Electrical Circuit Theory and Technology John Bird 4th Edition

Electrical Circuit Theory and Technology

In Memory of Elizabeth

Electrical Circuit Theory and Technology

Fourth edition

John Bird, BSc(Hons), CEng, CSci, CMath, FIET, MIEE, FIIE, FIMA, FCollT

AMSTERDAM BOSTON HEIDELBERG LONDON NEW YORK OXFORDPARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO

Newnes is an imprint of Elsevier

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Contents

Preface xi

Part 1 Basic electrical engineeringprinciples 1

1 Units associated with basic electricalquantities 3

1.1 SI units 31.2 Charge 31.3 Force 41.4 Work 41.5 Power 41.6 Electrical potential and e.m.f. 51.7 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 16

3 Resistance variation 173.1 Resistor construction 173.2 Resistance and resistivity 173.3 Temperature coefficient of resistance 203.4 Resistor colour coding and ohmic values 22

4 Batteries and alternative sources of energy 254.1 Introduction to batteries 254.2 Some chemical effects of electricity 264.3 The simple cell 264.4 Corrosion 27

4.5 E.m.f. and internal resistance of a cell 274.6 Primary cells 304.7 Secondary cells 314.8 Cell capacity 334.9 Safe disposal of batteries 334.10 Fuel cells 334.11 Alternative and renewable energy sources 34

Revision Test 1 35

5 Series and parallel networks 365.1 Series circuits 365.2 Potential divider 375.3 Parallel networks 395.4 Current division 415.5 Loading effect 455.6 Potentiometers and rheostats 465.7 Relative and absolute voltages 485.8 Earth potential and short circuits 505.9 Wiring lamps in series and in parallel 50

6 Capacitors and capacitance 526.1 Introduction to capacitors 526.2 Electrostatic field 526.3 Electric field strength 536.4 Capacitance 546.5 Capacitors 546.6 Electric flux density 556.7 Permittivity 556.8 The parallel plate capacitor 566.9 Capacitors connected in parallel and series 586.10 Dielectric strength 626.11 Energy stored 626.12 Practical types of capacitor 636.13 Discharging capacitors 65

7 Magnetic circuits 667.1 Introduction to magnetism and

magnetic circuits 667.2 Magnetic fields 677.3 Magnetic flux and flux density 677.4 Magnetomotive force and magnetic

field strength 687.5 Permeability and BH curves 697.6 Reluctance 70

vi Contents

7.7 Composite series magnetic circuits 717.8 Comparison between electrical and

magnetic quantities 747.9 Hysteresis and hysteresis loss 75

Revision Test 2 76

8 Electromagnetism 778.1 Magnetic field due to an electric current 778.2 Electromagnets 798.3 Force on a current-carrying conductor 808.4 Principle of operation of a simple

d.c. motor 848.5 Principle of operation of a moving

coil-instrument 848.6 Force on a charge 85

9 Electromagnetic induction 869.1 Introduction to electromagnetic induction 869.2 Laws of electromagnetic induction 879.3 Rotation of a loop in a magnetic field 909.4 Inductance 919.5 Inductors 929.6 Energy stored 939.7 Inductance of a coil 939.8 Mutual inductance 95

10 Electrical measuring instruments andmeasurements 97

10.1 Introduction 9810.2 Analogue instruments 9810.3 Moving-iron instrument 9810.4 The moving-coil rectifier instrument 9910.5 Comparison of moving-coil, moving-iron

and moving-coil rectifier instruments 9910.6 Shunts and multipliers 10010.7 Electronic instruments 10110.8 The ohmmeter 10110.9 Multimeters 10210.10 Wattmeters 10210.11 Instrument loading effect 10210.12 The oscilloscope 10510.13 Virtual test and measuring instruments 10910.14 Virtual digital storage oscilloscopes 11010.15 Waveform harmonics 11310.16 Logarithmic ratios 11410.17 Null method of measurement 11610.18 Wheatstone bridge 11610.19 D.C. potentiometer 11710.20 A.C. bridges 11810.21 Measurement errors 118

11 Semiconductor diodes 12111.1 Types of material 12111.2 Semiconductor materials 12211.3 Conduction in semiconductor materials 12311.4 The p-n junction 12311.5 Forward and reverse bias 12511.6 Semiconductor diodes 12811.7 Characteristics and maximum ratings 12811.8 Rectification 12911.9 Zener diodes 12911.10 Silicon controlled rectifiers 13011.11 Light emitting diodes 13111.12 Varactor diodes 13111.13 Schottky diodes 131

12 Transistors 13312.1 Transistor classification 13312.2 Bipolar junction transistors (BJT) 13412.3 Transistor action 13412.4 Leakage current 13512.5 Bias and current flow 13612.6 Transistor operating configurations 13712.7 Bipolar transistor characteristics 13712.8 Transistor parameters 13812.9 Current gain 14012.10 Typical BJT characteristics and maximum

ratings 14012.11 Field effect transistors 14112.12 Field effect transistor characteristics 14212.13 Typical FET characteristics and maximum

ratings 14412.14 Transistor amplifiers 14412.15 Load lines 146

Revision Test 3 150

Main formulae for Part 1 151

Part 2 Electrical principles andtechnology 153

13 D.c. circuit theory 15513.1 Introduction 15513.2 Kirchhoffs laws 15513.3 The superposition theorem 15913.4 General d.c. circuit theory 16213.5 Thvenins theorem 16413.6 Constant-current source 16813.7 Nortons theorem 16913.8 Thvenin and Norton equivalent networks 17113.9 Maximum power transfer theorem 175

Contents vii

14 Alternating voltages and currents 17814.1 Introduction 17814.2 The a.c. generator 17814.3 Waveforms 17914.4 A.c. values 18014.5 Electrical safety insulation and fuses 18314.6 The equation of a sinusoidal waveform 18414.7 Combination of waveforms 18614.8 Rectification 19014.9 Smoothing of the rectified output waveform 191

Revision Test 4 193

15 Single-phase series a.c. circuits 19415.1 Purely resistive a.c. circuit 19515.2 Purely inductive a.c. circuit 19515.3 Purely capacitive a.c. circuit 19515.4 RL series a.c. circuit 19715.5 RC series a.c. circuit 20015.6 RLC series a.c. circuit 20215.7 Series resonance 20515.8 Q-factor 20615.9 Bandwidth and selectivity 20815.10 Power in a.c. circuits 20815.11 Power triangle and power factor 209

16 Single-phase parallel a.c. circuits 21216.1 Introduction 21216.2 RL parallel a.c. circuit 21216.3 RC parallel a.c. circuit 21316.4 LC parallel a.c. circuit 21516.5 LRC parallel a.c. circuit 21616.6 Parallel resonance and Q-factor 21916.7 Power factor improvement 223

17 D.c. transients 22917.1 Introduction 22917.2 Charging a capacitor 22917.3 Time constant for a CR circuit 23017.4 Transient curves for a CR circuit 23117.5 Discharging a capacitor 23417.6 Camera flash 23717.7 Current growth in an LR circuit 23717.8 Time constant for an LR circuit 23717.9 Transient curves for an LR circuit 23817.10 Current decay in an LR circuit 23917.11 Switching inductive circuits 24217.12 The effect of time constant on a

rectangular waveform 242

18 Operational amplifiers 24418.1 Introduction to operational amplifiers 24418.2 Some op amp parameters 246

18.3 Op amp inverting amplifier 24718.4 Op amp non-inverting amplifier 24918.5 Op amp voltage-follower 25018.6 Op amp summing amplifier 25018.7 Op amp voltage comparator 25118.8 Op amp integrator 25218.9 Op amp differential amplifier 25318.10 Digital to analogue (D/A) conversion 25418.11 Analogue to digital (A/D) conversion 255

Revision Test 5 257

19 Three-phase systems 25819.1 Introduction 25819.2 Three-phase supply 25819.3 Star connection 25919.4 Delta connection 26219.5 Power in three-phase systems 26319.6 Measurement of power in three-phase

systems 26519.7 Comparison of star and delta connections 26919.8 Advantages of three-phase systems 270

20 Transformers 27120.1 Introduction 27120.2 Transformer principle of operation 27220.3 Transformer no-load phasor diagram 27420.4 E.m.f. equation of a transformer 27520.5 Transformer on-load phasor diagram 27720.6 Transformer construction 27820.7 Equivalent circuit of a transformer 27920.8 Regulation of a transformer 28120.9 Transformer losses and efficiency 28120.10 Resistance matching 28420.11 Auto transformers 28620.12 Isolating transformers 28820.13 Three-phase transformers 28820.14 Current transformers 28920.15 Voltage transformers 290

Revision Test 6 291

21 D.c. machines 29221.1 Introduction 29221.2 The action of a commutator 29321.3 D.c. machine construction 29321.4 Shunt, series and compound windings 29421.5 E.m.f. generated in an armature winding 29421.6 D.c. generators 29621.7 Types of d.c. generator and their

characteristics 296

viii Contents

21.8 D.c. machine losses 30021.9 Efficiency of a d.c. generator 30121.10 D.c. motors 30121.11 Torque of a d.c. machine 30221.12 Types of d.c. motor and their

characteristics 30421.13 The efficiency of a d.c. motor 30821.14 D.c. motor starter 31021.15 Speed control of d.c. motors 31021.16 Motor cooling 313

22 Three-phase induction motors 31422.1 Introduction 31422.2 Production of a rotating magnetic field 31522.3 Synchronous speed 31622.4 Construction of a three-phase induction

motor 31722.5 Principle of operation of a three-phase

induction motor 31822.6 Slip 31822.7 Rotor e.m.f. and frequency 32022.8 Rotor impedance and current 32022.9 Rotor copper loss 32122.10 Induction motor losses and efficiency 32122.11 Torque equation for an induction motor 32322.12 Induction motor torquespeed

characteristics 32522.13 Starting methods for induction motors 32622.14 Advantages of squirrel-cage induction

motors 32822.15 Advantages of wound rotor induction

motor 32822.16 Double cage induction motor 32822.17 Uses of three-phase induction motors 328

Revision Test 7 329

Main formulae for Part 2 330

Part 3 Advanced circuit theoryand technology 333

23 Revision of complex numbers 33523.1 Introduction 33523.2 Operations involving Cartesian complex

numbers 33623.3 Complex equations 33823.4 The polar form of a complex number 33923.5 Multiplication and division using complex

numbers in polar form 340

23.6 De Moivres theorem powers and rootsof complex numbers 342

24 Application of complex numbers to seriesa.c. circuits 344

24.1 Introduction 34424.2 Series a.c. circuits 34424.3 Further worked problems on series

a.c. circuits 351

25 Application of complex numbers to parallela.c. networks 356

25.1 Introduction 35625.2 Admittance, conductance and susceptance 35625.3 Parallel a.c. networks 36025.4 Further worked problems on parallel

a.c. networks 363

26 Power in a.c. circuits 36726.1 Introduction 36726.2 Determination of power in a.c. circuits 36726.3 Power triangle and power factor 36926.4 Use of complex numbers for

determination of power 37026.5 Power factor improvement 375

Revision Test 8 380

27 A.c. bridges 38127.1 Introduction 38127.2 Balance conditions for an a.c. bridge 38127.3 Types of a.c. bridge circuit 38227.4 Worked problems on a.c. bridges 387

28 Series resonance and Q-factor 39128.1 Introduction 39128.2 Series resonance 39128.3 Q-factor 39428.4 Voltage magnification 39528.5 Q-factors in series 39828.6 Bandwidth 39928.7 Small deviations from the resonant

frequency 403

29 Parallel resonance and Q-factor 40529.1 Introduction 40529.2 The LRC parallel network 40629.3 Dynamic resistance 40629.4 The LRCR parallel network 40729.5 Q-factor in a parallel network 40729.6 Further worked problems on parallel

resonance and Q-factor 412

Revision Test 9 415

Contents ix

30 Introduction to network analysis 41630.1 Introduction 41630.2 Solution of simultaneous equations using

determinants 41730.3 Network analysis using Kirchhoffs laws 418

31 Mesh-current and nodal analysis 42531.1 Mesh-current analysis 42531.2 Nodal analysis 429

32 The superposition theorem 43632.1 Introduction 43632.2 Using the superposition theorem 43632.3 Further worked problems on the

superposition theorem 441

33 Thvenins and Nortons theorems 44533.1 Introduction 44533.2 Thvenins theorem 44533.3 Further worked problems on Thvenins

theorem 45233.4 Nortons theorem 45633.5 Thvenin and Norton equivalent networks 462

Revision Test 10 468

34 Delta-star and star-delta transformations 46934.1 Introduction 46934.2 Delta and star connections 46934.3 Delta-star transformation 46934.4 Star-delta transformation 478

35 Maximum power transfer theorems andimpedance matching 481

35.1 Maximum power transfer theorems 48135.2 Impedance matching 487


36 Complex waveforms 49136.1 Introduction 49136.2 The general equation for a complex

waveform 49236.3 Harmonic synthesis 49336.4 Fourier series of periodic and non-periodic

functions 50036.5 Even and odd functions and Fourier series

over any range 50536.6 Rms value, mean value and the form

factor of a complex wave 50936.7 Power associated with complex waves 51336.8 Harmonics in single-phase circuits 51536.9 Further worked problems on harmonics

in single-phase circuits 519

36.10 Resonance due to harmonics 52336.11 Sources of harmonics 525

37 A numerical method of harmonic analysis 52937.1 Introduction 52937.2 Harmonic analysis on data given in tabular

or graphical form 52937.3 Complex waveform considerations 532

38 Magnetic materials 53638.1 Revision of terms and units used with

magnetic circuits 53638.2 Magnetic properties of materials 53738.3 Hysteresis and hysteresis loss 53938.4 Eddy current loss 54238.5 Separation of hysteresis and eddy current

losses 54538.6 Non-permanent magnetic materials 54738.7 Permanent magnetic materials 549


39 Dielectrics and dielectric loss 55139.1 Electric fields, capacitance and permittivity 55139.2 Polarization 55139.3 Dielectric strength 55239.4 Thermal effects 55339.5 Mechanical properties 55339.6 Types of practical capacitor 55339.7 Liquid dielectrics and gas insulation 55339.8 Dielectric loss and loss angle 554

40 Field theory 55740.1 Field plotting by curvilinear squares 55740.2 Capacitance between concentric cylinders 56140.3 Capacitance of an isolated twin line 56640.4 Energy stored in an electric field 56940.5 Induced e.m.f. and inductance 57140.6 Inductance of a concentric cylinder (or

coaxial cable) 57140.7 Inductance of an isolated twin line 57440.8 Energy stored in an electromagnetic field 576

41 Attenuators 57941.1 Introduction 57941.2 Characteristic impedance 58041.3 Logarithmic ratios 58141.4 Symmetrical T- and -attenuators 58341.5 Insertion loss 58941.6 Asymmetrical T- and -sections 592

x Contents

41.7 The L-section attenuator 59541.8 Two-port networks in cascade 59741.9 ABCD parameters 60041.10 ABCD parameters for networks 60341.11 Characteristic impedance in terms of

ABCD parameters 609


42 Filter networks 61242.1 Introduction 61242.2 Basic types of filter sections 61242.3 The characteristic impedance and the

attenuation of filter sections 61442.4 Ladder networks 61642.5 Low-pass filter sections 61742.6 High-pass filter sections 62342.7 Propagation coefficient and time delay in

filter sections 62842.8 m-derived filter sections 63442.9 Practical composite filters 639

43 Magnetically coupled circuits 64343.1 Introduction 64343.2 Self-inductance 67343.3 Mutual inductance 64343.4 Coupling coefficient 64543.5 Coils connected in series 64643.6 Coupled circuits 64943.7 Dot rule for coupled circuits 654

44 Transmission lines 66144.1 Introduction 66144.2 Transmission line primary constants 66144.3 Phase delay, wavelength and velocity of

propagation 66344.4 Current and voltage relationships 66444.5 Characteristic impedance and

propagation coefficient in terms of theprimary constants 666

44.6 Distortion on transmission lines 67044.7 Wave reflection and the reflection

coefficient 67244.8 Standing waves and the standing

wave ratio 675

45 Transients and Laplace transforms 68045.1 Introduction 68045.2 Response of RC series circuit to a step

input 680

45.3 Response of RL series circuit to a stepinput 683

45.4 LRC series circuit response 68645.5 Introduction to Laplace transforms 68945.6 Inverse Laplace transforms and the

solution of differential equations 69345.7 Laplace transform analysis directly from

the circuit diagram 69945.8 LRC series circuit using Laplace

transforms 70945.9 Initial conditions 712


Main formulae for Part 3: Advanced circuittheory and technology 717

Part 4 General reference 723

Standard electrical quantities their symbolsand units 725

Greek alphabet 728

Common prefixes 729

Resistor colour coding and ohmic values 730

Index 731

On the Website

Some practical laboratory experiments

1 Ohms law 22 Series-parallel 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.booksite.elsevier.com/newnes/bird

Preface

Electrical Circuit Theory and Technology 4th Edi-tion provides coverage for a wide range of coursesthat contain electrical principles, circuit theory andtechnology in their syllabuses, from introductory todegree level.

New topics included in this edition include resistorconstruction and colour code, more information on bat-teries, dangers of constant high current with insulation,loading effect of instruments, earth point and causes ofshort circuits, potentiometers and rheostats, and rectifiersmoothing. In addition, freely available on the websiteare 10 straightforward laboratory experiments whichmay be downloaded and edited.

A free Internet download of a sample (over 700) of the1000 further problems contained in the book is availableto tutors see next page.

Also, a free Internet download of a PowerPoint pre-sentation of all 1100 illustrations contained in the textis available see next page.

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 would needto know. An introduction to units, electrical circuits,resistance variation, batteries and alternative sourcesof energy, series and parallel circuits, capacitors andcapacitance, magnetic circuits, electromagnetism, elec-tromagnetic induction, electrical measuring instrumentsand measurements, semiconductor diodes and transis-tors 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 suitable

for Degree, Foundation Degree, Higher National Cer-tificate/Diploma and City & Guilds courses in electricaland electronic/telecommunications engineering. Thetwo earlier sections of the book will provide a valuablereference/revision for students at this level. Complexnumbers and their application to series and parallelnetworks, power in a.c. circuits, a.c. bridges, seriesand parallel resonance and Q-factor, network analysisinvolving Kirchhoffs laws, mesh and nodal analysis,the superposition theorem, Thevenins and Nortonstheorems, delta-star and star-delta transforms, maxi-mum power transfer theorems and impedance matching,complex waveforms, Fourier series, harmonic analy-sis, magnetic materials, dielectrics and dielectric loss,field theory, attenuators, filter networks, magneticallycoupled circuits, transmission line theory and tran-sients and Laplace transforms are all included in thissection.

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 learningobjectives are listed.

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

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 CircuitTheory and Technology 4th Edition contains some 800detailed worked problems, together with over 1000further problems (with answers), arranged within 177exercises that appear every few pages throughout thetext. Over 1100 line diagrams further enhance theunderstanding of the theory.

Fourteen Revision Tests have been included, inter-spersed within the text every few chapters. For example,Revision Test 1 tests understanding of Chapters 1 to 4,Revision Test 2 tests understanding of Chapters 5 to 7,Revision Test 3 tests understanding of Chapters 8 to 12,and so on. These Revision Tests do not have answers

xii Preface

given since it is envisaged that lecturers/instructorscould set the Revision Tests for students to attempt aspart of their course structure. Lecturers/ instructors mayobtain a complementary set of solutions of the Revi-sion Tests in an Instructors Manual available fromthe publishers via the Internet see below.

Learning by Example is at the heart of ElectricalCircuit Theory and Technology 4th Edition.

JOHN BIRDRoyal Naval School of Marine Engineering,

HMS Sultan,formerly University of Portsmouthand Highbury College, Portsmouth

Free web downloads

Sample of worked solutions to exercises

Within the text are some 1000 further prob-lems arranged within 177 exercises. A sampleof over 700 worked solutions has been prepared

and is available to lecturers only at http://www.booksite.elsevier.com/newnes/bird

Laboratory experiments

Ten practical laboratory experiments are included.It may be that tutors will want to edit theseexperiments to suit their own equipment/com-ponent availability. Go to http://www.booksite.elsevier.com/newnes/bird

Instructors manual

This provides full worked solutions and markscheme for all 14 Revision Tests in this book.The material is available to lecturers only andis available at http://www.booksite.elsevier.com/newnes/bird

Illustrations

Lecturers can download electronic files for all 1100illustrations in this fourth edition. The materialis available to lecturers only and is available athttp://www.booksite.elsevier.com/newnes/bird

Part 1

Basic electrical engineeringprinciples

1 Units associated with basic electrical quantities 32 An introduction to electric circuits 83 Resistance variation 174 Batteries and alternative sources of energy 25

Revision Test 1 355 Series and parallel networks 366 Capacitors and capacitance 527 Magnetic circuits 66

Revision Test 2 768 Electromagnetism 779 Electromagnetic induction 86

10 Electrical measuring instruments and measurements 9711 Semiconductor diodes 12112 Transistors 133

Revision Test 3 150Main formulae for Part 1 151

This page intentionally left blank

Chapter 1

Units associated with basicelectrical quantities

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 4.

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 point

DOI: 10.1016/B978-1-85617-770-2.00001-X

Part

1

4 Electrical Circuit Theory and Technology

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)

in an electric circuit when a current of one ampere ismaintained 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

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 m

s2=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) where onejoule is one newton metre. The joule is defined as thework 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 is onejoule per second. Power is defined as the rate of doingwork 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

Part

1

Units associated with basic electrical quantities 5

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 J25 s

= 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=forcedistance andforce=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

= 98 100 J20 s

= 4905 J/s=4905 W or 4.905 kW

Now try the following exercise

Exercise 1 Further problems on unitsassociated with basic electricalquantities.

(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? [600 N]

2. Find the accelerating force when a car havinga mass of 1.7 Mg increases its speed with aconstant acceleration of 3 m/s2 [5.1 kN]

3. A force of 40 N accelerates a mass at 5 m/s2.Determine the mass. [8 kg]

4. Determine the force acting downwardson a mass of 1500 g suspended on astring. [14.72 N]

5. A force of 4 N moves an object 200 cm in thedirection of the force. What amount of workis done? [8 J]

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? [12.5 kJ]

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. [4.5 W]

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.

[(a) 29.43 kNm (b) 981 W]

9. What quantity of electricity is carried by6.241021 electrons? [1000 C]

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

11. A current of 3 A flows for 5 minutes. Whatcharge is transferred? [900 C]

12. How long must a current of 0.1 A flow so asto transfer a charge of 30 C? [5 minutes]

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

[(a) 1 nF (b) 20 000 pF (c) 5 MHz(d) 0.047 M (e) 320A]

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

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.

Part

1


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). 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= 1

10siemen=0.1 S

(b) G= 1R= 1

5103 S=0.2103 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 an elec-tric circuit and the voltage across the circuit is V volts,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

1 kWh = 1000 watt hour= 1000 3600 watt seconds or joules= 3 600 000 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=voltagecurrent.Hence

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= 1000

250=4 A

Hence the current taken from the supply is 4 A.


Exercise 2 Further problems on unitsassociated with basic electricalquantities

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

[(a) 0.1 S (b) 0.5 mS (c) 500 S]

2. A conductor has a conductance of 50S. Whatis its resistance? [20 k]

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?

[1 kW]

4. 450 J of energy are converted into heat in1 minute. What power is dissipated? [7.5 W]

5. A current of 10 A flows through a conductorand 10 W is dissipated. What p.d. exists acrossthe ends of the conductor? [1 V]

Part

1

Units associated with basic electrical quantities 7

6. A battery of e.m.f. 12 V supplies a currentof 5 A for 2 minutes. How much energy issupplied in this time? [7.2 kJ]

7. A dc electric motor consumes 36 MJ when con-nected to a 250 V supply for 1 hour. Find thepower rating of the motor and the current takenfrom the supply. [10 kW, 40 A]

1.9 Summary of terms, units andtheir 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 ms1

Acceleration a metres per m/s2 orsecond ms2squared

Force F newton N

Electrical Q coulomb Ccharge orquantity

Electric current I ampere A

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 725.

Chapter 2

An introduction toelectric circuits


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 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 = V

Ror R= V

I 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

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

DOI: 10.1016/B978-1-85617-770-2.00002-1

Part

1

An introduction to electric circuits 9

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 quantity ofelectricity

All atoms consist of protons, neutrons and electrons.The protons, which have positive electrical charges, andthe neutrons, which have no electrical charge, are con-tained within the nucleus. Removed from the nucleusare minute negatively charged particles called electrons.Atoms of different materials differ from one another byhaving different numbers of protons, neutrons and elec-trons. An equal number of protons and electrons existwithin an atom and it is said to be electrically balanced,as the positive and negative charges cancel each otherout. When there are more than two electrons in an atomthe electrons are arranged into shells at various distancesfrom 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 thatare loosely connected to the nucleus and can easily movethrough the material from one atom to another.

Insulators are materials whose electrons are heldfirmly to their nucleus.

The unit used to measure the quantity of elec-trical charge Q is called the coulomb C (where 1coulomb=6.241018 electrons).

If the drift of electrons in a conductor takes place atthe rate of one coulomb per second the resulting currentis said to be a current of one ampere.Thus, 1 ampere=1 coulomb per second or 1 A=1 C/sHence, 1 coulomb=1 ampere second or 1 C=1 AsGenerally, 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 coulombs

Problem 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= 0.24

15 103 =0.24 103

15= 240

15= 16 A

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

Part

1


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


Exercise 3 Further problems on electriccurrent and quantity of charge

1. In what time would a current of 10 A transfera charge of 50 C? [5 s]

2. A current of 6 A flows for 10 minutes. Whatcharge is transferred? [3600 C]

3. How long must a current of 100 mA flow so asto transfer a charge of 80 C? [13 min 20 s]

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.

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 through

it 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 measuring resis-tance.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 acrossa 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.

Part

1


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 of onerevolution of the revolving object, and the duration of theview very short, the object will appear to be stationary.(See Chapter 10 for more detail about electrical mea-suring 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. againstcurrent. 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.

V

AR1

R2

l

Figure 2.3

p.d.

00 ll

(b)(a)

p.d.

Figure 2.4

If the resistor R1 in Figure 2.3 is replaced by acomponent such as a lamp then the graph shown in

Figure 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= 20

0.8= 200

8= 25

2.7 Multiples and sub-multiples

Currents, voltages and resistances can often be verylarge or very small. Thus multiples and sub-multiplesof 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 729.

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=2000Current I =10 mA

=10103 A or 10103

or10

1000A

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

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= 501 000 000

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

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= 12

50 103 =12 103

50

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

Resistance R = VI= 100

5 103 =100 103

5

= 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= 120

50 103

= 1200.05

= 12 0005

=2400 or 2.4 k

(b) Resistance R= 120200 106 =

120

0.0002

= 1 200 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.

Figure 2.5

For resistor A,

R= VI= 20 A

20 mA= 20

0.02= 2000

2=1000 or 1 k

For resistor B,

R= VI= 16 V

5 mA= 16

0.005= 16 000

5=3200 or 3.2 k

Part

1



Exercise 4 Further problemson Ohms law

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. [7]

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. [(a) 0.25 A (b) 960]

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

[2 m, 5 m]

Figure 2.6

4. Determine the p.d. which must be applied to a5 k resistor such that a current of 6 mA mayflow. [30 V]

5. A 20 V source of e.m.f. is connected across acircuit having a resistance of 400. Calculatethe current flowing. [50 mA]

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.

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= 250

0.4= 2500

4=625

Problem 10. Calculate the power dissipated whena current of 4 mA flows through a resistance of5 k.

Power P= I 2 R= (4103)2(5103)=161065103=80103=0.08 W or 80 mW

Part

1


Alternatively, since I =4103 and R=5103then from Ohms law,voltage V = IR=41035103=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= 240

30=8 A

Power, 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 by thecoil.

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

(b) Power dissipated by coil, P= I 2 R= 52100=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= 240

960= 24

96= 1

4A 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 the

home 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= 12

40=0.3 A

Power consumed, P=VI = (12)(0.3)=3.6 WEnergy dissipated=powertime= (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 kJProblem 16. Electrical equipment in an officetakes a current of 13 A from a 240 V supply.Estimate the cost per week of electricity if theequipment 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=powertime= (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

Part

1


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

=powertime=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=powertimeEnergy used by one 3 kW fire in 20 hours=3 kW20 h=60 kWh

Hence weekly energy used by two 3 kW fires=260=120 kWh

Energy 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


Exercise 5 Further problemson power andenergy

1. The hot resistance of a 250 V filament lampis 625. Determine the current taken by thelamp and its power rating. [0.4 A, 100 W]

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

[(a) 2 k (b) 0.5 M]

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.

[20, 2.88 kW, 57.6 kWh]

4. Determine the power dissipated when a cur-rent of 10 mA flows through an appliancehaving a resistance of 8 k. [0.8 W]

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

[9.5 W]

6. A current of 4 A flows through a conductor and10 W is dissipated. What p.d. exists across theends of the conductor? [2.5 V]

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.

[(a) 0.5 W (b) 1.33 W (c) 40 W]

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 kJ]

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 12.82 p.

[22.82]

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 12.50 p per unit.

[3 kW, 90 kWh, 11.25]

Part

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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 nocurrent 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 shorttime at switch on. If a fuse is fitted to withstand thislarge current there would be no protection against faultswhich cause the current to rise slightly above the nor-mal value. Therefore special anti-surge fuses are fitted.These can stand 10 times the rated current for 10 milli-seconds. If the surge lasts longer than this the fuse willblow.

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

= 4.17 A

Hence a 5 A fuse is most appropriate

(b) For the fire,

current I = PV

= 3000240

= 30024

= 12.5 A

Hence a 13 A fuse is most appropriate


Exercise 6 Further problemon fuses

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.

[3 A, 5 A]

2.12 Insulation and the dangers ofconstant high current flow

The use of insulation materials on electrical equipment,whilst being necessary, also has the effect of prevent-ing heat loss, i.e. the heat is not able to dissipate, thuscreating possible danger of fire. In addition, the insulat-ing material has a maximum temperature rating this isheat it can withstand without being damaged. The cur-rent rating for all equipment and electrical componentsis therefore limited to keep the heat generated withinsafe limits. In addition, the maximum voltage presentneeds to be considered when choosing insulation.

Chapter 3

Resistance variationAt the end of this chapter you should be able to:

recognise three common methods of resistor construction appreciate that electrical resistance depends on four factors appreciate that resistance R= l

a, where is the resistivity

recognize typical values of resistivity and its unit perform calculations using R= l

a 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

3.1 Resistor construction

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

(i) 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.

(ii) Metal oxide resistors

With a metal oxide resistor a thin coating of platinumis deposited on a glass plate; it is then fired and a thintrack etched out. It is then totally enclosed in an outertube.

Metal oxide resistors are used in electronic equip-ment.

(iii) Carbon resistors

This type of resistor is made from a mixture of carbonblack resin binder and a refractory powder that is pressedinto shape and heated in a kiln to form a solid rodof standard length and width. The resistive value ispredetermined by the ratio of the mixture. Metal endconnections are crimped onto the rod to act as connect-ing points for electrical circuitry. This type of resistoris small and mass-produced cheaply; it has limitedaccuracy and a low power rating.

Carbon resistors are used in electronic equipment.

3.2 Resistance and resistivity

The resistance of an electrical conductor depends on4 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.

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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 insertinga constant of proportionality into this relationship thetype of material used may be taken into account. Theconstant of proportionality is known as the resistivityof the material 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 unit

cube of the material measured between opposite facesof the cube.

Resistivity varies with temperature and some typ-ical values of resistivities measured at about roomtemperature 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 lHence, 6005 m or 600=(k)(5), where k is thecoefficient of proportionality. Hence,

k = 6005

= 120

When the length l is 8 m, then resistance

R= kl=(120)(8)=960

(b) When the resistance is 420, 420= kl, fromwhich,

length l = 420k

= 420120

= 3.5 m

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=5 mm2

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=0.06

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Resistance variation 19

If 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 2000

100

= 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= 36 mm2= 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


Exercise 7 Further problems on resistanceand resistivity

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.

[(a) 8.75 (b) 5 m]

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. [(a) 5 (b) 0.625 mm2]

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.

[0.32]

4. Find the resistance of 800 m of copper cableof cross-sectional area 20 mm2. Take the resis-tivity of copper as 0.02m. [0.8]

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 0.03106 m. [1.5 mm2]

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.

[0.026m]

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7. Find the resistance of 1 km of copper cablehaving a diameter of 10 mm if the resistivityof copper is 0.017106m. [0.216]

3.3 Temperature coefficient ofresistance

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

The temperature coefficient of resistance of amaterial is the increase in the resistance of a 1 resistorof that material when it is subjected to a rise of tem-perature of 1C. The symbol used for the temperaturecoefficient 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 ofresistance 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 0C

R = 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.

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

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

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

= 271+ 0.133 =

27

1.133

= 23.83

Problem 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.

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1


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.


Exercise 8 Further problems ontemperature coefficient ofresistance

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. [69]

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.

[24.69]

3. The temperature coefficient of resistance forcarbon at 0C is0.00048/C. What is the sig-nificance of the minus sign? A carbon resistorhas a resistance of 500 at 0C. Determine itsresistance at 50C. [488]

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. [26.4]

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.

[70C]

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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.

[64.8]

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.

[5.952]

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

from Table 3.1, which means that the value of the resistoris 33102=0.33

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 of:brown-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 codingbrown-black-brown-silver lie?

From Table 3.1, brown-black-brown-silver indicates1010, i.e. 100, with a tolerance of 10%

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1


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 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 M

Hence 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.

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%

Table 3.2

Resistance Marked as:Value

0.47 R47

1 1R0

4.7 4R7

47 47R

100 100R

1 k 1K0

10 k 10 K

10 M 10 M

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: 68 KK

Now try the following exercises

Exercise 9 Further problems on resistorcolour coding and ohmic values

1. Determine the value and tolerance of a resistorhaving a colour coding of: blue-grey-orange-red [68 k2%]

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

[4.720%]3. Determine the value and tolerance of a resistor

having a colour coding of: blue-white-black-black-gold [6905%]

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

[green-brown-orange-red]

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

[brown-black-green-silver]

6. Determine the range of values expected for aresistor with colour coding: red-black-green-silver [1.8 M to 2.2 M]

7. Determine the range of values expected fora resistor with colour coding: yellow-black-orange-brown [39.6 k to 40.4 k]

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

[(a) 0.222% (b) 4.7 k1%]9. Determine the letter and digit code for a

resistor having a value of 100 k5%[100 KJ]

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

[6 M8 M]

Chapter 4

Batteries and alternativesources 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 the terms e.m.f., E , and internal resistance, r, of a cell perform calculations using V =E Ir determine the total e.m.f. and total internal resistance for cells connected in series and in parallel distinguish between primary and secondary cells explain the construction and practical applications of the Leclanch, mercury, leadacid and alkaline cells list the advantages and disadvantages of alkaline cells over leadacid cells understand the term cell capacity and state its unit understand the importance of safe battery disposal appreciate advantages of fuel cells and their likely future applications understand the implications of alternative energy sources and state five examples

4.1 Introduction to batteries

A battery is a device that converts chemical energyto electricity. If an appliance is placed between itsterminals the current generated will power the device.Batteries are an indispensable item for many electronicdevices and are essential for devices that require powerwhen no mains power is available. For example, withoutthe battery, there would be no mobile phones or laptopcomputers.

The battery is now over 200 years old and batteriesare found almost everywhere in consumer and industrial

products. Some practical examples where batteries areused include:

in laptops, in cameras, in mobile phones, incars, in watches and clocks, for security equip-ment, in electronic meters, for smoke alarms, formeters used to read gas, water and electricityconsumption at home, to power a camera for anendoscope looking internally at the body, and fortransponders used for toll collection on highwaysthroughout the world

Batteries tend to be split into two categories primary,which are not designed to be electrically re-charged,

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i.e. are disposable (see Section 4.6), and secondarybatteries, which are designed to be re-charged, suchas those used in mobile phones (see Section 4.7).

In more recent years it has been necessary to designbatteries with reduced size, but with increased lifespanand capacity.

If an application requires small size and high powerthen the 1.5 V battery is used. If longer lifetime isrequired then the 3 to 3.6 V battery is used. In the 1970sthe 1.5 V manganese battery was gradually replaced bythe alkaline battery. Silver oxide batteries were grad-ually introduced in the 1960s and are still the preferredtechnology for watch batteries today.

Lithium-ion batteries were introduced in the 1970sbecause of the need for longer lifetime applications.Indeed, some such batteries have been known to lastwell over 10 years before replacement, a characteris-tic that means that these batteries are still very muchin demand today for digital cameras, and sometimesfor watches and computer clocks. Lithium batteriesare capable of delivering high currents but tend to beexpensive.

More types of batteries and their uses are listed inTable 4.2 on page 32.

4.2 Some chemical effects ofelectricity

A material must contain charged particles to be able toconduct electric current. In solids, the current is carriedby electrons. Copper, lead, aluminium, iron and carbonare some examples of solid conductors. In liquids andgases, the current is carried by the part of a moleculewhich has acquired an electric charge, called ions.These can possess a positive or negative charge, andexamples include hydrogen ion H+, copper ion Cu++and hydroxyl ion OH. Distilled water contains noions and is a poor conductor of electricity, whereas saltwater contains ions and is a fairly good conductor ofelectricity.

Electrolysis is the decomposition of a liquid com-pound by the passage of electric current through it.Practical applications of electrolysis include the elec-troplating of metals (see below), the refining of copperand the extraction of aluminium from its ore.

An electrolyte is a compound which will undergoelectrolysis. Examples include salt water, copper sul-phate and sulphuric acid.

The electrodes are the two conductors carrying cur-rent to the electrolyte. The positive-connected electrode

Copperelectrode(anode)

RA

1 2I

Zinc electrode(cathode)

Dilutesulphuric acid(electrolyte)

Figure 4.1

is called the anode and the negative-connected electrodethe cathode.

When two copper wires connected to a battery areplaced in a beaker containing a salt water solution, cur-rent will flow through the solution. Air bubbles appeararound the wires as the water is changed into hydrogenand oxygen by electrolysis.

Electroplating uses the principle of electrolysis toapply a thin coat of one metal to another metal. Somepractical applications include the tin-plating of steel,silver-plating of nickel alloys and chromium-plating ofsteel. If two copper electrodes connected to a batteryare placed in a beaker containing copper sulphate as theelectrolyte it is found that the cathode (i.e. the electrodeconnected to the negative terminal of the battery) gainscopper whilst the anode loses copper.

4.3 The simple cell

The purpose of an electric cell is to convert chemicalenergy into electrical energy.

A simple cell comprises two dissimilar conductors(electrodes) in an electrolyte. Such a cell is shown inFigure 4.1, comprising copper and zinc electrodes. Anelectric current is found to flow between the electrodes.Other possible electrode pairs exist, including zincleadand zinciron. The electrode potential (i.e. the p.d. mea-sured between the electrodes) varies for each pair ofmetals. By knowing the e.m.f. of each metal with respectto some standard electrode, the e.m.f. of any pair of met-als may be determined. The standard used is the hydro-gen electrode. The electrochemical series is a wayof listing elements in order of electrical potential, andTable 4.1 shows a number of elements in such a series.

In a simple cell two faults exist those due topolarization and local action.

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Batteries and alternative sources of energy 27

Table 4.1 Part of the electro-chemical series

Potassium

sodium

aluminium

zinc

iron

lead

hydrogen

copper

silver

carbon

Polarization

If the simple cell shown in Figure 4.1 is left connectedfor some time, the current I decreases fairly rapidly.This is because of the formation of a film of hydrogenbubbles on the copper anode. This effect is known asthe polarization of the cell. The hydrogen prevents fullcontact between the copper electrode and the electrolyteand this increases the internal resistance of the cell. Theeffect can be overcome by using a chemical depolariz-ing agent or depolarizer, such as potassium dichromatewhich removes the hydrogen bubbles as they form. Thisallows the cell to deliver a steady current.

Local action

When commercial zinc is placed in d

Electrical Circuit Theory and Technology John Bird 4th Edition

Engineering

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