practice Exercise 1 Charge, force, workand power (Answers on page 743)

(Take g=9.81 m/s2 where appropriate)1. What force is required to give a mass of 20 kg

an acceleration of 30 m/s2?

2. Find the accelerating force when a car hav-ing a mass of 1.7 Mg increases its speed witha constant acceleration of 3 m/s2.

3. A force of 40 N accelerates a mass at 5 m/s2.Determine the mass.

4. Determine the force acting downwards on amass of 1500 g suspended on a string.

5. A force of 4 N moves an object 200 cm in thedirection of the force. What amount of workis done?

6. A force of 2.5 kN is required to lift a load.How much work is done if the load is liftedthrough 500 cm?

7. An electromagnet exerts a force of 12 N andmoves a soft iron armature through a dis-tance of 1.5 cm in 40 ms. Find the powerconsumed.

8. A mass of 500 kg is raised to a height of 6 min 30 s. Find (a) the work done and (b) thepower developed.

9. What quantity of electricity is carried by6.241021 electrons?

10. In what time would a current of 1 A transfera charge of 30 C?

11. A current of 3 A flows for 5 minutes. Whatcharge is transferred?

12. How long must a current of 0.1 A flow so asto transfer a charge of 30 C?

13. Rewrite the following as indicated:

(a) 1000 pF= . . . . . . . . . nF(b) 0.02F= . . . . . . . . .. pF(c) 5000 kHz= . . . . . . . . . MHz(d) 47 k= . . . . . . .. M(e) 0.32 mA= . . . . . . . A


Part

1


1.6 Electrical potential and e.m.f.

The unit of electric potential is the volt (V), where onevolt is one joule per coulomb. One volt is defined as thedifference in potential between two points in a conductorwhich, when carrying a current of one ampere, dissipatesa power of one watt, i.e.

volts = wattsamperes

= joules/secondamperes

= joulesampere seconds

= joulescoulombs

(The volt is named after the Italian physicist AlessandroVolta.)

A change in electric potential between two points inan electric circuit is called a potential difference. Theelectromotive force (e.m.f.) provided by a source ofenergy such as a battery or a generator is measured involts.

1.7 Resistance and conductance

The unit of electric resistance is the ohm(), whereone ohm is one volt per ampere. It is defined as theresistance between two points in a conductor when aconstant electric potential of one volt applied at the twopoints produces a current flow of one ampere in theconductor. Thus,

resistance in ohms, R= VI

where V is the potential difference across the two pointsin volts and I is the current flowing between the twopoints in amperes.

The reciprocal of resistance is called conductanceand is measured in siemens (S), named after the Germaninventor and industrialist Ernst Siemen. Thus,

conductance in siemens, G= 1R

where R is the resistance in ohms.

Problem 6. Find the conductance of a conductorof resistance (a) 10, (b) 5 k and (c) 100 m.

(a) Conductance G = 1R

= 110

siemen=0.1 S

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(b) G = 1R

= 15103 S=0.210

3 S=0.2 mS

(c) G = 1R

= 1100103 S=

103

100S=10 S

1.8 Electrical power and energy

When a direct current of I amperes is flowing in anelectric circuit and the voltage across the circuit is Vvolts, then

power in watts, P = VIElectrical energy = Power time

= VIt joulesAlthough the unit of energy is the joule, when deal-ing with large amounts of energy the unit used is thekilowatt hour (kWh) where

1kWh = 1000 watt hour= 1000 3600 watt seconds or joules= 3600 000 J

Problem 7. A source e.m.f. of 5 V supplies acurrent of 3 A for 10 minutes. How much energy isprovided in this time?

Energy=power time and power=voltagecurrentHence

Energy=VIt =53 (1060)=9000 Ws or J=9 kJ

Problem 8. An electric heater consumes 1.8 MJwhen connected to a 250 V supply for 30 minutes.Find the power rating of the heater and the currenttaken from the supply.

Energy=power time, hencepower = energy

time

= 1.8 106 J

30 60 s =1000 J/s=1000 W

i.e. Power rating of heater=1 kW

Power P =VI, thus I = PV

= 1000250

=4 A

Hence the current taken from the supply is 4 A

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Part

1

Units associated with basic electrical quantities 7


Practice Exercise 2 E.m.f., resistance,conductance, power and energy (Answers onpage 743)

1. Find the conductance of a resistor of resistance(a) 10, (b) 2 k, (c) 2 m.

2. A conductor has a conductance of 50S. Whatis its resistance?

3. An e.m.f. of 250 V is connected across aresistance and the current flowing through theresistance is 4 A. What is the power developed?

4. 450 J of energy are converted into heat in1 minute. What power is dissipated?

5. A current of 10 A flows through a conductorand 10 W is dissipated. What p.d. exists acrossthe ends of the conductor?

6. A battery of e.m.f. 12 V supplies a currentof 5 A for 2 minutes. How much energy issupplied in this time?

7. A d.c. electric motor consumes 36 MJ whenconnected to a 250 V supply for 1 hour. Findthe power rating of the motor and the currenttaken from the supply.

1.9 Summary of terms, unitsand their symbols

Quantity Quantity Unit Unitsymbol symbol

Length l metre m

Mass m kilogram kg

Time t second s

Velocity v metres per m/s orsecond m s1

Acceleration a metres per m/s2 orsecond m s2squared

Force F newton N

Electrical Q coulomb Ccharge orquantity

Electric I ampere A

current

Resistance R ohm

Conductance G siemen S

Electromotive E volt Vforce

Potential V volt Vdifference

Work W joule J

Energy E (or W) joule J

Power P watt W

As progress is made through Electrical Circuit Theoryand Technology many more terms will be met. A fulllist of electrical quantities, together with their symbolsand units are given in Part 4, page 737.

For fully worked solutions to each of the problems in Practice Exercises 1 and 2 in this chapter,go to the website:

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Chapter 2

An introduction toelectric circuits

Why it is important to understand: An introduction to electric circuits

Electric circuits are a part of the basic fabric of modern technology. A circuit consists of electrical elementsconnected together, and we can use symbols to draw circuits. Engineers use electrical circuits to solveproblems that are important in modern society, such as in the generation, transmission and consumptionof electrical power and energy. The outstanding characteristics of electricity compared with other powersources are its mobility and flexibility. The elements in an electric circuit include sources of energy,resistors, capacitors, inductors, and so on. Analysis of electric circuits means determining the unknownquantities such as voltage, current and power associated with one or more elements in the circuit. Basicelectric circuit analysis and laws are explained in this chapter and knowledge of these are essential in thesolution of engineering problems.


recognize common electrical circuit diagram symbols understand that electric current is the rate of movement of charge and is measured in amperes appreciate that the unit of charge is the coulomb calculate charge or quantity of electricity Q from Q = It understand that a potential difference (p.d.) between two points in a circuit is required for current to flow appreciate that the unit of p.d. is the volt understand that resistance opposes current flow and is measured in ohms appreciate what an ammeter, a voltmeter, an ohmmeter, a multimeter, an oscilloscope, a wattmeter, a bridge

megger, a tachometer and stroboscope measure

distinguish between linear and non-linear devices

state Ohms law as V = IR or I = VR

or R = VI

use Ohms law in calculations, including multiples and sub-multiples of units describe a conductor and an insulator, giving examples of each

appreciate that electrical power P is given by P = VI = I 2R = V2

Rwatts


Part

1

An introduction to electric circuits 9

calculate electrical power define electrical energy and state its unit calculate electrical energy state the three main effects of an electric current, giving practical examples of each explain the importance of fuses in electrical circuits appreciate the dangers of constant high current flow with insulation materials

2.1 Standard symbols for electricalcomponents

Symbols are used for components in electrical circuitdiagrams and some of the more common ones are shownin Figure 2.1.

Conductor

Cell

Switch

Ammeter Voltmeter Indicator lamp

Filament lamp Fuse

Battery of 3 cells Alternative symbolfor battery

Variable resistor

Two conductorscrossing but not

joined

Two conductorsjoined together

A V

Fixed resistorPower supply

Figure 2.1

2.2 Electric current and quantityof electricity

All atoms consist of protons, neutrons and electrons.The protons, which have positive electrical charges,and the neutrons, which have no electrical charge,are contained within the nucleus. Removed from the

nucleus are minute negatively charged particles calledelectrons. Atoms of different materials differ from oneanother by having different numbers of protons, neu-trons and electrons. An equal number of protons andelectrons exist within an atom and it is said to be elec-trically balanced, as the positive and negative chargescancel each other out. When there are more than twoelectrons in an atom the electrons are arranged intoshells at various distances from the nucleus.

All atoms are bound together by powerful forces ofattraction existing between the nucleus and its elec-trons. Electrons in the outer shell of an atom, however,are attracted to their nucleus less powerfully than areelectrons whose shells are nearer the nucleus.

It is possible for an atom to lose an electron; theatom, which is now called an ion, is not now electri-cally balanced, but is positively charged and is thus ableto attract an electron to itself from another atom. Elec-trons that move from one atom to another are called freeelectrons and such random motion can continue indef-initely. However, if an electric pressure or voltage isapplied across any material there is a tendency for elec-trons to move in a particular direction. This movementof free electrons, known as drift, constitutes an electriccurrent flow. Thus current is the rate of movement ofcharge.Conductors are materials that contain electrons that areloosely connected to the nucleus and can easily movethrough the material from one atom to another.Insulators are materials whose electrons are held firmlyto their nucleus.The unit used to measure the quantity of electri-cal charge Q is called the coulomb C (where 1coulomb=6.241018 electrons).If the drift of electrons in a conductor takes place at therate of one coulomb per second the resulting current issaid to be a current of one ampere.Thus, 1 ampere=1 coulomb per second or 1 A=1 C/s.

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Part

1


Hence, 1 coulomb=1 ampere second or 1C=1 As.Generally, if I is the current in amperes and t thetime in seconds during which the current flows, thenI t represents the quantity of electrical charge incoulombs, i.e.

quantity of electrical charge transferred,

Q = I t coulombsProblem 1. What current must flow if 0.24coulombs is to be transferred in 15 ms?

Since the quantity of electricity, Q = It, then

I = Qt

= 0.2415 103 =

0.24 10315

= 24015

= 16 A

Problem 2. If a current of 10 A flows for fourminutes, find the quantity of electricity transferred.

Quantity of electricity, Q = It coulombsI =10 A; t =460=240 sHence Q =10240=2400 C


Practice Exercise 3 Electric current andcharge (Answers on page 743)

1. In what time would a current of 10 A transfera charge of 50 C?

2. A current of 6 A flows for 10 minutes. Whatcharge is transferred?

3. How long must a current of 100 mA flow so asto transfer a charge of 80 C?

2.3 Potential difference andresistance

For a continuous current to flow between two points ina circuit a potential difference (p.d.) or voltage, V, isrequired between them; a complete conducting path isnecessary to and from the source of electrical energy.The unit of p.d. is the volt, V (named in honour of theItalian physicist Alessandro Volta).

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Figure 2.2 shows a cell connected across a filamentlamp. Current flow, by convention, is considered as flow-ing from the positive terminal of the cell, around thecircuit to the negative terminal.

Currentflow

A

V

1

Figure 2.2

The flow of electric current is subject to friction. Thisfriction, or opposition, is called resistance, R, and is theproperty of a conductor that limits current. The unit ofresistance is the ohm; 1 ohm is defined as the resistancewhich will have a current of 1 ampere flowing throughit when 1 volt is connected across it, i.e.

resistance R = potential differencecurrent

2.4 Basic electrical measuringinstruments

An ammeter is an instrument used to measure cur-rent and must be connected in series with the circuit.Figure 2.2 shows an ammeter connected in series withthe lamp to measure the current flowing through it. Sinceall the current in the circuit passes through the ammeterit must have a very low resistance.A voltmeter is an instrument used to measure p.d. andmust be connected in parallel with the part of the cir-cuit whose p.d. is required. In Figure 2.2, a voltmeter isconnected in parallel with the lamp to measure the p.d.across it. To avoid a significant current flowing throughit a voltmeter must have a very high resistance.An ohmmeter is an instrument for measuringresistance.A multimeter, or universal instrument, may be used tomeasure voltage, current and resistance. An Avometerand fluke are typical examples.The oscilloscope may be used to observe waveformsand to measure voltages and currents. The display ofan oscilloscope involves a spot of light moving across

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Part

1


a screen. The amount by which the spot is deflectedfrom its initial position depends on the p.d. applied tothe terminals of the oscilloscope and the range selected.The displacement is calibrated in volts per cm. Forexample, if the spot is deflected 3 cm and the volts/cmswitch is on 10 V/cm then the magnitude of the p.d. is3 cm10 V/cm, i.e. 30 V.A wattmeter is an instrument for the measurement ofpower in an electrical circuit.A BM80 or a 420 MIT megger or a bridge meggermay be used to measure both continuity and insula-tion resistance. Continuity testing is the measurementof the resistance of a cable to discover if the cable iscontinuous, i.e. that it has no breaks or high-resistancejoints. Insulation resistance testing is the measurementof resistance of the insulation between cables, individ-ual cables to earth or metal plugs and sockets, and so on.An insulation resistance in excess of 1 M is normallyacceptable.A tachometer is an instrument that indicates the speed,usually in revolutions per minute, at which an engineshaft is rotating.A stroboscope is a device for viewing a rotating objectat regularly recurring intervals, by means of either (a) arotating or vibrating shutter, or (b) a suitably designedlamp which flashes periodically. If the period betweensuccessive views is exactly the same as the time ofone revolution of the revolving object, and the dura-tion of the view very short, the object will appear to bestationary.See Chapter 10 for more detail about electrical measur-ing instruments and measurements.

2.5 Linear and non-linear devices

Figure 2.3 shows a circuit in which current I can bevaried by the variable resistor R2. For various settingsof R2, the current flowing in resistor R1, displayed onthe ammeter, and the p.d. across R1, displayed on thevoltmeter, are noted and a graph is plotted of p.d. against

V

AR1

R2

l

Figure 2.3

p.d.

00 ll

(b)(a)

p.d.

Figure 2.4

current. The result is shown in Figure 2.4(a), where thestraight line graph passing through the origin indicatesthat current is directly proportional to the p.d. Sincethe gradient i.e. (p.d./current) is constant, resistance R1is constant. A resistor is thus an example of a lineardevice.

If the resistor R1 in Figure 2.3 is replaced by acomponent such as a lamp, then the graph shown inFigure 2.4(b) results when values of p.d. are noted forvarious current readings. Since the gradient is changing,the lamp is an example of a non-linear device.

2.6 Ohms law

Ohms law states that the current I flowing in a circuitis directly proportional to the applied voltage V andinversely proportional to the resistance R, provided thetemperature remains constant. Thus,

I = VR

or V = IR or R = VI

For a practical laboratory experiment on Ohms law,see the website.

Problem 3. The current flowing through a resistoris 0.8 A when a p.d. of 20 V is applied. Determinethe value of the resistance.

From Ohms law,

resistance R = VI

= 200.8

= 2008

= 25

2.7 Multiples and sub-multiples

Currents, voltages and resistances can often be verylarge or very small. Thus multiples and sub-multiples

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Part

1


Table 2.1

Prefix Name Meaning Example

M mega multiply by 1 000 000 2 M= 2 000 000 ohms(i.e.106)

k kilo multiply by 1000 10 kV=10 000 volts(i.e.103)

m milli divide by 1000 25 mA= 251000

A = 0.025 amperes(i.e.103)

micro divide by 1 000 000 50V= 501000 000

V = 0.000 05 volts(i.e.106)

of units are often used, as stated in Chapter 1. The mostcommon ones, with an example of each, are listed inTable 2.1.A more extensive list of common prefixes are given onpage 741.

Problem 4. Determine the p.d. which must beapplied to a 2 k resistor in order that a current of10 mA may flow.

Resistance R =2 k=2103 =2000

Current I =10 mA

=10103 A or 10103

or10

1000A

=0.01 AFrom Ohms law, potential difference,V = IR=(0.01) (2000)=20 V

Problem 5. A coil has a current of 50 mA flowingthrough it when the applied voltage is 12 V. What isthe resistance of the coil?

Resistance, R = VI

= 1250 103 =

12 10350

= 12 00050

= 240

Problem 6. A 100 V battery is connected across aresistor and causes a current of 5 mA to flow.Determine the resistance of the resistor. If the

voltage is now reduced to 25 V, what will be thenew value of the current flowing?

Resistance R = VI

= 1005 103 =

100 1035

= 20 103 =20 k

Current when voltage is reduced to 25 V,

I = VR

= 2520 103 =

25

20 103 = 1.25 mA

Problem 7. What is the resistance of a coil whichdraws a current of (a) 50 mA and (b) 200A from a120 V supply?

(a) Resistance R = VI

= 12050 103

= 1200.05

= 12 0005

=2400 or 2.4 k

(b) Resistance R = 120200 106 =

120

0.0002

= 1200 0002

= 600 000 or 600 kor 0.6 M

Problem 8. The current/voltage relationship fortwo resistors A and B is as shown in Figure 2.5.Determine the value of the resistance of eachresistor.

Part

1


Figure 2.5

For resistor A,

R = VI

= 20 A20 mA

= 200.02

= 20002

=1000 or 1k

For resistor B,

R = VI

= 16V5mA

= 160.005

= 160005

=3200 or 3.2 k


Practice Exercise 4 Ohms law (Answers onpage 743)

1. The current flowing through a heating elementis 5 A when a p.d. of 35 V is applied across it.Find the resistance of the element.

2. A 60 W electric light bulb is connected to a240 V supply. Determine (a) the current flow-ing in the bulb and (b) the resistance of thebulb.

3. Graphs of current against voltage for tworesistors, P and Q, are shown in Figure 2.6.Determine the value of each resistor.

Figure 2.6

4. Determine the p.d. which must be applied to a5 k resistor such that a current of 6 mA mayflow.

5. A 20 V source of e.m.f. is connected across acircuit having a resistance of 400. Calculatethe current flowing.

2.8 Conductors and insulators

A conductor is a material having a low resistance whichallows electric current to flow in it. All metals are con-ductors and some examples include copper, aluminium,brass, platinum, silver, gold and carbon.An insulator is a material having a high resistancewhich does not allow electric current to flow in it.Some examples of insulators include plastic, rubber,glass, porcelain, air, paper, cork, mica, ceramics andcertain oils.

2.9 Electrical power and energy

Electrical power

Power P in an electrical circuit is given by the prod-uct of potential difference V and current I , as stated inChapter 1. The unit of power is the watt, W. Hence

P = V I watts (1)From Ohms law, V = IRSubstituting for V in equation (1) gives:

P = (IR) I

i.e. P = I2R watts

Also, from Ohms law, I = VR

Substituting for I in equation (1) gives:

P = V VR

i.e. P = V2

Rwatts

There are thus three possible formulae which may beused for calculating power.

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25

20

Part

1


Problem 9. A 100 W electric light bulb isconnected to a 250 V supply. Determine (a) thecurrent flowing in the bulb, and (b) the resistance ofthe bulb.

Power P =V I , from which current I = PV

(a) Current I = 100250

= 1025

= 25

=0.4 A

(b) Resistance R = VI

= 2500.4

= 25004

=625

Problem 10. Calculate the power dissipated whena current of 4 mA flows through a resistanceof 5 k.

Power P = I 2 R = (4103)2(5103)=16106 5103 =80103

=0.08 W or 80 mWAlternatively, since I =4103 and R =5103,then from Ohms law,voltage V = IR=4103 5103 =20 VHence, power P =V I =204103 =80 mW

Problem 11. An electric kettle has a resistance of30. What current will flow when it is connectedto a 240 V supply? Find also the power rating of thekettle.

Current, I = VR

= 24030

=8 APower, P =VI =2408 =1920 W

=1.92 kW=power rating of kettle

Problem 12. A current of 5 A flows in thewinding of an electric motor, the resistance of thewinding being 100. Determine (a) the p.d. acrossthe winding, and (b) the power dissipated bythe coil.

(a) Potential difference across winding,V = IR=5100=500 V

(b) Power dissipated by coil, P = I 2 R =52 100=2500 W or 2.5 kW

(Alternatively, P =V I =5005=2500 Wor 2.5 kW)

Problem 13. The hot resistance of a 240 Vfilament lamp is 960. Find the current taken bythe lamp and its power rating.

From Ohms law,

current I = VR

= 240960

= 2496

= 14

A or 0.25 A

Power rating P =VI = (240)(

1

4

)=60 W

Electrical energy

Electrical energy = power timeIf the power is measured in watts and the time in secondsthen the unit of energy is watt-seconds or joules. If thepower is measured in kilowatts and the time in hoursthen the unit of energy is kilowatt-hours, often calledthe unit of electricity. The electricity meter in thehome records the number of kilowatt-hours used and isthus an energy meter.

Problem 14. A 12 V battery is connected across aload having a resistance of 40. Determine thecurrent flowing in the load, the power consumedand the energy dissipated in 2 minutes.

Current I = VR

= 1240

=0.3 APower consumed, P =VI = (12)(0.3)=3.6 WEnergy dissipated

=power time= (3.6 W)(260 s)=432 J (since 1 J=1 Ws)

Problem 15. A source of e.m.f. of 15 V supplies acurrent of 2 A for six minutes. How much energy isprovided in this time?

Energy=power time, and power=voltagecurrentHence energy=VIt =152 (660)

=10 800 Ws or J=10.8 kJ

Problem 16. Electrical equipment in an officetakes a current of 13 A from a 240 V supply.Estimate the cost per week of electricity if the

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Part

1


equipment is used for 30 hours each week and1 kWh of energy costs 13.56 p.

Power=VI watts=24013=3120 W=3.12 kWEnergy used per week

=power time= (3.12 kW)(30 h)=93.6 kWh

Cost at 13.56 p per kWh=93.613.56=1269.216 pHence weekly cost of electricity=12.69

Problem 17. An electric heater consumes 3.6 MJwhen connected to a 250 V supply for 40 minutes.Find the power rating of the heater and the currenttaken from the supply.

Power= energytime

= 3.6 106

40 60J

s(or W)=1500 W

i.e. Power rating of heater=1.5 kWPower P =VI, thus I = P

V= 1500

250=6 A

Hence the current taken from the supply is 6 A

Problem 18. Determine the power dissipated bythe element of an electric fire of resistance 20when a current of 10 A flows through it. If the fire ison for 6 hours determine the energy used and thecost if 1 unit of electricity costs 13 p.

Power P = I 2 R = 102 20 = 100 20 = 2000 Wor 2 kW (Alternatively, from Ohms law,V = IR = 10 20 = 200 V, hencepower P = V I = 200 10 = 2000 W=2 kW)Energy used in 6 hours

=power time=2 kW6 h=12 kWh

1 unit of electricity=1 kWhHence the number of units used is 12Cost of energy=1213=1.56

Problem 19. A business uses two 3 kW fires foran average of 20 hours each per week, and six150 W lights for 30 hours each per week. If the costof electricity is 14.25 p per unit, determine theweekly cost of electricity to the business.

Energy=power timeEnergy used by one 3 kW fire in 20 hours

=3 kW20 h=60 kWhHence weekly energy used by two 3 kW fires

=260=120 kWhEnergy used by one 150 W light for 30 hours

=150 W 30 h=4500 Wh=4.5 kWh

Hence weekly energy used by six 150 W lamps=64.5 =27 kWh

Total energy used per week=120+27=147 kWh1 unit of electricity=1 kWh of energyThus weekly cost of energy at

14.25 p per kWh=14.25 147=2094.75 p=20.95


Practice Exercise 5 Power and energy(Answers on page 743)

1. The hot resistance of a 250 V filament lampis 625. Determine the current taken by thelamp and its power rating.

2. Determine the resistance of a coil connectedto a 150 V supply when a current of (a) 75 mA,(b) 300A flows through it.

3. Determine the resistance of an electric firewhich takes a current of 12 A from a 240 Vsupply. Find also the power rating of the fireand the energy used in 20 h.

4. Determine the power dissipated when a cur-rent of 10 mA flows through an appliancehaving a resistance of 8 k.

5. 85.5 J of energy are converted into heat in nineseconds. What power is dissipated?

6. A current of 4 A flows through a conductor and10 W is dissipated. What p.d. exists across theends of the conductor?

7. Find the power dissipated when:(a) a current of 5 mA flows through a resis-

tance of 20 k

(b) a voltage of 400 V is applied across a120 k resistor

(c) a voltage applied to a resistor is 10 kV andthe current flow is 4 mA.

Part

1


8. A battery of e.m.f. 15 V supplies a current of2 A for 5 min. How much energy is supplied inthis time?

9. In a household during a particular week three2 kW fires are used on average 25 h each andeight 100 W light bulbs are used on average35 h each. Determine the cost of electricity forthe week if 1 unit of electricity costs 15 p.

10. Calculate the power dissipated by the elementof an electric fire of resistance 30 when acurrent of 10 A flows in it. If the fire is on for30 hours in a week determine the energy used.Determine also the weekly cost of energy ifelectricity costs 13.5 p per unit.

2.10 Main effects of electric current

The three main effects of an electric current are:

(a) magnetic effect

(b) chemical effect

(c) heating effect.

Some practical applications of the effects of an electriccurrent include:

Magnetic effect: bells, relays, motors, generators,transformers, telephones, car-ignition

and lifting magnets (see Chapter 8)

Chemical effect: primary and secondary cells and

electroplating (see Chapter 4)

Heating effect: cookers, water heaters, electric fires,

irons, furnaces, kettles and

soldering irons

2.11 Fuses

If there is a fault in a piece of equipment then excessivecurrent may flow. This will cause overheating and possi-bly a fire; fuses protect against this happening. Currentfrom the supply to the equipment flows through the fuse.The fuse is a piece of wire which can carry a stated cur-rent; if the current rises above this value it will melt. Ifthe fuse melts (blows) then there is an open circuit and

no current can then flow thus protecting the equipmentby isolating it from the power supply.

The fuse must be able to carry slightly more than thenormal operating current of the equipment to allow fortolerances and small current surges. With some equip-ment there is a very large surge of current for a short timeat switch on. If a fuse is fitted to withstand this large cur-rent there would be no protection against faults whichcause the current to rise slightly above the normal value.Therefore special anti-surge fuses are fitted. These canstand 10 times the rated current for 10 milliseconds. Ifthe surge lasts longer than this the fuse will blow.A circuit diagram symbol for a fuse is shown inFigure 2.1 on page 9.

Problem 20. If 5 A, 10 A and 13 A fuses areavailable, state which is most appropriate for thefollowing appliances, which are both connected to a240 V supply(a) electric toaster having a power rating of 1 kW(b) electric fire having a power rating of 3 kW.

Power P =VI, from which current I = PV

(a) For the toaster,

current I = PV

= 1000240

= 10024

= 4.17A

Hence a 5 A fuse is most appropriate

(b) For the fire,

current I = PV

= 3000240

= 30024

= 12.5A

Hence a 13 A fuse is most appropriate


Practice Exercise 6 Fuses (Answers onpage 743)

1. A television set having a power rating of 120 Wand electric lawn-mower of power rating 1 kWare both connected to a 240 V supply. If 3 A,5 A and 10 A fuses are available state which isthe most appropriate for each appliance.

Part

1


2.12 Insulation and the dangersof constant high current flow

The use of insulation materials on electrical equip-ment, whilst being necessary, also has the effect ofpreventing heat loss, i.e. the heat is not able to dissi-pate, thus creating possible danger of fire. In addition,

the insulating material has a maximum temperaturerating this is heat it can withstand without being dam-aged. The current rating for all equipment and electricalcomponents is therefore limited to keep the heat gen-erated within safe limits. In addition, the maximumvoltage present needs to be considered when choosinginsulation.

For fully worked solutions to each of the problems in Practice Exercises 3 to 6 in this chapter,go to the website:



Chapter 3

Resistance variationWhy it is important to understand: Resistance variation

An electron travelling through the wires and loads of an electric circuit encounters resistance. Resistanceis the hindrance to the flow of charge. The flow of charge through wires is often compared to the flow ofwater through pipes. The resistance to the flow of charge in an electric circuit is analogous to the frictionaleffects between water and the pipe surfaces as well as the resistance offered by obstacles that are presentin its path. It is this resistance that hinders the water flow and reduces both its flow rate and its drift speed.Like the resistance to water flow, the total amount of resistance to charge flow within a wire of an electriccircuit is affected by some clearly identifiable variables. Factors which affect resistance are length, cross-sectional area and type of material. The value of a resistor also changes with changing temperature, butthis is not as we might expect, mainly due to a change in the dimensions of the component as it expandsor contracts. It is due mainly to a change in the resistivity of the material caused by the changing activityof the atoms that make up the resistor. Resistance variation due to length, cross-sectional area, type ofmaterial and temperature variation are explained in this chapter, with calculations to aid understanding.In addition, the resistor colour coding/ohmic values are explained.


recognize three common methods of resistor construction appreciate that electrical resistance depends on four factors

appreciate that resistance R = la

, where is the resistivity

recognize typical values of resistivity and its unit

perform calculations using R = la

define the temperature coefficient of resistance, recognize typical values for perform calculations using R = R0(1+ ) determine the resistance and tolerance of a fixed resistor from its colour code determine the resistance and tolerance of a fixed resistor from its letter and digit code


Part

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Resistance variation 19

3.1 Resistor construction

There is a wide range of resistor types. Four of the mostcommon methods of construction are:

(i) Surface Mount Technology (SMT)

Many modern circuits use SMT resistors. Their manu-facture involves depositing a film of resistive materialsuch as tin oxide on a tiny ceramic chip.The edges of theresistor are then accurately ground or cut with a laser togive a precise resistance across the ends of the device.Tolerances may be as low as 0.02% and SMT resistorsnormally have very low power dissipation. Their mainadvantage is that very high component density can beachieved.

(ii) Wire wound resistors

A length of wire such, as nichrome or manganin, whoseresistive value per unit length is known, is cut to thedesired value and wound around a ceramic former priorto being lacquered for protection. This type of resistorhas a large physical size, which is a disadvantage; how-ever, they can be made with a high degree of accuracy,and can have a high power rating.

Wire wound resistors are used in power circuits andmotor starters.

(iii) Metal film resistors

Metal film resistors are made from small rods ofceramic coated with metal, such as a nickel alloy. Thevalue of resistance is controlled firstly by the thicknessof the coating layer (the thicker the layer, the lowerthe value of resistance), and secondly by cutting a finespiral groove along the rod using a laser or diamondcutter to cut the metal coating into a long spiral strip,which forms the resistor.

Metal film resistors are low tolerance, precise resis-tors (1% or less) and are used in electronic circuits.

(iv) Carbon film resistors

Carbon film resistors have a similar construction tometal film resistors but generally with wider tolerance,typically 5%. They are inexpensive, in common use,and are used in electronic circuits.

3.2 Resistance and resistivity

The resistance of an electrical conductor depends onfour factors, these being: (a) the length of the conductor,(b) the cross-sectional area of the conductor, (c) the typeof material and (d) the temperature of the material.

Resistance, R, is directly proportional to length, l, ofa conductor, i.e. R l. Thus, for example, if the length ofa piece of wire is doubled, then the resistance is doubled.

Resistance, R, is inversely proportional to cross-sectional area, a, of a conductor, i.e. R 1/a. Thus, forexample, if the cross-sectional area of a piece of wire isdoubled then the resistance is halved.

Since R l and R 1/a then R l/a. By inserting aconstant of proportionality into this relationship the typeof material used may be taken into account. The con-stant of proportionality is known as the resistivity of thematerial and is given the symbol (Greek rho). Thus,

resistance R = la

ohms

is measured in ohm metres (m)The value of the resistivity is that resistance of a unitcube of the material measured between opposite facesof the cube.Resistivity varies with temperature and some typicalvalues of resistivities measured at about room tempera-ture are given below:

Copper 1.7108m (or 0.017m)Aluminium 2.6108m (or 0.026m)Carbon (graphite) 10108m (or 0.10m)Glass 11010m (or 104 m)Mica 11013m (or 107 m)

Note that good conductors of electricity have a low valueof resistivity and good insulators have a high value ofresistivity.

Problem 1. The resistance of a 5 m length of wireis 600. Determine (a) the resistance of an 8 mlength of the same wire, and (b) the length of thesame wire when the resistance is 420.

(a) Resistance, R, is directly proportional to length, l,i.e. R l.Hence, 6005 m or 600=(k)(5), where k is thecoefficient of proportionality. Hence,

k = 6005

= 120When the length l is 8 m, then resistance

R =kl= (120)(8)=960(b) When the resistance is 420, 420= kl, from

which,

length l = 420k

= 420120

= 3.5 m

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Problem 2. A piece of wire of cross-sectionalarea 2 mm2 has a resistance of 300. Find (a) theresistance of a wire of the same length and materialif the cross-sectional area is 5 mm2, (b) thecross-sectional area of a wire of the same lengthand material of resistance 750.

Resistance, R, is inversely proportional to cross-

sectional area, a, i.e. R 1a

.

Hence 300 12 mm2

or 300 = (k)(

1

2

)

from which the coefficient of proportionality,k =3002=600(a) When the cross-sectional area a =5mm2

then R =(k)(

1

5

)= (600)

(1

5

)=120

(Note that resistance has decreased as the cross-sectional area is increased.)

(b) When the resistance is 750 then 750=(k)(1/a),from which cross-sectional area,

a = k750

= 600750

= 0.8 mm2

Problem 3. A wire of length 8 m andcross-sectional area 3 mm2 has a resistance of0.16. If the wire is drawn out until itscross-sectional area is 1 mm2, determine theresistance of the wire.

Resistance, R, is directly proportional to length, l, andinversely proportional to the cross-sectional area, a, i.e.

R la

or R =k(

l

a

), where k is the coefficient of

proportionality.

Since R =0.16, l =8 and a =3, then 0.16=(k)(

8

3

)

from which

k =0.16 38

=0.06If the cross-sectional area is reduced to 13 of its originalarea then the length must be tripled to 38, i.e. 24 m

New resistance, R = k(

l

a

)= 0.06

(24

1

)= 1.44

Problem 4. Calculate the resistance of a 2 kmlength of aluminium overhead power cable if thecross-sectional area of the cable is 100 mm2. Takethe resistivity of aluminium to be 0.03106m.

Length l =2 km=2000 m;area, a =100 mm2 =100106 m2;resistivity = 0.03106m

Resistance, R = la

= (0.03 106m)(2000 m)

(100 106 m2)

= 0.03 2000100

= 0.6Problem 5. Calculate the cross-sectional area, inmm2, of a piece of copper wire, 40 m in length andhaving a resistance of 0.25. Take the resistivity ofcopper as 0.02106m.

Resistance, R = la

hence cross-sectional area a = lR

= (0.02 106m)(40 m)

0.25

= 3.2 106 m2

= (3.2 106) 106 mm2 = 3.2 mm2

Problem 6. The resistance of 1.5 km of wire ofcross-sectional area 0.17 mm2 is 150. Determinethe resistivity of the wire.

Resistance, R = la

hence, resistivity, = Ral

= (150)(0.17 106 m2)

(1500 m)

= 0.017106 m or 0.017m

Problem 7. Determine the resistance of 1200 mof copper cable having a diameter of 12 mm if theresistivity of copper is 1.7108m.

Cross-sectional area of cable, a = r2 = ( 122 )2= 36mm2

= 36 106 m2

Resistance, R = la

= (1.7 108m)(1200 m)

(36 106 m2)

= 1.7 1200 106

108 36 =1.7 12

36

= 0.180

Part

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Practice Exercise 7 Resistance andresistivity (Answers on page 743)

1. The resistance of a 2 m length of cable is 2.5.Determine (a) the resistance of a 7 m length ofthe same cable and (b) the length of the samewire when the resistance is 6.25.

2. Some wire of cross-sectional area 1 mm2 hasa resistance of 20. Determine (a) the resis-tance of a wire of the same length and mate-rial if the cross-sectional area is 4 mm2, and(b) the cross-sectional area of a wire of thesame length and material if the resistance is32.

3. Some wire of length 5 m and cross-sectionalarea 2 mm2 has a resistance of 0.08. If thewire is drawn out until its cross-sectional areais 1 mm2, determine the resistance of the wire.

4. Find the resistance of 800 m of copper cableof cross-sectional area 20 mm2. Take the resis-tivity of copper as 0.02m.

5. Calculate the cross-sectional area, in mm2, ofa piece of aluminium wire 100 m long and hav-ing a resistance of 2. Take the resistivity ofaluminium as 0.03106m.

6. (a) What does the resistivity of a materialmean?

(b) The resistance of 500 m of wire of cross-sectional area 2.6 mm2 is 5. Determinethe resistivity of the wire in m.

7. Find the resistance of 1 km of copper cablehaving a diameter of 10 mm if the resistivityof copper is 0.017106m.

3.3 Temperature coefficientof resistance

In general, as the temperature of a material increases,most conductors increase in resistance, insulatorsdecrease in resistance, whilst the resistances of somespecial alloys remain almost constant.

The temperature coefficient of resistance of amaterial is the increase in the resistance of a 1

resistor of that material when it is subjected to a rise oftemperature of 1C. The symbol used for the tempera-ture coefficient of resistance is (Greek alpha). Thus, ifsome copper wire of resistance 1 is heated through1C and its resistance is then measured as 1.0043then =0.0043/C for copper. The units are usu-ally expressed only as per C, i.e. =0.0043/Cfor copper. If the 1 resistor of copper is heatedthrough 100C then the resistance at 100C would be1+1000.0043=1.43.Some typical values of temperature coefficient of resis-tance measured at 0C are given below:

Copper 0.0043/C Aluminium 0.0038/C

Nickel 0.0062/C Carbon 0.000 48/CConstantan 0 Eureka 0.000 01/C

(Note that the negative sign for carbon indicates that itsresistance falls with increase of temperature.)If the resistance of a material at 0C is known, theresistance at any other temperature can be determinedfrom:

R = R0(1 + 0 )

where R0 = resistance at 0CR = resistance at temperature C0 = temperature coefficient of resistance at 0C.

Problem 8. A coil of copper wire has a resistanceof 100 when its temperature is 0C. Determine itsresistance at 70C if the temperature coefficient ofresistance of copper at 0C is 0.0043/C.

Resistance R = R0(1+0 )

Hence resistance at 70C, R70 = 100[1+(0.0043)(70)]= 100[1 + 0.301]= 100(1.301)= 130.1

Problem 9. An aluminium cable has a resistanceof 27 at a temperature of 35C. Determine itsresistance at 0C. Take the temperature coefficientof resistance at 0C to be 0.0038/C.

Part

1


Resistance at C, R = R0(1+0 )Hence resistance at 0C, R0 = R

(1 +0)= 27

[1 + (0.0038)(35)]= 27

1 + 0.133 =27

1.133

= 23.83Problem 10. A carbon resistor has a resistance of1 k at 0C. Determine its resistance at 80C.Assume that the temperature coefficient ofresistance for carbon at 0C is 0.0005/C.

Resistance at temperature C, R = R0(1+0 )i.e. R = 1000[1 + (0.0005)(80)]

= 1000[1 0.040] = 1000(0.96)= 960

If the resistance of a material at room temperature(approximately 20C), R20, and the temperature coef-ficient of resistance at 20C, 20, are known, then theresistance R at temperature C is given by:

R = R20[1 +20( 20)]

Problem 11. A coil of copper wire has aresistance of 10 at 20C. If the temperaturecoefficient of resistance of copper at 20C is0.004/C, determine the resistance of the coil whenthe temperature rises to 100C.

Resistance at C, R = R20[1+20(20)]Hence resistance at 100C,R100 =10[1+(0.004)(10020)]

=10[1+(0.004)(80)]=10[1+0.32]=10(1.32)=13.2

Problem 12. The resistance of a coil ofaluminium wire at 18C is 200. The temperatureof the wire is increased and the resistance rises to240. If the temperature coefficient of resistance ofaluminium is 0.0039/C at 18C, determine thetemperature to which the coil has risen.

Let the temperature rise to

Resistance at C, R = R18[1+18(18)]i.e. 240=200[1+(0.0039)(18)]

240=200+(200)(0.0039)(18)240200=0.78(18)40=0.78(18)

40

0.78= 18

51.28= 18, from which,=51.28+18=69.28C

Hence the temperature of the coil increases to69.28C

If the resistance at 0C is not known, but is known atsome other temperature 1, then the resistance at anytemperature can be found as follows:

R1 = R0(1 +01) and R2 = R0(1 +02)Dividing one equation by the other gives:

R1R2

= 1 +011 +02

where R2 = resistance at temperature 2

Problem 13. Some copper wire has a resistanceof 200 at 20C. A current is passed through thewire and the temperature rises to 90C. Determinethe resistance of the wire at 90C, correct to thenearest ohm, assuming that the temperaturecoefficient of resistance is 0.004/C at 0C.

R20 =200, 0 =0.004/CR20R90

= [1 +0(20)][1 +0(90)]

Hence R90 = R20[1 + 900][1 + 200]

= 200[1 + 90(0.004)][1 + 20(0.004)]

= 200[1 + 0.36][1 + 0.08]

= 200(1.36)(1.08)

= 251.85

i.e. the resistance of the wire at 90C is 252

Part

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Practice Exercise 8 Temperature coefficientof resistance (Answers on page 743)

1. A coil of aluminium wire has a resistance of50 when its temperature is 0C. Determineits resistance at 100C if the temperature coef-ficient of resistance of aluminium at 0C is0.0038/C.

2. A copper cable has a resistance of 30 ata temperature of 50C. Determine its resis-tance at 0C. Take the temperature coefficientof resistance of copper at 0C as 0.0043/C.

3. The temperature coefficient of resistance forcarbon at 0C is 0.00048/C. What is the sig-nificance of the minus sign? A carbon resistorhas a resistance of 500 at 0C. Determine itsresistance at 50C.

4. A coil of copper wire has a resistance of 20at 18C. If the temperature coefficient of resis-tance of copper at 18C is 0.004/C, determinethe resistance of the coil when the temperaturerises to 98C.

5. The resistance of a coil of nickel wire at20C is 100. The temperature of the wireis increased and the resistance rises to 130.If the temperature coefficient of resistanceof nickel is 0.006/C at 20C, determine thetemperature to which the coil has risen.

6. Some aluminium wire has a resistance of 50at 20C. The wire is heated to a tempera-ture of 100C. Determine the resistance of thewire at 100C, assuming that the temperaturecoefficient of resistance at 0C is 0.004/C.

7. A copper cable is 1.2 km long and has across-sectional area of 5 mm2. Find its resis-tance at 80C if at 20C the resistivity ofcopper is 0.02106m and its temperaturecoefficient of resistance is 0.004/C.

3.4 Resistor colour coding andohmic values

(a) Colour code for fixed resistors

The colour code for fixed resistors is given in Table 3.1

Table 3.1

Colour Significant Multiplier Tolerancefigures

Silver 102 10%

Gold 101 5%Black 0 1

Brown 1 10 1%Red 2 102 2%Orange 3 103

Yellow 4 104

Green 5 105 0.5%Blue 6 106 0.25%Violet 7 107 0.1%Grey 8 108

White 9 109

None 20%

(i) For a four-band fixed resistor (i.e. resistancevalues with two significant figures):yellow-violet-orange-red indicates 47 k with atolerance of 2%(Note that the first band is the one nearest the endof the resistor.)

(ii) For a five-band fixed resistor (i.e. resistancevalues with three significant figures): red-yellow-white-orange-brown indicates 249 kwith a toler-ance of 1%(Note that the fifth band is 1.5 to 2 times wider thanthe other bands.)

Problem 14. Determine the value and toleranceof a resistor having a colour coding of:orange-orange-silver-brown.

The first two bands, i.e. orange-orange, give 33 fromTable 3.1.The third band, silver, indicates a multiplier of 102 fromTable 3.1, which means that the value of the resistor is33102 =0.33

Part

1


The fourth band, i.e. brown, indicates a toleranceof 1% from Table 3.1. Hence a colour coding oforange-orange-silver-brown represents a resistor ofvalue 0.33 with a tolerance of 1%

Problem 15. Determine the value and toleranceof a resistor having a colour coding ofbrown-black-brown.

The first two bands, i.e. brown-black, give 10 fromTable 3.1.

The third band, brown, indicates a multiplier of 10from Table 3.1, which means that the value of the resistoris 1010=100

There is no fourth band colour in this case; hence,from Table 3.1, the tolerance is 20%. Hence a colourcoding of brown-black-brown represents a resistor ofvalue 100 with a tolerance of 20%

Problem 16. Between what two values should aresistor with colour coding brown-black-brown-silver lie?

From Table 3.1, brown-black-brown-silver indicates1010, i.e. 100, with a tolerance of 10%This means that the value could lie between

(100 10% of 100)and (100 + 10% of 100)

i.e. brown-black-brown-silver indicates any valuebetween 90 and 110

Problem 17. Determine the colour coding for a47 k resistor having a tolerance of 5%

From Table 3.1, 47 k=47103 has a colour codingof yellow-violet-orange. With a tolerance of 5%, thefourth band will be gold.Hence 47 k5% has a colour coding of:yellow-violet-orange-gold.

Problem 18. Determine the value and toleranceof a resistor having a colour coding of orange-green-red-yellow-brown.

Orange-green-red-yellow-brown is a five-band fixedresistor and from Table 3.1 indicates: 352104 witha tolerance of 1%

352 104= 3.52 106, i.e. 3.52 MHence orange-green-red-yellow-brown indicates3.52 M1%(b) Letter and digit code for resistors

Another way of indicating the value of resistors is theletter and digit code shown in Table 3.2.

Table 3.2

Resistance value Marked as

0.47 R47

1 1R0

4.7 4R7

47 47R

100 100R

1 k 1K0

10 k 10 K

10 M 10 M

Tolerance is indicated as follows: F =1%,G =2%, J =5%, K =10% and M =20%

Thus, for example,

R33M = 0.33 20%4R7K = 4.7 10%390RJ = 390 5%

Problem 19. Determine the value of a resistormarked as 6K8F

From Table 3.2, 6K8F is equivalent to: 6.8 k1%

Problem 20. Determine the value of a resistormarked as 4M7M

From Table 3.2, 4M7M is equivalent to: 4.7 M20%

Problem 21. Determine the letter and digit codefor a resistor having a value of 68 k10%

From Table 3.2, 68 k10% has a letter and digit codeof: 68KK

Part

1


Now try the following Practice Exercises

Practice Exercise 9 Resistor colour codingand ohmic values (Answers on page 743)

1. Determine the value and tolerance of a resistorhaving a colour coding of: blue-grey-orange-red.

2. Determine the value and tolerance of a resis-tor having a colour coding of: yellow-violet-gold.

3. Determine the value and tolerance of a resistorhaving a colour coding of: blue-white-black-black-gold.

4. Determine the colour coding for a 51 k four-band resistor having a tolerance of 2%

5. Determine the colour coding for a 1 M four-band resistor having a tolerance of 10%

6. Determine the range of values expected for aresistor with colour coding: red-black-green-silver.

7. Determine the range of values expected fora resistor with colour coding: yellow-black-orange-brown.

8. Determine the value of a resistor marked as(a) R22G (b) 4K7F

9. Determine the letter and digit code for aresistor having a value of 100 k5%

10. Determine the letter and digit code for aresistor having a value of 6.8 M20%

For fully worked solutions to each of the problems in Practice Exercises 7 to 9 in this chapter,go to the website:



Chapter 4

Batteries and alternativesources of energy

Why it is important to understand: Batteries and alternative sources of energy

Batteries store electricity in a chemical form, inside a closed-energy system. They can be re-chargedand re-used as a power source in small appliances, machinery and remote locations. Batteries can stored.c. electrical energy produced by renewable sources such as solar, wind and hydro power in chemicalform. Because renewable energy-charging sources are often intermittent in their nature, batteries provideenergy storage in order to provide a relatively constant supply of power to electrical loads regardless ofwhether the sun is shining or the wind is blowing. In an off-grid photovoltaic (PV) system, for example,battery storage provides a way to power common household appliances regardless of the time of day or thecurrent weather conditions. In a grid-tie with battery backup PV system, batteries provide uninterruptedpower in case of utility power failure. Energy causes movement; every time something moves, energy isbeing used. Energy moves cars, makes machines run, heats ovens, and lights our homes. One form ofenergy can be changed into another form. When petrol is burned in a vehicle engine, the energy storedin petrol is changed into heat energy. When we stand in the sun, light energy is changed into heat. Whena torch or flashlight is turned on, chemical energy stored in the battery is changed into light and heat. Tofind energy, look for motion, heat, light, sound, chemical reactions or electricity. The sun is the source ofall energy. The suns energy is stored in coal, petroleum, natural gas, food, water and wind. While thereare two types of energy, renewable and non-renewable, most of the energy we use comes from burningnon-renewable fuels coal, petroleum or oil or natural gas. These supply the majority of our energyneeds because we have designed ways to transform their energy on a large scale to meet consumer needs.Regardless of the energy source, the energy contained in them is changed into a more useful form ofelectricity. This chapter explains the increasingly important area of battery use and briefly looks at somealternative sources of energy.


list practical applications of batteries understand electrolysis and its applications, including electroplating appreciate the purpose and construction of a simple cell explain polarization and local action explain corrosion and its effects define th

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