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Electrical and Electronic Principles and Technology

To Sue

Electrical and Electronic Principles
and Technology
Third edition
John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE,
FIIE, FIMA, FCollT

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD
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Newnes is an imprint of Elsevier

Newnes is an imprint of Elsevier
Linacre House, Jordan Hill, Oxford OX2 8DP, UK
30 Corporate Drive, Suite 400, Burlington, MA 01803, USA
First edition 2000 previously published as Electrical Principles and Technology for Engineering
Reprinted 2001
Second edition 2003
Reprinted 2004, 2005, 2006
Third edition 2007
Copyright © 2000, 2003, 2007, John Bird
...
All rights reserved
The right of John Bird to be identified as the author of this work
has been asserted in accordance with the Copyright, Designs
and Patents Act 1988
Permissions may be sought directly from Elsevier’s Science & Technology Rights
Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333;
email: permissions@elsevier
...
Alternatively you can submit your request online by
visiting the Elsevier web site at http://elsevier
...
Because of rapid advances in the medical sciences, in particular, independent
verification of diagnoses and drug dosages should be made
British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library
Library of Congress Cataloguing in Publication Data
A catalogue record for this book is available from the Library of Congress
ISBN: 978-0-75-068556-6
For information on all Newnes publications
visit our website at www
...
elsevier
...
charontec
...
1 SI units
1
...
3 Force
1
...
5 Power
1
...
m
...

1
...
8 Electrical power and energy
1
...
1 Electrical/electronic system
block diagrams
2
...
3 Electric current and quantity of
electricity
2
...
5 Basic electrical measuring
instruments
2
...
7 Ohm’s law
2
...
9 Conductors and insulators
2
...
11 Main effects of electric current
2
...
1 Resistance and resistivity
3
...
3 Resistor colour coding and
ohmic values

xi

4

1

3
3
4
4
4
4
5
5
6

Revision Test 1

7
9

5

9
10
11
11
12
12
13
13
14
15
17
17
20
20
22
24

Batteries and alternative sources of energy
4
...
2 Some chemical effects of
electricity
4
...
4 Corrosion
4
...
m
...
and internal resistance
of a cell
4
...
7 Secondary cells
4
...
9 Safe disposal of batteries
4
...
11 Alternative and renewable
energy sources

6

28
28
29
29
30
30
33
34
36
36
36
37
40

Series and parallel networks
5
...
2 Potential divider
5
...
4 Current division
5
...
6 Wiring lamps in series and in
parallel

41
41
42
44
47

Capacitors and capacitance
6
...
2 Electrostatic field
6
...
4 Capacitance
6
...
6 Electric flux density
6
...
8 The parallel plate capacitor
6
...
10 Dielectric strength
6
...
12 Practical types of capacitor
6
...
1 Introduction to magnetism
and magnetic circuits
7
...
3 Magnetic flux and flux density
7
...
5 Permeability and B–H curves
7
...
7 Composite series magnetic
circuits
7
...
9 Hysteresis and hysteresis loss
Revision Test 2
8

9

Electromagnetism
8
...
2 Electromagnets
8
...
4 Principle of operation of a
simple d
...
motor
8
...
6 Force on a charge
Electromagnetic induction
9
...
2 Laws of electromagnetic
induction
9
...
4 Inductance
9
...
6 Energy stored
9
...
8 Mutual inductance

10 Electrical measuring instruments and
measurements
10
...
2 Analogue instruments
10
...
4 The moving-coil rectifier
instrument
10
...
6
10
...
8
10
...
10
10
...
12
10
...
C
...
C
...
1 Types of material
11
...
3 Conduction in semiconductor
materials
11
...
5 Forward and reverse bias
11
...
7 Characteristics and maximum
ratings
11
...
9 Zener diodes
11
...
11 Light emitting diodes
11
...
13 Schottky diodes

140
140
141

10
...
15
10
...
17
10
...
19
10
...
21
10
...
1 Transistor classification
12
...
3 Transistor action
12
...
5 Bias and current flow
12
...
7 Bipolar transistor
characteristics
12
...
9 Current gain
12
...
11 Field effect transistors
12
...
13 Typical FET characteristics and
maximum ratings
12
...
15 Load lines
Revision Test 3
Formulae for basic electrical and electronic
engineering principles

Section 2 Further Electrical and
Electronic Principles

163

15
...
8
15
...
10
15
...
C
...
1 Introduction
13
...
3 The superposition theorem
13
...
c
...
5 Thévenin’s theorem
13
...
7 Norton’s theorem
13
...
9 Maximum power transfer
theorem

14 Alternating voltages and currents
14
...
2 The a
...
generator
14
...
4 A
...
values
14
...
6 Combination of waveforms
14
...
8 Smoothing of the rectified
output waveform
Revision Test 4
15 Single-phase series a
...
circuits
15
...
c
...
2 Purely inductive a
...
circuit
15
...
c
...
4 R–L series a
...
circuit
15
...
c
...
6 R–L–C series a
...
circuit

179
179
179
183
186
188
193
193

18

197
200
205
205
205
206
207
211
213
217
218
221
222
222
222
223
225
228
230

19

Series resonance
Q-factor
Bandwidth and selectivity
Power in a
...
circuits
Power triangle and power
factor

234
235
237
237
238

Single-phase parallel a
...
circuits
16
...
2 R–L parallel a
...
circuit
16
...
c
...
4 L–C parallel circuit
16
...
c
...
6 Parallel resonance and
Q-factor
16
...
1 Introduction
17
...
3 Low-pass filters
17
...
5 Band-pass filters
17
...
C
...
1 Introduction
18
...
3 Time constant for a C–R circuit
18
...
5 Discharging a capacitor
18
...
7 Current growth in an
L–R circuit
18
...
9 Transient curves for an
L–R circuit
18
...
11 Switching inductive circuits
18
...
1 Introduction to operational
amplifiers
19
...
3 Op amp inverting amplifier
19
...
5
19
...
7
19
...
9
19
...
11 Analogue to digital (A/D)
conversion
Revision Test 5
Formulae for further electrical and
electronic engineering principles

Section 3 Electrical Power Technology

295
296
297
297
298
300
301
305

306

309

20 Three-phase systems
20
...
2 Three-phase supply
20
...
4 Delta connection
20
...
6 Measurement of power in
three-phase systems
20
...
8 Advantages of three-phase
systems

311
311
311
312
315
317

21 Transformers
21
...
2 Transformer principle of
operation
21
...
4 E
...
f
...
5 Transformer on-load phasor
diagram
21
...
7 Equivalent circuit of a
transformer
21
...
9 Transformer losses and
efficiency
21
...
11 Auto transformers
21
...
13 Three-phase transformers

327
327

319
324
324

328
330
331
333
335
335
337
338
341
343
345
345

21
...
15 Voltage transformers
Revision Test 6

22

346
348
351

D
...
machines
22
...
2 The action of a commutator
22
...
C
...
4 Shunt, series and compound
windings
22
...
m
...
generated in an armature
winding
22
...
C
...
7 Types of d
...
generator and their
characteristics
22
...
C
...
9 Efficiency of a d
...
generator
22
...
C
...
11 Torque of a d
...
motor
22
...
c
...
13 The efficiency of a d
...
motor
22
...
C
...
15 Speed control of d
...
motors
22
...
1 Introduction
23
...
3 Synchronous speed
23
...
5 Principle of operation of a threephase induction motor
23
...
7 Rotor e
...
f
...
8 Rotor impedance and current
23
...
10 Induction motor losses and
efficiency
23
...
12 Induction motor torque-speed
characteristics
23
...
14 Advantages of squirrel-cage
induction motors

378
378

354
354
356
356
360
361
362
363
365
369
371
371
374

379
380
381
382
382
383
384
385
385
387
390
391
391

Contents ix
23
...
16 Double cage induction motor
23
...
c
...
c
...
c
...

New topics included in this edition are a complete
update on semiconductor diodes and transistors, and
additional material on batteries, fuel cells and alternative and renewable energies, relative and absolute
voltages, self and mutual inductance, virtual test and
measuring instruments
...

A new feature of this third edition is that a free Internet download (lecturers only) is available of a sample
of solutions (some 400) of the 530 further problems
contained in the book — see below
...

The third edition of this textbook provides coverage
of the following syllabuses:
(i) ‘Electrical and Electronic Principles’ (Unit 5,
BTEC National Certificate and National
Diploma) — see chapters 1–10, 11(part), 14,
16, 18(part), 21(part), 22(part)
...

(iii) Parts of the following BTEC National units:
Electrical Applications, Three Phase Systems,
Principles and Applications of Electronic Devices
and Circuits, Aircraft Electrical Machines, and
Telecommunications Principles
...


(v) ‘Electrical and Electronic Principles’, Units of the
City & Guilds Level 3 Certificate in Engineering
(2800)
...

(vii) Any introductory/Access/Foundation course
involving Electrical and Electronic Engineering
Principles
...

Section 2, comprising chapters 13 to 19, involves
Further Electrical and Electronic Principles, with
chapters on d
...
circuit theorems, alternating voltages and currents, single-phase series and parallel
networks, filter networks, d
...
transients and operational
amplifiers
...
c
...

Each topic considered in the text is presented in a
way that assumes in the reader little previous knowledge of that topic
...
The theory is
kept to a minimum, for problem solving is extensively
used to establish and exemplify the theory
...


xii Preface
Electrical and Electronic Principles and Technology, 3rd Edition contains 400 worked problems, together with 340 multi-choice questions (with
answers at the back of the book)
...
Over 500 line diagrams
further enhance the understanding of the theory
...

At regular intervals throughout the text are seven
Revision Tests to check understanding
...
These Revision Tests do

not have answers given since it is envisaged that lecturers/instructors could set the Tests for students to attempt
as part of their course structure
...

I am very grateful to Mike Tooley for his help in
updating chapters on Semiconductor diodes, Transistors, and Measuring instruments and measurements
...
‘Learning by
Example’ is at the heart of Electrical and Electronic
Principles and Technology, 3rd Edition
...

Solutions Manual
Within the text are some 530 further problems
arranged within 145 Exercises
...

Instructor’s Manual
This manual provides full worked solutions and
mark scheme for all 7 Revision tests in this book
...
To access
the lecturer support material, please go to
http://textbooks
...
com and search for the
book
...
If you do not
have an account for the textbook website already,
you will need to register and request access to the
book’s subject area
...


This page intentionally left blank

Section 1

Basic Electrical and
Electronic Engineering
Principles

This page intentionally left blank

Chapter 1

Units associated with basic
electrical 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
...
f
...
1

Derived SI units use combinations of basic units and
there are many of them
...
This was introduced in 1960
and is now adopted by the majority of countries as the
official system of measurement
...
The six most common multiples, with their
meaning, are listed below:
Prefix

Name

Meaning

M

mega

multiply by 1 000 000 (i
...
×106 )

k

kilo

multiply by 1000 (i
...
×103 )

second, s

m

milli

divide by 1000 (i
...
×10−3 )

electric current

ampere, A

μ

micro

divide by 1 000 000 (i
...
×10−6 )

thermodynamic temperature

kelvin, K

n

nano

divide by 1 000 000 000
(i
...
×10−9 )

luminous intensity

candela, cd

p

pico

amount of substance

mole, mol

divide by 1 000 000 000 000
(i
...
×10−12 )

Quantity

Unit

length

metre, m

mass

kilogram, kg

time

Section 1

4 Electrical and Electronic Principles and Technology
1
...
4 Work

The unit of charge is the coulomb (C) where one
coulomb is one ampere second
...
24 ×
1018 electrons)
...
Thus,
charge, in coulombs Q = It
where I is the current in amperes and t is the time in
seconds
...
If a current of 5 A flows for 2 minutes,
find the quantity of electricity transferred
...
3

I = 5 A, t = 2 × 60 = 120 s
Q = 5 × 120 = 600 C

The unit of work or energy is the joule (J) where one
joule is one newton metre
...
Thus
work done on a body, in joules, W = Fs
where F is the force in newtons and s is the distance in
metres moved by the body in the direction of the force
...


1
...
Power is defined as the rate of doing
work or transferring energy
...
Thus,
power, in watts,

Force

The unit of force is the newton (N) where one newton
is one kilogram metre per second squared
...
Thus,
force, in newtons

F = ma

where m is the mass in kilograms and a is the acceleration in metres per second squared
...
81 m/s2
...
A mass of 5000 g is accelerated at
2 m/s2 by a force
...

Force = mass × acceleration
= 5 kg × 2 m/s2 = 10 kg m/s2 = 10 N
...
Find the force acting vertically
downwards on a mass of 200 g attached to a wire
...
2 kg and acceleration due to gravity,
g = 9
...
2 kg × 9
...
962 N

P=

energy, in joules, W = Pt

Problem 4
...
How much work is done if the
machine is moved 20 m and what average power is
utilized if the movement takes 25 s?
Work done = force × distance
= 200 N × 20 m
= 4 000 Nm or 4 kJ
work done
Power =
time taken
=

4000 J
= 160 J/s = 160 W
25 s

Problem 5
...
What is (a) the work done
and (b) the power developed?

(a) Work done = force × distance
and force = mass × acceleration

Hence,
distance of 1
...
Find the power
consumed
...
5 W]

work done = (1000 kg × 9
...
1 kNm or 98
...
905 kW

Power =

Now try the following exercise
Exercise 1 Further problems on charge,
force, work and power
(Take g = 9
...
A mass of 500 kg is raised to a height of 6 m
in 30 s
...

[(a) 29
...
Rewrite the following as indicated:
(a) 1000 pF =
...
02 μF =
...
MHz
(d) 47 k =
...
32 mA =
...
047 M (e) 320 μA]

1
...
24 × 1021 electrons?
[1000 C]
2
...
A current of 3 A flows for 5 minutes
...
How long must a current of 0
...
6

The unit of electric potential is the volt (V), where one
volt is one joule per coulomb
...
e
...
What force is required to give a mass of 20 kg
an acceleration of 30 m/s2 ?
[600 N]
6
...
7 Mg increases its speed with a
constant acceleration of 3 m/s2
...
1 kN]
5 m/s2
...
A force of 40 N accelerates a mass at
Determine the mass
...
Determine the force acting downwards on
a mass of 1500 g suspended on a string
...
72 N]
9
...
What amount of work
is done?
[8 J]
10
...
5 kN is required to lift a load
...
5 kJ]
11
...
m
...


volts =

A change in electric potential between two points in
an electric circuit is called a potential difference
...
m
...
) provided by a source of
energy such as a battery or a generator is measured in
volts
...
7

Resistance and conductance

The unit of electric resistance is the ohm( ), where
one ohm is one volt per ampere
...
Thus,
resistance, in ohms

R=

V
I

Section 1

Units associated with basic electrical quantities 5

Section 1

6 Electrical and Electronic Principles and Technology
where V is the potential difference across the two points,
in volts, and I is the current flowing between the two
points, in amperes
...
Thus
conductance, in siemens G =

1
R

where R is the resistance in ohms
...
Find the conductance of a conductor
of resistance: (a) 10 (b) 5 k (c) 100 m
...
1 S
R 10

(b) G =

1
1
=
S = 0
...
2 mS
R 5 × 103

(c) G =

1
103
1
S=
=
S = 10 S
−3
R 100 × 10
100

1
...
A source e
...
f
...
How much energy is
provided in this time?

Energy = power × time, and power = voltage × current
...
An electric heater consumes 1
...

Find the power rating of the heater and the current
taken from the supply
...
8 × 106 J
=
time
30 × 60 s
= 1000 J/s = 1000 W

i
...
power rating of heater = 1 kW
Power P = VI, thus I =

P 1000
=
=4A
V
250

Hence the current taken from the supply is 4 A
...
m
...
, resistance, conductance, power and
energy
1
...
1 S (b) 0
...
A conductor has a conductance of 50 μS
...
An e
...
f
...
What is the power
developed?
[1 kW]
4
...
What power is dissipated?
[7
...
A current of 10 A flows through a conductor
and 10 W is dissipated
...
d
...
A battery of e
...
f
...
How much energy is
supplied in this time?
[7
...
A d
...
electric motor consumes 36 MJ when
connected to a 250 V supply for 1 hour
...

[10 kW, 40 A]

1
...
Complete the following:
Force =
...
What do you understand by the term ‘potential difference’?
4
...
Name the units used to measure:
(a) the quantity of electricity
(b) resistance
(c) conductance
6
...
Define electrical energy and state its unit

C

8
...
What is electromotive force?

Electric current I

ampere

Resistance

R

ohm

Conductance

G

siemen

S

Electromotive
force

E

volt

V

Potential
difference

V

volt

V

Work

W

joule

Energy

E (or W)

joule

J

Power

P

watt

W

10
...
c
...
What does ‘SI units’ mean?

11
...
m
...

(d) p
...

12
...
A resistance of 50 k
(a) 20 S
(c) 0
...
02 S
(d) 20 kS

2
...
03 A
(d) 1 J = 1 N/m

Section 1

Units associated with basic electrical quantities 7

Section 1

8 Electrical and Electronic Principles and Technology
3
...
4 W (b) 20 W (c) 40 W (d) 200 W
4
...
The value of the force required is:
(a) 2
...
24 N
5
...

The current flowing is:
(a) 120 A (b) 480 A (c) 2 A (d) 8 A
6
...
The energy consumed by the
resistor is:
(a) 0
...
02 kWh
7
...
Electromotive force is provided by:
(a) resistance’s
(b) a conducting path
(c) an electric current
(d) an electrical supply source

9
...
In order that work may be done:
(a) a supply of energy is required
(b) the circuit must have a switch
(c) coal must be burnt
(d) two wires are necessary
11
...
The unit of current is the:
(a) volt
(b) coulomb
(c) joule
(d) ampere

Chapter 2

An introduction to
electric circuits
At the end of this chapter you should be able to:
• appreciate that engineering systems may be represented by block diagrams
• 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
...
is the volt

•
•
•
•
•

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

understand that resistance opposes current flow and is measured in ohms
appreciate what an ammeter, a voltmeter, an ohmmeter, a multimeter and an oscilloscope measure

distinguish between linear and non-linear devices
state Ohm’s law as V = IR or I = V /R or R = V /I
• use Ohm’s 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 2 R = V 2 /R watts

explain the importance of fuses in electrical circuits

2
...

Figure 2
...


10 Electrical and Electronic Principles and Technology

Section 1

A
...
Supply

Microphone

Loudspeaker
Amplifier

Thermostat
Error
Heating
+
system
Temperature
−
command
Actual
temperature

Enclosure
Temperature
of enclosure

Figure 2
...
1

A sub-system is a part of a system which performs
an identified function within the whole system; the
amplifier in Fig
...
1 is an example of a sub-system
...
2
...

The illustration in Fig
...
1 is called a block diagram
and electrical/electronic systems, which can often be
quite complicated, can be better understood when broken down in this way
...

As another example of an engineering system,
Fig
...
2 illustrates a temperature control system containing a heat source (such as a gas boiler), a fuel controller
(such as an electrical solenoid valve), a thermostat and
a source of electrical energy
...
2
...
2
...


network could comprise a file server, coaxial cable, network adapters, several computers and a laser printer; an
electromechanical system is another example, where a
car electrical system could comprise a battery, a starter
motor, an ignition coil, a contact breaker and a distributor
...


2
...
2
...


Conductor

Gas
boiler

Solenoid

Two conductors
crossing but not
joined

Two conductors
joined together

Fixed resistor

Alternative symbol
for fixed resistor

Variable resistor

Cell

240 V

Battery of 3 cells

Alternative symbol
for battery

Switch

Filament lamp

Fuse

A

V

Ammeter

Voltmeter

Fuel
supply

Thermostat
Set temperature
Radiators
Enclosed space

Figure 2
...
A communications system is an example, where a local area

Figure 2
...
3

Electric current and quantity of
electricity

All atoms consist of protons, neutrons and electrons
...
Removed from the nucleus
are minute negatively charged particles called electrons
...
An equal number of protons and electrons exist
within an atom and it is said to be electrically balanced,
as the positive and negative charges cancel each other
out
...

All atoms are bound together by powerful forces of
attraction existing between the nucleus and its electrons
...

It is possible for an atom to lose an electron; the atom,
which is now called an ion, is not now electrically balanced, but is positively charged and is thus able to attract
an electron to itself from another atom
...

However, if an electric pressure or voltage is applied
across any material there is a tendency for electrons to
move in a particular direction
...
Thus current is the rate of movement of charge
...

Insulators are materials whose electrons are held
firmly to their nucleus
...
24 × 1018 electrons)
If the drift of electrons in a conductor takes place at
the rate of one coulomb per second the resulting current
is said to be a current of one ampere
...
e
...
What current must flow if 0
...
24
0
...
If a current of 10 A flows for four
minutes, find the quantity of electricity transferred
...
I = 10 A and
t = 4 × 60 = 240 s
...
In what time would a current of 10 A transfer
a charge of 50 C?
[5 s]
2
...
What
charge is transferred?
[3600 C]
3
...
4 Potential difference and
resistance
For a continuous current to flow between two points in
a circuit a potential difference (p
...
) or voltage, V, is
required between them; a complete conducting path is
necessary to and from the source of electrical energy
...
d
...

Figure 2
...
Current flow, by convention, is considered as
flowing from the positive terminal of the cell, around
the circuit to the negative terminal
...
d
...
e
...

(See Chapter 10 for more detail about electrical
measuring instruments and measurements
...
6
V

Figure 2
...
This
friction, or opposition, is called resistance R and is the
property of a conductor that limits current
...
e
...
6 shows a circuit in which current I can be
varied by the variable resistor R2
...
d
...
d
...
The result is shown in Fig
...
7(a) where the
straight line graph passing through the origin indicates
that current is directly proportional to the p
...
Since the
gradient, i
...
(p
...
)/(current) is constant, resistance R1
is constant
...

V

2
...
Figure 2
...
Since
all the current in the circuit passes through the ammeter
it must have a very low resistance
...
d
...
d
...
In Fig
...
5, a voltmeter is
connected in parallel with the lamp to measure the p
...

across it
...

An ohmmeter is an instrument for measuring
resistance
...
An ‘Avometer’
and ‘Fluke’ are typical examples
...
The display of
an oscilloscope involves a spot of light moving across
a screen
...
d
...

The displacement is calibrated in ‘volts per cm’
...
6

p
...


p
...


0

l
(a)

0

l
(b)

Figure 2
...
2
...
2
...
d
...
Since the gradient is changing, the lamp is an
example of a non-linear device
...
7 Ohm’s law

From Ohm’s law, potential difference,

Ohm’s law states that the current I flowing in a circuit
is directly proportional to the applied voltage V and
inversely proportional to the resistance R, provided the
temperature remains constant
...
01 A
3
1000
10

V = IR = (0
...
A coil has a current of 50 mA
flowing through it when the applied voltage is
12 V
...
The current flowing through a resistor
is 0
...
d
...
Determine
the value of the resistance
...
8
8

2
...
Thus multiples and sub-multiples
of units are often used, as stated in Chapter 1
...
1
...
Determine the p
...
which must be
applied to a 2 k resistor in order that a current of
10 mA may flow
...
A 100 V battery is connected across a
resistor and causes a current of 5 mA to flow
...
If the
voltage is now reduced to 25 V, what will be the
new value of the current flowing?
Resistance R =

V
100 × 103
100
=
=
−3
I
5 × 10
5
= 20 × 103 = 20 k

Current when voltage is reduced to 25 V,
I=

= 2 × 103 = 2000

Current I = 10 mA = 10 × 10−3 A
Table 2
...
25 mA
R
20 × 103
20

Name

Meaning

Example

M

mega

multiply by 1 000 000
(i
...
×106 )

2 M = 2 000 000 ohms

k

kilo

multiply by 1000
(i
...
×103 )

10 kV = 10 000 volts

m

milli

divide by 1000
(i
...
×10−3 )

25 mA =

μ

micro

divide by 1 000 000
(i
...
×10−6 )

50 μV =

25
A
1000
= 0
...
000 05 volts

Section 1

An introduction to electric circuits 13

Problem 7
...
05
5

= 2400
(b) Resistance R =
=

or 2
...
0002
1 200 000
= 600 000
2
or 600 k

Exercise 6

Further problems on Ohm’s law

2
...
Determine (a) the current flowing in the bulb and (b) the resistance of the
bulb
...
25 A (b) 960 ]
3
...
2
...

[2 m , 5 m ]

or 0
...
The current/voltage relationship for
two resistors A and B is as shown in Fig
...
8
Determine the value of the resistance of each
resistor
...
The current flowing through a heating element
is 5 A when a p
...
of 35 V is applied across it
...

[7 ]

V
120
(a) Resistance R = =
I
50 × 10−3

6
Q
4
2

20
Current/mA

Section 1

14 Electrical and Electronic Principles and Technology

Resistor A
0

15

4

8

12
16
Voltage/μV

20

24

10

Figure 2
...
Determine the p
...
which must be applied to a
5 k resistor such that a current of 6 mA may
flow
...
A 20 V source of e
...
f
...
Calculate
the current flowing
...
8

For resistor A,
R=

20 V
20
2000
V
=
=
=
I
20 mA
0
...
005
5
= 3200

or 3
...
9 Conductors and insulators
A conductor is a material having a low resistance which
allows electric current to flow in it
...

An insulator is a material having a high resistance
which does not allow electric current to flow in it
...


2
...
The unit of power is the watt, W
...
Substituting for V in equation (1) gives:

= 80 mW
Problem 11
...
What current will flow when it is connected
to a 240 V supply? Find also the power rating of the
kettle
...
e
...
Substituting for I in
equation (1) gives:
V
P=V×
R
V2
i
...

P=
watts
R
There are thus three possible formulae which may be
used for calculating power
...
92 kW = power rating of kettle
Problem 12
...
Determine (a) the p
...
across the
winding, and (b) the power dissipated by the coil
...
A 100 W electric light bulb is
connected to a 250 V supply
...

Power P = V × I, from which, current I =
(a)

Resistance R =

Potential difference across winding,
V = IR = 5 × 100 = 500 V

(b)

Power dissipated by coil,
P = I 2 R = 52 × 100
= 2500 W or 2
...
4 A
250 25 5

(b)

= 2500 W or 2
...
4
4

Problem 13
...
Find the current taken by
the lamp and its power rating
...
Calculate the power dissipated when
a current of 4 mA flows through a resistance of
5k
...
08 W or 80 mW

V
240
=
= 8A
R
30

From Ohm’s law,
V
240
=
R
960
1
24
= A or 0
...
If the
power is measured in kilowatts and the time in hours
then the unit of energy is kilowatt-hours, often called
the ‘unit of electricity’
...

Problem 14
...
Determine the
current flowing in the load, the power consumed
and the energy dissipated in 2 minutes
...
3 A
R
40
Power consumed, P = VI = (12)(0
...
6 W
...
6 W)(2 × 60 s)
= 432 J (since 1 J = 1 Ws)
Problem 15
...
m
...
of 15 V supplies a
current of 2 A for 6 minutes
...
5p per kWh = 93
...
5 = 1170p
...
70
Problem 17
...
6 MJ
when connected to a 250 V supply for 40 minutes
...


Power =

energy 3
...
e
...
5 kW
...

thus

I=

Problem 18
...
If the fire is
on for 6 hours determine the energy used and the
cost if 1 unit of electricity costs 13p
...

(Alternatively, from Ohm’s law,

Energy = power × time, and power = voltage × current
...
8 kJ
Problem 16
...

Estimate the cost per week of electricity if the
equipment is used for 30 hours each week and
1 kWh of energy costs 12
...

Power = VI watts = 240 × 13
= 3120 W = 3
...
12 kW) × (30 h)
= 93
...

Energy used in 6 hours
= power × time = 2 kW × 6 h = 12 kWh
...
Cost of energy = 12 ×13 = £1
...
A business uses two 3 kW fires for
an average of 20 hours each per week, and six
150 W lights for 30 hours each per week
...

Energy = power × time
...


Hence weekly energy used by two 3 kWfires
= 2 × 60 = 120 kWh
...
5 kWh
...
5 = 27 kWh
...

1 unit of electricity = 1 kWh of energy
...
58
...
The hot resistance of a 250 V filament lamp
is 625
...

[0
...
Determine the resistance of a coil connected
to a 150 V supply when a current of (a) 75 mA
(b) 300 μA flows through it
...
5 M ]

9
...
c
...

Find the power rating of the motor and the
current taken from the supply
...
A p
...
of 500 V is applied across the winding of an electric motor and the resistance
of the winding is 50
...

[5 kW]
11
...
Determine the cost of electricity
for the week if 1 unit of electricity costs 15 p
...
70]
12
...
If the fire is on for
30 hours in a week determine the energy used
...
5p per unit
...
15]

3
...
Find also the power rating of the fire
and the energy used in 20 h
...
88 kW, 57
...
Determine the power dissipated when a current of 10 mA flows through an appliance
having a resistance of 8 k
...
8 W]

(a) magnetic effect
(b) chemical effect
(c) heating effect

5
...
5 J of energy are converted into heat in 9 s
...
5 W]

Some practical applications of the effects of an electric
current include:

6
...
What p
...
exists
across the ends of the conductor?
[2
...
Find the power dissipated when:
(a) a current of 5 mA flows through a resistance of 20 k
(b) a voltage of 400 V is applied across a
120 k resistor
(c) a voltage applied to a resistor is 10 kV
and the current flow is 4 m
[(a) 0
...
33 W (c) 40 W]
8
...
m
...
15 V supplies a current of
2 A for 5 min
...
11

2
...
This will cause overheating and

Section 1

An introduction to electric circuits 17

Section 1

18 Electrical and Electronic Principles and Technology
possibly a fire; fuses protect against this happening
...
The fuse is a piece of wire which can carry a
stated current; if the current rises above this value it will
melt
...
The
fuse must be able to carry slightly more than the normal
operating current of the equipment to allow for tolerances and small current surges
...
If a fuse is fitted to withstand this large current there would be no protection against faults which
cause the current to rise slightly above the normal value
...
These can
stand 10 times the rated current for 10 milliseconds
...

A circuit diagram symbol for a fuse is shown in
Fig
...
4 on page 10
...
If 5 A, 10 A and 13 A fuses are
available, state which is most appropriate for the
following appliances which are both connected to a
240 V supply: (a) Electric toaster having a power
rating of 1 kW (b) Electric fire having a power
rating of 3 kW
...
17 A
V
240
24
Hence a 5 A fuse is most appropriate
current I =

(b) For the fire,
P
3000
300
=
=
= 12
...
A television set having a power rating of 120 W
and electric lawnmower of power rating 1 kW
are both connected to a 250 V supply
...
[3 A, 5 A]

Exercise 9 Short answer questions on the
introduction to electric circuits
1
...
State the unit of
(a) current
(b) potential difference
(c) resistance
3
...
What is a multimeter?
5
...
Give one example of
(a) a linear device
(b) a non-linear device
7
...
What is a conductor? Give four examples
9
...
Complete the following statement:
‘An ammeter has a
...
with the load’
11
...
resistance and must be
connected
...
State the unit of electrical power
...
State two units used for electrical energy
14
...
What is the function of a fuse in an electrical circuit?

Exercise 10

Multi-choice problems on the
introduction to electric circuits
(Answers on page 398)

1
...
06
(c) 1000 minutes

(b) 0
...
6 s

2
...
1 coulomb is
transferred in 10 ms is:
(a) 1 A
(b) 10 A
(c) 10 mA
(d) 100 mA
3
...
d
...
1 V

4
...
The power dissipated by a resistor of 4
when a current of 5 A passes through it is:
(a) 6
...
Which of the following statements is true?
(a) Electric current is measured in volts
(b) 200 k resistance is equivalent to 2 M
(c) An ammeter has a low resistance and
must be connected in parallel with a
circuit
(d) An electrical insulator has a high
esistance
7
...
The energy consumed by the
resistor is:
(a) 0
...
7 kWh
(c) 9 kWh
(d) 27 kWh

8
...
Voltage drop is the:
(a) maximum potential
(b) difference in potential between two
points
(c) voltage produced by a source
(d) voltage at the end of a circuit
10
...
The largest number of 100 W electric light
bulbs which can be operated from a 240 V
supply fitted with a 13 A fuse is:
(a) 2 (b) 7 (c) 31 (d) 18
12
...
5 kW heater in
5 minutes is:
(a) 5 J
(b) 450 J
(c) 7500 J
(d) 450 000 J
13
...
1

Resistance and resistivity

The resistance of an electrical conductor depends on
four factors, these being: (a) the length of the conductor,
(b) the cross-sectional area of the conductor, (c) the
type of material and (d) the temperature of the material
...
e
...
Thus, for example, if the length of
a piece of wire is doubled, then the resistance is doubled
...
e
...
Thus,
for example, if the cross-sectional area of a piece of
wire is doubled then the resistance is halved
...
By inserting
a constant of proportionality into this relationship the
type of material used may be taken into account
...
Thus,
ρl
ohms
a
ρ is measured in ohm metres ( m)
...

resistance

R=

Resistivity varies with temperature and some typical values of resistivities measured at about room
temperature are given below:
Copper 1
...
6 × 10

m (or 0
...
026 μ m)

Carbon (graphite) 10 × 10−8
Glass 1 × 10

m (0
...

Problem 1
...
Determine (a) the resistance of an 8 m
length of the same wire, and (b) the length of the
same wire when the resistance is 420
...
e
...
Hence, 600 ∝ 5 m or 600 = (k)(5),

where k is the coefficient of proportionality
...
5 m
k
120

Problem 2
...
Find (a) the
resistance of a wire of the same length and material
if the cross-sectional area is 5 mm2 , (b) the
cross-sectional area of a wire of the same length
and material of resistance 750
...
e
...
16, l = 8 and a = 3, then 0
...
16 × 3/8 = 0
...
e
...
03 × 10−6 m)(2000 m)
=
(100 × 10−6 m2 )
= 0
...
02 × 10−6 m)(40 m)
=
R
0
...
2 × 10−6 m2
= (3
...
2 mm2

from which
cross-sectional area, a =

0
...
Calculate the cross-sectional area, in
mm2 , of a piece of copper wire, 40 m in length and
having a resistance of 0
...
Take the resistivity of
copper as 0
...


a=

1
a

= 1
...
)
(b) When the resistance is 750 then

24
1

Length l = 2 km = 2000 m,
area a = 100 mm2 = 100 × 10−6 m2
and resistivity ρ = 0
...


from which, the coefficient of proportionality,
k = 300 × 2 = 600

R = (k)( 1 ) = (600)( 1 ) = 120
5
5

= 0
...
Calculate the resistance of a 2 km
length of aluminium overhead power cable if the
cross-sectional area of the cable is 100 mm2
...
03 × 10−6 m
...
8 mm

2

Problem 3
...
16
...

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

R ∝ l/a or R = k(l/a), where k is the coefficient of
proportionality
...
The resistance of 1
...
17 mm2 is 150
...

Resistance, R = ρl/a hence
Ra
resistivity ρ =
l
(150 )(0
...
017 × 10−6
or 0
...
Determine the resistance of 1200 m
of copper cable having a diameter of 12 mm if the
resistivity of copper is 1
...


5
...
Take the resistivity of
aluminium as 0
...

[1
...
Find the resistance of 800 m of copper cable of
cross-sectional area 20 mm2
...
02 μ m
...
8 ]

ρl
a

=

(1
...
7 × 1200 × 106
108 × 36π

=

1
...
180

Now try the following exercise
Exercise 11 Further problems on resistance
and resistivity
1
...
5
...
25
...
75 (b) 5 m]
2
...

Determine (a) the resistance of a wire of the
same length and material if the cross-sectional
area is 4 mm2 , and (b) the cross-sectional area
of a wire of the same length and material if the
resistance is 32
...
625 mm2 ]
3
...
08
...

[0
...
The resistance of 500 m of wire of crosssectional area 2
...
Determine the
resistivity of the wire in μ m
...
026 μ m]
7
...
017 × 10−6 m
...
216 ]

3
...

The temperature coefficient of resistance of a material is the increase in the resistance of a 1 resistor
of that material when it is subjected to a rise of temperature of 1◦ C
...
Thus, if
some copper wire of resistance 1 is heated through
1◦ C and its resistance is then measured as 1
...
0043 / ◦ C for copper
...
e
...
0043/◦ C
for copper
...
0043 = 1
...
Some typical values of
temperature coefficient of resistance measured at 0◦ C
are given below:
Copper
Nickel
Constantan
Aluminium
Carbon
Eureka

0
...
0062/◦ C
0
0
...
00048/◦ C
0
...
)

If the resistance of a material at 0◦ C is known the
resistance at any other temperature can be determined
from:

i
...

Rθ = 1000[1 + (−0
...
040] = 1000(0
...
A coil of copper wire has a resistance
of 100 when its temperature is 0◦ C
...
0043/◦ C
...
Hence resistance at
100◦ C,
R100 = 100[1 + (0
...
301]

Problem 11
...
If the temperature
coefficient of resistance of copper at 20◦ C is
0
...

Resistance at θ ◦ C,
Rθ = R20 [1 + α20 (θ − 20)]
Hence resistance at 100◦ C,
R100 = 10[1 + (0
...
301) = 130
...
004)(80)]
Problem 9
...
Determine its
resistance at 0◦ C
...
0038/◦ C
...
Hence resistance
at 0◦ C,
R0 =

Rθ
27
=
(1 + α0 θ) [1 + (0
...
133
27
=
= 23
...
133

Problem 10
...
Determine its resistance at 80◦ C
...
0005/◦ C
...
32]
= 10(1
...
2
Problem 12
...
The temperature
of the wire is increased and the resistance rises to
240
...
0039/◦ C at 18◦ C determine the
temperature to which the coil has risen
...
Resistance at θ ◦ C,
Rθ = R18 [1 + α18 (θ − 18)]
i
...


240 = 200[1 + (0
...
0039)(θ − 18)
240 − 200 = 0
...
78(θ − 18)
40
= θ − 18
0
...
28 = θ − 18, from which,
θ = 51
...
28◦ C

Section 1

Resistance variation 23

Section 1

24 Electrical and Electronic Principles and Technology
Hence the temperature of the coil increases to
69
...
Some copper wire has a resistance
of 200 at 20◦ C
...
Determine
the resistance of the wire at 90◦ C, correct to the
nearest ohm, assuming that the temperature
coefficient of resistance is 0
...


and

R20 = 200 , α0 = 0
...
004)]
[1 + 20(0
...
36]
[1 + 0
...
36)
= 251
...
08)

i
...
the resistance of the wire at 90◦ C is 252 , correct
to the nearest ohm

its resistance at 100◦ C if the temperature coefficient of resistance of aluminium at 0◦ C is
0
...
A copper cable has a resistance of 30 at
a temperature of 50◦ C
...
Take the temperature coefficient
of resistance of copper at 0◦ C as 0
...
69 ]
3
...
00048/◦ C
...
Determine
its resistance at 50◦ C
...
A coil of copper wire has a resistance of 20
at 18◦ C
...
004/◦ C, determine
the resistance of the coil when the temperature
rises to 98◦ C
[26
...
The resistance of a coil of nickel wire at
20◦ C is 100
...

If the temperature coefficient of resistance of
nickel is 0
...

[70◦ C]
6
...
The wire is heated to a temperature
of 100◦ C
...
004/◦ C
...
8 ]
7
...
2 km long and has a
cross-sectional area of 5 mm2
...
02 × 10−6 m and its temperature
coefficient of resistance is 0
...

[5
...
A coil of aluminium wire has a resistance of
50 when its temperature is 0◦ C
...
3 Resistor colour coding and
ohmic values
(a) Colour code for fixed resistors
The colour code for fixed resistors is given in Table 3
...
1
Colour

Significant
Figures

Multiplier

Tolerance

Silver

–

10−2

±10%

Gold

–

10−1

±5%

Black

0

1

–

Brown

1

10

±1%

Red

2

102

±2%

Orange

3

103

–

Yellow

4

104

–

Green

5

105

±0
...
25%

Violet

7

107

±0
...
e
...
1
...
33 with a tolerance of ±1%

(i) For a four-band fixed resistor (i
...
resistance
values with two significant figures):
yellow-violet-orange-red indicates 47 k with a
tolerance of ±2%
(Note that the first band is the one nearest the end
of the resistor)
(ii) For a five-band fixed resistor (i
...
resistance
values with three significant figures): red-yellowwhite-orange-brown indicates 249 k with a tolerance of ±1%
(Note that the fifth band is 1
...
Determine the value and tolerance
of a resistor having a colour coding of:
orange-orange-silver-brown
...
e
...
1
...
1, which means that the value of the
resistor is 33 × 10−2 = 0
...
Determine the value and tolerance
of a resistor having a colour coding of:
brown-black-brown
...
e
...
1
...
1, which means that the value of the
resistor is 10 × 10 = 100
There is no fourth band colour in this case; hence,
from Table 3
...
Hence a colour
coding of brown-black-brown represents a resistor of
value 100 with a tolerance of ±20%
Problem 16
...
1, brown-black-brown-silver indicates
10 × 10, i
...
100 , with a tolerance of ±10%
This means that the value could lie between
(100 − 10% of 100)
and

(100 + 10% of 100)

i
...
brown-black-brown-silver indicates any value
between 90 and 110
Problem 17
...

From Table 3
...
With a tolerance of ±5%, the
fourth band will be gold
...


Problem 18
...


Section 1

Resistance variation 25

Section 1

26 Electrical and Electronic Principles and Technology
orange-green-red-yellow-brown is a five-band fixed
resistor and from Table 3
...
52 × 106 , i
...
3
...
52 M ± 1%

From Table 3
...
7 M ± 20%

Problem 21
...


indicates

(b) Letter and digit code for resistors

From Table 3
...
2
...
2
Marked as:

1
...
47

R47

1

1R0

4
...
Determine the value and tolerance of a resistor having a colour coding of: yellow-violetgold
[4
...
Determine the colour coding for a 51 k fourband resistor having a tolerance of ±2%
[green-brown-orange-red]

Tolerance is indicated as follows: F = ±1%,
G = ±2%, J = ±5%, K = ±10% and M = ±20%
Thus, for example,
R33M = 0
...
Determine the value and tolerance of a resistor having a colour coding of: blue-whiteblack-black-gold
[690 ± 5%]

± 20%

4R7K = 4
...
Determine the value of a resistor
marked as 6K8F
...
2, 6K8F is equivalent to: 6
...
Determine the value of a resistor
marked as 4M7M
...
Determine the colour coding for a 1 M fourband resistor having a tolerance of ±10%
[brown-black-green-silver]
6
...
8 M to 2
...
Determine the range of values expected for
a resistor with colour coding: yellow-blackorange-brown
[39
...
4 k ]
8
...
22 ± 2% (b) 4
...
Determine the letter and digit code for a
resistor having a value of 100 k ± 5%
[100 KJ]
10
...
8 M ± 20%
[6 M8 M]

Exercise 14 Short answer questions on
resistance variation
1
...
If the length of a piece of wire of constant
cross-sectional area is halved, the resistance
of the wire is
...
If the cross-sectional area of a certain length
of cable is trebled, the resistance of the cable
is
...
What is resistivity? State its unit and the
symbol used
...
Complete the following:
Good conductors of electricity have a
...
value of resistivity
6
...

7
...

8
...
The resistance of a 2 km length of cable of
cross-sectional area 2 mm2 and resistivity of
2 × 10−8 m is:
(a) 0
...
02 m
(d) 200
4
...
If its resistance is 0
...
The symbol for the unit of temperature coefficient of resistance is:
(a)
/◦ C
(b)
◦C
(c)
(d)
/ ◦C
6
...

If the temperature coefficient of resistance for
the wire is 0
...
4
(b) 1
...
A nickel coil has a resistance of 13 at 50◦ C
...
006/◦ C, the resistance at 0◦ C is:
(a) 16
...
3
(d) 0
...
Explain briefly the letter and digit code for
resistors

8
...
A resistor marked as 4K7G indicates a value of:
(a) 47 ± 20%
(b) 4
...
47 ± 10%
(d) 4
...
The unit of resistivity is:
(a) ohms
(b) ohm millimetre
(c) ohm metre
(d) ohm/metre
2
...
Its new resistance is:
(a) 100
(b) 200
(c) 50
(d) 400

Section 1

Resistance variation 27

Chapter 4

Batteries and alternative
sources of energy
At the end of this chapter you should be able to:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

list practical applications of batteries
understand electrolysis and its applications, including electroplating
appreciate the purpose and construction of a simple cell
explain polarisation and local action
explain corrosion and its effects
define the terms e
...
f
...
m
...
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 Leclanché, mercury, lead–acid and alkaline cells
list the advantages and disadvantages of alkaline cells over lead–acid 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
...
If an appliance is placed between its
terminals the current generated will power the device
...
For example, without
the battery, there would be no mobile phones or laptop
computers
...
Some practical examples where batteries are
used include:
in laptops, in cameras, in mobile phones, in
cars, in watches and clocks, for security equipment, in electronic meters, for smoke alarms,
for meters used to read gas, water and electricity
consumption at home, to power a camera for an
endoscope looking internally at the body, and for
transponders used for toll collection on highways
throughout the world
Batteries tend to be split into two categories – primary,
which are not designed to be electrically re-charged,

Batteries and alternative sources of energy 29

4
...
In solids, the current is carried by electrons
...
In liquids and gases, the current is carried by the part of a
molecule which has acquired an electric charge, called
ions
...
Distilled water contains
no ions and is a poor conductor of electricity, whereas
salt water contains ions and is a fairly good conductor
of electricity
...

Practical applications of electrolysis include the electroplating of metals (see below), the refining of copper
and the extraction of aluminium from its ore
...
Examples include salt water, copper sulphate and sulphuric acid
...
The positive-connected electrode

R
A
I

ϩ

Copper
electrode
(anode)

Ϫ
Zinc electrode
(cathode)

Dilute
sulphuric acid
(electrolyte)

Figure 4
...

When two copper wires connected to a battery are
placed in a beaker containing a salt water solution, current will flow through the solution
...

Electroplating uses the principle of electrolysis to
apply a thin coat of one metal to another metal
...
If two copper electrodes connected to a battery
are placed in a beaker containing copper sulphate as the
electrolyte it is found that the cathode (i
...
the electrode
connected to the negative terminal of the battery) gains
copper whilst the anode loses copper
...
3 The simple cell
The purpose of an electric cell is to convert chemical
energy into electrical energy
...
Such a cell is shown in
Fig
...
1, comprising copper and zinc electrodes
...

Other possible electrode pairs exist, including zinc–lead
and zinc–iron
...
e
...
d
...
By knowing the e
...
f
...
m
...
of any pair of metals may be determined
...
The electrochemical series is a way
of listing elements in order of electrical potential, and
Table 4
...

In a simple cell two faults exist – those due to
polarisation and local action
...
e
...
6), and secondary
batteries, which are designed to be re-charged, such as
those used in mobile phones (see Section 4
...

In more recent years it has been necessary to design
batteries with reduced size, but with increased lifespan
and capacity
...
5 V battery is used
...
6 V battery is used
...
5 V manganese battery was gradually replaced by
the alkaline battery
...

Lithium-ion batteries were introduced in the 1970s
because of the need for longer lifetime applications
...
Lithium batteries
are capable of delivering high currents but tend to be
expensive
...
2 on page 35
...
1

Part of the electrochemical series

Potassium
sodium
aluminium
zinc
iron
lead
hydrogen

the zinc electrode in the cell shown in Fig
...
1 is
negative and the copper electrode is positive
...
m
...
produced
by the cell
...

(ii) When two metal electrodes are used in a simple
cell the one that is higher in the series tends to
dissolve in the electrolyte
...
4
...

This is because of the formation of a film of hydrogen
bubbles on the copper anode
...
The hydrogen prevents full
contact between the copper electrode and the electrolyte
and this increases the internal resistance of the cell
...
This
allows the cell to deliver a steady current
...

The reason for this is that impurities, such as traces of
iron, are present in the zinc which set up small primary
cells with the zinc
...
This action is known as local
action of the cell
...

When two metals are used in a simple cell the electrochemical series may be used to predict the behaviour
of the cell:
(i) The metal that is higher in the series acts as the
negative electrode, and vice-versa
...
4 Corrosion
Corrosion is the gradual destruction of a metal in a
damp atmosphere by means of simple cell action
...
Thus, if metals widely spaced
in the electrochemical series, are used in contact with
each other in the presence of an electrolyte, corrosion
will occur
...

The effects of corrosion include the weakening of
structures, the reduction of the life of components and
materials, the wastage of materials and the expense of
replacement
...
Also, iron may be galvanised, i
...
,
plated with zinc, the layer of zinc helping to prevent the
iron from corroding
...
5 E
...
f
...
m
...
), E, of a cell is the p
...

between its terminals when it is not connected to a load
(i
...
the cell is on ‘no load’)
...
m
...
of a cell is measured by using a high resistance voltmeter connected in parallel with the cell
...
For
example, if the resistance of a cell is 1 and that of
a voltmeter 1 M then the equivalent resistance of the
circuit is 1 M + 1 , i
...
approximately 1 M , hence
no current flows and the cell is not loaded
...
This is caused by the internal
resistance of the cell which is the opposition of the
material of the cell to the flow of current
...
Figure 4
...
m
...
E volts and
internal resistance, r, and XY represents the terminals
of the cell
...
2

When a load (shown as resistance R) is not connected,
no current flows and the terminal p
...
, V = E
...
The p
...
available at the cell terminals is less than the e
...
f
...
m
...
12 volts and internal resistance 0
...
d
...
4
...

When a current flows in the opposite direction to that
shown in Fig
...
2 the cell is said to be charging (V > E)
...

The cells in a battery may be connected in series or in
parallel
...
m
...
= sum of cell’s e
...
f
...
m
...
and internal
resistance:
Total e
...
f
...
m
...
of one cell
Total internal resistance of n cells
=

1
× internal resistance of one cell
n

Problem 1
...
2 and an e
...
f
...
2 V are
connected (a) in series, (b) in parallel
...
m
...
and the internal resistance of the batteries
so formed
...
01)
= 12 − 1 = 11 V

(a) When connected in series, total e
...
f

When different values of potential difference V across
a cell or power supply are measured for different values
of current I, a graph may be plotted as shown in Fig
...
3
...
m
...
E of the cell or power supply is the p
...

across its terminals on no load (i
...
when I = 0), then E
is as shown by the broken line
...
m
...

= 2
...
6 V
Total internal resistance
= sum of cell’s internal resistance
= 0
...
6
(b) When connected in parallel, total e
...
f

Terminal p
...
, V

E

= e
...
f
...
2 V

V

0

Figure 4
...
2 = 0
...
A cell has an internal resistance of
0
...
m
...
of 2
...
Calculate its terminal
p
...
if it delivers (a) 5 A (b) 50 A
...
4
...
m
...

total resistance
15
=
58 + 2
15
=
= 0
...
d
...
m
...
of
cell, I = current flowing and r = internal resistance
of cell
E = 2
...
02

I

Hence terminal p
...

V = 2
...
02) = 2
...
1 = 1
...
d
...
0 − 50(0
...
e
...
0 − 1
...
0 V

Thus the terminal p
...
decreases as the current
drawn increases
...
The p
...
at the terminals of a battery
is 25 V when no load is connected and 24 V when a
load taking 10 A is connected
...

When no load is connected the e
...
f
...
d
...
e
...
d
...
e
...
4
(b) P
...
at battery terminals, V = E − Ir
i
...
V = 15 − (0
...
5 V
Now try the following exercise
Exercise 16 Further problems on e
...
f
...
Twelve cells, each with an internal resistance
of 0
...
m
...
of 1
...
Determine the
e
...
f
...


24 = 25 − (10)r

Hence, rearranging, gives
10r = 25 − 24 = 1
and the internal resistance,
r=

1
= 0
...
Ten 1
...
2 , are connected in series
to a load of 58
...
d
...

(a) For ten cells, battery e
...
f
...
5 = 15 V,
and the total internal resistance, r = 10 × 0
...


[(a) 18 V, 2
...
5 V, 0
...
A cell has an internal resistance of 0
...
m
...
of 2
...
Calculate its terminal p
...
if
it delivers
(a) 1 A

(b) 20 A

(c) 50 A

[(a) 2
...
6 V (c) 0
...
The p
...
at the terminals of a battery is 16 V
when no load is connected and 14 V when a
load taking 8 A is connected
...

[0
...
A battery of e
...
f
...
2 supplies a load taking 10 A
...
d
...

[18 V, 1
...
Ten 2
...
1 are connected in series to a
load of 21
...
d
...
For the circuits shown in Fig
...
5 the resistors represent the internal resistance of the
batteries
...
m
...
across PQ
(ii) the total equivalent internal resistances
of the batteries
...
25 ]
4V

1Ω

5V

2Ω

1Ω

3V

are exhausted
...


Lechlanché cell
A typical dry Lechlanché cell is shown in Fig
...
6
...
m
...
of about 1
...
The
hydrogen film on the carbon electrode forms faster than
can be dissipated by the depolariser
...

The cell is the most commonly used of primary cells, is
cheap, requires little maintenance and has a shelf life of
about 2 years
...

(ammonium chloride, manganese dioxide
and powdered carbon)

Q

P

(a)
2V

2V

1Ω

ELECTROLYTE
(sal ammoniac, zinc chloride,
plaster of paris, water)

1Ω

ZINC CASE CATHODE
(negative terminal)

2V

1Ω

2V

1Ω

DRY LECLANCHÉ CELL

Figure 4
...
5
7
...
8 V when a
load taking 80 A is connected
...
What would be the
terminal voltage when a load taking 20 A is
connected?
[0
...
2 V]

A typical mercury cell is shown in Fig
...
7
...
m
...
of about 1
...
Its main advantages over the
Lechlanché cell is its smaller size and its long shelf
life
...

Steel cap cathode
(negative terminal)
Insulating gasket
Steel case anode
(positive terminal)
Zinc cylinder
Electrolyte
(potassium hydroxide)

4
...
7

Section 1

Batteries and alternative sources of energy 33

Section 1

34 Electrical and Electronic Principles and Technology
4
...
Examples of secondary cells include the lead–acid cell and the
nickel cadmium and nickel-metal cells
...


Lead–acid cell

(ii) the oxygen in the lead peroxide combines with
hydrogen in the electrolyte to form water
...

The terminal p
...
of a lead–acid cell when fully discharged is about 1
...
A cell is charged by connecting
a d
...
supply to its terminals, the positive terminal of
the cell being connected to the positive terminal of the
supply
...
During charging:
(i) the lead sulphate on the positive and negative
plates is converted back to lead peroxide and lead
respectively, and

A typical lead–acid cell is constructed of:
(i) A container made of glass, ebonite or plastic
...

The plates are interleaved as shown in the plan
view of Fig
...
8 to increase their effective crosssectional area and to minimize internal resistance
...
The relative
density of the electrolyte thus increases
...
The
colour of the negative plate when fully charged is grey
and when discharged is light grey
...
8

(iii) Separators made of glass, celluloid or wood
...

The relative density (or specific gravity) of a lead–acid
cell, which may be measured using a hydrometer, varies
between about 1
...
19 when discharged
...
d
...

When a cell supplies current to a load it is said to be
discharging
...
The tubes are assembled into nickel–steel
plates
...
In the nickel–
cadmium cell the negative plate is made of cadmium
...
The plates are separated by insulating rods and
assembled in steel containers which are then enclosed
in a non-metallic crate to insulate the cells from one
another
...
d
...
2 V
...
2
Type of battery

Common uses

Hazardous
component

Disposal recycling
options

Sulphuric acid
and lead

Recycle – most petrol
stations and garages
accept old car batteries,
and council waste facilities
have collection points
for lead acid batteries

Wet cell (i
...
a primary cell that has a liquid electrolyte)
Lead acid batteries

Electrical energy supply for vehicles
including cars, trucks, boats, tractors
and motorcycles
...
e
...
9 Safe disposal of batteries

(iv) For a given capacity is lighter in weight
(v) Can be left indefinitely in any state of charge or
discharge without damage
(vi) Is not self-discharging
Disadvantages of an nickel cadmium and nickel-metal
cells over a lead–acid cell include:
(i) Is relatively more expensive
(ii) Requires more cells for a given e
...
f
...
Practical examples include traction and marine
work, lighting in railway carriages, military portable
radios and for starting diesel and petrol engines
...
2, page 35
...
8

Cell capacity

The capacity of a cell is measured in ampere-hours
(Ah)
...
Typical discharge characteristics for a lead–acid
cell are shown in Fig
...
9

2
...
d
...
0
1
...
9

10

Battery disposal has become a topical subject in the
UK because of greater awareness of the dangers and
implications of depositing up to 300 million batteries
per annum – a waste stream of over 20 000 tonnes –
into landfill sites
...
Other batteries can be recycled for
their metal content
...
If
batteries containing heavy metals are disposed of incorrectly, the metals can leach out and pollute the soil and
groundwater, endangering humans and wildlife
...
e
...
Mercury can cause damage
to the human brain, spinal system, kidneys and liver
...
It is increasingly
important to correctly dispose of all types of batteries
...
2 lists types of batteries, their common uses,
their hazardous components and disposal recycling
options
...
From the Waste Electrical and Electronic
Equipment (WEEE) Regulations 2006, commencing
July 2007 all producers (manufacturers and importers)
of electrical and electronic equipment will be responsible for the cost of collection, treatment and recycling of
obligated WEEE generated in the UK
...
10

Fuel cells

A fuel cell is an electrochemical energy conversion
device, similar to a battery, but differing from the latter
in that it is designed for continuous replenishment of
the reactants consumed, i
...
it produces electricity from
an external source of fuel and oxygen, as opposed to
the limited energy storage capacity of a battery
...
e
...

Typical reactants used in a fuel cell are hydrogen
on the anode side and oxygen on the cathode side
(i
...
a hydrogen cell)
...
Virtually continuous longterm operation is feasible as long as these flows are
maintained
...
The only
by-product of a fuel cell operating on pure hydrogen is
water vapour
...
However, continued
research and development is likely to make fuel cell
vehicles available at market prices within a few years
...
A fuel cell running
on hydrogen can be compact, lightweight and has no
moving parts
...
11 Alternative and renewable
energy sources
Alternative energy refers to energy sources which
could replace coal, traditional gas and oil, all of which
increase the atmospheric carbon when burned as fuel
...

Coal, gas and oil are not renewable because, although
the fields may last for generations, their time span is
finite and will eventually run out
...
Solar energy is one of the most resourceful sources
of energy for the future
...

However, about one third of this energy is either
absorbed by the outer atmosphere or reflected back
into space
...
Solar panels on roofs
capture heat in water storage systems
...

2
...
The fins of a windmill rotate
in a vertical plane which is kept vertical to the wind
by means of a tail fin and as wind flow crosses the

blades of the windmill it is forced to rotate and can
be used to generate electricity (see chapter 9)
...
The average wind
velocity of Earth is around 9 m/s, and the power
that could be produced when a windmill is facing
a wind of 10 m
...
h
...
e
...
5 m/s) is around
50 watts
...
Hydroelectricity is achieved by the damming of
rivers and utilising the potential energy in the water
...
The
system has enormous initial costs but has relatively
low maintenance costs and provides power quite
cheaply
...
Tidal power utilises the natural motion of the tides
to fill reservoirs which are then slowly discharged
through electricity-producing turbines
...
Geothermal energy is obtained from the internal
heat of the planet and can be used to generate steam
to run a steam turbine which, in turn, generates electricity
...
Drilling 3 miles from the surface of the
Earth, a temperature of 100◦ C is encountered; this is
sufficient to boil water to run a steam-powered electric power plant
...
Fortunately, however,
volcanic features called geothermal hotspots are
found all around the world
...

Now try the following exercises

Exercise 17 Short answer questions on the
chemical effects of electricity
1
...
State five practical applications of batteries
3
...
What is electrolysis?
5
...
Conduction in electrolytes is due to
...
A positive-connected electrode is called the

...

8
...
The purpose of an electric cell is to convert

...


28
...
What is meant by (a) alternative energy
(b) renewable energy
30
...


10
...
What is the electrochemical series?
12
...
What is corrosion? Name two effects of corrosion and state how they may be prevented
14
...
m
...
of a cell? How
may the e
...
f
...
Define internal resistance
16
...
m
...
of E volts, an internal
resistance of r ohms and supplies a current I
amperes to a load, the terminal p
...
V volts
is given by: V =
...
Name the two main types of cells
18
...
A battery consists of:
(a) a cell
(c) a generator

(b) a circuit
(d) a number of cells

2
...
d
...
m
...
2 V and
internal resistance 0
...
5 V
(c) 1
...
5 V

3
...
m
...
of 2 V and internal resistance 0
...

The resulting battery will have:

22
...
m
...
of 2 V and an internal resistance
of 0
...
m
...
of 10 V and an internal resistance of 2
...
m
...
of 2 V and an internal resistance
of 0
...
m
...
of 10 V and an internal resistance of 0
...
In what unit is the capacity of a cell measured?

4
...
Why is safe disposal of batteries important?

(a) an e
...
f
...
5
(b) an e
...
f
...
5
(c) an e
...
f
...
1
(d) an e
...
f
...
1

19
...
Name two types of secondary cells
21
...
Name any six types of battery and state three
common applications for each
26
...
State the advantages of fuel cells

5
...
d
...
m
...

(d) A secondary cell may be recharged
after use
6
...
Five 2 V cells, each having an internal resistance of 0
...
The current flowing in the
circuit is:
(a) 10 A

(b) 1
...
5 A

(d)

2
A
3

8
...
d
...
Which of the following statements is true?
(a) The capacity of a cell is measured in volts
(b) A primary cell converts electrical energy
into chemical energy
(c) Galvanising iron helps to prevent corrosion
(d) A positive electrode is termed the cathode

10
...
d
...
m
...

the greater the e
...
f
...
d
...
The negative pole of a dry cell is made of:
(a)
(b)
(c)
(d)

carbon
copper
zinc
mercury

12
...
Which of the following statements is true?
(a) A zinc carbon battery is rechargeable and
is not classified as hazardous
(b) A nickel cadmium battery is not rechargeable and is classified as hazardous
(c) A lithium battery is used in watches and
is not rechargeable
(d) An alkaline manganese battery is used in
torches and is classified as hazardous

Section 1

Batteries and alternative sources of energy 39

Section 1

Revision test 1
This revision test covers the material contained in Chapters 1 to 4
...

1
...
Determine the power consumed
...
A d
...
motor consumes 47
...
Determine the power rating of the motor and the current
taken from the supply
...
A 100 W electric light bulb is connected to a 200 V
supply
...

(4)
4
...

(2)
5
...
Determine the power
dissipated by the element
...
5p per unit
...
Calculate the resistance of 1200 m of copper
cable of cross-sectional area 15 mm2
...
02 μ m
(5)
7
...
If the temperature coefficient

of resistance at 0◦ C is 0
...
(a) Determine the values of the resistors with the
following colour coding:
(i) red-red-orange-silver
(ii) orange-orange-black-blue-green
(b) What is the value of a resistor marked as
47 KK?
(6)
9
...
40 and an e
...
f
...
5 V are connected in
series to a load of 38
...
(a) Determine the current flowing in the circuit and the p
...
at the battery
terminals
...
d
...

(10)
10
...

(12)
11
...

(15)

Chapter 5

Series and parallel networks
At the end of this chapter you should be able to:
•
•
•
•
•

calculate unknown voltages, current and resistances in a series circuit
understand voltage division in a series circuit
calculate unknown voltages, currents and resistances in a parallel network
calculate unknown voltages, currents and resistances in series-parallel networks
understand current division in a two-branch parallel network

• understand and perform calculations on relative and absolute voltages
• describe the advantages and disadvantages of series and parallel connection of lamps

5
...
1 shows three resistors R1 , R2 and R3 connected
end to end, i
...
in series, with a battery source of V volts
...
d
...

R1

R3

V1

A

R2
V2

From Ohm’s law: V1 = IR1 , V2 = IR2 , V3 = IR3 and
V = IR where R is the total circuit resistance
...
Dividing
throughout by I gives
R = R1 + R2 + R3
Thus for a series circuit, the total resistance is
obtained by adding together the values of the separate
resistance’s
...
1

Problem 1
...
5
...
d
...


In a series circuit

R1

(a) the current I is the same in all parts of the circuit
and hence the same reading is found on each of
the ammeters shown, and

4A

(b) the sum of the voltages V1 , V2 and V3 is equal to
the total applied voltage, V ,
i
...


V = V1 + V2 + V3

V1

R2

V2
V

Figure 5
...
25
I
4
V1 5
(c) Resistance R1 =
= = 1
...
5
I
4
V3 6
Resistance R3 =
= = 1
...
25 + 0
...
5
= 3
...
For the circuit shown in Fig
...
3,
determine the p
...
across resistor R3
...
Find also the
value of resistor R2
...
4

Power dissipated in the 11

resistor,

P = I 2 R = (0
...
25)(11) = 2
...
2 Potential divider
The voltage distribution for the circuit shown in
Fig
...
5(a) is given by:

R3

V1 =
I

11 Ω

9Ω

R1
V and V2 =
R1 + R 2

R2
V
R1 + R 2

V3
R1

25 V

R2

Figure 5
...
d
...
25 A,
R
100
which is the current flowing in each resistor
4
V2
Resistance R2 =
=
= 16
I
0
...
A 12 V battery is connected in a
circuit having three series-connected resistors
having resistance’s of 4 , 9 and 11
...
d
...
Find also the power
dissipated in the 11 resistor
...
5
...
5 A,
R
24
which is the current in the 9 resistor
...
d
...
5 × 9 = 4
...
5

The circuit shown in Fig
...
5(b) is often referred to as a
potential divider circuit
...
Frequently the divider consists
of two resistors as shown in Fig
...
5(b), where
R2
R1 + R 2

VOUT =

VIN

Rx

V1
Iϭ3A

A potential divider is the simplest way of producing a
source of lower e
...
f
...
m
...
, and
is the basic operating mechanism of the potentiometer,
a measuring device for accurately measuring potential
differences (see page 131)
...
8

Value of unknown resistance,
Rx = 8 − 2 = 6

Problem 4
...
5
...
d
...
6

Figure 5
...
5
...
6 kWh

50 V
6Ω

V

Now try the following exercise
Figure 5
...
Two resistors are connected in series
across a 24 V supply and a current of 3 A flows in
the circuit
...
d
...
If the circuit is
connected for 50 hours, how much energy is used?

The circuit diagram is shown in Fig
...
8
(a) Total circuit resistance
R=

V
24
=
=8
I
3

Exercise 19 Further problems on series
circuits
1
...
d’s measured across three resistors connected in series are 5 V, 7 V and 10 V, and the
supply current is 2 A
...

[(a) 22 V (b) 11 (c) 2
...
5 , 5 ]
2
...
5
...
If the total circuit resistance
is 36 determine the supply current and the
value of resistors R1 , R2 and R3
[10 V, 0
...
3

R3

5V

Parallel networks

Figure 5
...
e
...


3V

18 V
I1

Figure 5
...
When the switch in the circuit in Fig
...
10 is
closed the reading on voltmeter 1 is 30 V and
that on voltmeter 2 is 10 V
...
5 ]

I2

R2
A2

I3

R3
A3

I

5Ω

Rx
A
V2

A

V

V1

Figure 5
...
10

4
...
5
...
e
...
11

5
...
If one
of the resistors has a value of 2
...
d
...
4 resistor
...
2 (b) 12 V]
6
...
6 A at 55 V
...
Find the value of the
stabilising resistor to be connected in series
...
77 ]
7
...
It is required
to reduce the current to 12 A
...

[(a) 4 (b) 48 V]

and

(b) the source p
...
, V volts, is the same across each of
the resistors
...
Since
V
V
V
V
I = I1 + I2 + I3 then
=
+
+
R
R1
R2
R3
Dividing throughout by V gives:
1
1
1
1
=
+
+
R
R1
R2
R3
This equation must be used when finding the total resistance R of a parallel circuit
...
e
...
For the circuit shown in Fig
...
13,
determine (a) the reading on the ammeter, and
(b) the value of resistor R2
...
For the circuit shown in Fig
...
15,
find (a) the value of the supply voltage V and
(b) the value of current I
...
13

P
...
across R1 is the same as the supply voltage V
Hence supply voltage, V = 8 × 5 = 40 V
(a) Reading on ammeter,
I=

V
40
=
= 2A
R3
20

(b) Current flowing through R2 = 11 − 8 − 2 = 1 A
...
Two resistors, of resistance 3 and
6 , are connected in parallel across a battery
having a voltage of 12 V
...

The circuit diagram is shown in Fig
...
14

Figure 5
...
d
...

Alternatively,
1
1
1
1
1+3+6
10
=
+
+
=
=
R
60 20 10
60
60
Hence total resistance
60
= 6 , and current
10
V
60
I= =
= 10 A
R
6

R=

Figure 5
...
Given four 1 resistors, state how
they must be connected to give an overall resistance
1
1
of (a) 4 (b) 1 (c) 1 1 (d) 2 2 , all four
3
resistors being connected in each case
...
5
...
e
...
Find the equivalent resistance for
the circuit shown in Fig
...
20

Figure 5
...
20

(b) Two in series, in parallel with another two in
series (see Fig
...
17), since 1 and 1 in series
gives 2 , and 2 in parallel with 2 gives
2×2
4
= =1
2+2
4

R3 , R4 and R5 are connected in parallel and their
equivalent resistance R is given by
1
1 1
1
6+3+1
10
= + +
=
=
R
3 6 18
18
18
hence R = (18/10) = 1
...
The circuit is now equivalent to four resistors in series and the equivalent circuit
resistance = 1 + 2
...
8 + 4 = 9

Figure 5
...
5
...
Resistances of 10 , 20 and 30
are connected (a) in series and (b) in parallel to a
240 V supply
...

(a) The series circuit is shown in Fig
...
21

Figure 5
...
e
...
5
...
21

The equivalent resistance
RT = 10 + 20 + 30 = 60
Supply current I =

Figure 5
...
5
...
21

(I)

Similarly,
given by:
1
1
1
1
6+3+2
11
=
+
+
=
=
RT
10 20 30
60
60
60
hence RT =
11
Supply current

current

V
I
=
R2
R2

R 1 R2
R1 + R 2

R1
R1 + R 2

=

(I)

Summarising, with reference to Fig
...
23
I1 =

V
240
240 × 11
I=
= 60 =
= 44 A
RT
60
11

I2 =

and

(Check:
I1 =

V
240
=
= 24 A,
R1
10

I2 =

V
240
=
= 12 A
R2
20

and I3 =

I2 =

R2
R1 + R 2
R1
R1 + R 2

(I)

(I)

Problem 12
...
5
...
d
...


V
240
=
= 8A
R3
30

For a parallel circuit I = I1 + I2 + I3
= 24 + 12 + 8 = 44 A, as above)

5
...
5
...
24

(a) The equivalent resistance Rx of R2 and R3 in
parallel is:
Rx =

6×2
= 1
...
5 + 1
...
23

I=

200
V
= 25 A
=
RT
8

Section 1

Series and parallel networks 47

Section 1

48 Electrical and Electronic Principles and Technology
(b) The current flowing through R1 and R4 is 25 A
...
5 kW,
(b) the current flowing in each of the four resistors
...
25 A
The current flowing through R3
=

R2
R2 + R 3

=

6
25 = 18
...
e
...
5
...
5
...
25

p
...
across R1 , i
...


Figure 5
...
e
...
The
equivalent resistance of R1 and R2 in parallel is
RT =

150
15 × 10
=
=6
15 + 10
25
The equivalent resistance of resistors R3 and Rx in
parallel is equal to 25 − 6 , i
...
19
...


Method 1
V1 = IR1 = (25)(2
...
5 V

The voltage V1 = IR, where R is 6 , from above, i
...

V1 = (10)(6) = 60 V
...
d
...
e
...
d
...
5) = 37
...
d
...

=
R3
38

p
...
across R4 , i
...

V4 = IR4 = (25)(4) = 100 V

Thus I4 = 5 A also, since I = 10 A
...
d
...
d
...
5 V

V2
190
= 38
=
I4
5

Method 2
Since the equivalent resistance of R3 and Rx in parallel
is 19 ,

Problem 13
...
5
...
e
...
Thus, in this case, since
RT = 19 and R3 = 38 , then Rx = 38 could have
been deduced on sight
...
6 A
10

Now try the following exercise

I=

15
(10)
15 + 10

=

R1
R1 + R 2

From Fig
...
28(b),

From Fig
...
27

=

Current I2 =

Figure 5
...
For the arrangement shown in
Fig
...
27, find the current Ix
...
Resistances of 4 and 12 are connected
in parallel across a 9 V battery
...

[(a) 3 (b) 3 A (c) 2
...
75 A]
2
...
5
...
5 A, 2
...
27

Commencing at the right-hand side of the arrangement
shown in Fig
...
27, the circuit is gradually reduced in
stages as shown in Fig
...
28(a)–(d)
...
5
...
25

Figure 5
...
Find the equivalent resistance when the following resistances are connected (a) in series
(b) in parallel (i) 3 and 2 (ii) 20 k and
40 k (iii) 4 , 8 and 16 (iv) 800 , 4 k
and 1500
[(a) (i) 5
(ii) 60 k
(iii) 28
(iv) 6
...
2
(ii) 13
...
29 (iv) 461
...
31

4
...
5
...
For the circuit shown in Fig
...
32, find
(a) V1 , (b) V2 , without calculating the current
flowing
...
Resistors of 20 , 20 and 30 are connected in parallel
...
If the complete circuit expends a power of 0
...

[2
...
Find the equivalent resistance between terminals C and D of the circuit shown in
Fig
...
30(b)
[27
...
32

9
...
5
...
5 A, I3 = 1 2 A, I4 = 5 A
3
6
I5 = 3 A, I6 = 2 A, V1 = 20 V, V2 = 5 V,
V3 = 6 V]

Figure 5
...
30

7
...
5
...

[(a) 1
...
Find the current I in Fig
...
34

Figure 5
...
8 A]

11
...
4 is connected in series with
another of 3
...
What resistance must be
placed across the one of 2
...
2 ]
12
...
A p
...
of 10 V is
applied to the circuit
...
Find the p
...

required to send the same current through the
8 resistor
...
5 Relative and absolute voltages

an ‘absolute potential’
...
5
...
e
...
If the voltage is
negative w
...
t
...
r
...
earth, and is written
as VC = −30 V or VC = 30 V −ve
...
For the circuit shown in Fig
...
36,
calculate (a) the voltage drop across the 4 k
resistor, (b) the current through the 5 k resistor,
(c) the power developed in the 1
...
r
...
earth, and (e) the
absolute voltage at point X
...
r
...
) any
other point in the circuit
...
5
...
The total resistance,
RT = 30 + 50 + 5 + 15 = 100
200
current, I =
= 2A
100

30 Ω

A

50 Ω

C

X
1
...
36

(a) Total circuit resistance, RT = [(1 + 4)k
lel with 5 k ] in series with 1
...
e
...
This is known as a
‘relative voltage’
...
5
...
r
...
B is I × 50, i
...
2 × 50 = 100 V and
is written as VAB = 100 V
...
35

It must also be indicated whether the voltage at A w
...
t
...
Point A is nearer to the
positive terminal than B so is written as VAB = 100 V or
VAB = +100 V or VAB = 100 V +ve
...

If the voltage at B w
...
t
...

If the reference point is changed to the earth point
then any voltage taken w
...
t
...
5 = 4 k
5+5

Total circuit current, IT =

V
24
=
= 6 mA
RT 4 × 103

By current division, current in top branch
5
=
× 6 = 3 mA
5+1+4
Hence, volt drop across 4 k resistor
= 3 × 10−3 × 4 × 103 = 12 V
(b) Current through the 5 k
=

resistor
1+4
× 6 = 3 mA
5+1+4

(c) Power in the 1
...
5 × 103 ) = 54 mW
(d) The voltage at the earth point is 0 volts
...
Since
moving from the earth point to point X is moving

Section 1

Series and parallel networks 51

Section 1

52 Electrical and Electronic Principles and Technology
towards the negative terminal of the voltage source,
the voltage at point X w
...
t
...
r
...
earth’, hence the absolute
voltage at point X is −12 V
...


8Ω
B

Now try the following exercise

30 V

Exercise 21 Further problems on relative
and absolute voltages
1
...
5
...

[(a) +40 V, +29
...
4 V,
+16 V (c) −5
...
6 Wiring lamps in series and in
parallel
Series connection
Figure 5
...


B

5Ω

100 V

Figure 5
...
37

2
...
5
...
r
...
earth, and (e) the absolute voltage
at point X
...
68 V (b) 0
...
8 mW
(d) +2
...
88 V]
8Ω

X

7Ω

5Ω

18 Ω
30 Ω

12 V

Figure 5
...
In the bridge circuit of Fig
...
39 calculate (a)
the absolute voltages at points A and B, and
(b) the voltage at A relative to B
...
40

(i) Each lamp has only (240/3) V, i
...
80 V across it
and thus each lamp glows dimly
...
e
...

(iii) If a lamp is removed from the circuit or if a lamp
develops a fault (i
...
an open circuit) or if the
switch is opened, then the circuit is broken, no
current flows, and the remaining lamps will not
light up
...

The series connection of lamps is usually limited to
decorative lighting such as for Christmas tree lights
...
41 shows three similar lamps, each rated at
240 V, connected in parallel across a 240 V supply
...
If four identical lamps are connected in parallel
and the combined resistance is 100 , find the
resistance of one lamp
...
Three identical filament lamps are connected
(a) in series, (b) in parallel across a 210 V supply
...
d
...

[(a) 70 V (b) 210 V]

Figure 5
...

(ii) If any lamp is removed from the circuit or develops a fault (open circuit) or a switch is opened,
the remaining lamps are unaffected
...


Exercise 23 Short answer questions on
series and parallel networks
1
...


2
...


3
...
If three identical lamps are
connected in parallel and the combined resistance is
150 , find the resistance of one lamp
...
Show that for three resistors R1 , R2 and R3
connected in parallel the equivalent resistance
R is given by
1
1
1
1
+
=
+
R
R1
R2
R3

Let the resistance of one lamp be R, then
1
1
1
1
3
= + + =
150
R R R
R
from which, R = 3 × 150 = 450

Problem 17
...
State
(a) the voltage across each lamp, and (b) the effect
of lamp C failing
...
Explain the potential divider circuit
6
...
Compare the merits of wiring lamps in
(a) series (b) parallel

Exercise 24
(a) Since each lamp is identical and they are connected
in series there is 150/3 V, i
...
50 V across each
...
e
...


Multi-choice questions on
series and parallel networks
(Answers on page 398)

1
...
If two 4 resistors are connected in parallel
the effective resistance of the circuit is:
(a) 8
(b) 4
(c) 2
(d) 1

7
...
5
...
With the switch
closed the ammeter reading will indicate:
(a)

3
...
5
...
42

4
...
d
...
The equivalent resistance when a resistor
1
of 1 is connected in parallel with a 4
3
resistance is:
1
1
3
(a) 7
(b) 7
(c) 12
(d) 4

2Ω

1
3

A
6Ω

6V

Figure 5
...
The total resistance of two resistors R1 and
R2 when connected in parallel is given by:
1
1
(a) R1 + R2
(b)
+
R 1 R2
(c)

R1 + R2
R 1 R2

(c) 3 A

(d) 4 3 A
5

(d)

R1 R2
R1 + R2

11
...
5
...
005
(b) 5
(c) 125
(d) 200
R
A
V

10 Ω

Figure 5
...
A 10 resistor is connected in parallel with
a 15 resistor and the combination in series
with a 12 resistor
...
With the switch in Fig
...
43 closed the
ammeter reading will indicate:
(b)

1
4

9
...
333

A

(a) 108 A

(c)

Chapter 6

Capacitors and capacitance
At the end of this chapter you should be able to:
• appreciate some applications of capacitors
• describe an electrostatic field
• appreciate Coulomb’s law
• define electric field strength E and state its unit
• define capacitance and state its unit
• describe a capacitor and draw the circuit diagram symbol
• perform simple calculations involving C = Q/V and Q = It
• define electric flux density D and state its unit
• define permittivity, distinguishing between ε0 , εr and ε
• perform simple calculations involving
D=

Q
,
A

E=

V
d

and

D
= ε0 εr
E

• understand that for a parallel plate capacitor,
C=

ε0 εr A(n − 1)
d

• perform calculations involving capacitors connected in parallel and in series
• define dielectric strength and state its unit
1
• state that the energy stored in a capacitor is given by W = 2 CV 2 joules
• describe practical types of capacitor
• understand the precautions needed when discharging capacitors

6
...
Next to the resistor, the capacitor is the
most commonly encountered component in electrical
circuits
...
For example, capacitors

are used to smooth rectified a
...
outputs, they are
used in telecommunication equipment – such as radio
receivers – for tuning to the required frequency, they are
used in time delay circuits, in electrical filters, in oscillator circuits, and in magnetic resonance imaging (MRI)
in medical body scanners, to name but a few practical
applications
...
2

Electrostatic field

Figure 6
...
If an electron that
has a negative charge is placed between the plates,
a force will act on the electron tending to push it away
from the negative plate B towards the positive plate,
A
...

Any region such as that shown between the plates in
Fig
...
1, in which an electric charge experiences a force,
is called an electrostatic field
...
In Fig
...
1, the direction of the force
is from the positive plate to the negative plate
...
The closeness of the lines is an indication of the
field strength
...
d
...


+

(a)

+

−

(b)

Figure 6
...
1

Figure 6
...
6
...
Electric lines
of force (often called electric flux lines) are continuous
and start and finish on point charges; also, the lines cannot cross each other
...

This is because lines of force from the charged body
terminate on its surface
...
However,
it should be remembered that they are only aids to the
imagination
...
e
...
This is known as
Coulomb’s law
...
6 μC is given by:
force=k

6
...
6 × 10−6 )2
≈ (9 × 109 )
d2
(16 × 10−3 )2
= 90 newtons

Electric field strength

Figure 6
...
They are connected to
opposite terminals of a battery of voltage V volts
...
3

is therefore an electric field in the space between the
plates
...
6
...
Over the area in which there is negligible
fringing,

Fixed capacitor

Variable capacitor

Figure 6
...
Electric field
strength is also called potential gradient
...


6
...

Thus the presence of the field indicates the presence of
equal positive and negative electric charges on the two
plates of Fig
...
3
...
The property
of this pair of plates which determines how much charge
corresponds to a given p
...
between the plates is called
their capacitance:
capacitance C =

Q
V

The unit of capacitance is the farad F (or more usually
μF = 10−6 F or pF = 10−12 F), which is defined as the
capacitance when a p
...
of one volt appears across the
plates when charged with one coulomb
...
(a) Determine the p
...
across a 4 μF
capacitor when charged with 5 mC (b) Find the
charge on a 50 pF capacitor when the voltage
applied to it is 2 kV
...

Since C =

Q
5 × 10−3
Q
then V = =
V
C
4 × 10−6
=

5 × 106
5000
=
3
4 × 10
4

Hence p
...
V = 1250 V or 1
...
5 Capacitors

5×2
= 0
...
1 μC
Every system of electrical conductors possesses capacitance
...
In these
examples the capacitance is undesirable but has to
be accepted, minimised or compensated for
...

Devices specially constructed to possess capacitance
are called capacitors (or condensers, as they used to
be called)
...
A capacitor has the ability to store
a quantity of static electricity
...
6
...
A direct current of 4 A flows into a
previously uncharged 20 μF capacitor for 3 ms
...
d
...

I = 4 A, C = 20 μF = 20 × 10−6 F and
t = 3 ms = 3 × 10−3 s
...

V=

Q 4 × 3 × 10−3
=
C
20 × 10−6
=

12 × 106
= 0
...
d
...
A 5 μF capacitor is charged so that
the p
...
between its plates is 800 V
...

C = 5 μF = 5 × 10−6 F, V = 800 V and
I = 2 mA = 2 × 10−3 A
...
Thus,
t=

4 × 10−3
Q
= 2s
=
I
2 × 10−3

Hence, the capacitor can provide an average
discharge current of 2 mA for 2 s
...
Find the charge on a 10 μF capacitor when the
applied voltage is 250 V
...
5 mC]
2
...

[2 kV]
3
...
4 kV
...

[2
...
For how long must a charging current of 2 A be
fed to a 5 μF capacitor to raise the p
...
between
its plates by 500 V
...
25 ms]
5
...
Determine
the p
...
between the plates
...
A 16 μF capacitor is charged at a constant
current of 4 μA for 2 min
...
d
...

[30 V, 480 μC]
7
...
5 ms when
the p
...
between the plates is 2 kV
...

[7
...
6

Electric flux density

Unit flux is defined as emanating from a positive charge
of 1 coulomb
...

Electric flux density D is the amount of flux passing
through a defined area A that is perpendicular to the
direction of the flux:
Q
electric flux density, D = coulombs/metre2
A
Electric flux density is also called charge density, σ
...
7 Permittivity
At any point in an electric field, the electric field strength
E maintains the electric flux and produces a particular
value of electric flux density D at that point
...
e
...
The value of ε0 is 8
...

When an insulating medium, such as mica, paper,
plastic or ceramic, is introduced into the region of an
electric field the ratio of D/E is modified:
D
= ε0 εr
E
where εr , the relative permittivity of the insulating
material, indicates its insulating power compared with
that of vacuum:
relative permittivity,
εr =

flux density in material
flux density in vacuum

εr has no unit
...
00;
polythene, 2
...

The product ε0 εr is called the absolute permittivity,
ε, i
...

ε = ε0 εr
The insulating medium separating charged surfaces is
called a dielectric
...
They are therefore used to separate conductors at different potentials,
such as capacitor plates or electric power lines
...
Two parallel rectangular plates
measuring 20 cm by 40 cm carry an electric charge
of 0
...
Calculate the electric flux density
...
25 kV determine the electric field
strength
...
2 μC = 0
...
2 × 10−6
0
...
5 μC/m2
800

Voltage V = 0
...

Electric field strength
250
V
= 50 kV/m
=
E=
d
5 × 10−3

(a) For air: εr = 1 and

D
= ε0 εr
E

Hence electric flux density
D = Eε0 εr
= (250 × 103 × 8
...
213 μC/m2
(b) For polythene, εr = 2
...
85 × 10−12 × 2
...
089 μC/m2
Now try the following exercise
Exercise 26 Further problems on electric
field strength, electric flux
density and permittivity
(Where appropriate take ε0 as 8
...
The flux density between two plates
separated by mica of relative permittivity 5 is
2 μC/m2
...

Flux density D = 2 μC/m2 = 2 × 10−6 C/m2 ,
ε0 = 8
...

D/E = ε0 εr , hence voltage gradient,
E=

D
2 × 10−6
=
V/m
ε0 εr 8
...
2 kV/m

Problem 6
...
d
...
8 mm apart
...
3
Electric field strength
E=

V
200
= 250 kV/m
=
d
0
...
A capacitor uses a dielectric 0
...
What is the electric field
strength across the dielectric at this voltage?
[750 kV/m]
2
...

If the effective area of each plate is 5 cm2 find
the electric flux density of the electric field
...
A charge of 1
...
Calculate the electric flux density
...
5 kV determine the
electric field strength
...
5 μC/m2 , 50 kV/m]
4
...
Given that the
area of each plate is 50 cm2 , calculate the electric flux density in the dielectric separating
the plates
...
The electric flux density between two plates
separated by polystyrene of relative permittivity 2
...
Find the voltage gradient
between the plates
...
Two parallel plates having a p
...
of 250 V
between them are spaced 1 mm apart
...
Find also
the electric flux density when the dielectric
between the plates is (a) air and (b) mica of
relative permittivity 5
...
213 μC/m2
(b) 11
...
8 The parallel plate capacitor
For a parallel-plate capacitor, as shown in Fig
...
5(a),
experiments show that capacitance C is proportional to
the area A of a plate, inversely proportional to the plate
spacing d (i
...
the dielectric thickness) and depends on
the nature of the dielectric:
Capacitance, C =

ε0 εrA
farads
d

Figure 6
...
1 mm = 0
...
85 × 10−12 F/m and εr = 100
Capacitance,
C=

where ε0 = 8
...
(a) A ceramic capacitor has an
effective plate area of 4 cm2 separated by 0
...
Calculate
the capacitance of the capacitor in picofarads
...
2 μC what will be the p
...
between the
plates?

8
...
85 × 10−12 × 100 × 4 × 10−4
F
0
...
6
...

Ten plates are shown, forming nine capacitors
with a capacitance nine times that of one pair of
plates
...
Thus capacitance

ε0 εr A
farads
d

8
...
2 × 10−6
Q
=
V = 339 V
C
3540 × 10−12

Problem 8
...
If the
capacitance of the capacitor is 4425 pF determine
the effective thickness of the paper if its relative
permittivity is 2
...

A = 800 cm2 = 800 × 10−4 m2 = 0
...
85 × 10−12 F/m

and εr = 2
...
Since
ε0 εA A
ε 0 εr A
then d =
d
C
−12 × 2
...
08
8
...
0004 m

C=
i
...


Hence, the thickness of the paper is 0
...

Problem 9
...
2 mm thick
...

n = 19 thus n − 1 = 18, A = 75 × 75 = 5625 mm2 =
5625 × 10−6 m2 , εr = 5, ε0 = 8
...
2 mm = 0
...
Capacitance,
ε0 εr A(n − 1)
d
8
...
2 × 10−3
= 0
...
4 nF

C=

Now try the following exercise
Exercise 27 Further problems on parallel
plate capacitors
(Where appropriate take ε0 as 8
...
A capacitor consists of two parallel plates each
of area 0
...
1 mm in air
...
[885 pF]
2
...
2 m2
...
885 mm]
3
...
75 mm
thick having a relative permittivity of 2
...
14 pF]
4
...
102 mm thick with a relative permittivity
of 6
...
A parallel plate capacitor is made from
25 plates, each 70 mm by 120 mm interleaved with mica of relative permittivity
5
...

[2
...
A capacitor is constructed with parallel plates
and has a value of 50 pF
...
The capacitance of a parallel plate capacitor
is 1000 pF
...
40 mm
...

[1
...
The charge on the square plates of a multiplate capacitor is 80 μC when the potential
between them is 5 kV
...
102 mm and relative permittivity
4
...

[40 mm]
9
...
d
...
The dielectric is
to be polythene (εr = 2
...
Find (a) the thickness of the
polythene needed, and (b) the area of a plate
...
005 mm (b) 10
...
9 Capacitors connected in parallel
and series
(a) Capacitors connected in parallel
Figure 6
...

When the charging current I reaches point A it divides,
some flowing into C1 , some flowing into C2 and some
into C3
...
The conductor between
plates ‘b’ and ‘c’ is electrically isolated from the rest
of the circuit so that an equal but opposite charge of
+Q coulombs must appear on plate ‘c’, which, in turn,
induces an equal and opposite charge of −Q coulombs
on plate ‘d’, and so on
...
In a series circuit:

C1

C2

Q3

Section 1

Q1

C3

l

l

V = V1 + V2 + V3
V
Total charge, Q T = Q1 + Q2 + Q3

Since V =

Q
Q
Q
Q
Q
+
+
then
=
C
C
C1
C2
C3

where C is the total equivalent circuit capacitance, i
...


Figure 6
...
The capacitors each store
a charge and these are shown as Q1 , Q2 and Q3
respectively
...

Therefore CV = C1 V + C2 V + C3 V where C is the total
equivalent circuit capacitance, i
...

C = C1 + C2 + C3
It follows that for n parallel-connected capacitors,
C = C1 + C2 + C3 · · · · · · + Cn
i
...
the equivalent capacitance of a group of parallelconnected capacitors is the sum of the capacitances
of the individual capacitors
...


1
1
1
1
=
+
+
C
C1
C2
C3
It follows that for n series-connected capacitors:
1
1
1
1
1
=
+
+
+ ···+
C
C1
C2
C3
Cn
i
...
for series-connected capacitors, the reciprocal of the
equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances
...

For the special case of two capacitors in series:

Hence

C=

(b) Capacitors connected in series
Figure 6
...
Let the
p
...
across the individual capacitors be V1 , V2 and V3
respectively as shown
...
e
...
Calculate the equivalent capacitance
of two capacitors of 6 μF and 4 μF connected (a) in
parallel and (b) in series
...
Thus

Figure 6
...
This induces an equal but opposite charge

C1 C2
C1 + C 2

C=

24
6×4
=
= 2
...
What capacitance must be
connected in series with a 30 μF capacitor for the
equivalent capacitance to be 12 μF?
Let C = 12 μF (the equivalent capacitance),
C1 = 30 μF and C2 be the unknown capacitance
...
Capacitance’s of 3 μF, 6 μF and
12 μF are connected in series across a 350 V supply
...
d
...

The circuit diagram is shown in Fig
...
8
...
Capacitance’s of 1 μF, 3 μF, 5 μF
and 6 μF are connected in parallel to a direct
voltage supply of 100 V
...

(a) The equivalent capacitance C for four capacitors
in parallel is given by:
C = C 1 + C2 + C3 + C4
(b) Total charge QT = CV where C is the equivalent
circuit capacitance i
...

QT = 15 × 10

× 100 = 1
...
5 mC
(c) The charge on the 1 μF capacitor
Q1 = C1 V = 1 × 10−6 × 100 = 0
...
3 mC
The charge on the 5 μF capacitor
Q3 = C3 V =

(a) The equivalent circuit capacitance C for three
capacitors in series is given by:
1
1
1
1
=
+
+
C C1
C2
C3
1
1 1
1
4+2+1
7
i
...

= + +
=
=
C 3 6 12
12
12
Hence the equivalent circuit capacitance

i
...
C = 1 + 3 + 5 + 6 = 15 μF

−6

Figure 6
...
5 mC
The charge on the 6 μF capacitor
Q4 = C4 V = 6 × 10−6 × 100 = 0
...
1 + 0
...
5 + 0
...
5 mC = QT ]

C=

5
12
= 1 μF or 1
...
6 mC

QT =

Since the capacitors are connected in series
0
...

(c) The voltage across the 3 μF capacitor,
V1 =
=

Q
C1
0
...
6 × 10−3
= 100 V
6 × 10−6

Section 1

Capacitors and capacitance 63

Section 1

64 Electrical and Electronic Principles and Technology
The voltage across the 12 μF capacitor,
V3 =
=

Q
C3
0
...

V1 + V2 + V3 = 200 + 100 + 50 = 350 V = supply
voltage]
In practice, capacitors are rarely connected in series
unless they are of the same capacitance
...
e
...
d
...
e
...

Problem 14
...
6
...


(b) The charge on each of the capacitors shown in
Fig
...
10 will be the same since they are connected
in series
...

Then
i
...


Q = C 1 V 1 = C2 V 2
5V1 = 15V2
V1 = 3V2

Also

(1)

V1 + V2 = 240 V

Hence 3V2 + V2 = 240 V from equation (1)
Thus

V2 = 60 V and V1 = 180 V

Hence the voltage across QR is 60 V
(c) The charge on the 15 μF capacitor is
C2 V2 = 15 × 10−6 × 60 = 0
...
36 mC
The charge on the 3 μF capacitor is
3 × 10−6 × 180 = 0
...
9

(a) 2 μF in parallel with 3 μF gives an equivalent
capacitance of 2 μF + 3 μF = 5 μF
...
6
...


Exercise 28 Further problems on capacitors
in parallel and series
1
...
Determine
the equivalent capacitance in each case
...
5 μF]
2
...
10

The equivalent capacitance of 5 μF in series with
15 μF is given by
75
5 × 15
μF =
μF = 3
...
What value of capacitance would be obtained
if capacitors of 0
...
10 μF are
connected (a) in series and (b) in parallel
[(a) 0
...
25 μF]
4
...
Find
the total equivalent circuit capacitance
...
2 μF?
[2
...
4 μF]
5
...
02 μF, 0
...
10 μF
(iii) 50 pF and 450 pF
(iv) 0
...
0102 μF

1
...
0125 μF

45 pF

(iv)

196
...
17 μF

6
...
6
...
5 μF capacitor
...
2 μF (b) 100 V]

Figure 6
...
Three 12 μF capacitors are connected in
series across a 750 V supply
...
d
...

[(a) 4 μF (b) 3 mC (c) 250 V]
8
...
d
...

[(a) 150 V, 90 V (b) 0
...
In Fig
...
12 capacitors P, Q and R are identical and the total equivalent capacitance of
the circuit is 3 μF
...
2 μF each]

Figure 6
...
Capacitances of 4 μF, 8 μF and 16 μF are
connected in parallel across a 200 V supply
...

[(a) 28 μF (b) 5
...
8 mC, 1
...
2 mC]
11
...
The capacitances of P, Q and R
are 4 μF, 12 μF and 8 μF respectively
...
c
...
d
...

[(a) 5
...
4 mC on P,
1
...
6 mC on R]
12
...
6
...
857 μF, (b) 1
...
85 μC on each]
2μF

2μF
2μF

2μF

2μF
2μF

C1

2μF
50 V

Figure 6
...
10

Dielectric strength

The maximum amount of field strength that a dielectric can withstand is called the dielectric strength of the
material
...
A capacitor is to be constructed so
that its capacitance is 0
...
d
...
25 kV across its terminals
...
Find (a) the
thickness of the mica needed, and (b) the area of a
plate assuming a two-plate construction
...


(a) Energy stored
W=

(b) Power =

Energy stored

(b) Capacitance,
C=

ε0 εr A
d

1
CV 2
2
2W
V2 =
C
2W
p
...
V =
=
C
W=

hence

i
...


energy
0
...
A 12 μF capacitor is required to
store 4 J of energy
...
d
...


(a) Dielectric strength,
V
E=
d
V
1
...
025 mm

1
1
CV 2 joules = × 3 × 10−6 × 4002
2
2
3
= × 16 × 10−2 = 0
...
5 V
3

Problem 18
...

If the energy stored is 1
...

1
Energy stored W = 2 CV 2 and C = Q/V
...
2 × 10−6 × 0
...
85 × 10−12 × 6
= 0
...
6 cm2

from which

Q = 10 mC = 10 × 10−3 C

6
...
2 J

and
(a) Voltage
V=

2 × 1
...
24 kV or 240 V
Q
10 × 10−3

(b) Capacitance
Problem 16
...


C=

Q
10 × 10−3
10 × 106
μF
=
F=
V
240
240 × 103
= 41
...
Variable air capacitors
...
The set of moving plates
rotate on a spindle as shown by the end view of
Fig
...
14
...
85 × 10−12 F/m)
1
...
Find (a) the capacitance and (b) the energy stored
...
02 μF (b) 0
...
Find the energy stored in a 10 μF capacitor
when charged to 2 kV
...
A 3300 pF capacitor is required to store 0
...
Find the p
...
to which the capacitor
must be charged
...
A capacitor is charged with 8 mC
...
4 J find (a) the voltage and (b) the
capacitance
...
A capacitor, consisting of two metal plates
each of area 50 cm2 and spaced 0
...
Calculate (a) the energy stored, (b) the electric flux
density and (c) the potential gradient
...
593 μJ (b) 5
...
A bakelite capacitor is to be constructed to have
a capacitance of 0
...
Allowing a safe value of field stress of 25 MV/m find
(a) the thickness of bakelite required, (b) the
area of plate required if the relative permittivity of bakelite is 5, (c) the maximum energy
stored by the capacitor and (d) the average
power developed if this energy is dissipated
in a time of 20 μs
...
04 mm (b) 361
...
02 J (d) 1 kW]

6
...
The main types
include: variable air, mica, paper, ceramic, plastic,
titanium oxide and electrolytic
...
14

As the moving plates are rotated through half a
revolution, the meshing, and therefore the capacitance, varies from a minimum to a maximum
value
...

The maximum value of such capacitors is between
500 pF and 1000 pF
...
Mica capacitors
...
6
...


Mica sheets

Metal foil
(lead or aluminium)

Figure 6
...
Mica is easily
obtained in thin sheets and is a good insulator
...
2 μF
...
The mica
is coated on both sides with a thin layer of silver
which forms the plates
...
Such capacitors have
a constant capacitance with change of temperature,
a high working voltage rating and a long service life
and are used in high frequency circuits with fixed
values of capacitance up to about 1000 pF
...
Paper capacitors
...
6
...

Metal end cap for
connection to
metal foil

5
...
Some plastic materials such as
polystyrene and Teflon can be used as dielectrics
...
Plastic capacitors operate well under conditions of high temperature, provide a precise value of capacitance, a very
long service life and high reliability
...
16

The whole is usually impregnated with oil or
wax to exclude moisture, and then placed in a plastic or aluminium container for protection
...
The maximum value of this type
of capacitor is between 500 pF and 10 μF
...

Connection

Conducting
coating

Figure 6
...
19

Ceramic
tube

Conducting
coating
(e
...
silver)

Figure 6
...
Ceramic capacitors
...
For high values, a
tube of ceramic material is used as shown in the
cross section of Fig
...
17
...
6
...
6
...
Certain ceramic materials have a very high permittivity and this enables
capacitors of high capacitance to be made which
are of small physical size with a high working
voltage rating
...
1 μF and may be used in
high frequency electronic circuits subject to a wide
range of temperatures
...
Titanium oxide capacitors have a very high capacitance with a small physical size when used at a low
temperature
...
Electrolytic capacitors
...
The finished capacitor is usually assembled in an aluminium container
and hermetically sealed
...

This oxide layer is very thin and forms the dielectric
...
) Such
capacitors must always be used on d
...
and must
be connected with the correct polarity; if this is
not done the capacitor will be destroyed since the
oxide layer will be destroyed
...
These capacitors possess a much larger
capacitance than other types of capacitors of similar dimensions due to the oxide film being only a
few microns thick
...
c
...


12
...
Name two practical examples where capacitance is present, although undesirable
14
...


6
...
Thus precautions must be
taken to ensure that the capacitor is automatically discharged after the supply is switched off
...

Now try the following exercises

15
...
What is the capacitance of the capacitor?
16
...
The equivalent capacitance is
...
Three 3 μF capacitors are connected in series
...

18
...
Name three factors upon which capacitance
depends

Exercise 30 Short answer questions on
capacitors and capacitance

20
...
Define ‘permittivity of free space’

1
...
State five practical applications of capacitors
3
...
Complete the statements:
Like charges
...


22
...
State the formula used to determine the
energy stored by a capacitor
24
...
How can an ‘electric field’ be established
between two parallel metal plates?

25
...
What is capacitance?

26
...
State the unit of capacitance
8
...
Complete the statements:
(a) 1 μF =
...
State three advantages and one disadvantage
of mica capacitors
28
...
F

10
...
Complete the statement:
Electric flux density D =

······
······

29
...
What main advantages do plastic capacitors
possess?
31
...
What is the main disadvantage of electrolytic
capacitors?

Section 1

Capacitors and capacitance 69

Section 1

70 Electrical and Electronic Principles and Technology
33
...
What safety precautions should be taken
when a capacitor is disconnected from a
supply?

Exercise 31

Multi-choice questions on
capacitors and capacitance
(Answers on page 398)

1
...
The capacitance of a capacitor is the ratio
(a) charge to p
...
between plates
(b) p
...
between plates to plate spacing
(c) p
...
between plates to thickness of
dielectric
(d) p
...
between plates to charge
3
...
d
...
The charge on a 10 pF capacitor when the
voltage applied to it is 10 kV is
(a) 100 μC
(b) 0
...
1 μC
(d) 0
...
Four 2 μF capacitors are connected in parallel
...
5 μF
(c) 2 μF
(d) 6 μF
6
...

The equivalent capacitance is
(a) 8 μF
(b) 0
...
State which of the following is false
...
Which of the following statement is false?
(a) An air capacitor is normally a variable type
(b) A paper capacitor generally has a shorter
service life than most other types of
capacitor
(c) An electrolytic capacitor must be used
only on a
...
supplies
(d) Plastic capacitors generally operate satisfactorily under conditions of high temperature
9
...
25 mJ
(b) 0
...
25 J
(d) 1
...
The capacitance of a variable air capacitor is
at maximum when
(a) the movable plates half overlap the fixed
plates
(b) the movable plates are most widely
separated from the fixed plates
(c) both sets of plates are exactly meshed
(d) the movable plates are closer to one side
of the fixed plate than to the other
11
...

The capacitance of the capacitor is:
(a) 2 × 109 F
(b) 0
...
5 mF
(d) 0
...
m
...

l
S=
=
μ0 μr A
• perform calculations on composite series magnetic circuits
• compare electrical and magnetic quantities
• appreciate how a hysteresis loop is obtained and that hysteresis loss is proportional to its area

7
...

The association between electricity and magnetism is

a fairly recent finding in comparison with the very first
understanding of basic magnetism
...
For example, they are used in motors and
generators, telephones, relays, loudspeakers, computer hard drives and floppy disks, anti-lock brakes,
cameras, fishing reels, electronic ignition systems, keyboards, t
...
and radio components and in transmission
equipment
...


7
...
A permanent magnet will position itself in a north and south
direction when freely suspended
...

The area around a magnet is called the magnetic field
and it is in this area that the effects of the magnetic force
produced by the magnet can be detected
...
Michael Faraday suggested
that the magnetic field could be represented pictorially,
by imagining the field to consist of lines of magnetic
flux, which enables investigation of the distribution and
density of the field to be carried out
...
A bar magnet is placed
on a flat surface covered by, say, cardboard, upon which
is sprinkled some iron filings
...
7
...
If a number of magnets of different strength are used, it is found that the stronger the
field the closer are the lines of magnetic flux and vice
versa
...
The strength of
the magnetic field decreases as we move away from
the magnet
...


If a compass is placed in the magnetic field in various
positions, the direction of the lines of flux may be determined by noting the direction of the compass pointer
...
The direction of a line of flux is from the north pole to the south
pole on the outside of the magnet and is then assumed to
continue through the magnet back to the point at which
it emerged at the north pole
...

The laws of magnetic attraction and repulsion can be
demonstrated by using two bar magnets
...
7
...

Lines of flux are imagined to contract and the magnets
try to pull together
...
In Fig
...
2(b), with similar poles
adjacent (i
...
two north poles), repulsion occurs, i
...

the two north poles try to push each other apart, since
magnetic flux lines running side by side in the same
direction repel
...
2

7
...
1

Magnetic flux and flux density

Magnetic flux is the amount of magnetic field (or
the number of lines of force) produced by a magnetic
source
...
The unit of magnetic flux is the weber, Wb
...
The unit of
magnetic flux density is the tesla, T , where
1 T = 1 Wb/m2
...
A magnetic pole face has a
rectangular section having dimensions 200 mm by
100 mm
...

Flux = 150 μWb = 150 × 10−6 Wb
Cross sectional area A = 200 × 100 = 20 000 mm2
= 20 000 × 10−6 m2
...
0075 T or 7
...
The maximum working flux density
of a lifting electromagnet is 1
...
If the
total magnetic flux produced is 353 mWb,
determine the radius of the pole face
...
8 T and
flux = 353 mWb = 353 × 10−3 Wb
...
e
...
1961 m2
1
...
Hence πr 2 √ 0
...
1961/π and radius r = (0
...
250 m
i
...
the radius of the pole face is 250 mm
...
4 Magnetomotive force and
magnetic field strength
Magnetomotive force (m
...
f
...
m
...
Fm = NI amperes
where N is the number of conductors (or turns) and I is
the current in amperes
...
m
...
However since ‘turns’
have no dimensions, the S
...
unit of m
...
f
...


Magnetic field strength (or magnetising force),
H=

NI
ampere per metre
l

where l is the mean length of the flux path in metres
...
m
...
= NI = Hl amperes
Problem 3
...
If the coil is uniformly wound
around the circuit and has 750 turns, find the
current in the coil
...
Since H = NI/l, then
I=

8000 × π × 30 × 10−2
Hl
=
N
750

Thus, current I = 10
...
What is the flux density in a magnetic field of
cross-sectional area 20 cm2 having a flux of
3 mWb?
[1
...
Determine the total flux emerging from a magnetic pole face having dimensions 5 cm by
6 cm, if the flux density is 0
...
7 mWb]
3
...
9 T and the effective area
of a pole face is circular in cross-section
...

[32 cm]
4
...
Calculate (a) the magnetomotive force, and (b) the magnetic field strength
...
An electromagnet of square cross-section produces a flux density of 0
...
If the magnetic

Section 1

Magnetic circuits 73

Section 1

74 Electrical and Electronic Principles and Technology
flux is 720 μWb find the dimensions of the
electromagnet cross-section
...
Find the magnetic field strength applied to a
magnetic circuit of mean length 50 cm when
a coil of 400 turns is applied to it carrying a
current of 1
...
A solenoid 20 cm long is wound with 500 turns
of wire
...

[1 A]
8
...
If the coil has 500 turns find
the current in the coil
...
85 A]

7
...
e
...
This constant is μ0 , the permeability of free space (or the magnetic space constant)
and is equal to 4π × 10−7 H/m, i
...
for air, or any
non-magnetic medium, the ratio
B
= μ0
H
(Although all non-magnetic materials, including air,
exhibit slight magnetic properties, these can effectively
be neglected
...
From its
definition, μr for a vacuum is 1
...
For non-magnetic materials
this is a straight line
...
7
...
3

The relative permeability of a ferromagnetic material is proportional to the slope of the B–H curve and
thus varies with the magnetic field strength
...
A flux density of 1
...
Find the relative permeability of the
steel under these conditions
...
e
...
2
=
= 764
μ0 H
(4π × 10−7 )(1250)

Problem 5
...
m
...
required to produce a flux density of
0
...


For air: B = μ0 H (since μr = 1)

(a) H =

Magnetic field strength,
H=

0
...
m
...
= Hl = 198 940 × 12 × 10−3 = 2387 A
Problem 6
...
The
ring has a mean circumference of 40 cm and a
uniform cross-sectional area of 4 cm2
...


Problem 8
...
Determine the m
...
f
...
3 mWb in the ring
...


300 × 5
NI
=
l
40 × 10−2

Flux density B =

= 3750 A/m
(b) For a non-magnetic material μr = 1, thus flux
density B = μ0 H
i
...
2 m and
0
...
3 T
10 × 10−4

From the magnetisation curve for cast iron on
page 74, when B = 0
...
m
...
= Hl = 1000 × 0
...
Such a solution is shown below in Table 7
...


B = 4π × 10−7 × 3750
Problem 9
...


= 4
...
4
μr =
=
= 200
μ0 H
(4π × 10−7 )(1592)

A = 10 cm2 = 10 × 10−4 m2 ,
= 0
...


(a) Magnetic field strength
H=

2000 × 0
...
712 × 10−3 )(4 × 10−4 )
= 1
...
An iron ring of mean diameter 10 cm
is uniformly wound with 2000 turns of wire
...
25 A is passed through the coil a flux
density of 0
...
Find (a) the
magnetising force and (b) the relative permeability
of the iron under these conditions
...
25 A and B = 0
...
2
...
7
...
The
curve demonstrates the change that occurs in the relative
permeability as the magnetising force increases
...
1
Part of
circuit

Material

(Wb)

Ring

Cast iron

0
...
3

A

(T)

H from
graph

l(m)

m
...
f
...
2

200

Section 1

Magnetic circuits 75

Section 1

76 Electrical and Electronic Principles and Technology
Table 7
...
04

0
...
17

0
...
41

0
...
60

0
...
73

0
...
79

H(A/m)

200

400

500

1000

1500

2000

3000

4000

5000

6000

7000

159

259

271

239

218

195

159

135

116

101

90

μr =

107 B
×
4π
H

mr B
300 1
...
Find the
flux density now in the solenoid
...
05 mT (b) 1
...
0
B

200 0
...
6
100 0
...
2

0

1000 2000 3000 4000 5000
H (A/m)

6000 7000

Figure 7
...
Find the magnetic field strength and the magnetomotive force needed to produce a flux
density of 0
...

[(a) 262 600 A/m (b) 3939 A]
2
...
If the flux required in the air-gap
is 0
...
m
...
necessary
...
(a) Determine the flux density produced in an
air-cored solenoid due to a uniform magnetic
field strength of 8000 A/m (b) Iron having a
relative permeability of 150 at 8000 A/m is

4
...
084 × 10−4 H/m
...
Find the relative permeability of a piece of silicon iron if a flux density of 1
...

[1478]
6
...
When
a current of 0
...
5 T is set up in the steel
...

[1000]
7
...
Find the current required in a
coil of 1200 turns wound on the ring to produce
a flux of 0
...
(Use the magnetisation
curve for cast steel shown on page 74)
[0
...
(a) A uniform mild steel ring has a diameter
of 50 mm and a cross-sectional area of 1 cm2
...
m
...
necessary to produce a
flux of 50 μWb in the ring
...
25 A]
9
...

From your graph determine (a) the value of μr
when the magnetic field strength is 1200 A/m,
and (b) the value of the magnetic field strength
when μr is 500
...
6

Reluctance

Reluctance S (or RM ) is the ‘magnetic resistance’ of
a magnetic circuit to the presence of magnetic flux
...

Ferromagnetic materials have a low reluctance and
can be used as magnetic screens to prevent magnetic
fields affecting materials within the screen
...
Determine the reluctance of a piece
of mumetal of length 150 mm and cross-sectional
area 1800 mm2 when the relative permeability is
4000
...

Reluctance,
S=

l
μ0 μr A

150 × 10−3
(4π × 10−7 )(4000)(1800 × 10−6 )
= 16 580/H

=

Absolute permeability,
μ = μ0 μr = (4π × 10−7 )(4000)
= 5
...
A mild steel ring has a radius of
50 mm and a cross-sectional area of 400 mm2
...
5 A flows in a coil wound uniformly
around the ring and the flux produced is 0
...

If the relative permeability at this value of current is
200 find (a) the reluctance of the mild steel and
(b) the number of turns on the coil
...
5 A, = 0
...
125 × 106 /H

(b) S =

m
...
f
...
m
...
= S

i
...


NI = S

Hence, number of terms
N=

S
3
...
1 × 10−3
=
I
0
...
Part of a magnetic circuit is made from steel
of length 120 mm, cross sectional area 15 cm2
and relative permeability 800
...

[(a) 79 580 /H (b) 1 mH/m]
2
...
2 mm2
...
40 A flows in a coil
wound uniformly around the circuit and the
flux produced is 200 μWb
...

[(a) 466 000 /H (b) 233]

7
...

Problem 12
...
5 cm2
...
4 A
flows
...


Section 1

Magnetic circuits 77

Section 1

78 Electrical and Electronic Principles and Technology
From the B–H curve for silicon iron on page 74, when
B = 1
...
Hence the m
...
f
...
4 = 660 A

For the 6 cm long path:
Reluctance S1 =
=

l1
μ 0 μ r A1
6 × 10−2
(4π × 10−7 )(750)(1 × 10−4 )

= 6
...
5 × 10−4 )
= 4
...
e
...
4 T (This assumes no leakage or fringing
occurring)
...
4
=
= 1 114 000 A/m
μ0
4π × 10−7

Hence the m
...
f
...

Total m
...
f
...
6 mWb
= 660 + 2228 = 2888 A
...
3 at top of next page
...
366 + 4
...
61 × 105/H
m
...
f
...
e
...
m
...

NI
=
S
S

=

S=

Problem 14
...
5 shows a ring formed with
two different materials — cast steel and mild steel
...
4
= 7
...
61 × 105

Flux density in the 2 cm path,
B=

A

=

7
...
51 T
0
...
5

Problem 13
...
If
the mean length of the silicon iron path is 40 cm
calculate the magnetomotive force to produce a flux
of 0
...
The magnetisation curve for silicon is
shown on page 74
...
The total m
...
f
...
m
...
’s of each part
...
7 × 10−3
= 1
...
5 mm2

Find the total m
...
f
...
Determine also
the total circuit reluctance
...
3
Material

m
...
f
...
7 × 10−3

5 × 10−4

1
...
4

0
...
4

1
...
m
...

= Hl(A)

(Wb)

660

= 1 114 000

Table 7
...
0

1400

400 × 10−3

560

500 × 10−6

312
...
6

4800

300 × 10−3

1440

Total:

Part of
circuit

2000 A

(Wb)

A tabular solution is shown in Table 7
...

m
...
f
...
A section through a magnetic
circuit of uniform cross-sectional area 2 cm2 is
shown in Fig
...
6
...
The air gap is 1 mm wide and the
coil has 5000 turns
...
Determine the current in the
coil to produce a flux density of 0
...


Reluctance of core S1 =

l1
and
μ0 μr A1

since B = μ0 μr H, then μr =
S1 =
μ0
=

l1
B
μ0 H

B

...
8)(2 × 10−4 )

For the air gap:
Reluctance, S2 =

l2
μ 0 μr A2

=

l2
(since μr = 1 for air)
μ 0 A2

=

1 × 10−3
(4π × 10−7 )(2 × 10−4 )

= 3 979 000/H
Total circuit reluctance
S = S1 + S2 = 1 172 000 + 3 979 000

Figure 7
...
80 T, H = 750 A/m
(from page 74)
...
80 × 2 × 10−4 = 1
...
m
...


thus
m
...
f
...

[0
...
6 × 10−4 )
=
N
5000
= 0
...
A magnetic circuit of cross-sectional area
0
...
With a 100 turn coil
carrying 2 A, find the value of flux existing in
the circuit
...
195 mWb]
2
...
Determine
the m
...
f
...
8 mWb in the ring
...
(b) If a radial air
gap 1
...
m
...
now necessary to maintain the
same flux in the ring
...
7

5
...
5 cm2
...
m
...
required to cause a flux of
0
...
Find also
the total reluctance of the circuit
...

[550 A, 1
...
Figure 7
...
When each of the air gaps are 1
...
m
...
required to produce a
flux density of 0
...
Use the
B–H curves shown on page 74
...
A closed magnetic circuit made of silicon
iron consists of a 40 mm long path of crosssectional area 90 mm2 and a 15 mm long path
of cross-sectional area 70 mm2
...
39 A flows
...

[1
...
For the magnetic circuit shown in Fig
...
7 find
the current I in the coil needed to produce a
flux of 0
...
The silicon

Figure 7
...
8

Comparison between electrical
and magnetic quantities

Electrical circuit

Magnetic circuit

e
...
f
...
m
...
Fm (A)

current I (A)

flux

resistance R ( )

reluctance S (H−1 )

I=

E
R

R=

ρl
A

7
...
m
...

S

S=

l
μ 0 μr A

Hysteresis and hysteresis loss

Hysteresis loop
Let a ferromagnetic material which is completely
demagnetised, i
...
one in which B = H = 0 be subjected
to increasing values of magnetic field strength H and
the corresponding flux density B measured
...
7
...
At a particular value of H, shown as
Oy, it becomes difficult to increase the flux density any
further
...
Thus by is
the saturation flux density
...
When H is reduced
to zero, flux remains in the iron
...
7
...

When H is increased in the opposite direction, the flux
density decreases until, at a value shown as Od, the
flux density has been reduced to zero
...
e
...

Further increase of H in the reverse direction causes
the flux density to increase in the reverse direction until
saturation is reached, as shown by curve de
...

It is seen from Fig
...
9 that the flux density
changes lag behind the changes in the magnetic field
strength
...
The closed
figure bcdefgb is called the hysteresis loop (or the
B/H loop)
...
e
...
This energy appears as heat in the
specimen and is called the hysteresis loss
...

The area of a hysteresis loop varies with the type of
material
...

Figure 7
...


Figure 7
...
c
...
Thus a hysteresis loop
with a large area (as with hard steel) is often unsuitable
since the energy loss would be considerable
...


Section 1

Magnetic circuits 81

Section 1

82 Electrical and Electronic Principles and Technology
10
...

magnetic field strength
11
...
The value of the permeability of free space
is
...
What is a magnetisation curve?
14
...
and the unit
of reluctance is
...
Make a comparison between magnetic and
electrical quantities
16
...
10

Now try the following exercise
Exercise 36 Short answer questions on
magnetic circuits
1
...
What is a permanent magnet?

17
...
State the units of (a) remanence (b) coercive
force
19
...
Complete the statement: magnetic materials
have a
...
reluctance
21
...
Sketch the pattern of the magnetic field associated with a bar magnet
...

Now try the following exercise
4
...
The symbol for magnetic flux is
...

6
...
The symbol for magnetic flux density is
...

8
...
m
...
is
...
m
...
is the
...
Another name for the magnetising force is

...
and its unit is
...
The unit of magnetic flux density is the:
(a) weber
(b) weber per metre
(c) ampere per metre (d) tesla
2
...
The cross-sectional area of the core is:
(a) 0
...
02 m2
2
(c) 20 m
(d) 50 m2

3
...
2 T
(b) 2 T
(c) 20 T
(d) 20 mT
Questions 4 to 8 refer to the following data:
A coil of 100 turns is wound uniformly on a
wooden ring
...
The current in the coil is 1 A
...
The magnetomotive force is:
(a) 1 A
(b) 10 A
(c) 100 A
(d) 1000 A
5
...
85 × 10−10 T
−7 T
(c) 4π × 10
(d) 40π μT
7
...
04π μWb
(b) 0
...
85 μWb
(d) 4π μWb
8
...
5
(c)
×109 H−1
π

(b)

1000 H−1

(d)

108 −1
H
8
...
Which of the following statements is false?
(a) For non-magnetic materials reluctance
is high
(b) Energy loss due to hysteresis is greater
for harder magnetic materials than for
softer magnetic materials
(c) The remanence of a ferrous material is
measured in ampere/metre
(d) Absolute permeability is measured in
henrys per metre

10
...
The reluctance of the
circuit is 2 × 106 H
...
5 μWb
11
...
From the following list, match the magnetic quantities
with their equivalent electrical quantities
...
m
...

(d) flux
(e) m
...
f
...
The effect of an air gap in a magnetic circuit
is to:
(a) increase the reluctance
(b) reduce the flux density
(c) divide the flux
(d) reduce the magnetomotive force

13
...
With this arrangement, the magnets will:
(a) attract each other
(b) have no effect on each other
(c) repel each other
(d) lose their magnetism

Section 1

Magnetic circuits 83

Section 1

Revision test 2
This revision test covers the material contained in Chapters 5 to 7
...

1
...
If a 10 V supply voltage is connected
across the arrangement determine the current flowing through and the p
...
across the 7 resistor
...

(6)
2
...
R2
...
d
...
Assume electrical energy costs 14p
per unit
...

(7)
5
...
This arrangement is then connected
in series with a 10 μF capacitor
...
d
...
Find (a) the
equivalent capacitance of the circuit, (b) the voltage
across the 10 μF capacitor, and (c) the charge on
each capacitor
...
The charge on the plates of a capacitor is 8 mC when
the potential between them is 4 kV
...

(2)

6
...
The ring has a uniform
cross-sectional area of 200 mm2 and a mean circumference of 500 mm
...

(5)

4
...
48 μC
...
Calculate (a) the electric flux
density (b) the electric field strength, and (c) the

7
...
If the mean length
of the mild steel path is 300 mm, calculate the magnetomotive force to produce a flux of 0
...

(Use the B–H curve on page 74)
(8)

R1 = 2
...
1

R5 = 11 Ω

Chapter 8

Electromagnetism
At the end of this chapter you should be able to:
• understand that magnetic fields are produced by electric currents
• apply the screw rule to determine direction of magnetic field
• recognize that the magnetic field around a solenoid is similar to a magnet
• apply the screw rule or grip rule to a solenoid to determine magnetic field direction
• recognize and describe practical applications of an electromagnet, i
...
electric bell, relay, lifting magnet,
telephone receiver
• appreciate factors upon which the force F on a current-carrying conductor depends
• perform calculations using F = BIl and F = BIl sin θ
• recognize that a loudspeaker is a practical application of force F
• use Fleming’s left-hand rule to pre-determine direction of force in a current carrying conductor
• describe the principle of operation of a simple d
...
motor
• describe the principle of operation and construction of a moving coil instrument
• appreciate that force F on a charge in a magnetic field is given by F = QvB
• perform calculations using F = QvB

Wire

8
...

Let a piece of wire be arranged to pass vertically
through a horizontal sheet of cardboard on which is
placed some iron filings, as shown in Fig
...
1(a)
...
By placing a compass in different positions the lines of flux are
seen to have a definite direction as shown in Fig
...
1(b)
...
The effect on both the iron
filings and the compass needle disappears when the current is switched off
...
1

by the electric current
...
If the current is increased the strength of
the field increases and, as for the permanent magnet,

Section 1

86 Electrical and Electronic Principles and Technology
field round a straight conductor is in the form of concentric cylinders as shown in Fig
...
2, the field direction
depending on the direction of the current flow
...
8
...

In Fig
...
1, the effect of only a small part of the magnetic field is shown
...
e
...
This may be thought of
as the feathered end of the shaft of an arrow
...
8
...


Direction of
current flow

(ii) Current flowing towards the viewer, i
...
out of the
paper, is indicated by
...
See Fig
...
3(b)
...

For example, with current flowing away from the
viewer (Fig
...
3(a)) a right-hand thread screw driven
into the paper has to be rotated clockwise
...

A magnetic field set up by a long coil, or solenoid, is
shown in Fig
...
4(a) and is seen to be similar to that of a
bar magnet
...
8
...
The direction of the magnetic
field produced by the current I in the solenoid may be
found by either of two methods, i
...
the screw rule or
the grip rule
...
2

(a) Current flowing away
from viewer

(b) Current flowing
towards viewer

(a) The screw rule states that if a normal right-hand
thread screw is placed along the axis of the solenoid

Figure 8
...
4

S

N

l

ϩ

Ϫ

(b) Magnetic field of an iron cored solenoid

and is screwed in the direction of the current
it moves in the direction of the magnetic field
inside the solenoid
...

Thus in Figs
...

(b) The grip rule states that if the coil is gripped
with the right hand, with the fingers pointing in
the direction of the current, then the thumb, outstretched parallel to the axis of the solenoid, points
in the direction of the magnetic field inside the
solenoid
...
2

Electromagnets

The solenoid is very important in electromagnetic theory since the magnetic field inside the solenoid is
practically uniform for a particular current, and is
also versatile, inasmuch that a variation of the current
can alter the strength of the magnetic field
...


(i) Electric bell
Problem 1
...
5 shows a coil of wire wound
on an iron core connected to a battery
...


There are various types of electric bell, including the
single-stroke bell, the trembler bell, the buzzer and a
continuously ringing bell, but all depend on the attraction exerted by an electromagnet on a soft iron armature
...
8
...
Since the iron-cored coil is energised
the soft iron armature is attracted to the electromagnet
...

When the circuit is broken the coil becomes demagnetised and the spring steel strip pulls the armature back to
its original position
...


Figure 8
...
8
...
8
...
The polarity of the field
is determined either by the screw rule or by the grip rule
...


Electromagnet

Soft iron
armature

Striker
S
Gong

Figure 8
...
6

A relay is similar to an electric bell except that contacts
are opened or closed by operation instead of a gong
being struck
...
8
...
When
the coil is energised the hinged soft iron armature is

Section 1

Electromagnetism 87

88 Electrical and Electronic Principles and Technology

Section 1

To electrical
circuit

(iv) Telephone receiver
Fixed
contacts

Electromagnet

Iron
core
Hinged
armature

Supply to coil

Whereas a transmitter or microphone changes sound
waves into corresponding electrical signals, a telephone
receiver converts the electrical waves back into sound
waves
...
8
...
A thin, flexible diaphragm of magnetic
material is held in position near to the magnetic poles
but not touching them
...
The vibration produces sound
variations corresponding to those transmitted
...
8
Diaphragm

attracted to the electromagnet and pushes against two
fixed contacts so that they are connected together, thus
closing some other electrical circuit
...
A
typical robust lifting magnet, capable of exerting large
attractive forces, is shown in the elevation and plan view
of Fig
...
9 where a coil, C, is wound round a central core,
P, of the iron casting
...

The load, Q, which must be of magnetic material is lifted
when the coils are energised, the magnetic flux paths,
M, being shown by the broken lines
...
10

8
...
The force on the current-carrying conductor
in a magnetic field depends upon:
(a) the flux density of the field, B teslas
(b) the strength of the current, I amperes,
(c) the length of the conductor perpendicular to the
magnetic field, l metres, and
(d) the directions of the field and the current
...
9

When the conductor and the field are at an angle θ ◦ to
each other then:
Force F = BIl sin θ newtons

Since when the magnetic field, current and conductor
are mutually at right angles, F = BIl, the magnetic flux
density B may be defined by B = (F)/(Il), i
...
the flux
density is 1 T if the force exerted on 1 m of a conductor
when the conductor carries a current of 1 A is 1 N
...
12

Loudspeaker
A simple application of the above force is the moving
coil loudspeaker
...

Figure 8
...
A moving coil,
called the voice or speech coil, is suspended from the
end of a paper or plastic cone so that it lies in the gap
...

The cone acts as a piston, transferring this force to the
air, and producing the required sound waves
...
8
...
9)(20)(0
...
e
...
7 N
If the current-carrying conductor shown in Fig
...
3 (a) is
placed in the magnetic field shown in Fig
...
13(a), then
the two fields interact and cause a force to be exerted
on the conductor as shown in Fig
...
13(b) The field is
strengthened above the conductor and weakened below,
thus tending to move the conductor downwards
...
4) and the moving-coil instrument (see
Section 8
...


Soft iron pole pieces

N

S

Voice coil
Cone

(a)

S

Permanent
magnet

N

N
S

S

Direction of motion
of conductor
(b)

Figure 8
...
11

Problem 2
...
9 T
...
Determine also the value of the force
if the conductor is inclined at an angle of 30◦ to the
direction of the field
...
9 T, I = 20 A and l = 30 cm = 0
...
9)(20)(0
...
8
...
e
...
4 N
...
8
...

Summarising:
First finger – Field
SeCond finger – Current
ThuMb – Motion

Section 1

Electromagnetism 89

90 Electrical and Electronic Principles and Technology
Motion

Section 1

Force F = BIl,
Magnetic
field

Figure 8
...
Determine the current required in a
400 mm length of conductor of an electric motor,
when the conductor is situated at right-angles to a
magnetic field of flux density 1
...
92 N is to be exerted on the conductor
...
92 N, l = 400 mm = 0
...
2 T
...
92
= 4A
(1
...
4)

B=

A

hence

force F =

Il
A
(0
...
35) newtons
π(0
...
155 N

i
...


Current

and

Problem 5
...
8
...
8
...
8
...
8
...
8
...

S
N

S

S

N

N
(a)

(b)

(c)

(d)

Figure 8
...
e
...
The current is
flowing towards the viewer, and using the screw
rule, the direction of the field is anticlockwise
...
8
...


If the current flows downwards, the direction of its
magnetic field due to the current alone will be clockwise when viewed from above
...
e
...
e
...
Hence the force on
the conductor will be from back to front (i
...
toward
the viewer)
...


(b) Using a similar method to part (a) it is seen that
the force on the conductor is to the right — see
Fig
...
16(b)
...
A conductor 350 mm long carries a
current of 10 A and is at right-angles to a magnetic
field lying between two circular pole faces each of
radius 60 mm
...
5 mWb, calculate the magnitude of the
force exerted on the conductor
...
A coil is wound on a rectangular
former of width 24 mm and length 30 mm
...
8 T, the
axis being perpendicular to the field
...


l = 350 mm = 0
...
06)2 m2 and = 0
...
5×10−3 Wb

(c) Using Fleming’s left-hand rule, or by sketching as
in Fig
...
16(c), it is seen that the current is toward
the viewer, i
...
out of the paper
...
8
...


conductor is situated at right-angles to the
magnetic field of flux density 1
...
20 N is to be exerted on the conductor
...
0 A]
3
...
Calculate the flux
density of the magnetic field if a current of
15 A in the conductor produces a force on it of
3
...

[0
...
A conductor 300 mm long carries a current of
13 A and is at right-angles to a magnetic field
between two circular pole faces, each of diameter 80 mm
...
75 mWb calculate the force exerted
on the conductor
...
582 N]

Figure 8
...
8 T, length of conductor lying
at right-angles to field l = 30 mm = 30 × 10−3 m
and current I = 50 mA = 50 × 10−3 A
...
8 × 50 × 10−3 × 30 × 10−3
= 1
...
0012 N
(b) When there are 300 turns on the coil there are
effectively 300 parallel conductors each carrying
a current of 50 mA
...
Hence force on coil side,
F = 300 BIl = 300 × 0
...
36 N
Now try the following exercise
Exercise 38 Further problems on the force
on a current-carrying
conductor
1
...
5 T
...
What is the force when the
conductor and field are at an angle of 45◦ ?
[21
...
8 N]
2
...
c
...
(a) A 400 mm length of conductor carrying
a current of 25 A is situated at right-angles
to a magnetic field between two poles of an
electric motor
...
If the force exerted on the conductor
is 80 N and the total flux between the pole
faces is 1
...

(b) If the conductor in part (a) is vertical, the
current flowing downwards and the direction
of the magnetic field is from left to right, what
is the direction of the 80 N force?
[(a) 14
...
A coil is wound uniformly on a former having a width of 18 mm and a length of 25 mm
...
75 T, the axis being perpendicular to
the field
...

[(a) 2
...
9 N]

8
...
c
...
8
...
When current flows in the coil
a magnetic field is set up around the coil which interacts
with the magnetic field produced by the magnets
...
This causes
a torque and the coil rotates anticlockwise
...
8
...
If the current
is not reversed and the coil rotates past this position the
forces acting on it change direction and it rotates in the
opposite direction thus never making more than half a
revolution
...
17

Figure 8
...
This is the principle of operation of a d
...
motor
which is thus a device that takes in electrical energy and
converts it into mechanical energy
...
5

Principle of operation of a
moving-coil instrument

A moving-coil instrument operates on the motor principle
...
If the flux density B is made constant
(by using permanent magnets) and the conductor is a
fixed length (say, a coil) then the force will depend only
on the current flowing in the conductor
...
8
...
(The air-gap is kept
as small as possible, although for clarity it is shown
exaggerated in Fig
...
18) The coil is supported by steel
pivots, resting in jewel bearings, on a cylindrical iron
core
...
e
...
Current flowing in the coil
produces forces as shown in Fig
...
18(b), the directions
being obtained by Fleming’s left-hand rule
...
e
...
Since force is proportional to current
the scale is linear
...
The
moving-coil instrument will measure only direct current or voltage and the terminals are marked positive
and negative to ensure that the current passes through
the coil in the correct direction to deflect the pointer ‘up
the scale’
...


8
...
An electron in a television tube has a
charge of 1
...
5 μT
...


From above, force F = QvB newtons, where
Q = charge in coulombs = 1
...
5 × 10−6 T
...
6 × 10−19 × 3 × 107 × 18
...
6 × 3 × 18
...
8 × 10−18 = 8
...
Determine the speed of a 10−19 C charge travelling perpendicular to a field of flux density
10−7 T, if the force on the charge is 10−20 N
[106 m/s]

Exercise 40 Short answer questions on electromagnetism
1
...
rule
...
Sketch the magnetic field pattern associated
with a solenoid connected to a battery and
wound on an iron bar
...

3
...

4
...

5
...
Name
them
...
The direction of the force on a conductor in
a magnetic field may be predetermined using
Fleming’s
...

7
...

8
...
19 shows a simplified diagram of
a section through the coil of a moving-coil
instrument
...


Now try the following exercises
Exercise 39 Further problems on the force
on a charge
1
...
19

Section 1

Electromagnetism 93

Section 1

94 Electrical and Electronic Principles and Technology
9
...
c
...

10
...
Briefly explain the principle
of operation of such an instrument
...
A conductor carries a current of 10 A at
right-angles to a magnetic field having a flux
density of 500 mT
...
Figure 8
...
If the current flows into the
coil at C, the coil will:
(a) commence to rotate anti-clockwise
(b) commence to rotate clockwise
(c) remain in the vertical position
(d) experience a force towards the
north pole

2
...
21

3
...
8
...
The force on an electron travelling at 107 m/s
in a magnetic field of density 10 μT is
1
...
The electron has a charge of:
(a) 1
...
6 × 10−15 C
−19 C
(c) 1
...
6 × 10−25 C
7
...
20

4
...
8
...
A relay can be used to:
(a) decrease the current in a circuit
(b) control a circuit more readily
(c) increase the current in a circuit
(d) control a circuit from a distance

9
...
The magnetic field due to a current-carrying
conductor takes the form of:
(a) rectangles
(b) concentric circles
(c) wavy lines
(d) straight lines radiating outwards

Section 1

Electromagnetism 95

Chapter 9

Electromagnetic induction
At the end of this chapter you should be able to:
• understand how an e
...
f
...
m
...
, E = Blv or E = Blv sin θ
• calculate induced e
...
f
...
m
...
given N, t, L, change of flux or change of current
• appreciate factors which affect the inductance of an inductor
• draw the circuit diagram symbols for inductors
1
• calculate the energy stored in an inductor using W = 2 LI 2 joules

N
N2
and L =
I
S
dI1
N1 N2
• calculate mutual inductance using E2 = −M
and M =
dt
S

• calculate inductance L of a coil, given L =

9
...
m
...
) is produced in the conductor
...
m
...

produced causes an electric current to flow round the circuit
...
m
...
(and thus current) is ‘induced’ in

the conductor as a result of its movement across the magnetic field
...

Figure 9
...

(a) When the magnet is moved at constant speed
towards the coil (Fig
...
1(a)), a deflection is noted

S

N

9
...
m
...
is set up whenever the magnetic
field linking that circuit changes
...
m
...
in any circuit is proportional to the rate of change of the
magnetic flux linking the circuit
...
1

on the galvanometer showing that a current has
been produced in the coil
...
9
...

(d) When the coil is moved at the same speed as
in (a) and the magnet held stationary the same
galvanometer deflection is noted
...
m
...
is always such that
it tends to set up a current opposing the motion or the
change of flux responsible for inducing that e
...
f
...
9
...
If the first
finger points in the direction of the magnetic field and the
thumb points in the direction of motion of the conductor
relative to the magnetic field, then the second finger will
point in the direction of the induced e
...
f
...
m
...


(e) When the relative speed is, say, doubled, the
galvanometer deflection is doubled
...

(g) When the number of turns of wire of the coil is
increased, a greater gal vanometer deflection is
noted
...
m
...
1(c) shows the magnetic field associated
with the magnet
...
It is the relative movement of the
magnetic flux and the coil that causes an e
...
f
...
This effect is
known as electromagnetic induction
...
2 evolved from
experiments such as those described above
...
2

Section 1

Electromagnetic induction 97

Section 1

98 Electrical and Electronic Principles and Technology
In a generator, conductors forming an electric circuit
are made to move through a magnetic field
...
m
...
is induced in the conductors and
thus a source of e
...
f
...
A generator converts
mechanical energy into electrical energy
...
c
...

The induced e
...
f
...
9
...
5
E
=
= 0
...
At what velocity must a conductor
75 mm long cut a magnetic field of flux density
0
...
m
...
of 9 V is to be induced in it?
Assume the conductor, the field and the direction of
motion are mutually perpendicular
...
m
...
E = Blv, hence velocity v = E/Bl
Thus
v=

N
l

v

9 × 103
0
...
3

where B, the flux density, is measured in teslas, l, the
length of conductor in the magnetic field, is measured
in metres, and v, the conductor velocity, is measured in
metres per second
...
A conductor 300 mm long moves at a
uniform speed of 4 m/s at right-angles to a uniform
magnetic field of flux density 1
...
Determine the
current flowing in the conductor when (a) its ends
are open-circuited, (b) its ends are connected to a
load of 20 resistance
...
6)(75 × 10−3 )

Problem 3
...
If the flux leaving a pole
face is 5 μWb, find the magnitude of the induced
e
...
f
...

v = 15 m/s, length of conductor in magnetic field,
l = 2 cm = 0
...
02)(15)(1)
=
4 × 10−4
= 3
...
m
...
induced in it but this e
...
f
...
Induced e
...
f
...
25)

300
(4) = 1
...
5 V has been
induced
...
75 sin 60◦ = 3
...
75 sin 30◦ = 1
...
The wing span of a metal aeroplane is
36 m
...
m
...
induced between its wing tips
...


Induced e
...
f
...
e
...
m
...

i
...
towards the viewer or out of the paper, as
shown in Fig
...
5(b)

(400)(1000)
3600
4000
=
m/s
36

=

Hence
E = Blv = (40 × 10−6 )(36)

4000
36

= 0
...
The diagrams shown in Fig
...
4
represents the generation of e
...
f’s
...
9
...
m
...
in Fig
...
4(b), (iii) the polarity of the
magnetic system in Fig
...
4(c)

N
S
S

N
(b)

(a)

(c)

Figure 9
...
5

The direction of the e
...
f
...
m
...
may be obtained by either Lenz’s law or
Fleming’s Right-hand rule (i
...
GeneRator rule)
...
9
...
Hence the force on
the conductor is to the right
...
m
...
is
always such as to oppose the effect producing it
...

(ii) Using Fleming’s right-hand rule:
First finger – Field,
i
...
N → S, or right to left;

(iii) The polarity of the magnetic system of Fig
...
4(c)
is shown in Fig
...
5(c) and is obtained using
Fleming’s right-hand rule
...
m
...

1
...
2 T
...
m
...
induced in
the conductor
...
135 V]
2
...
6 T to induce
in it an e
...
f
...
8 V
[25 m/s]
3
...
2 T
...

[(a) 0 (b) 0
...
A straight conductor 500 mm long is moved
with constant velocity at right angles both
to its length and to a uniform magnetic field
...
m
...
induced in the conductor
is 2
...
If the
conductor forms part of a closed circuit of
total resistance 5 ohms, calculate the force on
the conductor
...
25 N]
5
...
Assuming the
back axle of the car is 1
...
m
...
generated in the
axle due to motion
...
56 mV]
6
...
5 cm
...
m
...
in each case
...
9 V (c) 24 V]
7
...
85 T magnetic field
...

[(a) 10
...
408 N]

l
N

S
+

−

Figure 9
...

The induced e
...
f
...

Therefore the total e
...
f
...
m
...
E for the loop conductor is now given by:
E = 2NBlv sin θ
Problem 6
...
4 T, the longer side of the coil actually cutting this
flux
...
(a) Calculate the maximum
generated e
...
f
...
m
...
E = 2NBLv sin θ
where number of turns, N = 80, flux density,
B = 1
...
12 m,
velocity, v = ωr =

9
...
6 shows a view of a looped conductor whose
sides are moving across a magnetic field
...
By definition, the induced e
...
f
...


1200
× 2π rad/s
60

0
...
6π m/s,
and for maximum e
...
f
...
m
...
induced,
E = 2NBlv sin θ
= 2 × 80 × 1
...
12 × 1
...
1 volts
(b) Since

E = 2NBlv sin θ

90 = 2 × 80 × 1
...
12 × v × 1
90
from which, v =
= 3
...
4 × 0
...
348
v = ωr hence, angular velocity, ω = =
0
...
7 rad/s
83
...

Since E = 2NBlv sin θ, then with N, B, l and θ
being constant, E ∝ v
If from (a), 135
...
88 rev/min
135
...
88 = 799 rev/min
Now try the following exercise
Exercise 43 Further problems on induced
e
...
f
...
A rectangular coil of sides 8 cm by 6 cm is rotating in a magnetic field such that the longer sides
cut the magnetic field
...
m
...
if there are 60 turns on the coil,
the flux density is 1
...

[72
...
A generating coil on a former 100 mm long has
120 turns and rotates in a 1
...

Calculate the maximum e
...
f
...

[47
...
If the coils in problems 1 and 2 generates 60 V,
calculate (a) the new speed for each coil, and
(b) the flux density required if the speed is
unchanged
...
33 T, 1
...
4

Inductance

Inductance is the name given to the property of a circuit
whereby there is an e
...
f
...

When the e
...
f
...
m
...
is induced
in a circuit by a change of flux due to current changing
in an adjacent circuit, the property is called mutual
inductance, M
...

A circuit has an inductance of one henry when an
e
...
f
...
m
...
in a coil of N turns,
d
E = −N
volts
dt
where d is the change in flux in Webers, and dt is the
time taken for the flux to change in seconds (i
...
ddt is
the rate of change of flux)
...
m
...
in a coil of inductance L henrys,
dI
E = −L volts
dt
where dI is the change in current in amperes and dt is the
time taken for the current to change in seconds (i
...
dI
dt
is the rate of change of current)
...
Determine the e
...
f
...


Induced e
...
f
...
A flux of 400 μWb passing through a
150-turn coil is reversed in 40 ms
...
m
...
induced
...

Induced e
...
f
...
m
...
induced, E = −3 volts
Problem 9
...
m
...
induced in a coil
of inductance 12 H by a current changing at the rate
of 4 A/s
...
m
...
E = −L

dI
= −(12)(4)
dt
= −48 volts

Problem 10
...
m
...
of 1
...
Determine the inductance of the coil
...
5 kV = 1500 V
|E| = L

inductance, L =

dI
dt

Now try the following exercise
Exercise 44 Further problems on
inductance
1
...
m
...
induced in a coil of 200 turns
when there is a change of flux of 30 mWb
linking with it in 40 ms
...
An e
...
f
...
Find the time, in milliseconds, in
which the flux makes the change
...
An ignition coil having 10 000 turns has an
e
...
f
...
What rate of
change of flux is required for this to happen?
[0
...
A flux of 0
...
Find the magnitude of the average e
...
f
...

[3
...
Calculate the e
...
f
...
An average e
...
f
...
Calculate the time taken for the
current to reverse
...
15 H and change in current, dI = 6 − (−6) = 12 A (since the current is
reversed)
...
15)(12)
=
|E|
40
= 0
...
5 Inductors
A component called an inductor is used when the property of inductance is required in a circuit
...
Factors which
affect the inductance of an inductor include:
(i) the number of turns of wire — the more turns the
higher the inductance
(ii) the cross-sectional area of the coil of wire —
the greater the cross-sectional area the higher the
inductance
(iii) the presence of a magnetic core — when the coil
is wound on an iron core the same current sets
up a more concentrated magnetic field and the
inductance is increased
(iv) the way the turns are arranged — a short thick
coil of wire has a higher inductance than a long
thin one
...
9
...
9
...


1
W = LI2 joules
2
Problem 12
...
How much energy is stored in
the magnetic field of the inductor?

Energy stored,
Wire

Coil of
wire

(a)

1
1
W = LI 2 = (8)(3)2 = 36 joules
2
2

(b)

Figure 9
...
8

An iron-cored inductor is often called a choke since,
when used in a
...
circuits, it has a choking effect,
limiting the current flowing through it
...
To reduce
inductance to a minimum the wire may be bent back on
itself, as shown in Fig
...
9, so that the magnetising effect
of one conductor is neutralised by that of the adjacent
conductor
...
Standard
resistors may be non-inductively wound in this manner
...
An inductor of 20 H has a current of 2
...
Find the energy stored in the magnetic
field of the inductor
...
5 J]
2
...
18 mJ]
3
...
Calculate the inductance
of the coil
...
7

Inductance of a coil

If a current changing from 0 to I amperes, produces
a flux change from 0 to
webers, then dI = I and
d =
...
3,
Figure 9
...
6 Energy stored
An inductor possesses an ability to store energy
...
m
...
E =

N
t

from which, inductance of coil,
L=

N
henrys
I

=

LI
t

Section 1

104 Electrical and Electronic Principles and Technology
Since E = −L

d
dI
= −N
dt
dt

then L = N

d
dt

dt
dI

(Alternatively,
E = −N

d
i
...
L = N
dI
From chapter 7, m
...
f
...
m
...

S

i
...


L=

N d(N I)
since m
...
f
...
e
...
m
...
When a current of 1
...
If the
coil inductance is 0
...

For a coil, L =

N
I

N=

LI

Thus

Problem 13
...


=

(0
...
5)
= 10 000 turns
90 × 10−6

Problem 16
...
Calculate the flux linking
the coil and the e
...
f
...


For a coil, inductance
N
L=
I

=

(800)(5 × 10−3 )
4

Coil inductance, L =
=1H
flux

Problem 14
...
Calculate (a) the inductance of the coil,
(b) the energy stored in the magnetic field, and
(c) the average e
...
f
...


=

N
I

from which,

LI
(3)(2)
=
= 8 × 10−3 = 8 mWb
N
750

Induced e
...
f
...
5 H
I
3
(b) Energy stored,
1
1
W = LI 2 = (12
...
25 J
2
2
(c) Induced e
...
f
...
5)
dt
150 × 10−3
= −250 V

E = −N

d
8 × 10−3
= −(750)
dt
20 × 10−3
= −300 V)

Problem 17
...
If
when carrying a current of 0
...
m
...
if
the current is reduced to zero in 80 ms
...
9
...

c
...
a = 400 mm2

=
d mm
0
12

I = 0
...
Calculate the coil inductance when a current
of 5 A in a coil of 1000 turns produces a flux
of 8 mWb linking with the coil
...
6 H]
6
...
5 H
...
8 A]

Figure 9
...
e
...
A coil of 2500 turns has a flux of 10 mWb
linking with it when carrying a current of 2 A
...
m
...

induced in the coil when the current collapses
to zero in 20 ms
...
5 H, 1
...
56 H
dI
(0
...
m
...
, E = −L = −(2
...
A flux of 30 mWb links with a 1200 turn
coil when a current of 5 A is passing through
the coil
...
m
...
induced if
the current is reduced to zero in 0
...
2 H (b) 90 J (c) 180 V]
2
...
m
...
of 2 kV is induced in a coil when a
current of 5 A collapses uniformly to zero in
10 ms
...

[4 H]
3
...
m
...
of 60 V is induced in a coil
of inductance 160 mH when a current of 7
...
Calculate the time taken for the
current to reverse
...
When a current of 2 A flows in a coil, the flux
linking with the coil is 80 μWb
...
5 H, calculate the number of
turns of the coil
...
A coil of 1200 turns has a flux of 15 mWb
linking with it when carrying a current of 4 A
...
m
...

induced in the coil when the current collapses
to zero in 25 ms
[4
...
A coil has 300 turns and an inductance of
4
...
How many turns would be needed to
produce a 0
...
A steady current of 5 A when flowing in a
coil of 1000 turns produces a magnetic flux
of 500 μWb
...
The current of 5 A is then reversed in
12
...
Calculate the e
...
f
...

[0
...
An iron ring has a cross-sectional area of
500 mm2 and a mean length of 300 mm
...
Calculate (a) the current
required to set up a flux of 500 μWb in the
coil, (b) the inductance of the system, and
(c) the induced e
...
f
...

[(a) 1
...
51 mH (c) −50 V]

9
...
m
...
in the second coil,
E2 = −M

dI1
volts
dt

Section 1

Electromagnetic induction 105

Section 1

106 Electrical and Electronic Principles and Technology
where M is the mutual inductance between two coils,
in henrys, and (dI1 /dt) is the rate of change of current
in the first coil
...

If the fluxes 1 and 2 are produced from currents I1
and I2 in coils A and B respectively, then the reluctance
could be expressed as:
S=

I1 N1
1

=

I2 N2
2

If the flux in coils A and B are the same and produced from the current I1 in coil A only, assuming
100% coupling, then the mutual inductance can be
expressed as:
M=
Multiplying by

N1
N1

S=

Thus, mutual inductance, M =

Problem 20
...
2 H
...
m
...
in the second coil, (b) the change of
flux linked with the second coil if it is wound with
500 turns
...
m
...


1

N2 1 N1
I1 N 1

dI1
dt
= −(0
...
m
...
|E2 | = MdI1 /dt, i
...
1
...
Thus
mutual inductance,

10 − 4
10 × 10−3

= −120 V

(b) Induced e
...
f
...
Calculate the mutual inductance
between two coils when a current changing at
200 A/s in one coil induces an e
...
f
...
5 V in the
other
...
72
=
=
= 40 A/s
dt
M
0
...
m
...
|E2 | = M

|E2 |dt
d
, hence d =
dt
N

Thus the change of flux,
d

=

(120)(10 × 10−3 )
= 2
...
In the device shown in Fig
...
11,
when the current in the primary coil of 1000 turns
increases linearly from 1 A to 6 A in 200 ms, an
e
...
f
...
Determine
(a) the mutual inductance of the two coils, (b) the
reluctance of the former, and (c) the self-inductance
of the primary coil
...
5
= 0
...
5 mH
200
NP =1000

Problem 19
...
Calculate the steady rate of change
of current in one coil to induce an e
...
f
...
72 V
in the other
...
11

NS = 480

(a) ES = M

dIp
from which,
dt

mutual inductance, M =

ES
=
dIP
dt

15
6−1
200 × 10−3

15
=
= 0
...
60

= 800 000 A/Wb or 800 kA/Wb
(c) Primary self-inductance, LP =

2
NP (1000)2
=
S
800 000

is reversed in 0
...
m
...
induced in the other coil, (b) the number of turns on the other coil if the flux change
linking with the other coil is 5 mWb
[(a) −0
...
When the current in the primary coil of 400
turns of a magnetic circuit increases linearly
from 10 mA to 35 mA in 100 ms, an e
...
f
...

Determine (a) the mutual inductance of the
two coils, (b) the reluctance of the former,
and (c) the self-inductance of the secondary
coil
[(a) 0
...
18 H]

= 1
...
The mutual inductance between two coils is
150 mH
...
m
...

induced in one coil when the current in the
other is increasing at a rate of 30 A/s
...
5 V]
2
...
m
...
of 80 mV in the other
...
6 mH]
3
...
75 H
...
m
...
induced
in one coil when a current of 2
...
The mutual inductance between two coils is
240 mH
...
m
...
induced in the other coil, (b) the change
of flux linked with the other coil if it is wound
with 400 turns
...
4 mWb]
5
...
06 H exists between
two coils
...
What is electromagnetic induction?
2
...
State Lenz’s law
4
...
The direction of an induced e
...
f
...
rule
6
...
m
...
E induced in a moving conductor
may be calculated using the formula E = Blv
...
The total e
...
f
...

8
...
State and define the unit of inductance
10
...
m
...
E is given by E =
...
If a current of I amperes flowing in a coil of
N turns produces a flux of webers, the coil
inductance L is given by L =
...
The energy W stored by an inductor is given
by W =
...
If the number of turns of a coil is N and its
reluctance is S, then the inductance, L, is
given by: L =
...
What is mutual inductance? State its symbol
15
...
The e
...
f
...
volts
16
...

The mutual inductance, M, is given by:
M =
...
A current changing at a rate of 5 A/s in a coil
of inductance 5 H induces an e
...
f
...
A bar magnet is moved at a steady speed
of 1
...
The magnet is now withdrawn along the
same path at 0
...
The deflection of the
galvanometer is in the:
(a) same direction as previously, with the
magnitude of the deflection doubled
(b) opposite direction as previously, with
the magnitude of the deflection halved

(c) same direction as previously, with the
magnitude of the deflection halved
(d) opposite direction as previously,
with the magnitude of the deflection
doubled
3
...
m
...
is:
(a) 1 V
(b) 4 V
(c) 100 V
(d) 400 V
4
...

The coil inductance is:
(a) 106 H
(b) 1 H
(d) 1 mH
(c) 1 μH
5
...
m
...
of 1 V is induced in a conductor
moving at 10 cm/s in a magnetic field of 0
...

The effective length of the conductor in the
magnetic field is:
(a) 20 cm
(b) 5 m
(c) 20 m
(d) 50 m
6
...
m
...

(b) An induced e
...
f
...
m
...
is
always such as to oppose the effect
producing it
(d) The induced e
...
f
...
The effect of inductance occurs in an electrical circuit when:
(a) the resistance is changing
(b) the flux is changing
(c) the current is changing
8
...
The mutual inductance between two coils,
when a current changing at 20 A/s in one coil
induces an e
...
f
...
5 H
(b) 200 mH
(c) 0
...
A strong permanent magnet is plunged into
a coil and left in the coil
...
Self-inductance occurs when:
(a) the current is changing
(b) the circuit is changing
(c) the flux is changing
(d) the resistance is changing
12
...
m
...
of a chemical cell
(b) the e
...
f
...
c
...
c
...
c
...
c
...
In order to detect electrical quantities
such as current, voltage, resistance or power, it is necessary to transform an electrical quantity or condition into
a visible indication
...

The digital instrument has, in the main, become
the instrument of choice in recent years; in particular, computer-based instruments are rapidly replacing
items of conventional test equipment, with the virtual
storage test instrument, the digital storage oscilloscope, being the most common
...


10
...
A mechanical
force is produced by the current or voltage which
causes the pointer to deflect from its zero position
...
1

Figure 10
...
The ammeter shown has a f
...
d
...


10
...
10
...
When
current flows in the solenoid, a pivoted soft-iron
disc is attracted towards the solenoid and the
movement causes a pointer to move across a scale
...
10
...
When current passes through
the solenoid, the two pieces of iron are magnetized in the same direction and therefore repel each
other
...
The
force moving the pointer is, in each type, proportional to I 2 and because of this the direction of

(b) A controlling device
...
It also
prevents the pointer always going to the maximum
deflection
...

(c) A damping device
...
There are
three main types of damping used — eddy-current
damping, air-friction damping and fluid-friction
damping
...
A linear scale is shown in Fig
...
1(a),
where the divisions or graduations are evenly spaced
...
e
...
s
...
) of 100 V
...
10
...
2

Section 1

112 Electrical and Electronic Principles and Technology
current does not matter
...
c
...
c
...


10
...
c
...
10
...
The average
value of the full wave rectified current is 0
...

However, a meter being used to measure a
...
is usually calibrated in r
...
s
...
For sinusoidal quantities
the indication is (0
...
637Im ) i
...
1
...
Rectifier instruments have scales calibrated
in r
...
s
...
c
...


There is no difference between the basic instrument
used to measure current and voltage since both use a
milliammeter as their basic part
...
s
...
for currents of only a few
milliamperes
...
Such a diverting resistor is called
a shunt
...
10
...

Hence Ia ra = Is Rs
...
10
...
From
Fig
...
4(b),
V = Va + VM = Ira + IRM
Thus the value of the multiplier,
RM =

Figure 10
...
5 Comparison of moving-coil,
moving-iron and moving-coil
rectifier instruments
See Table at top of next page
...


Figure 10
...
A moving-coil instrument gives a
f
...
d
...
Calculate the value of the shunt to be
connected in parallel with the meter to enable it to
be used as an ammeter for measuring currents up
to 50 A
The circuit diagram is shown in Fig
...
5, where
ra = resistance of instrument = 25 , Rs = resistance

10
...

A voltmeter, which measures p
...
, has a high resistance (ideally infinite) and must be connected in parallel
with the part of the circuit whose p
...
is required
...
5

Type of instrument

Moving-coil

Moving-iron

Moving-coil rectifier

Suitable for measuring

Direct current and
voltage

Direct and alternating
currents and voltage
(reading in rms value)

Alternating current
and voltage (reads
average value but
scale is adjusted to
give rms value for
sinusoidal waveforms)

Scale

Linear

Non-linear

Linear

Method of control

Hairsprings

Hairsprings

Hairsprings

Method of damping

Eddy current

Air

Eddy current

Frequency limits

—

20–200 Hz

20–100 kHz

Advantages

1
...
High sensitivity
3
...
Low power
consumption

1
...

3
...


Robust construction
Relatively cheap
Measures dc and ac
In frequency range
20–100 Hz reads
rms correctly regardless
of supply wave-form

1
...
High sensitivity
3
...
Lower power
consumption
5
...
Only suitable for dc
2
...
Easily damaged

1
...
Affected by stray
magnetic fields
3
...
Liable to temperature errors
5
...
More expensive
than moving iron type
2
...
04 A, Is = current flowing
in shunt and I = total circuit current required to give
f
...
d
...

Since I = Ia + Is then Is = I − Ia
Is = 50 − 0
...
96 A
...
e
...
04)(25)
= 0
...
96
= 20
...
02 m needs to be connected in parallel with the
instrument
...
A moving-coil instrument having a
resistance of 10 , gives a f
...
d
...
Calculate the value of the multiplier to be
connected in series with the instrument so that it
can be used as a voltmeter for measuring p
...
s
...
10
...
State the mode of
resistance connection in each case
...
10 m in parallel
(b) 4
...
6

of multiplier I = total permissible instrument current =
8 mA = 0
...
d
...
s
...
= 100 V
V = Va + VM = Ira + IRM
i
...

100 = (0
...
008)RM
or 100 − 0
...
008 RM , thus
RM =

99
...
008

= 12
...
49 k needs to be connected in series with the
instrument
...
A moving-coil instrument gives f
...
d
...
Neglecting the resistance
of the instrument, calculate the approximate
value of series resistance needed to enable the
instrument to measure up to (a) 20 V (b) 100 V
(c) 250 V
[(a) 2 k (b) 10 k (c) 25 k ]
2
...
s
...
of
4 mA
...
s
...
should be (a) 15 mA
(b) 20 A (c) 100 A
[(a) 18
...
00 m (c) 2
...
A moving-coil instrument having a resistance
of 20 , gives a f
...
d
...

Calculate the value of the multiplier to be connected in series with the instrument so that it
can be used as a voltmeter for measuring p
...
’s
up to 200 V
[39
...
A moving-coil instrument has a f
...
d
...
Calculate the values
of resistance required to enable the instrument to be used (a) as a 0–10 A ammeter, and

5
...
Calculate the value of resistance
that converts the movement into (a) an ammeter with a maximum deflection of 50 A (b) a
voltmeter with a range 0–250 V
[(a) 12
...
63 k in series]

10
...
c
...

The digital voltmeter (DVM) is one which provides
a digital display of the voltage being measured
...
22) and a very high input
resistance, constant on all ranges
...
c
...
c
...
c
...

Instruments for a
...
measurements are generally calibrated with a sinusoidal alternating waveform to indicate r
...
s
...
Some instruments, such as the movingiron and electro-dynamic instruments, give a true r
...
s
...
With other instruments the indication is
either scaled up from the mean value (such as with the
rectified moving-coil instrument) or scaled down from
the peak value
...
15), and whenever a quantity is non-sinusoidal, errors in instrument readings
can occur if the instrument has been calibrated for
sine waves only
...


10
...
A simple ohmmeter circuit is shown in

Fig
...
7(a)
...
In the ohmmeter this energy is supplied by a self-contained source
of voltage, such as a battery
...
s
...
on the
milliammeter
...
Thus f
...
d
...
When terminals XX are open
circuited no current flows and R (=E/O) is infinity, ∞
...
These instruments
measure d
...
currents and voltages, resistance and continuity, a
...
(r
...
s
...


10
...
Fig
...
8 shows typical connections
of a wattmeter used for measuring power supplied to a
load
...


Figure 10
...
A cramped (non-linear) scale results and is ‘back
to front’, as shown in Fig
...
7(b)
...
An ohmmeter designed for measuring low
values of resistance is called a continuity tester
...
e
...
g
...


10
...
If a battery
is incorporated then resistance can also be measured
...
An ‘Avometer’ is a typical example
...
Only one measurement can be
performed at a time
...
c
...
c
...


Figure 10
...
11 Instrument ‘loading’ effect
Some measuring instruments depend for their operation
on power taken from the circuit in which measurements
are being made
...
e
...

The resistance of voltmeters may be calculated since
each have a stated sensitivity (or ‘figure of merit’), often
stated in ‘k per volt’ of f
...
d
...
In
a
...
circuits the impedance of the instrument varies with
frequency and thus the loading effect of the instrument
can change
...
Calculate the power dissipated by the
voltmeter and by resistor R in Fig
...
9 when
(a) R = 250 (b) R = 2 M
...
10

(b) Actual ammeter reading

Figure 10
...
s
...

Hence, Rv = (10 k /V) × (200 V) = 2000 k =
2 M
...


= V /(R + ra ) = 10/(500 + 50)
= 18
...

Thus the ammeter itself has caused the circuit
conditions to change from 20 mA to 18
...

(c) Power dissipated in the ammeter
= I 2 ra = (18
...
53 mW
...
18 × 10−3 )2 (500) = 165
...


When R = 250 , current in resistor,
IR =

V
100
=
= 0
...
4) = 40 W
...


Problem 5
...
s
...
of 100 V
and a sensitivity of 1
...
10
...
Determine
(a) the value of voltage V1 with the voltmeter not
connected, and (b) the voltage indicated by the
voltmeter when connected between A and B
...
In this case the higher load
resistance reduced the power dissipated such that
the voltmeter is using as much power as the load
...
An ammeter has a f
...
d
...
The ammeter is used to
measure the current in a load of resistance 500
when the supply voltage is 10 V
...


Figure 10
...
s
...

and sensitivity 1
...
6 k /V = 160 k
...
10
...
10
...


40
100 = 40 V
40 + 60

i
...


40 × 160
k
200

= 32 k

Figure 10
...
10
...
Thus the voltage indicated on the
voltmeter is
32
100 V = 34
...
The error is reduced by
using a voltmeter with a higher sensitivity
...
(a) A current of 20 A flows through a
load having a resistance of 2
...
(b) A wattmeter,
whose current coil has a resistance of 0
...
10
...
Determine the
wattmeter reading
...
Calculate: (a) the approximate value of current
(neglecting the ammeter resistance), (b) the
actual current in the circuit, (c) the power dissipated in the ammeter, (d) the power dissipated
in the 1 k resistor
...
250 A
(b) 0
...
832 W (d) 56
...
(a) A current of 15 A flows through a load having a resistance of 4
...
(b) A wattmeter, whose
current coil has a resistance of 0
...
10
...
Determine the wattmeter
reading assuming the current in the load is still
15 A
...
5 W]
3
...
6 k resistor
...
6 k resistor? The p
...
across the 1
...
s
...

250 V and sensitivity 100 /V
...

[160 V; 156
...
A 240 V supply is connected across a load
resistance R
...
s
...
of 300 V and a figure of
merit (i
...
sensitivity) of 8 k /V
...

Comment on the results obtained
...
6 mW]

Figure 10
...
01 = 2
...
The wattmeter
reading is thus I 2 RT = (20)2 (2
...
12 The oscilloscope
The oscilloscope is basically a graph-displaying
device – it draws a graph of an electrical signal
...
From the graph it is possible to:
• determine the time and voltage values of a signal

Now try the following exercise
Exercise 51 Further problems on
instrument ‘loading’ effects
1
...
c
...
c
...
14

• tell how much of the signal is noise and whether the
noise is changing with time
Oscilloscopes are used by everyone from television
repair technicians to physicists
...

The usefulness of an oscilloscope is not limited to the
world of electronics
...
e
...
An automobile engineer uses
an oscilloscope to measure engine vibrations; a medical
researcher uses an oscilloscope to measure brain waves,
and so on
...
An analogue oscilloscope works by directly
applying a voltage being measured to an electron beam
moving across the oscilloscope screen
...
This gives an immediate
picture of the waveform
...
11, page 301) to convert the voltage
being measured into digital information
...

For many applications either an analogue or digital
oscilloscope is appropriate
...

Analogue oscilloscopes are often preferred when it
is important to display rapidly varying signals in ‘real
time’ (i
...
as they occur)
...
They can process the
digital waveform data or send the data to a computer
for processing
...
Digital storage
oscilloscopes are explained in Section 10
...


Analogue oscilloscopes
When an oscilloscope probe is connected to a circuit, the
voltage signal travels through the probe to the vertical
system of the oscilloscope
...
14 shows a simple
block diagram that shows how an analogue oscilloscope
displays a measured signal
...
Next, the signal travels directly to the vertical deflection plates of the
cathode ray tube (CRT)
...
(An electron
beam hitting phosphor inside the CRT creates the glowing dot
...

The signal also travels to the trigger system to start or
trigger a ‘horizontal sweep’
...

Also, adjusting the focus and intensity controls enable
a sharp, visible display to be created
...
With no voltage applied to the Y plates the
position of the spot trace on the screen is noted
...
For example, in
Fig
...
15(a), with no voltage applied to the Y
plates, the spot trace is in the centre of the screen
(initial position) and then the spot trace moves
2
...
c
...
With the ‘volts/cm’ switch on
10 volts/cm the magnitude of the direct voltage is
2
...
e
...

(ii) With alternating voltage measurements, let
a sinusoidal waveform be displayed on an
oscilloscope screen as shown in Fig
...
15(b)
...
Triggering
the horizontal system causes the horizontal time base
to move the glowing dot across the screen from left
to right within a specific time interval
...
At higher speeds, the dot
may sweep across the screen up to 500 000 times each
second
...
e
...
e
...

The trigger is necessary to stabilise a repeating signal
...

In conclusion, to use an analogue oscilloscope, three
basic settings to accommodate an incoming signal need
to be adjusted:

Figure 10
...
e
...
02 s
...
02
If the ‘volts/cm’ switch is on, say, 20 volts/cm
then the amplitude or peak value of the sinewave
shown is 20 volts/cm × 2 cm, i
...
40 V
...
m
...
voltage = √ = 28
...
m
...
voltage =

Double beam oscilloscopes are useful whenever two
signals are to be compared simultaneously
...
r
...

demands reasonable skill in adjustment and use
...


Digital oscilloscopes
Some of the systems that make up digital oscilloscopes are the same as those in analogue oscilloscopes;
however, digital oscilloscopes contain additional data
processing systems – as shown in the block diagram of
Fig
...
16
...


Section 1

120 Electrical and Electronic Principles and Technology
Display
Acquisition System
Vertical System

Attenuator

Processing
Analogue to
Digital
Convertor

Vertical
Amplifier

Probe

Memory

Digital
Display
System

Horizontal System

Trigger System

Sample
Clock

Clock Time Base

Figure 10
...
Next,
the analogue to digital converter (ADC) in the acquisition system samples the signal at discrete points in
time and converts the signals’ voltage at these points to
digital values called sample points
...
The rate at which the clock ‘ticks’ is called
the sample rate and is measured in samples per second
...
More than one sample point
may make up one waveform point
...
The number of waveform points used to
make a waveform record is called a record length
...
The display receives these record points after
being stored in memory
...
Pre-trigger may be available, allowing events to be seen before the trigger
point
...

Problem 7
...
10
...
The ‘time/cm’ (or timebase
control) switch is on 100 μs/cm and the ‘volts/cm’
(or signal amplitude control) switch is on 20 V/cm

Figure 10
...
10
...
20 assume that the squares shown
are 1 cm by 1 cm)
(a) The width of one complete cycle is 5
...
Hence
the periodic time,
T = 5
...
52 ms
...
92 kHz
...
52 × 10−3

(c) The peak-to-peak height of the display is 3
...
6 cm × 20 V/cm = 72 V

Problem 8
...
10
...
2 V/cm
...


(b) The peak-to-peak height of the waveform is 5 cm
...

1
(c) Amplitude = 2 × 25 V = 12
...
e
...
5 V, and r
...
s
...
18

(a) The width of one complete cycle is 3
...
Hence
the periodic time,
T = 3
...

(b) Frequency, f =

peak voltage
12
...
84 V
√
2
2

Problem 10
...
10
...
m
...
values, (c) their phase
difference
...


1
1
=
= 5
...

T 0
...
4 cm hence the magnitude
of the pulse voltage
= 3
...
2 V/cm = 0
...

Problem 9
...
10
...
If the
‘time/cm’ switch is on 500 μs/cm and the ‘volts/cm’
switch is on 5 V/cm, find, for the waveform, (a) the
frequency, (b) the peak-to-peak voltage, (c) the
amplitude, (d) the r
...
s
...


Figure 10
...
Hence the periodic time, T , of each
waveform is 5 cm × 100 μs/cm, i
...
0
...

Frequency of each waveform,
f =

(b) The peak value of waveform A is
2 cm × 2 V/cm = 4 V, hence the r
...
s
...
83 V

Figure 10
...
Hence
the periodic time, T is 4 cm × 500 μs/cm, i
...
2 ms
...
5 × 10−3

1
1
=
= 500 Hz
T
2 × 10−3

The peak value of waveform B is
2
...
m
...
value of
waveform B
√
=5/( 2) = 3
...
e
...
The phase
angle φ = 0
...
5 cm × 72◦ /cm = 36◦
...
m
...

voltage
...
14 Hz (b) 220 V (c) 77
...
For the square voltage waveform displayed on
an oscilloscope shown in Fig
...
21, find (a)
its frequency, (b) its peak-to-peak voltage
...
7 Hz (b) 176 V]

Figure 10
...
13 Virtual test and measuring
instruments

Figure 10
...
For the pulse waveform shown in Fig
...
22,
find (a) its frequency, (b) the magnitude of the
pulse voltage
...
56 Hz (b) 8
...
Probably the most commonly available virtual test instrument is the digital
storage oscilloscope (DSO)
...
In addition, the ability to save waveforms and
captured measurement data for future analysis or for
comparison purposes can be extremely valuable, particularly where evidence of conformance with standards or
specifications is required
...
The functions
and available measurements from such an instrument
usually includes:
• real time or stored waveform display

Figure 10
...
For the sinusoidal waveform shown in
Fig
...
23, determine (a) its frequency,

• digital display of frequency and/or periodic time
• accurate measurement of phase angle

• frequency spectrum display and analysis
• data logging (stored waveform data can be exported
in formats that are compatible with conventional
spreadsheet packages, e
...
as
...
g
...
jpg or
...

Virtual instruments can take various forms including:
• internal hardware in the form of a conventional PCI
expansion card
• external hardware unit which is connected to the
PC by means of either a conventional 25-pin parallel port connector or by means of a serial USB
connector
The software (and any necessary drivers) is invariably supplied on CD-ROM or can be downloaded from
the manufacturer’s web site
...


of between 10K and 100K samples per second
...

High-speed DSOs are rapidly replacing CRT-based
oscilloscopes
...

Additional features available with a computer-based
instrument include the ability to capture transient signals (as with a conventional digital storage ‘scope’) and
save waveforms for future analysis
...


Upper frequency limit
The upper signal frequency limit of a DSO is determined
primarily by the rate at which it can sample an incoming signal
...
14 Virtual digital storage
oscilloscopes
Several types of virtual DSO are currently available
...
A high-speed DSO is designed
for examining waveforms that are rapidly changing
...
Similarly, a high-resolution
DSO is useful for displaying waveforms with a high
degree of precision but it may not be suitable for examining fast waveforms
...

Low-cost DSO are primarily designed for low frequency signals (typically signals up to around 20 kHz)
and are usually able to sample their signals at rates

Typical sampling rate

Low-cost DSO

20K to 100K per second

High-speed DSO

100M to 1000M per second

High-resolution DSO

20M to 100M per second

In order to display waveforms with reasonable accuracy it is normally suggested that the sampling rate
should be at least twice and preferably more than five
times the highest signal frequency
...

The ‘five times rule’ merits a little explanation
...
Unfortunately, this no longer applies
in the case of a DSO where we need to sample at an even
faster rate if we are to accurately display the signal
...
Hence
the sampling rate should be at least five times that of

Section 1

Electrical measuring instruments and measurements 123

Section 1

124 Electrical and Electronic Principles and Technology
highest signal frequency in order to display a waveform
reasonably faithfully
...
Here
the sampling rate may be shared between the two channels
...
In such a case the upper
frequency limit would not be 4 MHz but only a mere
2 MHz
...
The relationship is as
follows:
x = 2n
where x is the number of discrete voltage levels and n is
the number of bits
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...

It is worth noting that most manufacturers define the
bandwidth of an instrument as the frequency at which
a sine wave input signal will fall to 0
...
e
...
To put this into context,
at the cut-off frequency the displayed trace will be in
error by a whopping 29%!

Buffer memory capacity
A DSO stores its captured waveform samples in a buffer
memory
...

The relationship between sampling rate and buffer
memory capacity is important
...

To put this into context, it’s worth considering a simple example
...
This signal will occur
in a time frame of 1 ms
...

To reconstruct the square wave we would need a minimum of about five samples per cycle so a minimum
sampling rate would be 5 × 10 MHz = 50M samples per
second
...
If each sample uses 16-bits
we would require 100 kbyte of extremely fast memory
...
So, for example, on the 1 V range an 8-bit DSO
is able to detect a voltage change of one two hundred
and fifty sixth of a volt or (1/256) V or about 4 mV
...
4% of full-scale
...
24 depicts a PicoScope software display showing multiple windows providing conventional
oscilloscope waveform display, spectrum analyser display, frequency display, and voltmeter display
...
26

528 ns
...


Autoranging

Figure 10
...
In Fig
...
25, the
peak value of the (nominal 10 V peak) waveform is
measured at precisely 9625 mV (9
...
The time to
reach the peak value (from 0 V) is measured as 246
...
2467 ms)
...
25

The addition of a second time cursor makes it possible to measure the time accurately between two events
...
10
...
The elapsed time between these two events is

Autoranging is another very useful feature that is often
provided with a virtual DSO
...


High-resolution DSO
High-resolution DSOs are used for precision applications where it is necessary to faithfully reproduce a
waveform and also to be able to perform an accurate
analysis of noise floor and harmonic content
...

Unlike the low-cost DSO, which typically has 8-bit
resolution and poor d
...
accuracy, these units are usually
accurate to better than 1% and have either 12-bit or
16-bit resolution
...

The increased resolution also allows the instrument to
be used as a spectrum analyser with very wide dynamic
range (up to 100 dB)
...

Bandwidth alone is not enough to ensure that a DSO
can accurately capture a high frequency signal
...

This response is sometimes referred to as a Maximally
Flat Envelope Delay (MFED)
...


Section 1

126 Electrical and Electronic Principles and Technology
It is important to remember that, if the input signal is
not a pure sine wave it will contain a number of higher
frequency harmonics
...
Thus,
to display a 1 MHz square wave accurately you need
to take into account the fact that there will be signal
components present at 3 MHz, 5 MHz, 7 MHz, 9 MHz,
11 MHz, and so on
...


Spectrum analysis
The technique of Fast Fourier Transformation (FFT)
calculated using software algorithms using data captured by a virtual DSO has made it possible to produce
frequency spectrum displays
...

Figure 10
...
Here the virtual DSO has been set to capture
samples at a rate of 4096 per second within a frequency
range of d
...
to 12
...
The display clearly shows
the second harmonic (at a level of −50 dB or −70 dB
relative to the fundamental), plus further harmonics at
3 kHz, 5 kHz and 7 kHz (all of which are greater than
75 dB down on the fundamental)
...
28

(a) The signal x is at a frequency of 3553 kHz
...
Thus, x is the third harmonic of
the signal ‘o’
(b) The signal at ‘o’ has an amplitude of +17
...
08 dB
...
46) − (−4
...
54 dB
(c) The amplitude of the second harmonic (shown
at approximately 2270 kHz) = −5 dB

10
...
27

Problem 11
...
28 shows the frequency
spectrum of a signal at 1184 kHz displayed by a
high-speed virtual DSO
...
This is a waveform which
varies sinusoidally with time t, has a frequency f ,
and a maximum value Vm
...
If
the waveform is not sinusoidal it is called a complex wave, and, whatever its shape, it may be
split up mathematically into components called
the fundamental and a number of harmonics
...
The fundamental (or first harmonic) is sinusoidal and has
the supply frequency, f ; the other harmonics are
also sine waves having frequencies which are integer multiples of f
...

(ii) A complex waveform comprising the sum of
the fundamental and a third harmonic of about

half the amplitude of the fundamental is shown
in Fig
...
29(a), both waveforms being initially
in phase with each other
...
In Fig
...
29(b), the third harmonic is
shown having an initial phase displacement from
the fundamental
...
10
...


second harmonic is shown with an initial phase
displacement from the fundamental and the positive and negative half cycles are dissimilar
...
10
...
The negative half cycle,
if reversed, appears as a mirror image of the
positive cycle about point B
...
10
...
The
positive and negative half cycles are seen to be
dissimilar
...
10
...


10
...
By definition, if the ratio
of two powers P1 and P2 is to be expressed in decibel
(dB) units then the number of decibels, X, is given by:
P2
P1

X = 10 lg

Figure 10
...
10
...
If further
even harmonics of appropriate amplitudes are
added a good approximation to a triangular wave
results
...
10
...
In Fig
...
29(d) the

dB

(1)

Thus, when the power ratio, P2 /P1 = 1 then the
decibel power ratio = 10 lg 1 = 0, when the power
ratio, P2 /P1 = 100 then the decibel power ratio =
10 lg 100 = +20 (i
...
a power gain), and when the power
ratio, P2 /P1 = 1/100 then the decibel power ratio =
10 lg 1/100 = −20 (i
...
a power loss or attenuation)
...
Power, P, is given by P = I 2 R or
P = V 2 /R
...
e
...
30

(from the laws of logarithms)
...
It is therefore necessary to state a reference
level to measure a number of decibels above or below
that reference
...

A voltmeter can be re-scaled to indicate the power
level directly in decibels
...
The
reference voltage V is then obtained from

i
...


V2
P=
,
R
V2
1 × 10−3 =
600

(a)

When

P2
= 3,
P1
X = 10 lg(3) = 10(0
...
77 dB

(b)

When

P2
= 20,
P1
X = 10 lg(20) = 10(1
...
0 dB

(c)

When

P2
= 400,
P1
X = 10 lg(400) = 10(2
...
0 dB

from which, V = 0
...
In general, the number
of dBm,
X = 20 lg

From above, the power ratio in decibels, X, is given by:
X = 10 lg(P2 /P1 )

V
0
...
05,
P1
20
X = 10 lg(0
...
30)

Thus V = 0
...
2
0
...
77 dBm and
V = 0
...
90
0
...
3 dBm, and so on
...
10
...
Errors are introduced with dB meters when
the circuit impedance is not 600
...
The ratio of two powers is (a) 3
(b) 20 (c) 4 (d) 1/20
...


= −13
...

Problem 13
...
Find the
decibel current ratio assuming the input and load
resistances of the system are equal
...
60)
= 12 dB gain

Problem 14
...
Determine the
power loss in decibels
...


P2
6
=
= 0
...
06)
P1
= 10(−1
...
22 dB
Hence the decibel power loss, or attenuation, is
12
...


19 = 10 lg

from which
If P1 = input power and P2 = output power then

Thus

1
...
4
P1
P2
= 79
...
2 = 15
...
9 =

Similarly for the second stage,
Problem 15
...
Find its output power
...
62
P1
and for the third stage,

Decibel power ratio = 10 lg(P2 /P1 ) where
P1 = input power = 8 mW, and P2 = output power
...
e
...
4 = lg
P1
P2
101
...
12 =
P1

Output power, P2 = 25
...
12)(8)
= 201 mW or 0
...
Determine, in decibels, the ratio of
output power to input power of a 3 stage
communications system, the stages having gains of
12 dB, 15 dB and −8 dB
...


The decibel ratio may be used to find the overall
power ratio of a chain simply by adding the decibel power ratios together
...
1585
P1
The overall power ratio is thus
15
...
62 × 0
...
4]
Problem 17
...
If the voltage gain is 27 dB,
calculate the value of the input voltage assuming
that the amplifier input resistance and load
resistance are equal
...
Hence
4
27
= lg
20
V1
4
i
...

1
...
35 =
V1
4
from which
V1 = 1
...
39
= 0
...
179 V
...
The ratio of two powers is (a) 3 (b) 10 (c) 20
(d) 10 000
...

[(a) 4
...
The ratio of two powers is (a) 10 (b) 1 (c) 40
3
1
(d) 100
...

[(a) −10 dB (b) −4
...
02 dB (d) −20 dB]

3
...
Determine the
decibel current ratio of output to input current
assuming input and output resistances of the
system are equal
...
98 dB]
4
...
Determine the power
loss in decibels
...
An amplifier has a gain of 24 dB and its input
power is 10 mW
...

[2
...
Determine, in decibels, the ratio of the output
power to input power of a four stage system,
the stages having gains of 10 dB, 8 dB, −5 dB
and 7 dB
...

[20 dB, 100]
7
...

If the voltage gain is 25 dB calculate the value
of the input voltage assuming that the amplifier
input resistance and load resistance are equal
...
39 mV]

9
...
Determine
the voltage at (a) 0 dB (b) 1
...
5 V?
[(a) 0
...
921 V
(c) 0
...
807 dB]

10
...

The method assumes:
(i) if there is any deflection at all, then some current
is flowing;
(ii) if there is no deflection, then no current flows (i
...

a null condition)
...
A sensitive milliammeter or microammeter with centre zero
position setting is called a galvanometer
...
18), in the d
...
potentiometer (see
Section 10
...
c
...
20)

10
...
31 shows a Wheatstone bridge circuit which
compares an unknown resistance Rx with others of
known values, i
...
R1 and R2 , which have fixed values,
and R3 , which is variable
...
No current then

8
...
8 dB, −12
...
8 dB
...
If a voltage of 15 mV is
applied to the input of the system, determine
the value of the output voltage
...
5 dB, 39
...
31

Electrical measuring instruments and measurements 131

R1 Rx = R2 R3 i
...
Rx =

Section 1

flows through the meter, VA = VB , and the bridge is said
to be ‘balanced’
...
In a Wheatstone bridge ABCD, a
galvanometer is connected between A and C, and a
battery between B and D
...
When the
bridge is balanced, the resistance between B and C
is 100 , that between C and D is 10 and that
between D and A is 400
...


The Wheatstone bridge is shown in Fig
...
32 where
Rx is the unknown resistance
...
33

(i
...
the galvanometer deflection is zero), shown as
length l1
...
m
...
E2 (see Fig
...
33(b)) and again balance is obtained (shown as l2 )
...
c
...

Such devices may be constructed in the form of a
resistive element carrying a sliding contact which
is adjusted by a rotary or linear movement of the
control knob
...
32

Hence, the unknown resistance, Rx = 4 k
...
In a d
...
potentiometer, balance is
obtained at a length of 400 mm when using a
standard cell of 1
...
Determine the e
...
f
...
0186 V, l1 = 400 mm and l2 = 650 mm
With reference to Fig
...
33,
E1
l1
=
E2
l2

10
...
C
...
c
...
m
...
’s and p
...
s
...
m
...
or p
...
In Fig
...
33(a),
using a standard cell of known e
...
f
...
655 volts
= (1
...
c
...
In a Wheatstone bridge PQRS, a galvanometer
is connected between Q and S and a voltage
source between P and R
...
When the
bridge is balanced, the resistance between Q
and R is 200 , that between R and S is 10
and that between S and P is 150
...
c
...
A commercial
or universal bridge is one which can be used to measure resistance, inductance or capacitance
...
c
...
e
...

A Maxwell-Wien bridge for measuring the inductance L and resistance r of an inductor is shown in
Fig
...
35
...
Balance is obtained in a d
...
potentiometer
at a length of 31
...
0186 volts
...
m
...
of a
dry cell if balance is obtained with a length of
46
...
525 V]

10
...
C
...
35

A Wheatstone bridge type circuit, shown in Fig
...
34,
may be used in a
...
circuits to determine unknown values of inductance and capacitance, as well as resistance
...
Thus
Z1 Z2 = Z3 Z4
Using complex quantities, Z1 = R1 , Z2 = R2 ,
Z3 =

product
R3 (−jXC )
i
...

R3 − jXC
sum

and Z4 = r + jXL
...
34

R1 R2 =
When the potential differences across Z3 and Zx (or
across Z1 and Z2 ) are equal in magnitude and phase,
then the current flowing through the galvanometer, G,
is zero
...
e
...
e
...


Equating the real parts gives:

can be adjusted to give a direct indication of Q-factor
...
21 following
...
e
...
c
...
A Maxwell bridge circuit ABCD has the following arm impedances: AB, 250 resistance; BC, 15 μF capacitor in parallel with
a 10 k resistor; CD, 400 resistor; DA,
unknown inductor having inductance L and
resistance R
...

[1
...
21 Q-meter
(3)

Problem 20
...
c
...
10
...
5 μF

The Q-factor for a series L–C–R circuit is the voltage
magnification at resonance, i
...

Q-factor =

=

From equation (2) above, inductance
L = R1 R2 C = (400)(400)(7
...
2 H
From equation (3) above, resistance,
r=

(400)(400)
R1 R2
=
= 32
R3
5000

From equation (2),
L
R2 =
R1 C
and from equation (3),
R1
R2
r
R1 L
L
R3 =
=
r R1 C
Cr

If the frequency is constant then R3 ∝ L/r ∝ ωL/r ∝
Q-factor (see Chapters 15 and 16)
...


The simplified circuit of a Q-meter, used for measuring
Q-factor, is shown in Fig
...
36
...
The frequency is
then varied until resonance causes voltage Vc to reach
a maximum value
...

Then
Vc
Vc
Q-factor =
=
Vr
Ir
In a practical Q-meter, Vr is maintained constant and
the electronic voltmeter can be calibrated to indicate
the Q-factor directly
...
In this way inductance L
may be calculated using

R3 =

Hence

voltage across capacitor
supply voltage

fr =

Since

1
√

2π LC
2πfL
Q=
R

then R may be calculated
...
396 × 10−3 )
100

= 9
...
36

Q-meters operate at various frequencies and instruments exist with frequency ranges from 1 kHz to
50 MHz
...
The accuracy of a
Q-meter is approximately ±5%
...
When connected to a Q-meter an
inductor is made to resonate at 400 kHz
...
Determine (a) the inductance, and (b) the
resistance of the inductor
...
The circuit diagram of a Q-meter is shown in
Fig
...
36
(a) At resonance,
fr =

1
√

2π LC
for a series L–C–R circuit
...
396 mH

Exercise 56 Further problem on the
Q-meter
1
...
If the capacitance of the Q-meter
capacitor is set to 300 pF determine (a) the
inductance L, and (b) the resistance R of the
inductor
...
351 mH (b) 10
...
22

Measurement errors

Errors are always introduced when using instruments to
measure electrical quantities
...

(i) Errors in the limitations of the instrument
The calibration accuracy of an instrument depends
on the precision with which it is constructed
...
For
example, industrial grade instruments have an accuracy
of ±2% of f
...
d
...
s
...
of 100 V
and it indicates 40 V say, then the actual voltage may
be anywhere between 40 ± (2% of 100), or 40 ± 2, i
...

between 38 V and 42 V
...
A typical graph is
shown in Fig
...
37 where it is seen that the accuracy
varies over the scale length
...
s
...
accuracy would tend to have an accuracy which is
much better than ±2% f
...
d
...


Problem 22
...
4% is measured as 2
...
5%
...


Figure 10
...

With linear scales the values of the sub-divisions are
reasonably easy to determine; non-linear scale graduations are more difficult to estimate
...
When reading
a meter scale it should be viewed from an angle perpendicular to the surface of the scale at the location of
the pointer; a meter scale should not be viewed ‘at an
angle’
...

(iii) Errors due to the instrument disturbing the
circuit
Any instrument connected into a circuit will affect
that circuit to some extent
...
Incorrect positioning of instruments in a
circuit can be a source of errors
...
10
...
Assuming ‘perfect’ instruments,
the resistance should be given by the voltmeter reading
divided by the ammeter reading (i
...
R = V /I)
...
10
...
10
...
Hence the voltmeter
reading divided by the ammeter reading will not give
the true value of the resistance R for either method of
connection
...
5 × 10−3 )(5 × 103 ) = 12
...
The
maximum possible error is 0
...
5% = 0
...

Hence the voltage, V = 12
...
9% of 12
...
9% of 12
...
9/100 × 12
...
1125 V = 0
...

Hence the voltage V may also be expressed as
12
...
11 volts (i
...
a voltage lying between 12
...
61 V)
...
The current I flowing in a resistor R
is measured by a 0–10 A ammeter which gives an
indication of 6
...
The voltage V across the
resistor is measured by a 0–50 V voltmeter, which
gives an indication of 36
...
Determine the
resistance of the resistor, and its accuracy of
measurement if both instruments have a limit of
error of 2% of f
...
d
...


Resistance,
R=

V
36
...
84
I
6
...
0 V and expressed
as a percentage of the voltmeter reading gives
±1
× 100% = ±2
...
5
Current error is ±2% of 10 A = ±0
...
2
× 100% = ±3
...
25
Maximum relative error = sum of errors = 2
...
2% = ±5
...
94% of 5
...
347
...
84
(rounding off)

Figure 10
...
94% or 5
...
35

Section 1

Electrical measuring instruments and measurements 135

Section 1

136 Electrical and Electronic Principles and Technology
Problem 24
...
0%; BC: R2 = 100 ± 0
...
5 ± 0
...

Determine the value of the unknown resistance and
its accuracy of measurement
...
10
...
e
...
5)
=
= 43
...
The voltage across a resistor is measured
by a 75 V f
...
d
...
The current flowing in the resistor is measured by a 20 A f
...
d
...
5 A
...
s
...

[4
...
08% or 4
...
25 ]
3
...
5%; RS, unknown resistance; SP,
273
...
1%
...

[27
...
6% or 27
...
71 ]

Exercise 58 Short answer questions on
electrical measuring
instruments and
measurements

Figure 10
...
e
...
0% + 0
...
2% = 1
...

Hence

resistor is 6 k ± 0
...
Determine the current flowing in the resistor and its accuracy
of measurement
...
25 mA ± 1
...
25 ± 0
...
25

1
...
25 = 0
...
7%

(rounding off)
...
25 ± 0
...
The p
...
across a resistor is measured as 37
...
5%
...
What is the main difference between an
analogue and a digital type of measuring
instrument?
2
...
Complete the following statements:
(a) An ammeter has a
...
with the circuit
(b) A voltmeter has a
...
with the circuit
4
...
What effect does the connection of (a) a shunt
(b) a multiplier have on a milliammeter?
6
...
Name two advantages of electronic measuring instruments compared with moving coil
or moving iron instruments
8
...
Name a type of ohmmeter used for measuring (a) low resistance values (b) high
resistance values
10
...
When may a rectifier instrument be used in
preference to either a moving coil or moving
iron instrument?
12
...
r
...
is capable
of measuring
13
...
What is a feature of waveforms containing
the fundamental and odd harmonics?
15
...
What does a power level unit of dBm
indicate?
17
...
Sketch a Wheatstone bridge circuit used for
measuring an unknown resistance in a d
...

circuit and state the balance condition
19
...
c
...
d
...
Which of the following would apply to a
moving coil instrument?
(a) An uneven scale, measuring d
...

(b) An even scale, measuring a
...

(c) An uneven scale, measuring a
...

(d) An even scale, measuring d
...

2
...
In question 1, which would refer to a moving
coil rectifier instrument?
4
...
Fig
...
40 shows a scale of a multi-range
ammeter
...
6 A
(c) 14 A
(d) 8
...
Name five types of a
...
bridge used for measuring unknown inductance, capacitance or
resistance
21
...
State the name of an a
...
bridge used for
measuring inductance
23
...
Why do instrument errors occur when
measuring complex waveforms?
25
...
State three main areas where errors are most
likely to occur in measurements

Figure 10
...
r
...

screen
...
The
‘variable’ switch is on 100 μs/cm and the

Section 1

Electrical measuring instruments and measurements 137

Section 1

138 Electrical and Electronic Principles and Technology
‘volts/cm’ switch is on 10 V/cm
...
4 ms
(d) 35
...
5 V
(i) 2
...
7 V

11
...
Peak-to-peak value of waveform Q
13
...
Frequency of both waveforms

6
...
Determine the periodic time of the waveform
8
...
R
...
s
...
R
...
s
...
Determine the frequency of the waveform
10
...
m
...
value of the waveform
Fig
...
41 shows double-beam c
...
o
...
For the quantities stated in questions 11 to 17, select the correct answer from
the following:
(a) 30 V (b) 0
...
Phase displacement of waveform Q relative
to waveform P
18
...
The decibel
power ratio of output power to input power
is:
(a) 9
(b) 9
...
9
(d) 19
...
The input and output voltages of a system
are 500 μV and 500 mV respectively
...
41

(q)

30
√ V
2

20
...
The decibel
ratio of output to input current (assuming the
input and load resistances are equal) is:
(a) 15
...
6
(d) 7
...
Which of the following statements is false?
(a) The Schering bridge is normally used
for measuring unknown capacitances
(b) A
...
electronic measuring instruments
can handle a much wider range of frequency than the moving coil instrument
(c) A complex waveform is one which is
non-sinusoidal
(d) A square wave normally contains the
fundamental and even harmonics

22
...
s
...
of 100 V, a sensitivity
of 1 k /V and an accuracy of ±2% of f
...
d
...
Which of the following
statements is false?
(a) Voltage reading is 50 ± 2 V
(b) Voltmeter resistance is 100 k
(c) Voltage reading is 50 V ± 2%
(d) Voltage reading is 50 V ± 4%

23
...


Insulators:

11
...
The classification depends on the
value of resistivity of the material
...
Some typical approximate values at normal
room temperatures are:

Conductors:
Aluminium
Brass (70 Cu/30 Zn)
Copper (pure annealed)
Steel (mild)

2
...
7×10−8
15 × 10−8

m
m
m
m

Semiconductors: (at 27◦ C)
Silicon
Germanium

2
...
45 m

Glass
Mica
PVC
Rubber (pure)

≥ 1010 m
≥ 1011 m
≥ 1013 m
1012 to 1014 m

In general, over a limited range of temperatures, the
resistance of a conductor increases with temperature
increase, the resistance of insulators remains approximately constant with variation of temperature and the
resistance of semiconductor materials decreases as the
temperature increases
...
11
...

As the temperature of semiconductor materials is
raised above room temperature, the resistivity is reduced
and ultimately a point is reached where they effectively become conductors
...
As

Semiconductor diodes 141

Insulator
Semiconductor
Valence shell

15

Electrons

t
Temperature °C

Nucleus

Figure 11
...


11
...
Electrons each carry a single unit
of negative electric charge while protons each exhibit
a single unit of positive charge
...
For example, if an atom
has eleven electrons, it will also contain eleven protons
...

Electrons are in constant motion as they orbit around
the nucleus of the atom
...
The maximum number of electrons present
in the first shell is two, in the second shell eight, and
in the third, fourth and fifth shells it is 18, 32 and
50, respectively
...

It is important to note that the movement of electrons
between atoms only involves those present in the outer
valence shell
...
11
...
If, however, the valence shell does
not have its full complement of electrons, the electrons
can be easily detached from their orbital bonds, and the
material has the properties associated with an electrical
conductor
...
If, however, an atom of a different element (i
...


Figure 11
...
11
...
These free electrons become available for use
as charge carriers and they can be made to move through
the lattice by applying an external potential difference
to the material
...
3

Similarly, if the impurity element introduced into the
pure silicon lattice has three electrons in its valence
shell, the absence of the fourth electron needed for
proper covalent bonding will produce a number of
spaces into which electrons can fit (see Fig
...
4)
...
Once again, current will
flow when an external potential difference is applied to
the material
...
4

Regardless of whether the impurity element produces surplus electrons or holes, the material will no
longer behave as an insulator, neither will it have the
properties that we normally associate with a metallic
conductor
...
Examples of
semiconductor materials include silicon (Si), germanium (Ge), gallium arsenide (GaAs), and indium
arsenide (InAs)
...
The amount of
impurity added usually varies from 1 part impurity in
105 parts semiconductor material to 1 part impurity to
108 parts semiconductor material, depending on the
resistivity required
...

The process of introducing an atom of another (impurity) element into the lattice of an otherwise pure
material is called doping
...
e
...
e
...
If,
however, the pure material is doped with an impurity having three electrons in its valence shell (i
...
a
trivalent impurity) it will become a p-type (i
...
positive type) semiconductor material
...


In semiconductor materials, there are very few charge
carriers per unit volume free to conduct
...
e
...


11
...
The ‘fifth’ valency
electron is not rigidly bonded and is free to conduct,
the impurity atom donating a charge carrier
...
One of the four bonds
associated with the semiconductor material is deficient
by one electron and this deficiency is called a hole
...
11
...
In this diagram, an electron moves
from A to B, giving the appearance that the hole moves
from B to A
...


Possible electron
movement
A
B

1

2

3
C

Hole
Impurity
atom

Figure 11
...
In
order to examine the charge situation, assume that separate blocks of p-type and n-type materials are pushed
together
...

At the junction, the donated electrons in the n-type
material, called majority carriers, diffuse into the
p-type material (diffusion is from an area of high density
to an area of lower density) and the acceptor holes in the
p-type material diffuse into the n-type material as shown
by the arrows in Fig
...
6
...
The p-type material
has gained electrons and becomes negatively charged
with respect to the n-type material and hence tends to
retain holes
...
The
area in the region of the junction becomes depleted of
holes and electrons due to electron-hole recombination,
and is called a depletion layer, as shown in Fig
...
7
...
6

Problem 1
...

(a) Silicon or germanium with no doping atoms added
are called intrinsic semiconductors
...
4 The p-n junction

Figure 11
...

This is called thermal generation of electronhole pairs
...
This positive charge
may attract another electron released from another
atom, creating a hole elsewhere
...

(b) When additional mobile electrons are introduced
by doping a semiconductor material with pentavalent atoms (atoms having five valency electrons),
these mobile electrons are called majority carriers
...

For p-type materials, the additional holes are
introduced by doping with trivalent atoms (atoms
having three valency electrons)
...
The relatively
few mobile electrons in the p-type material produced by intrinsic action are called minority
carriers
...

There are more free electrons in n-type material
than holes and more holes in p-type material than
electrons
...
This process is
called diffusion
...
Explain briefly why a junction
between p-type and n-type materials creates a
contact potential
...
When a p-n junction is
formed, the resistive property is replaced by a rectifying property, that is, current passes more easily in one
direction than the other
...

The total number of positive and negative charges are
equal
...

Again, the total number of positive and negative
charges are equal and the material is neither positively
nor negatively charged
...
Also, some of
the mobile holes in the p-type material diffuse into the
n-type material
...
This region, called
the depletion layer, acts as an insulator and is in the
order of 0
...
Since the n-type material has
lost electrons, it becomes positively charged
...


11
...
8

n-type material as is shown in Fig
...
9, the p-n junction is reverse biased
...
Thus, in theory, no current flows
...
This process is called electron-hole
generation by thermal excitation
...
11
...
The applied voltage opposes the
contact potential, and, in effect, closes the depletion
layer
...
An increase in the applied voltage
above that required to narrow the depletion layer (about
0
...
6 V for silicon), results in
a rapid rise in the current flow
...
9

The electrons in the p-type material and holes in
the n-type material caused by thermal excitation, are
called minority carriers and these will be attracted by
the applied voltage
...

Graphs depicting the current-voltage relationship for
forward and reverse biased p-n junctions, for both
germanium and silicon, are shown in Fig
...
10
...
The movement of minority carriers results in a small constant current flowing
...
The voltage at which this occurs is called the
breakdown voltage
...


Figure 11
...
Sketch the forward and reverse
characteristics of a silicon p-n junction diode and
describe the shapes of the characteristics drawn
...
The forward characteristic of a diode
is shown in Fig
...
11
...
4 V is applied, (b) the voltage
dropped across the diode when a forward current of
9 mA is flowing in it, (c) the resistance of the diode
when the forward voltage is 0
...

10
9 mA

(b)

8
Forward current (mA)

A typical characteristic for a silicon p-n junction is
shown in Fig
...
10
...
Due to like charges repelling, the holes
in the p-type material drift towards the junction
...
The width of the depletion layer and size of
the contact potential are reduced
...
6 V, very little current flows
...
6 V, majority carriers begin to cross the junction in large numbers and current starts to flow
...
6 V, the current
increases exponentially (see Fig
...
10)
...
The
holes in the p-type material are attracted towards the
negative terminal and the electrons in the n-type material are attracted towards the positive terminal (unlike
charges attract)
...


6

(c)

6 mA

4

2

1
...
4 V

0
0
...
2
0
...
6
Forward voltage (V)

0
...
8

Figure 11
...
11
...
4 V, current flowing, I = 1
...
67 V

Section 1

Semiconductor diodes 145

(c) From the graph, when V = 0
...

Thus, resistance of the diode,
V
0
...
1 × 103 = 100
=
I
6 × 10−3
(d) The onset of conduction occurs at approximately
0
...
This suggests that the diode is a Ge type
...
Corresponding readings of current, I,
and voltage, V , for a semiconductor device are
given in the table:
Vf (V) 0 0
...
2 0
...
4 0
...
6 0
...
8
0
0
0
1
9 24 50
If (mA) 0 0
Plot the I/V characteristic for the device and
identify the type of device
...
11
...
Since
the device begins to conduct when a potential of approximately 0
...


(a) From Fig
...
12, when the forward voltage is
0
...
76 V
Now try the following exercise
Exercise 60 Further problems on
semiconductor materials
and p-n junctions
1
...

2
...

3
...

Explain the effect these impurities have on
the form of conduction in silicon
...
With the aid of simple sketches, explain how
pure germanium can be treated in such a
way that conduction is predominantly due to
(a) electrons and (b) holes
...
Explain the terms given below when used
in semiconductor terminology: (a) covalent
bond, (b) trivalent impurity, (c) pentavalent
impurity, (d) electron-hole pair generation
...
Explain briefly why although both p-type
and n-type materials have resistive properties
when separate, they have rectifying properties when a junction between them exists
...
0

0
...
4

0
...
8

0
...
76 V
Forward voltage (V)

Figure 11
...
For the characteristic of Fig
...
12,
determine for the device (a) the forward current
when the forward voltage is 0
...


7
...
With the aid of diagrams
explain this statement and also how the
direction and magnitude of the applied
voltage affects the depletion layer
...
State briefly what you understand by
the terms: (a) reverse bias, (b) forward
bias, (c) contact potential, (d) diffusion,
(e) minority carrier conduction
...
Explain briefly the action of a p-n junction
diode: (a) on open-circuit, (b) when provided
with a forward bias, and (c) when provided
with a reverse bias
...

10
...
Explain
the change in the charge situation when compared with that in isolated p-type and n-type
materials
...

11
...
11
...

(a) What type of diode is this? Give reasons
...
5 V
...
(d) Determine the resistance of
the diode when the forward voltage is 0
...

[(a) Ge (b) 17 mA (c) 0
...
This property allows diodes to be used in
applications that require a circuit to behave differently
according to the direction of current flowing in it
...

A semiconductor diode is an encapsulated p-n junction fitted with connecting leads or tags for connection
to external circuitry
...
The connection
to the p-type material is referred to as the anode while
that to the n-type material is called the cathode
...
These include rectifier diodes for
use in power supplies, Zener diodes for use as voltage
reference sources, light emitting diodes, and varactor
diodes
...
14 shows the symbols used to represent diodes in electronic circuit diagrams, where ‘a’ is
the anode and ‘k’ the cathode
...
0

0
...
4

0
...
8

mt 2
(e) Triac

(d) Bridge rectifier

Forward voltage (V)

Figure 11
...
6 Semiconductor diodes
When a junction is formed between p-type and n-type
semiconductor materials, the resulting device is called
a semiconductor diode
...
14

k

a
(g) Photodiode

k

a
(h) Varactor diode

Section 1

Semiconductor diodes 147

Section 1

148 Electrical and Electronic Principles and Technology
11
...
Rectifier diodes need
to be able to cope with high values of reverse voltage
and large values of forward current, and consistency of
characteristics is of secondary importance in such applications
...
1 summarises the characteristics of
some common semiconductor diodes
...
This limit is
based on the physical size and construction of the diode
...
6 V and a reverse
breakdown voltage of 200 V
...
Typical values of maximum repetitive reverse voltage (VRRM ) or
peak inverse voltage (PIV) range from about 50 V to
over 500 V
...
If this voltage is exceeded the junction
may break down and the diode may suffer permanent
damage
...
8

Rectification

The process of obtaining unidirectional currents and
voltages from alternating currents and voltages is called
Table 11
...
Semiconductor diodes are commonly
used to convert alternating current (a
...
) to direct current
(d
...
), in which case they are referred to as rectifiers
...
Four diodes are connected as a bridge
rectifier – see Fig
...
14(d) – and are often used as a
full-wave rectifier
...

For methods of half-wave and full-wave rectification,
see Section 14
...


11
...

A similar effect, called avalanche breakdown, occurs
in less heavily doped diodes
...
For avalanche diodes, this breakdown
voltage usually occurs at voltages above 6 V
...
The symbol for a Zener diode is shown in
Fig
...
14(b) whilst a typical Zener diode characteristic
is shown in Fig
...
15
...
16
Figure 11
...
When a diode
is undergoing reverse breakdown and provided its maximum ratings are not exceeded, the voltage appearing
across it will remain substantially constant (equal to the
nominal Zener voltage) regardless of the current flowing
...

Zener diodes are available in various families
(according to their general characteristics, encapsulations and power ratings) with reverse breakdown
(Zener) voltages in the range 2
...


Problem 7
...
11
...
Use the characteristic to
determine (a) the current flowing in the diode when
a reverse voltage of 30 V is applied, (b) the voltage
dropped across the diode when a reverse current of
5 mA is flowing in it, (c) the voltage rating for the
Zener diode, and (d) the power dissipated in the
Zener diode when a reverse voltage of 30 V appears
across it
...
5 mA
(b) When I = −5 mA, the voltage dropped across the
diode, V = −27
...
5 × 10−3 ) = 0
...
10 Silicon controlled rectifiers
Silicon controlled rectifiers (or thyristors) are threeterminal devices which can be used for switching
and a
...
power control
...
In the off state, the silicon controlled
rectifier exhibits negligible leakage current, while in the
on state the device exhibits very low resistance
...


Section 1

150 Electrical and Electronic Principles and Technology
Once switched into the conducting state, the silicon controlled rectifier will remain conducting (i
...
it
is latched in the on state) until the forward current is
removed from the device
...
c
...
Where the device is used with an alternating
supply, the device will automatically become reset
whenever the main supply reverses
...

Like their conventional silicon diode counterparts,
silicon controlled rectifiers have anode and cathode
connections; control is applied by means of a gate terminal, g
...
11
...

In normal use, a silicon controlled rectifier (SCR)
is triggered into the conducting (on) state by means of
the application of a current pulse to the gate terminal –
see Fig
...
17
...
Triggering can become erratic when insufficient gate current is
available or when the gate current changes slowly
...
The viewing angle for round LEDs
tends to be in the region of 20◦ to 40◦ , whereas for
rectangular types this is increased to around 100◦
...
The symbol for an LED is shown in
Fig
...
14(f)
...
12 Varactor diodes
It was shown earlier that when a diode is operated in
the reverse biased condition, the width of the depletion
region increases as the applied voltage increases
...
11
...
The typical variation of capacitance
provided by a varactor is from about 50 pF to 10 pF as
the reverse voltage is increased from 2 V to 20 V
...
11
...


Controlled load, RL

RG
Gate trigger
pulse

AC or DC
supply
SCR

Figure 11
...
5 V to control a current of up to 5 A
...
18

11
...
11

Schottky diodes

Light emitting diodes

Light emitting diodes (LED) can be used as generalpurpose indicators and, compared with conventional
filament lamps, operate from significantly smaller voltages and currents
...
Most LEDs will provide
a reasonable level of light output when a forward current
of between 5 mA and 20 mA is applied
...
Round LEDs
are commonly available in the 3 mm and 5 mm (0
...
4 operates well as a rectifier and switching
device at relatively low frequencies (i
...
50 Hz to
400 Hz) but its performance as a rectifier becomes seriously impaired at high frequencies due to the presence
of stored charge carriers in the junction
...

This problem becomes increasingly more problematic
as the frequency of the a
...
supply is increased and the
periodic time of the applied voltage becomes smaller
...
State TWO applications for Schottky diodes
...
19

To avoid these problems a diode that uses a metalsemiconductor contact rather than a p-n junction (see
Fig
...
19) is employed
...
35 V) and a
slightly reduced maximum reverse voltage rating (typically 50 V to 200 V)
...

Schottky diodes are also extensively used in the construction of integrated circuits designed for high-speed
digital logic applications
...
The graph shown in Fig
...
21 was obtained
during an experiment on a Zener diode
...

(b) Determine the reverse voltage for a reverse
current of −20 mA
...
5 V
...

[(a) 5
...
8 V (c) –5 mA (d) 195 mW]

Reverse voltage (V)
−8

−6

−4

−2

0
0

−10

−20

−30

Now try the following exercises

−40

Exercise 61 Further problems on
semiconductor diodes
1
...
11
...


Reverse current (mA)

Figure 11
...
20

2
...

3
...

4
...


1
...
A semiconductor has a resistivity in the order
of …… to …… m
3
...
Over a limited range, the resistance of an
insulator …… with increase in temperature
5
...
Over a limited range, the resistance of a
conductor …… with increase in temperature

applied before an appreciable current starts
to flow

7
...
When a silicon p-n junction is forward biased,
approximately …… mV must be applied
before an appreciable current starts to flow

8
...
When a p-n junction is reversed biased, the
thickness or width of the depletion layer ……

9
...
If the thickness or width of a depletion layer
decreases, then the p-n junction is ……
biased

10
...
Name two p-type impurities

29
...
Antimony is called …… impurity

30
...
Arsenic has …… valency electrons

31
...
When phosphorus is introduced into a semiconductor material, mobile …… result
15
...
Indium has …… valency electrons
17
...
What is a thyristor? State a typical practical application and sketch its circuit diagram
symbol
33
...
When a p-n junction is formed, the n-type
material acquires a …… charge due to losing
……

34
...
When a p-n junction is formed, the p-type
material acquires a …… charge due to losing
……

35
...
What is meant by contact potential in a p-n
junction?
21
...
In a p-n junction, what is diffusion?
23
...
To reverse bias a p-n junction, the positive
terminal of the battery is connected to the
…… material
25
...

1
...
Intrinsic semiconductor materials have:
(a) covalent bonds forming a tetrahedral
structure
(b) pentavalent atoms added
(c) conduction by means of doping
(d) a resistance which increases with
increase of temperature
3
...
Free electrons in a p-type material:
(a) are majority carriers
(b) take no part in conduction
(c) are minority carriers
(d) exist in the same numbers as holes
5
...

6
...
Trivalent impurities:
(a) have three valeney electrons
(b) introduce holes when added to a semiconductor material
(c) can be introduced to a semiconductor
material by adding antimony atoms to it
(d) increase the conductivity of a semiconductor material when added to it
8
...
When a germanium p-n junction diode is
forward biased:
(a) current starts to flow in an appreciable amount when the applied voltage is
about 600 mV
(b) the thickness or width of the depletion
layer is reduced
(c) the curve representing the current flow
is exponential
(d) the positive terminal of the battery is
connected to the p-type material
10
...
1 Transistor classification
Transistors fall into two main classes – bipolar and
field effect
...
Transistors are also classified according to the
application that they are designed for, as shown in Table
12
...
Note that these classifications can be combined so

Table 12
...


The symbols and simplified junction models for
n-p-n and p-n-p transistors are shown in Fig
...
3
...


12
...
The junctions are, in fact, produced in a single slice
of silicon by diffusing impurities through a photographically reduced mask
...

The construction of typical n-p-n and p-n-p transistors
is shown in Figs
...
1 and 12
...
In order to conduct the
heat away from the junction (important in medium- and
high-power applications) the collector is connected to
the metal case of the transistor
...
3
Oxide layer
(insulation)
n

p

n

12
...
12
...
1

(a) the majority carriers in the n-type emitter material
are electrons
Base

Emitter

Oxide layer
(insulation)
p

Metal case
(conductor)
Collector

Figure 12
...
Around
99
...
5% of the electrons
will recombine with holes in the narrow base region
...
12
...
4

The transistor action for an n-p-n device is shown
diagrammatically in Fig
...
5(a)
...
12
...
Around 99
...
5% of the holes will recombine with
electrons in the narrow base region
...
4

Leakage current

(a) n-p-n bipolar junction transistor
Emitter
IE

n

p

Holes

p

Collector
IC

Base
IB

(b) p-n-p bipolar junction transistor

Figure 12
...

The base-collector junction is forward biased to these
minority carriers
...
However, a small leakage current, ICBO , flows from the base
to the collector due to thermally generated minority carriers (electrons in the collector and holes in the base),

being present
...

With modern transistors, leakage current is usually
very small (typically less than 100 nA) and in most
applications it can be ignored
...
With reference to a p-n-p transistor,
explain briefly what is meant by the term ‘transistor
action’ and why a bipolar junction transistor is so
named
...
12
...

When the emitter terminal is made sufficiently positive
with respect to the base, the base-emitter junction is
forward biased to the majority carriers
...

The base region is relatively lightly doped with
donor atoms (electrons) and although some electronhole recombination’s take place, perhaps 0
...

The base-collector junction is reverse biased to electrons in the base region, but forward biased to holes in
the base region
...
05IE

Collector
n
IC
>0
...
05IE

junction towards the negative potential of the collector
terminal
...

The essence of transistor action is this current control by means of the base-emitter voltage
...
Also thermally generated electrons in
the emitter and collector regions are minority carriers
as are holes in the base region
...
12
...
It is because a transistor
makes use of both types of charge carriers (holes and
electrons) that they are called bipolar
...


12
...
e
...

The base region is, however, made very narrow so that
carriers are swept across it from emitter to collector so
that only a relatively small current flows in the base
...
The direction of conventional current flow is
from emitter to collector in the case of a p-n-p transistor,
and collector to emitter in the case of an n-p-n device,
as shown in Fig
...
7
...
12
...
95IE
<0
...


Collector
p
IC
ICBO

Problem 2
...
Determine the value of base current
...
6

Emitter current,
IE = IB + IC
from which, base current, IB = IE − IC
Hence, base current,
IB = 102 − 100 = 2 mA

Section 1

Transistors 157

158 Electrical and Electronic Principles and Technology
e

IC

c

IE
VBE
−

b

+

−

IB

VCB

IC
IB

+

(a) n-p-n bipolar junction transistor (BJT)
IE

e

IC

c

IE
VBE
+
−

IB

VCB

b
+

IB

IC

−

(b) p-n-p bipolar junction transistor (BJT)

12
...

Fig
...
9 shows a typical input characteristic (IB plotted against VBE ) for an n-p-n bipolar junction transistor
operating in common-emitter mode
...
12
...

The input characteristic shows that very little base
current flows until the base emitter voltage VBE exceeds
0
...
Thereafter, the base current increases rapidly —
this characteristic bears a close resemblance to the
forward part of the characteristic for a silicon diode
...
7
500

12
...
These three circuit configurations depend
upon which one of the three transistor connections is
made common to both the input and the output
...
12
...


Base current, IB (μA)

Section 1

IE

300

200

100

0

Output

0

Common
Input

0
...
4
0
...
8
Base-emitter voltage,VBE (V)

1
...
9
Input
Output

Common
(a) Common emitter

(b) Common collector
Output

Input

Common
(c) Common base

Figure 12
...
10 shows a typical set of output (collector)
characteristics (IC plotted against VCE ) for an n-p-n
bipolar transistor
...
Note the ‘knee’ in the characteristic below VCE = 2 V
...
For this reason (i
...
since the collector current does
not change very much as the collector-emitter voltage
changes) we often refer to this as a constant current
characteristic
...
11 shows a typical transfer characteristic
for an n-p-n bipolar junction transistor
...


Transistors 159
IC

18

+

16

B1
Collector current, IC (mA)

IB

IB = 100μA

14

−

μA

VR1
VBE mV

mA
TR1
c
b

+
B2

V
VCE VR2

−

e

12

Figure 12
...
Then VR2 is set at various values and
corresponding values of VCE and IC are noted
...
This is repeated
for, say, IB = 40 μA, IB = 60 μA, and so on
...
12
...


IB = 60μA

6
IB = 40μA

4
2
0

IB = 20μA
0

2

4

6

8

10

12

14

16

18

20

Collector-emitter voltage, VCE (V)

12
...
10

The transistor characteristics met in the previous section
provide us with some useful information that can help us
to model the behaviour of a transistor
...
12
...
c
...
c
...
11

The slope of this curve (i
...
the ratio of IC to IB ) is the
common-emitter current gain of the transistor which is
explored further in Section 12
...

A circuit that can be used for obtaining the commonemitter characteristics of an n-p-n BJT is shown in
Fig
...
12
...
This is repeated for various settings
of VR1 and plotting the values gives the typical input
characteristic of Fig
...
9
...
12
...
c
...
c
...
c
...
c
...


(c) From Fig
...
13, VBE changes by 0
...

Hence,
dynamic value of input resistance
VBE
0
...
Figure 12
...
When
the collector-emitter voltage is 10 V and the base
current is 80 μA, determine (a) the value of collector
current, (b) the static value of output resistance,
and (c) the dynamic value of output resistance
...
Figure 12
...
When
the base-emitter voltage is 0
...


IB = 100μA

14
12

(c)

10 10 mA

(a, b)

IB = 80μA

1
...
14

(a) From Fig
...
14, when VCE = 10 V and IB = 80 μA,
(i
...
point (a, b) on the graph), the

100

collector current, IC = 10 mA
0
...
12
...
2

0
...
6
0
...
8

1
...
13

(a) From Fig
...
13, when VBE = 0
...
65 V, IB = 250 μA,
the static value of input resistance
=

hence,

VBE
0
...
6 k
IB
250 × 10−6

(b) When VCE = 10 V and IB = 80 μA then IC = 10 mA
from part (a)
...
8 mA (shown as point (c) on the graph)
Hence,
the dynamic value of output resistance
VCE
12
=
=
= 6
...
8 × 10−3

Problem 5
...
15 shows the transfer
characteristic for an n-p-n silicon transistor
...
5 mA, determine (a) the value
of collector current, (b) the static value of current
gain, and (c) the dynamic value of current gain
...
65 mA

100

0

1

0

2
...
15

(a) From Fig
...
15, when IB = 2
...
5 mA, IC = 280 mA
hence,
the static value of current gain
IC 280 × 10−3
= =
= 112
IB
2
...
12
...

Hence,
the dynamic value of current gain
=

IC (460 − 110) × 10−3
350
= 96
=
=
−3
IB (4
...
75) × 10
3
...
9 Current gain
As stated earlier, the common-emitter current gain is
given by the ratio of collector current, IC , to base current,
IB
...
Note that hFE is found from corresponding
static values while hfe is found by measuring the slope
of the graph
...

It is worth noting that current gain (hfe ) varies with
collector current
...
Current gain also falls to very low values
for power transistors when operating at very high values of collector current
...
It is, therefore, important to design circuits
on the basis of the minimum value for hfe in order to
ensure successful operation with a variety of different
devices
...
A bipolar transistor has a
common-emitter current gain of 125
...


Common-emitter current gain, hFE =

IC
IB

from which, base current,
IB =

IC
50 × 10−3
=
= 400 μA
hFE
125

12
...
2 summarises the characteristics of some
typical bipolar junction transistors for different applications, where IC max is the maximum collector current, VCE max is the maximum collector-emitter voltage,
PTOT max is the maximum device power dissipation,
and hfe is the typical value of common-emitter current
gain
...
2 Transistor characteristics and maximum ratings
Device

Type

I C max
...


P TOT
max
...
Which of the bipolar transistors
listed in Table 12
...


(a) BF180, since this transistor is designed for use in
radio frequency (RF) applications
(b) 2N3055, since this is the only device in the list that
can operate at a sufficiently high power level
(c) 2N3904, since switching transistors are designed
for use in pulse and square wave applications

Now try the following exercise
Exercise 64

Further problems on bipolar
junction transistors

1
...

2
...

3
...

4
...
Explain the shape of
the characteristics
...
Sketch the typical input characteristic relating base current and the base-emitter voltage
for a transistor connected in common-emitter
configuration and explain its shape
...
With the aid of a circuit diagram, explain
how the input and output characteristic of
a common-emitter n-p-n transistor may be
produced
...
Define the term ‘current gain’ for a bipolar
junction transistor operating in commonemitter mode
...
A bipolar junction transistor operates with
a collector current of 1
...
What is the value of common-emitter current
gain for the transistor in problem 8?
[24]
10
...
16

VBE (V) 0 0
...
2 0
...
4 0
...
6 0
...
8
IB (μA) 0 0 0 0 1 3 19 57 130
Plot the IB /VBE characteristic for the device
and use it to determine (a) the value of
IB when VBE = 0
...
65 V, and
(c) the dynamic value of input resistance
when VBE = 0
...
5 μA (b) 20 k (c) 3 k ]
11
...
1 2
...
1 4
...
9 5
...
7 7
...

[98]

12
...
The gate-source junction of a junction gate field effect transistor (JFET)
is effectively a reverse-biased p-n junction
...
To keep things simple, we will consider only JFET
devices
...
16 shows the basic construction of
an n-channel JFET
...
The ends of the channel (in which conduction takes place) form electrodes known as the source
and drain
...
The effective resistance between the source and drain is thus determined by
the voltage present at the gate
...
12
...
)
JFETs offer a very much higher input resistance
when compared with bipolar transistors
...
5 k
...

As with bipolar transistors, the characteristics of a
FET are often presented in the form of a set of graphs
relating voltage and current present at the transistors,
terminals
...
12 Field effect transistor
characteristics
A typical mutual characteristic (ID plotted against
VGS ) for a small-signal general-purpose n-channel field
effect transistor operating in common-source mode is
shown in Fig
...
17
...
At a certain value of VGS
the drain current falls to zero and the device is said to
be cut-off
...
18 shows a typical family of output characteristics (ID plotted against VDS ) for a small-signal
general-purpose n-channel FET operating in common
source mode
...
What will the value of emitter
current be?
[1
...
In this mode, the
input voltage is applied to the gate and the output current
appears in the drain (the source is effectively common
to both the input and output circuits)
...
c
...
0 −4
...
0 −3
...
0 −2
...
0 −1
...
0 −0
...
c
...
17

20

VGS = 0 V

18
Drain current, ID (mA)

16

VGS = −1 V

14
12

VGS = −2 V

10
8

VGS = −3 V

6
4

VGS = −4 V

2
0

VGS = −5 V
0

2

ID
VGS

4 6 8 10 12 14 16 18 20
Drain-source voltage, VDS (V)

(Note that ID means ‘change of ID ’ and
‘change of VGS ’)

VGS means

The method for determining these parameters from
the relevant characteristic is illustrated in worked problem 8 below
...
For most small-signal devices, gfs , is quoted for
values of drain current between 1 mA and 10 mA
...
It is, therefore, important to design circuits
on the basis of the minimum value for gfs , in order
to ensure successful operation with a variety of different devices
...
12
...


Figure 12
...
You might also like to compare this characteristic with the output characteristic for a transistor
operating in common-emitter mode that you met earlier
in Fig
...
10
...
Also, note
how the curves become flattened above this value with
the drain current ID not changing very significantly for a
comparatively large change in drain-source voltage VDS
...
Because of their flatness, they are
often said to represent a constant current characteristic
...
19

Problem 8
...
20 shows the mutual
characteristic for a junction gate field effect
transistor
...
5 V,
determine (a) the value of drain current, (b) the
dynamic value of forward transconductance
...
25 × −0
...
025 A = −25 mA

20
18

(b) The new value of drain current = (100 − 25)
= 75 mA

14
12
10
12 mA

8
6

Drain current, ID (mA)

16

5 mA
4
2
...
5 V
−5
...
5 −4
...
5 −3
...
5 −2
...
5 −1
...
5

2
0

0

Gate-source voltage, VGS (V)

12
...
3 summarises the characteristics of some typical field effect transistors for different applications,
where ID max is the maximum drain current, VDS max
is the maximum drain-source voltage, PD max is the
maximum drain power dissipation, and gfs typ is the
typical value of forward transconductance for the transistor
...


Figure 12
...
12
...
5 V, the drain
current, ID = 5 mA
(b) From Fig
...
20
gfs =

ID
(14
...
5) × 10−3
=
VGS
2
...
e
...
8 mS
2
...
A field effect transistor operates with
a drain current of 100 mA and a gate source bias of
−1 V
...
25
...
1 V, determine (a) the
change in drain current, and (b) the new value of
drain current
...
1 V
and the resulting change in drain current can be
determined from:
gfs =

ID
VGS

Hence, the change in drain current,
ID = gfs ×

VGS

Problem 10
...
3 would be most suitable for each
of the following applications: (a) the input stage of
a radio receiver, (b) the output stage of a transmitter,
and (c) switching a load connected to a
high-voltage supply
...
14 Transistor amplifiers
Three basic circuit arrangements are used for transistor
amplifiers and these are based on the three circuit configurations that we met earlier (i
...
they depend upon which
one of the three transistor connections is made common
to both the input and the output)
...

Where field effect transistors are used, the corresponding configurations are common source, common
drain (or source follower) and common gate
...
12
...
22 exhibit quite different performance characteristics, as shown in Tables 12
...
5
respectively
...
3

FET characteristics and maximum ratings

Device

Type

I D max
...


P D max
...


Application

2N2819

n-chan
...
5 mS

General purpose

2N5457

n-chan
...
2 mS

General purpose

2N7000

n-chan
...


100 mA

30 V

360 mW

BSS84

p-chan
...
27 S

IRF830

n-chan
...
5 A

500 V

75 W

3
...


4
...
8 S

RF power amplifier

0
...
3 mS

RF amplifier
Low-power switching

+V

+V

RL

RL

TR

TR
Output

Input

Output
Input

Common, 0 V

Common, 0 V

(a) Common emitter

(a) Common source

+V

+V

TR

TR

Input
RL

Input

Output

RL

Common, 0 V

Output

Common, 0 V

(b) Common collector

(b) Common drain

+V

+V

RL

RL

TR
TR
Input

Output
Input
Common, 0 V

Common, 0 V

(c) Common base
Bipolar transistor amplifier circuit configurations

Figure 12
...
22

Table 12
...
21)
Common emitter
Common collector

Common base

Voltage gain

medium/high (40)

unity (1)

high (200)

Current gain

high (200)

high (200)

unity (1)

Power gain

very high (8000)

high (200)

high (200)

Input resistance

medium (2
...
5

Characteristics of FET amplifiers

Parameter

Field effect transistor amplifiers (see Figure 12
...
Other types of amplifier
are ‘non-linear’, in which case their input and output
waveforms will not necessarily be similar
...
It
is also worth noting that a linear amplifier will become
non-linear when the applied input signal exceeds a
threshold value
...

The optimum value of bias for linear (Class A)
amplifiers is that value which ensures that the active
devices are operated at the mid-point of their characteristics
...

Furthermore, the collector current will flow throughout
the complete cycle of an input signal (i
...
conduction will take place over an angle of 360◦ )
...

In order to ensure that a static value of collector
current flows in a transistor, a small current must be
applied to the base of the transistor
...
Figure 12
...


+Vcc
R2
R1
C1

C2
TR1

Output

current that would flow with the device totally saturated
(VCE = 0 V)
...

Figure 12
...
The quiescent point (or
operating point) is the point on the load line that corresponds to the conditions that exist when no-signal is
applied to the stage
...
12
...
This position ensures that the collector voltage can swing both
positively (above) and negatively (below) its quiescent
value (VCQ )
...
23
ICQ

The a
...
signal is applied to the base terminal of the
transistor via a coupling capacitor, C1
...
c
...
C2 couples the signal out of the stage and
also prevents d
...
current flow appearing at the output
terminals
...
15 Load lines
The a
...
performance of a transistor amplifier stage can
be predicted using a load line superimposed on the relevant set of output characteristics
...
One end of the load
line corresponds to the supply voltage (VCC ) while the
other end corresponds to the value of collector or drain

Input (base current)
signal
IB = 20 μA
IB = 10 μA

Lo

ad

VCQ

line
VCC VCE(V)

Output (collector current)
signal

Figure 12
...
c
...
The corresponding collector current signal can be determined by simply moving
up and down the load line
...
The characteristic curves shown in
Fig
...
25 relate to a transistor operating in
common-emitter mode
...
2 k and an

18 V supply, determine (a) the quiescent values of
collector voltage and current (VCQ and ICQ ), and
(b) the peak-peak output voltage that would be
produced by an input signal of 40 μA peak-peak
...
4

1
...
0

3
...
6

5
...
2
kΩ

IB = 40 μA

10
Operating point
8
ICQ = 7
...
An n-p-n transistor has the following
characteristics, which may be assumed to be linear
between the values of collector voltage stated
...
3 V

6

8

10 12
VCQ = 9
...
8 V

20

Collector-emitter voltage, VCE (V)

Figure 12
...
12
...
The two ends of the load line will correspond to VCC , the 18 V supply, on the collectoremitter voltage axis and 18 V/1
...

Next we locate the operating point (or quiescent point) from the point of intersection of the
IB = 30 μA characteristic and the load line
...
e
...
Hence, VCQ = 9
...
3 mA
(b) Next we can determine the maximum and minimum values of collector-emitter voltage by
locating the appropriate intercept points on
Fig
...
25
...
The maximum and minimum
values of VCE are, respectively, 14
...
3 V
...
8 V − 3
...
5 V peak-peak

The transistor is used as a common-emitter
amplifier with load resistor RL = 1
...
The signal input resistance
is 1 k
...


The characteristics are drawn as shown in Fig
...
26
...
83
5
4
3
...
3

50 μA

X

3
30 μA

2
1
3
...
4
0

1

2

3

4
...
6 V
pk–pk

Figure 12
...
2 k = 5
...

(a) The operating point (or quiescent point), X, is
located from the point of intersection of the

Section 1

Transistors 169

Section 1

170 Electrical and Electronic Principles and Technology
IB = 50 μA characteristic and the load line
...
e
...
Hence, VCQ = 3
...
3 mA
(b) The maximum and minimum values of collectoremitter voltage may be determined by locating
the appropriate intercept points on Fig
...
26
...
The maximum and minimum values of VCE are, respectively,
4
...
4 V
...
9 V − 1
...
5 V peak-peak
(c) Voltage gain =

change in collector voltage
change in base voltage

The change in collector voltage = 3
...

The input voltage swing is given by: ib Ri ,
where ib is the base current swing = (70 − 30) =
40 μA and Ri is the input resistance = 1 k
...

Thus, voltage gain =
=

change in collector voltage
change in base voltage
VC
3
...
5
VB 40 × 10−3

(d) Dynamic current gain, hfe =

IC
IB

From Figure 12
...
e
...
0 mA
peak to peak
...

Hence, the dynamic current gain,
hfe =

IC 3
...
5 × 75 = 6562
...
State whether the following statements are true
or false:
(a) The purpose of a transistor amplifier is
to increase the frequency of the input
signal
...

(c) The output characteristics of a transistor relate the collector current to the base
current
...

(e) In a common-emitter amplifier, the output voltage is shifted through 180◦ with
reference to the input voltage
...

(g) The dynamic current gain of a transistor
is always greater than the static current
gain
...
In relation to a simple transistor amplifier
stage, explain what is meant by the terms:
(a) Class-A (b) saturation (c) cut-off
(d) quiescent point
...
Sketch the circuit of a simple Class-A BJT
amplifier and explain the function of the
components
...
Explain, with the aid of a labelled sketch,
how a load line can be used to determine the
operating point of a simple Class-A transistor
amplifier
...
Sketch circuits showing how a JFET can
be connected as an amplifier in: (a) common source configuration, (b) common drain
configuration, (c) common gate configuration
...

6
...
12
...
If this device is used in a
common-emitter amplifier circuit operating

from a 12 V supply with a base bias of 60 μA
and a load resistor of 1 k , determine (a) the
quiescent values of collector-emitter voltage
and collector current, and (b) the peak-peak
collector voltage when an 80 μA peak-peak
signal current is applied
...
5 V]
20

Collector current, IC (mA)

9
...

IB = 20 μA 50 μA

IB = 120 μA

18
IB = 100 μA

14
12

VCE (v)

80 μA

1
...
0

1
...
0 1
...
0

IC (mA) 1
...
4

16

3
...
2 6
...
1

IB = 80 μA

10
IB = 60 μA

8
6

IB = 40 μA

4
IB = 20 μA

2
0
0

2

4

6

8

10 12 14 16 18 20

Collector-emitter voltage, VCE (V)

Figure 12
...
The output characteristics of a JFET are shown
in Fig
...
28
...

[(a) 12
...
1 mA (b) 5
...
75]

20

VGS = 0 V

18
16
Drain current, ID (mA)

8
...
Determine the power gain
...
c
...
The signal input resistance is 800
...

[(a) 5
...
7 mA (b) 5
...
In a p-n-p transistor the p-type material
regions are called the
...
, and
the n-type material region is called the
...
In an n-p-n transistor, the p-type material
region is called the
...
and the

...
In a p-n-p transistor, the base-emitter junction is ……biased and the base-collector
junction is
...
28

4
...
biased and the base-emitter
junction is
...
Majority charge carriers in the emitter of a
transistor pass into the base region
...
doped
6
...
biased
7
...
flow

14
...
State
the advantage of a JFET over a bipolar
transistor
15
...
Leakage current flows from
...

in an n-p-n transistor

16
...
The input characteristic of IB against VBE for
a transistor in common-emitter configuration
is similar in shape to that of a
...
Name and sketch three possible circuit
arrangements used for FETs

10
...

static input resistance =
and

...

dynamic input resistance =

...
From a transistor output characteristic,

...


...

12
...

static current gain =
and dynamic

...

current gain =

...
Complete the following statements that refer
to a transistor amplifier:
(a) An increase in base current causes collector current to
...

(c) Under no-signal conditions the power
supplied by the battery to an amplifier
equals the power dissipated in the load
plus the power dissipated in the
...
gradient
(e) The gradient of the load line depends
upon the value of
...

(g) The current gain of a common-emitter
amplifier is always greater than
...
of the load line

18
...
Explain briefly the purpose of all
the components you show in your diagram
19
...
c
...
What is the quiescent point on a load line?

Exercise 67

Multi-choice problems on
transistors
(Answers on page 398)

In Problems 1 to 10 select the correct answer from
those given
...
In normal operation, the junctions of a p-n-p
transistor are:
(a) both forward biased
(b) base-emitter forward biased and basecollector reverse biased
(c) both reverse biased
(d) base-collector forward biased and baseemitter reverse biased
2
...
The current flow across the base-emitter
junction of a p-n-p transistor
(a) mainly electrons
(b) equal numbers of holes and electrons
(c) mainly holes
(d) the leakage current
4
...
In normal operation an n-p-n transistor connected in common-base configuration has
(a) the emitter at a lower potential than the
base
(b) the collector at a lower potential than
the base
(c) the base at a lower potential than the
emitter
(d) the collector at a lower potential than
the emitter
6
...
If the per unit value of electrons which leave
the emitter and pass to the collector is 0
...
4 mA
(b) the collector current is approximately
3
...
4 mA
(d) the base current is approximately
3
...
The base region of a p-n-p transistor is
(a) very thin and heavily doped with holes
(b) very thin and heavily doped with
electrons

(c) very thin and lightly doped with holes
(d) very thin and lightly doped with
electrons
9
...
4 V
10
...

In questions 11 to 15, which refer to the amplifier
shown in Fig
...
29, select the correct answer from
those given
+ V cc
R1

RL
V0

Vi
R2

RE

Figure 12
...
If RL short-circuited:
(a) the amplifier signal output would fall to
zero
(b) the collector current would fall to zero
(c) the transistor would overload
12
...
A voltmeter connected across RE reads zero
...
A voltmeter connected across RL reads zero
...
If RE short-circuited:
(a) the load line would be unaffected
(b) the load line would be affected
In questions 16 to 20, which refer to the
output characteristics shown in Fig
...
30,
select the correct answer from those given
...
30

2

4

6

8

10

12

VCE(V)

16
...
5 k
17
...
The greatest permissible peak input current
would be about
(a) 30 μA
(b) 35 μA
(d) 80 μA
(c) 60 μA
19
...
2 V
(b) 6
...
8 V
(d) 13 V
20
...
The marks for each question are shown in brackets
at the end of each question
...
A conductor, 25 cm long is situated at right angles
to a magnetic field
...
5 N
...
An electron in a television tube has a charge of
1
...
Calculate
the force exerted on the electron in the field
...
A lorry is travelling at 100 km/h
...
98 m, find
the e
...
f
...
(4)
4
...
m
...
of 2
...
Calculate
the inductance of the coil
...
Two coils, P and Q, have a mutual inductance of
100 mH
...
m
...
induced in
coil Q, and (b) the flux change linked with coil Q
if it is wound with 200 turns
...
A moving coil instrument gives a f
...
d
...

Determine the value of resistance required to
enable the instrument to be used (a) as a 0–5 A
ammeter, and (b) as a 0–200 V voltmeter
...

(8)
7
...
Its input power
is 5 mW
...

(3)
8
...
RT3
...
Determine for the waveform (a) the
frequency (b) the peak-to-peak voltage (c) the
amplitude (d) the r
...
s
...

(7)
9
...
Briefly describe each of the following, drawing
their circuit diagram symbol and stating typical
applications: (a) zenor diode (b) silicon controlled
rectifier (c) light emitting diode (d) varactor diode
(e) Schottky diode
(20)

Figure RT3
...
The following values were obtained during an
experiment on a varactor diode
...
Label your axes
clearly and use your graph to determine (a) the
capacitance when the reverse voltage is −17
...
5 V to −22
...

(8)
12
...

(7)
13
...
Assume that
the characteristics are linear between the values of
collector voltage stated
...
0

7
...
0

7
...
0

7
...
6

0
...
5

2
...
6

5
...
5 k load and collector supply voltage of 8 V
...
2 k
...

(18)

Section 1

Revision Test 3

Formulae for basic electrical and electronic engineering principles

Magnetic Circuits:

General:
Charge Q = It
Work W = Fs

Force F = ma
Power P =

B=

W
t

S=

A
mmf

Fm = NI

Conductance G =

or
1
R

I=

V
R

or

R=

Resistance R =

Power P = VI = I 2 R =

V
I

ρl
a

V2
R

Electromagnetism:
F = BIl sin θ

F = QvB

Electromagnetic Induction:
E = Blv sin θ

Resistance at θ ◦ C, Rθ = R0 (1 + α0 θ)

B
= μ0 μr
H

l
μ0 μr A

=

Energy W = Pt
Ohm’s law V = IR

NI
l

H=

E =−N

d
dI
= −L
dt
dt

1
N
N2
dI1
N1 N2
W = LI 2 L =
=
E2 = −M
M=
2
I
S
dt
S

Terminal p
...
of source, V = E − Ir
Series circuit R = R1 + R2 + R3 + · · ·
Parallel network

1
1
1
1
=
+
+
+ ···
R R 1 R2 R 3

Measurements:
Shunt Rs =

Ia ra
Is

Multiplier RM =

Power in decibels = 10 log

P2
P1

= 20 log

I2
I1

= 20 log

V2
V1

Capacitors and Capacitance:
V
E=
d
D
= ε0 εr
E

Q
C=
V
C=

Q
Q = It D =
A
ε0 εr A(n − 1)
d

1
W = CV 2
2

Capacitors in parallel C = C1 + C2 + C3 + · · ·
Capacitors in series

1
1
1
1
+
+
+ ···
=
C C1 C2 C3

Wheatstone bridge RX =

R2 R3
R1

Potentiometer E2 = E1

l2
l1

V − Ira
I

Section 2

Further Electrical and
Electronic Principles

This page intentionally left blank

Chapter 13

D
...
circuit theory
At the end of this chapter you should be able to:
• state and use Kirchhoff’s laws to determine unknown currents and voltages in d
...
circuits
• understand the superposition theorem and apply it to find currents in d
...
circuits
• understand general d
...
circuit theory
• understand Thévenin’s theorem and apply a procedure to determine unknown currents in d
...
circuits
• recognize the circuit diagram symbols for ideal voltage and current sources
• understand Norton’s theorem and apply a procedure to determine unknown currents in d
...
circuits
• appreciate and use the equivalence of the Thévenin and Norton equivalent networks
• state the maximum power transfer theorem and use it to determine maximum power in a d
...
circuit

13
...
c
...
2 following)
...
These include:
(i)
(ii)
(iii)
(iv)

the superposition theorem (see Section 13
...
5),
Norton’s theorem (see Section 13
...
8)

13
...
13
...
At any junction in an electric circuit
the total current flowing towards that junction is
equal to the total current flowing away from the
junction, i
...
I = 0

I1 + I2 = I3 + I4 + I5
or

I1 + I2 − I3 − I4 − I5 = 0

Figure 13
...
In any closed loop in a network, the
algebraic sum of the voltage drops (i
...
products
of current and resistance) taken around the loop is
equal to the resultant e
...
f
...

Thus, referring to Fig
...
2:
E1 − E2 = IR1 + IR2 + IR3
(Note that if current flows away from the positive
terminal of a source, that source is considered by
convention to be positive
...
13
...


180 Electrical and Electronic Principles and Technology
Problem 2
...
13
...
2

Problem 1
...
13
...
(b) Determine the value of
e
...
f
...
13
...


Figure 13
...
3

(a) Applying Kirchhoff’s current law:
For junction B: 50 = 20 + I1
...


1
...
The directions
chosen are arbitrary, but it is usual, as a starting
point, to assume that current flows from the positive
terminals of the batteries
...
13
...
e
...

Hence
I 3 = −90 A
(i
...
in the opposite direction to that shown in
Fig
...
3(a))
For junction E: I4 + I3 = 15
i
...

I4 = 15 − (−90)
...

Hence
I 5 = 80 A
(b) Applying Kirchhoff’s voltage law and moving
clockwise around the loop of Fig
...
3(b) starting
at point A:

Figure 13
...
Divide the circuit into two loops and apply Kirchhoff’s voltage law to each
...
13
...
5)
+ (I)(1
...
5 + 1
...
e
...
e
...
e
...
13
...
C
...
e
...
e
...
Determine, using Kirchhoff’s laws,
each branch current for the network shown in
Fig
...
7

4I1 + 5I2 = 2

(2)

3
...
7

(3) − (4) gives: −7I2 = 2
hence I2 = −2/7 = −0
...
e
...
13
...
286) = 4

From (1)

1
...
13
...
It is
usual, although not essential, to follow conventional
current flow with current flowing from the positive
terminal of the source

6I1 = 4 + 1
...
144
= 0
...
857 + (−0
...
571 A
Note that a third loop is possible, as shown in
Fig
...
6, giving a third equation which can be used
as a check:

Figure 13
...
The network is divided into two loops as shown in
Fig
...
8
...
5I1 + 2I2

i
...


(1)

For loop 2:

[Check: 2I1 − I2 = 2(0
...
286) = 2]

E2 = I2 R2 − (I1 − I2 )R3
Note that since loop 2 is in the opposite direction
to current (I1 − I2 ), the volt drop across R3 (i
...

(I1 − I2 )(R3 )) is by convention negative
...
e
...
Solving Equations (1) and (2) to find I1 and I2 :
Figure 13
...
37 A
27
16 = 0
...
37)

Equations (1) and (2) are simultaneous equations with
two unknowns, I1 and I2
...
37)
I1 =
= 6
...
5

208I1 + 416I2 = 1456

(4)

592I2 = 592

(4) − (3) gives:

Current flowing in R3 = (I1 − I2 )

I2 = 1 A

= 6
...
37 = 0
...
For the bridge network shown in
Fig
...
9 determine the currents in each of the
resistors
...
9

the current flowing in the 32

Let the current in the 2 resistor be I1 , then by Kirchhoff’s current law, the current in the 14 resistor is
(I − I1 )
...
13
...
Then the current in the 11 resistor is (I1 − I2 ) and that in the 3 resistor is (I − I1 + I2 )
...
13
...
e
...
Find currents I3 , I4 and I6 in Fig
...
11
[I3 = 2 A, I4 = −1 A, I6 = 3 A]

Figure 13
...
13
...
e
...
11

(2)

D
...
circuit theory 183
2
...
13
...

[(a) I1 = 4 A, I2 = −1 A, I3 = 13 A
(b) I1 = 40 A, I2 = 60 A, I3 = 120 A
I4 = 100 A, I5 = −80 A]

3
...
13
...

[I1 = 0
...
5 A]
4Ω

0
...
5 V
0
...
5 A

6
...
13
...

[(a) 60
...
10 mA
(c) 45
...
20 mW]
7
...
13
...
26 A, I2 = 0
...
16 A,
I4 = 1
...
58 A]

Figure 13
...
Use Kirchhoff’s laws to find the current flowing in the 6 resistor of Fig
...
14 and the
power dissipated in the 4 resistor
...
162 A, 42
...
3 The superposition theorem
The superposition theorem states:
In any network made up of linear resistances and containing more than one source of e
...
f
...

The superposition theorem is demonstrated in the
following worked problems

Figure 13
...
Find the current flowing in the 3 resistor for
the network shown in Fig
...
15(a)
...
d
...

[2
...
410 V, 3
...
Figure 13
...
m
...
, each with their
internal resistance
...


Section 2

Figure 13
...
12

184 Electrical and Electronic Principles and Technology

Figure 13
...
16

Procedure:
1
...
13
...
Label the currents in each branch and their directions as shown in Fig
...
18(a) and determine their
values
...
333
From the equivalent circuit of Fig
...
18(b)
E2
2
=
= 0
...
333 + r2
1
...
13
...
17

2
...
13
...
(Note that the choice of current directions
depends on the battery polarity, which, by convention is taken as flowing from the positive battery
terminal as shown)
R in parallel with r2 gives an equivalent resistance
of (4 × 1)/(4 + 1) = 0
...
13
...
8
2 + 0
...
429 A

From Fig
...
17(a),
I2 =
and

1
1
I1 = (1
...
286 A
4+1
5

I3 =

4
4
I1 = (1
...
143 A
4+1
5
by current division

3
...
13
...
857) = 0
...
857) = 0
...
Superimpose Fig
...
18(a) on to Fig
...
17(a) as
shown in Fig
...
19

Figure 13
...
Determine the algebraic sum of the currents flowing
in each branch
...
e
...
429 − 0
...
858 A (discharging)
Resultant current flowing through source 2, i
...

I4 − I3 = 0
...
143
= −0
...
C
...
e
...
286 + 0
...
572 A

From Fig 13
...
667 A
3 + 1
...
8
From Fig 13
...
667) = 1
...
667) = 0
...
13
...
Removing source E1 gives the circuit of
Fig
...
23(a) (which is the same as Fig
...
23(b))

Figure 13
...
For the circuit shown in Fig
...
21,
find, using the superposition theorem, (a) the
current flowing in and the p
...
across the 18
resistor, (b) the current in the 8 V battery and (c) the
current in the 3 V battery
...
23

4
...
23(a) and 13
...
13
...
21

E2
3
=
= 0
...
571
4
...
13
...
Removing source E2 gives the circuit of
Fig
...
22(a)
2
...
13
...
656) = 0
...
656) = 0
...
Superimposing Fig
...
23(a) on to Fig
...
22(a)
gives the circuit in Fig
...
24
6
...
167 − 0
...
073 A
P
...
across the 18
Figure 13
...
073×18 = 1
...
Use the superposition theorem to find the current in each branch of the network shown in
Fig
...
27
[10 V battery discharges at 1
...
857 A
Current through 10 resistor is 0
...
24

(b) Resultant current in the 8 V battery
= I1 + I5 = 1
...
562
= 2
...
27

= I2 + I4 = 1
...
656
= 2
...
Use the superposition theorem to determine
the current in each branch of the arrangement
shown in Fig
...
28
[24 V battery charges at 1
...
280 A
Current in 20 resistor is 1
...
Use the superposition theorem to find currents
I1 , I2 and I3 of Fig
...
25
[I1 = 2 A, I2 = 3 A, I3 = 5 A]
Figure 13
...
4

Figure 13
...
Use the superposition theorem to find the current in the 8 resistor of Fig
...
26 [0
...
c
...
c
...
13
...


Figure 13
...
26

(ii) The open-circuit voltage, E, across terminals AB
in Fig
...
30(a) is the same as the voltage across

D
...
circuit theory 187
the 6 resistor
...
13
...
e
...
32

Figure 13
...
13
...
m
...
, V is
the battery terminal voltage and r is the internal resistance of the battery (as shown in Section 4
...
For the circuit shown in Fig
...
31(b),
V = E − (−I)r, i
...
V = E + Ir

Figure 13
...
31

(iv) The resistance ‘looking-in’ at terminals AB in
Fig
...
32(a) is obtained by reducing the circuit
in stages as shown in Figures 13
...

Hence the equivalent resistance acrossAB is 7
...
13
...
d
...
Redrawing the circuit gives
Fig
...
33(b), from which
E=

4
4+6

with zero ohms gives an equivalent resistance
of (20 × 0)/(20 + 0) i
...
0
...
13
...
13
...
From Fig
...
33(e), the equivalent
resistance across AB,
r=

6×4
+ 3 = 2
...
4
6+4

(vii) To find the voltage across AB in Fig
...
34:
Since the 20 V supply is across the 5 and 15
resistors in series then, by voltage division, the
voltage drop across AC,

× 10 = 4 V

(vi) If the 10 V battery in Fig
...
33(a) is removed
and replaced by a short-circuit, as shown in
Fig
...
33(c), then the 20 resistor may be
removed
...

12 + 3

188 Electrical and Electronic Principles and Technology
made in the branch, were introduced into the branch,
all other e
...
f
...

The procedure adopted when using Thévenin’s theorem is summarised below
...
e
...
m
...
):
(i) remove the resistance R from that branch,
Figure 13
...

VA = VC − VAC = +20 − 5 = 15 V

Section 2

and
VB = VC − VBC = +20 − 16 = 4 V
...


(ii) determine the open-circuit voltage, E, across the
break,
(iii) remove each source of e
...
f
...
13
...
e
...
13
...
13
...
From Fig
...
27(c), the
equivalent resistance across AB
5 × 15 12 × 3
=
+
5 + 15 12 + 3

E
R+r

= 3
...
4 = 6
...
36

Problem 7
...
37
Figure 13
...
5 and 13
...

However, these theorems can be used to analyse
part of a circuit and in much more complicated
networks the principle of replacing the supply
by a constant voltage source in series with a
resistance (or impedance) is very useful
...
5 Thévenin’s theorem
Thévenin’s theorem states:
The current in any branch of a network is that which
would result if an e
...
f
...
d
...
37

Following the above procedure:
(i) The 10 resistance is removed from the circuit
as shown in Fig
...
38(a)
(ii) There is no current flowing in the 5
current I1 is given by
I1 =

resistor and

10
10
= 1A
=
R1 + R 2
2+8

D
...
circuit theory 189
R3 = 5 Ω

Following the procedure:

10 V

A
I1

(i) The 0
...
13
...


R2 = 8 Ω

R =2Ω
1

B
(a)

R2 = 8 Ω

r

12 V
1Ω

4Ω

I1

E

1Ω

r

4Ω

B

B

B
(b)

(a)

(b)
I

I

A

A

E=8V

1Ω+5 Ω
= 6Ω 4 Ω

R = 10 Ω

r = 6
...
38

A

r

E=4
...
8 Ω
r=2
...
40

P
...
across R2 = I1 R2 = 1 × 8 = 8 V
...
d
...
e
...
m
...
gives the circuit of
Fig
...
38(b) Resistance,
R 1 R2
2×8
r = R3 +
=5+
R1 + R 2
2+8
= 5 + 1
...
6
(iv) The equivalent Thévenin’s circuit is shown in
Fig
...
38(c)
E
8
8
Current I =
=
=
R+r
10 + 6
...
6
= 0
...
13
...
482 A
...
For the network shown in Fig
...
39
determine the current in the 0
...


(ii) Current I1 =

12
12
=
= 1
...
d
...
2) = 4
...

Hence p
...
across AB, i
...
the open-circuit voltage
across AB, E = 4
...
m
...
gives the circuit
shown in Fig
...
40(b)
...
13
...
13
...
4
4+6
10

(iv) The equivalent Thévenin’s circuit is shown in
Fig
...
40(d), from which, current
I=

E
4
...
8
=
=
r+R
2
...
8
3
...
5 A = current in the 0
...
39

A

A

A

resistor

Problem 9
...
13
...
Find also the power dissipated in the
4 resistor
...
13
...
571 A
Figure 13
...
13
...
571)2 (4) = 1
...
Determine the current in the 5
resistance of the network shown in Fig
...
43 using
Thévenin’s theorem
...


r2 =1 Ω

E2 =12 V

r1 =
0
...
43

r2 =1 Ω
B

Following the procedure:

(b)
I
E=2

2
3

V

A
R =4 Ω

2
r=3Ω

(i) The 5 resistance is removed from the circuit as
shown in Fig
...
44(a)
A
E2 =12V

r1 =
0
...
d
...
8 V
r = 0
...
5 Ω

B
(a)

Figure 13
...
5 Ω
E1 =
4V

I = 0
...
44

(see Section 13
...
(Alternatively, p
...
across
AB, E = E2 + I1 r2 = 2 + 2 (1) = 2 2 V)
3
3
(iii) Removing the sources of e
...
f
...
13
...
4 A
0
...
5
P
...
across AB,

(ii) Current I1 =

E = E1 − I1 r1 = 4 − (6
...
5) = 0
...
4(iii))
...
4)(2) = 0
...
C
...
5 × 2
=
= 0
...
5 + 2
2
...
13
...

3
P
...
across 10 resistor

0
...
8
E
=
=
= 0
...
4 + 5
5
...
13
...
148)(5) = 0
...
4(v))
...
d
...

(iii) Removing the source of e
...
f
...
13
...
The 20
resistance may thus be removed as shown in
Fig
...
46(c) (see Section 13
...


From Section 13
...
e
...
74 3
...
52 A
0
...
5
Also from Fig
...
44(d),

i
...


0
...
74 = 4 − (IA )(0
...
74 12
...
37 A
Hence current IB =
2
2
[Check, from Fig
...
44(d), IA = IB + I, correct to 2
significant figures by Kirchhoff’s current law]

(c)
A

I
E = 16 V

R=3Ω

r = 5Ω

B
(d)

Problem 11
...
13
...
The voltage
source has negligible internal resistance
...
46

From Fig
...
46(c), resistance,
2 10 × 5
2 50
r=1 +
=1 +
=5
3 10 + 5
3 15
(iv) The equivalent Thévenin’s circuit is shown in
Fig
...
46(d), from which, current,

Figure 13
...
13
...
)
Following the procedure
(i) The 3 resistance is removed from the circuit as
shown in Fig
...
46(a)
...
A Wheatstone Bridge network is
shown in Fig
...
47
...
m
...
gives the circuit
shown in Fig
...
44(b), from which resistance

192 Electrical and Electronic Principles and Technology
Point C is at a potential of +54 V
...
31 V
...
31 = 45
...
Between
C and B is a voltage drop of 44
...
Hence
the voltage at point B is 54 − 44
...
53 V
...
(See
Section 13
...
Assume the source of e
...
f
...


(iii) Replacing the source of e
...
f
...
e
...
13
...
The circuit is redrawn and
simplified as shown in Fig
...
48(c) and (d), from
which the resistance between terminals A and B,

Figure 13
...
13
...
d
...
692 + 2
...
163

2
(54)
2 + 11

(iv) The equivalent Thévenin’s circuit is shown in
Fig
...
48(e), from which, current

= 8
...
d
...
47 V

C
R2 =14 Ω

R4=
11 Ω

R3 = 3 Ω

2Ω

14 Ω

A

B

B

11 Ω

3Ω
D

D
(a)
C 14 Ω

A
11 Ω D 3 Ω
(c)

Figure 13
...
13
...
16
= 1A
4
...
d
...
47 −
8
...
16 V

E=
54 V

E
r + R5

C
D

A

I

14 Ω

11 Ω

B
3Ω

(d)

r=
4
...
16 V
(e)

D
...
circuit theory 193
Now try the following exercise
Exercise 70 Further problems on
Thévenin’s theorem

5
...
13
...

[0
...
Use Thévenin’s theorem to find the current
flowing in the 14 resistor of the network
shown in Fig
...
49
...

[0
...
64 W]

Figure 13
...
49

2
...
13
...

[2
...
07 W]

Figure 13
...
m
...
in series with a resistance
...
5, the Thévenin constant-voltage source consisted of a constant e
...
f
...
However this is not the only form of representation
...
It may be shown that the two forms are
equivalent
...
An ideal constant-current generator is one with infinite internal resistance so that it
supplies the same current to all loads
...
13
...
7 Norton’s theorem
3
...

4
...
13
...
Find, using
Thévenin’s theorem, the current flowing in the
4 resistor
...
918 A]

Norton’s theorem states:
The current that flows in any branch of a network is
the same as that which would flow in the branch if it
were connected across a source of electrical energy, the
short-circuit current of which is equal to the current that
would flow in a short-circuit across the branch, and
the internal resistance of which is equal to the resistance which appears across the open-circuited branch
terminals
...
To determine the current flowing
in a resistance R of a branch AB of an active network:
(i) short-circuit branch AB

Figure 13
...
6 Constant-current source

194 Electrical and Electronic Principles and Technology
(iii) remove all sources of e
...
f
...
13
...
e
...
6 Ω
10 Ω
B
(c)

Figure 13
...
13
...
53

Problem 13
...
13
...
6
(5) = 0
...
6 + 5 + 10

as obtained previously in Problem 7 using
Thévenin’s theorem
...
Use Norton’s theorem to determine
the current I flowing in the 4 resistance shown in
Fig
...
56

Figure 13
...
13
...
56

(ii) Fig
...
55(b) is equivalent to Fig
...
55(a)
...
m
...
is removed from
Fig
...
55(a) the resistance ‘looking-in’ at a break
made between A and B is given by:
r=

2×8
= 1
...
13
...
13
...
C
...
4 Ω

12 V

0
...
59

(iii) If the sources of e
...
f
...
13
...
m
...
is removed the resistance
‘looking-in’ at a break made between A and B is
given by:

r=

I=

5Ω

2Ω

4V

(b)

Figure 13
...
571 A,

as obtained previously in problems 2, 5 and 9
using Kirchhoff’s laws and the theorems of superposition and Thévenin

Problem 15
...
13
...
Hence find the currents flowing
in the other two branches
...
5 × 2
= 0
...
5 + 2

(iv) From the Norton equivalent network shown in
Fig
...
59(b) the current in the 5 resistance is
given by:
I=

0
...
148 A,
0
...

The currents flowing in the other two branches are
obtained in the same way as in Problem 10
...
52 A and the
current flowing from the 12 V source is 6
...

Problem 16
...
13
...
The voltage source
has negligible internal resistance
...
58

Following the procedure:
(i) The 5 branch is short-circuited as shown in
Fig
...
59(a)
(ii) From Fig
...
59(a),
ISC = I1 − I2 =

12
4
−
= 8−6 = 2A
0
...
60

Following the procedure:
(i) The branch containing the 3 resistance is shortcircuited as shown in Fig
...
61(a)

Section 2

4V

I

A

I2

196 Electrical and Electronic Principles and Technology
5Ω

A

10 Ω

I SC

5Ω

A

20 Ω

24 V

B

I SC

5Ω

A

20 Ω

24 V

24 V

10 Ω

r
B

B
(a)

(b)

(c)

5Ω

A

I

A

ISC = 4
...
61

(ii) From the equivalent circuit shown in Fig
...
61(b),

Section 2

ISC =

24
= 4
...
m
...
is removed the resistance ‘looking-in’ at a break made between A and
B is obtained from Fig
...
61(c) and its equivalent
circuit shown in Fig
...
61(d) and is given by:
r=

10 × 5
50
1
=
=3
10 + 5
15
3

Following the procedure:
(i) The 2 resistance branch is short-circuited as
shown in Fig
...
63(a)
(ii) Fig
...
63(b) is equivalent to Fig
...
63(a)
...
13
...

6+4

6Ω

B

I SC
B

(a)

(b)

4Ω A 8Ω

I

A

I SC = 9 A

(4
...


B
(c)

B
(d)

Figure 13
...
Determine the current flowing in the
2 resistance in the network shown in Fig
...
62

(iii) If the 15 A current source is replaced by an opencircuit then from Fig
...
63(c) the resistance
‘looking-in’ at a break made between A and B is
given by (6 + 4) in parallel with (8 + 7) , i
...

r=

Figure 13
...
13
...
C
...
75 A
6+2

Figure 13
...
m
...
E in series with
a resistance r feeding a load resistance R

Now try the following exercise
Exercise 71 Further problems on Norton’s
theorem

2
...
Determine the current flowing in the 6 resistance of the network shown in Fig
...
64 by
using Norton’s theorem
...
5 mA]

Figure 13
...
13
...
e
...
13
...


Figure 13
...
8 Thévenin and Norton
equivalent networks
The Thévenin and Norton networks shown in Fig
...
65
are equivalent to each other
...
e
...
67

Thus the two representations shown in Fig
...
65 are
equivalent
...
Convert the circuit shown in
Fig
...
68 to an equivalent Norton network
...
65

If terminals AB in Fig
...
65(a) are short-circuited,
the short-circuit current is given by E/r
...
13
...
For the circuit shown in Fig
...
65(a) to
be equivalent to the circuit in Fig
...
65(b) the same
short-circuit current must flow
...


Figure 13
...
Repeat Problems 1–4 of Exercise 69, page 186,
by using Norton’s theorem

198 Electrical and Electronic Principles and Technology
If terminals AB in Fig
...
68 are short-circuited, the
short-circuit current ISC = 10/2 = 5 A
The resistance ‘looking-in’ at terminals AB is 2
...
13
...
(a) Convert the circuit to the left of
terminals AB in Fig
...
72 to an equivalent
Thévenin circuit by initially converting to a Norton
equivalent circuit
...
8 resistor
...
8 Ω
r2 = 2 Ω
B

Figure 13
...
72

Problem 19
...
13
...


Section 2

(a) For the branch containing the 12 V source, converting to a Norton equivalent circuit gives
ISC = 12/3 = 4 A and r1 = 3
...
Thus Fig
...
73(a) shows a network
equivalent to Fig
...
72
A

ISC2 =
12 A

ISC1 =
4A

Figure 13
...
13
...


r2 = 2 Ω

r1 =
3Ω

B

(a)
A
16 A

A
19
...
2 Ω

The resistance ‘looking-in’ at terminals AB is 3
...
13
...
2 Ω
B
(b)

B
(c)

Figure 13
...
71

From Fig
...
73(a) the total short-circuit current is 4 + 12 = 16 A and the total resistance is given by (3 × 2)/(3 + 2) = 1
...
Thus
Fig
...
73(a) simplifies to Fig
...
73(b)
...
13
...
2) = 19
...
2
...
13
...


D
...
circuit theory 199
(b) When the 1
...
13
...
2
1
...
8

= 6
...
Determine by successive
conversions between Thévenin and Norton
equivalent networks a Thévenin equivalent
circuit for terminals AB of Fig
...
74
...


Figure 13
...
8
1800 + 200
7
...
9 mA
2000

Figure 13
...
13
...
13
...
Combining the 5 mA and 2 mA current
sources gives the equivalent network of Fig
...
75(b)
where the short-circuit current for the original two
branches considered is 7 mA and the resistance is
(2 × 3)/(2 + 3) = 1
...
13
...
The open-circuit voltage across CD
is (7 × 10−3 )(1
...
4 V and the resistance
‘looking-in’ at CD is 1
...
The open-circuit voltage
across EF is (1 × 10−3 ) (600) = 0
...
6 k
...
13
...
13
...
Combining the two Thévenin
circuits gives E = 8
...
6 = 7
...
2 + 0
...
8 k
Thus the Thévenin equivalent circuit for terminals AB
of Fig
...
74 is as shown in Fig
...
75(d)

Exercise 72 Further problems on Thévenin
and Norton equivalent
networks
1
...
13
...

[(a) ISC = 25 A, r = 2
(b) ISC = 2 mA, r = 5 ]

Figure 13
...
Convert the networks shown in Fig
...
77 to
Thévenin equivalent circuits
[(a) E = 20 V, r = 4
(b) E = 12 mV, r = 3 ]

Section 2

I=

200 Electrical and Electronic Principles and Technology
Norton equivalent and hence determine the
current flowing in the 5 resistance
...
22 A]

Figure 13
...
(a) Convert the network to the left of terminals
AB in Fig
...
78 to an equivalent Thévenin
circuit by initially converting to a Norton
equivalent network
...
80

Section 2

13
...
78

The maximum power transfer theorem states:
The power transferred from a supply source to a load is
at its maximum when the resistance of the load is equal
to the internal resistance of the source
...
13
...


(b) Determine the current flowing in the 1
...
13
...
2 (b) 6 A]
4
...
13
...
Hence determine
the current flowing in a 6 resistor connected
between A and B
...
81

Typical practical applications of the maximum power
transfer theorem are found in stereo amplifier design,
seeking to maximise power delivered to speakers, and
in electric vehicle design, seeking to maximise power
delivered to drive a motor
...
79

5
...
13
...
The circuit diagram of Fig
...
82
shows dry cells of source e
...
f
...
5
...
5 steps, calculate the power
dissipated by the load in each case
...
C
...


Figure 13
...
13
...
From the
maximum power transfer theorem, for maximum power
dissipation, RL = r = 1
...
82

RL ( )

0

0
...
0

E
I=
r + RL

2
...
0 1
...
5

1
...
0

2
...
333 1
...
5 4
...
13
...
5 + 1
...
5) = 150 W = maximum
power dissipated
Problem 24
...
13
...


2
...
94 3
...
56 3
...
84

I=

3
...
5

5
...
091 1
...
923 0
...
8
Figure 13
...
57 3
...
41 3
...
20

A graph of RL against P is shown in Fig
...
83
...
60 W which occurs
when RL is 2
...
e
...


Using the procedure for Thévenin’s theorem:
(i) Resistance RL is removed from the circuit as
shown in Fig
...
86(a)
(ii) The p
...
across AB is the same as the p
...
across
the 12 resistor
...
A d
...
source has an open-circuit
voltage of 30 V and an internal resistance of 1
...

State the value of load resistance that gives
maximum power dissipation and determine the
value of this power
...
m
...
gives the circuit of
Fig
...
86(b), from which, resistance,
r=

36
12 × 3
=
= 2
...
5 = 2
...
e
...
4)2 (0) = 0 W
...
5 ,
current I = E/(r + RL ) = 6/(2
...
5) = 2 A and
P = I 2 RL = (2)2 (0
...
00 W
...
0 , current I = 6/(2
...
0) = 1
...
714)2 (1
...
94 W
...
Determine the maximum power dissipated by the
load
...
A voltage source comprising six 2 V cells, each
having an internal resistance of 0
...
Determine the
maximum power transferred to the load
...
86

(iv) The equivalent Thévenin’s circuit supplying terminals AB is shown in Fig
...
86(c), from which,
current, I =

E
r + RL

5
...
c
...
Calculate (a) the current
in the load, (b) internal resistance r, and (c)
voltage V
...
4
12
= 2
...
4 + 2
...
5)2 (2
...


Section 2

Thus current, I =

Now try the following exercises

Exercise 74 Short answer questions on
d
...
circuit theory
1
...
d
...
State Kirchhoff’s current law
3
...
A d
...
source has an open-circuit voltage of
20 V and an internal resistance of 2
...
Find the value of
this power
...
Determine the value of the load resistance
RL shown in Fig
...
87 that gives maximum
power dissipation and find the value of the
power
...
6 , P = 57
...
87

3
...
c
...
State, in your own words, the superposition
theorem
5
...
State, in your own words, Norton’s theorem
7
...
c
...
c
...
Which of the following statements is true:
For the junction in the network shown in
Fig
...
88:
(a) I5 − I4 = I3 − I2 + I1
(b) I1 + I2 + I3 = I4 + I5
(c) I2 + I3 + I5 = I1 + I4
(d) I1 − I2 − I3 − I4 + I5 = 0

D
...
circuit theory 203

5
...
13
...
4 A
(c) 0
...
88

2
...
13
...
For the circuit shown in Fig
...
91, current
I2 is:
(a) 2 A
(b) 14
...
5 A
(d) 0 A
7
...
13
...
31
(b) 7
...
0
(d) 6
...
For the circuit shown in Fig
...
90, the
internal resistance r is given by:
I
V −E
(a)
(b)
V −E
I
I
E −V
(c)
(d)
E −V
I

Figure 13
...
With reference to Fig
...
93, which of the
following statements is correct?
(a) VPQ = 2 V
(b) VPQ = 15 V
(c) When a load is connected between P and
Q, current would flow from Q to P
(d) VPQ = 20 V
R
3Ω

Figure 13
...
For the circuit shown in Fig
...
91, voltage V is:
(a) 12 V

(b) 2 V

11 Ω

P

Q
15 V
2Ω

(c) 10 V (d) 0 V

4Ω
S

Figure 13
...
91

9
...
13
...
20
(c) 4
...
29

Section 2

Figure 13
...
For the circuit shown in Fig
...
94, maximum power transfer from the source is
required
...
5

13
...
13
...
97

Section 2

Figure 13
...
The open-circuit voltage E across terminals
XY of Fig
...
95 is:
(a) 0 V
(b) 20 V
(c) 4 V
(d) 16 V

12
...
13
...
96

15
...
13
...
The current flowing in the branches of a d
...

circuit may be determined using:
(a) Kirchhoff’s laws
(b) Lenz’s law
(c) Faraday’s laws
(d) Fleming’s left-hand rule

Figure 13
...
For the circuit shown in Fig
...
97, current
I1 is:
(a) 25 A
(b) 4 A
(c) 0 A
(d) 20 A

RL

Chapter 14

Alternating voltages and
currents
At the end of this chapter you should be able to:
• appreciate why a
...
is used in preference to d
...

• describe the principle of operation of an a
...
generator
• distinguish between unidirectional and alternating waveforms
• define cycle, period or periodic time T and frequency f of a waveform
• perform calculations involving T = 1/f
• define instantaneous, peak, mean and r
...
s
...
m
...
values and form and peak factors for given waveforms
• understand and perform calculations on the general sinusoidal equation v = Vm sin (ωt ± φ)
• understand lagging and leading angles
• combine two sinusoidal waveforms (a) by plotting graphically, (b) by drawing phasors to scale and (c) by
calculation

14
...
It is easier and cheaper to generate alternating current (a
...
) than direct current (d
...
)
and a
...
is more conveniently distributed than d
...

since its voltage can be readily altered using transformers
...
c
...
c
...
7)
...
2 The a
...
generator
Let a single turn coil be free to rotate at constant angular
velocity symmetrically between the poles of a magnet
system as shown in Fig
...
1
...
m
...
is generated in the coil (from Faraday’s
laws) which varies in magnitude and reverses its direction at regular intervals
...
14
...
In positions (a), (e) and (i) the conductors
of the loop are effectively moving along the magnetic
field, no flux is cut and hence no e
...
f
...
In
position (c) maximum flux is cut and hence maximum
e
...
f
...
In position (g), maximum flux is cut

206 Electrical and Electronic Principles and Technology
of the coil, one cycle of alternating e
...
f
...
This is the principle of operation of the a
...

generator (i
...
the alternator)
...
3 Waveforms

Section 2

Figure 14
...
2

and hence maximum e
...
f
...
However,
using Fleming’s right-hand rule, the induced e
...
f
...
In positions (b), (d), (f) and (h) some
flux is cut and hence some e
...
f
...
If all such
positions of the coil are considered, in one revolution

Figure 14
...
Some typical waveforms are shown in Fig
...
3
...
e
...
Waveforms (c) to (g) are called
alternating waveforms since their quantities are continually changing in direction (i
...
alternately positive
and negative)
...
14
...
It is the shape of the waveform of e
...
f
...

One complete series of values is called a cycle (i
...

from O to P in Fig
...
3(g))
...

The number of cycles completed in one second is
called the frequency, f , of the supply and is measured
in hertz, Hz
...
02 s or 20 ms
f
50
(b) Periodic time T =

1
1
=
f
20 000
= 0
...
Determine the frequencies for
periodic times of (a) 4 ms (b) 4 μs
...
25 MHz
(a) Frequency f =

14
...
C
...
They are represented
by small letters, i, v, e, etc
...
14
...

The largest value reached in a half cycle is called the
peak value or the maximum value or the amplitude of
the waveform
...
(see Fig
...
3(f) and (g))
...
m
...
is shown in Fig
...
3(g) and is the difference
between the maximum and minimum values in a cycle
...

Average or mean value =

area under the curve
length of base

The area under the curve is found by approximate methods such as the trapezoidal rule, the mid-ordinate rule
or Simpson’s rule
...

For a sine wave:

Problem 3
...
What is its frequency?
Time for 1 cycle = (8/5) ms = 1
...

1
1
1000
Frequency f = =
=
T
1
...
6
10 000
=
= 625 Hz
16
Now try the following exercise

average value = 0
...
e
...
The effective value is called
the root mean square (r
...
s
...
For example, the domestic mains supply in Great
Britain is 240 V and is assumed to mean ‘240 V rms’
...
m
...
values are I, V, E, etc
...
14
...
m
...
value is given by:

Exercise 76 Further problems on
frequency and periodic time
1
...
5 Hz (b) 100 Hz
(c) 40 kHz
[(a) 0
...


2
...
2 s
[(a) 200 Hz (b) 20 kHz (c) 5 Hz]
3
...
What is its frequency?
[800 Hz]
Figure 14
...
Determine the periodic time for
frequencies of (a) 50 Hz and (b) 20 kHz
...
707 × maximum value
√
(i
...
1/ 2 × maximum value)
r
...
s
...
11
Peak factor =

maximum value
r
...
s
...
41
...


Problem 4
...
14
...
m
...
value
(iv) form factor and (v) peak factor
...
14
...
Thus
=

rms value =
=

v2 + v 2 + v 2 + v 2
1
2
3
4
4
252 + 752 + 1252 + 1752
4

= 114
...

For example, if twice the number of ordinates
as that chosen above are used, the r
...
s
...
6 V)
r
...
s
...
6
=
= 1
...
m
...
value
200
=
= 1
...
6
(b) Rectangular waveform (Fig
...
5(b))
...
Hence
Figure 14
...
14
...

(i) Time for 1 complete cycle = 20 ms = periodic
time, T
...
5 Hz
area under curve
(ii) Average value over
=
half a cycle
length of base
=

10 × (8 × 10−3 )
8 × 10−3

= 10 A

Alternating voltages and currents 209
2
2
2
2
i1 + i2 + i3 + i4
= 10 A,
4
however many intervals are chosen, since the
waveform is rectangular
...
m
...
value =

(iv) Form factor =

10
r
...
s
...
m
...
value
10

Problem 5
...

time t (ms)

0

current i (A)

0
...
0 1
...
0 2
...
0 3
...
0 4
...
0
5

0

Assuming the negative half cycle is identical in
shape to the positive half cycle, plot the waveform
and find (a) the frequency of the supply, (b) the
instantaneous values of current after 1
...
8 ms, (c) the peak or maximum value, (d) the
mean or average value, and (e) the r
...
s
...


The half cycle of alternating current is shown plotted in
Fig
...
6

Figure 14
...
5 ms gives:
area under
= (0
...
14
...
5 × 10−3 )(351)
(0
...
1 A

(a) Time for a half cycle = 5 ms; hence the time for
1 cycle, i
...
the periodic time, T = 10 ms or 0
...
m
...
01

(b) Instantaneous value of current after 1
...
14
...
Instantaneous value of current after 3
...
14
...
8 A
10

Problem 6
...
m
...
value of a
sinusoidal current of maximum value 20 A
...
m
...
value = 0
...
707 × 20 = 14
...
Determine the peak and mean values
for a 240 V mains supply
...
m
...
value of voltage V = 0
...

A 240 V mains supply means that 240 V is the r
...
s
...
5 V
0
...
707
= peak value

Mean value
VAV = 0
...
637 × 339
...
3 V

2
...
14
...
m
...
value
(iv) the form factor (v) the peak factor
...
15
250 Hz
1
...
09
250 Hz
2
...
50 A
1
...
0
18 A
1
...
0]

(iii) 2
...
56 A
(iii) 50 V

Problem 8
...
Determine its maximum value and its r
...
s
...


Section 2

For a sine wave, mean value = 0
...

Hence
maximum value =

mean value
150
=
0
...
637
= 235
...
m
...
value = 0
...
707 × 235
...
5 V
Figure 14
...
c
...
An alternating current varies with time over
half a cycle as follows:
Current (A)
time (ms)

0
0

0
...
0 4
...
4
1
2
3
4

Current (A) 8
...
5 1
...
4 0
...
Plot the
curve and determine:
(a) the frequency (b) the instantaneous values
at 3
...
8 ms (c) its mean value and
(d) its r
...
s
...
5 A, 3
...
8 A (d) 4
...
An alternating voltage is triangular in shape,
rising at a constant rate to a maximum of 300 V
in 8 ms and then falling to zero at a constant
rate in 4 ms
...
Calculate
(a) the mean voltage over half a cycle, and
(b) the r
...
s
...
An alternating e
...
f
...
m
...
(V)
time (ms)

0
0

E
...
f
...
0

E
...
f
...
0

45
1
...
0

155
4
...
5 9
...
5

Alternating voltages and currents 211
The negative half cycle is identical in shape
to the positive half cycle
...
75 ms (c) the times when the voltage is
125 V (d) the mean value, and (e) the r
...
s
...
67 Hz (b) 115 V
(c) 4 ms and 10
...
Any quantity which
varies sinusoidally can thus be represented as a phasor
...
To show this
a periodic function is represented by y = sin(ωt ± φ),
where φ is the phase (or angle) difference compared
with y = sin ωt
...
14
...
Phasors y1 and y2 are shown in
Fig
...
9(b) at the time when t = 0
...
Calculate the r
...
s
...
1 V]
6
...
9 V, 180
...
Plot a sine wave of peak value 10
...
Show
that the average value of the waveform is
6
...
m
...

value is 7
...
A sinusoidal current has a mean value of
15
...
Determine its maximum and r
...
s
...

[23
...
65 A]

14
...
14
...

A rotating vector is known as a phasor
...
8

After time t seconds the vector 0A has turned through
an angle ωt
...
e
...
9

In Fig
...
9(c), y4 = sin(ωt − φ) starts φ radians later
than y3 = sin ωt and is thus said to lag y3 by φ radians
...
14
...

Given the general sinusoidal voltage,
v = V m sin(ωt ± φ), then
(i)
(ii)
(iii)
(iv)
(v)
(vi)

Amplitude or maximum value = Vm
Peak to peak value = 2Vm
Angular velocity = ω rad/s
Periodic time, T = 2π/ω seconds
Frequency, f = ω/2π Hz (since ω = 2πf )
φ = angle of lag or lead (compared with
v = Vm sin ωt)

Problem 9
...
8 sin 314 t volts
...
m
...
voltage,
(b) the frequency and (c) the instantaneous value of
voltage when t = 4 ms
...
Comparing
v = 282
...
8 V
...
m
...

voltage = 0
...
707 × 282
...
A sinusoidal voltage has a maximum value of
120 V
...
m
...
and average values
...
8 V, 76
...
e
...

Hence frequency,
f =

314
= 50 Hz
2π

(c) When t = 4 ms,
v = 282
...
8 sin(1
...
9 V
180◦
Note that 1
...
256 ×
π
= 71
...
8 sin 71
...
9 V, as above
...

Periodic time T =

2π
2π
=
= 200π rad/s
...
01
v = Vm sin(ωt + φ) thus becomes

ω=

v = 40 sin(200πt + φ) volts
...
e
...
5
Hence

Section 2

Problem 10
...
25) volts
...
m
...

value, (d) the periodic time, (e) the frequency, and
(f) the phase angle (in degrees and minutes) relative
to 75 sin 200πt
...
25) with the general
expression v = Vm sin(ωt ± φ) gives:
(a) Amplitude, or peak value = 75 V
(b) Peak-to-peak value = 2 × 75 = 150 V
(c) The r
...
s
...
707 × maximum value
= 0
...
Hence periodic time,
2π
2π
1
T=
=
=
= 0
...
5) = −30◦
π
π
= −30 ×
rads = − rads
180
6
π
v = 40 sin 200πt −
V
6

Problem 12
...
c
...
36) amperes
...

(a) Peak value = 120 A
2π
ω
2π
=
(since ω = 100π)
100π
1
=
= 0
...
01

(f) Phase angle, φ = 0
...
25 rads = 0
...
32◦

Hence phase angle = 14
...
02

Phase angle = 0
...
36 ×

Problem 11
...
01 s and a peak value of 40 V
...
Express the
instantaneous voltage in the form
v = Vm sin(ωt ± φ)
...
63◦ leading
π

(b) When t = 0,
i = 120 sin(0 + 0
...
63◦ = 42
...
36

= 120 sin 2
...
63◦ )
= 31
...
36) thus
(60/120) = sin (100πt + 0
...
36) = sin−1 0
...
5236 rads
...
5236 − 0
...
521 ms
100π

(e) When the current is a maximum, i = 120 A
...
36)
1 = sin(100πt + 0
...
36) = sin−1 1 = 90◦
= (π/2) rads
= 1
...

Hence time,

t=

1
...
36
= 3
...
An alternating voltage is represented by
v = 20 sin 157
...
Find (a) the maximum
value (b) the frequency (c) the periodic time
...
04 s
(d) 157
...
Find the peak value, the r
...
s
...
63 V, 200 Hz, 5 ms, 0◦ ]
(b) i = 50 sin(100πt + 0
...
35 A, 50 Hz, 0
...
19◦ lead]

(c) e = 200 sin (628
...
41) volts
[200 V, 141
...
01 s,
23
...
A sinusoidal current has a peak value of 30 A
and a frequency of 60 Hz
...
Express the instantaneous
current i in the form i = Im sin ωt
[i = 30 sin 120πt A]
4
...
When
time t = 0, v = −75 volts
...

[v = 200 sin(100πt − 0
...
The voltage in an alternating current circuit at
any time t seconds is given by v = 60 sin 40 t
volts
...
496 ms (b) 91
...
The instantaneous value of voltage in an a
...

circuit at any time t seconds is given by
v = 100 sin (50πt − 0
...
Find:
(a) the peak-to-peak voltage, the frequency,
the periodic time and the phase angle
(b) the voltage when t = 0
(c) the voltage when t = 8 ms
(d) the times in the first cycle when the
voltage is 60 V
(e) the times in the first cycle when the
voltage is −40 V
(f) the first time when the voltage is a maximum
...

[(a) 200 V, 25 Hz, 0
...
97◦ lagging
(b) −49
...
96 V (d) 7
...
23 ms (e) 25
...
71 ms
(f) 13
...
6 Combination of waveforms
The resultant of the addition (or subtraction) of two
sinusoidal quantities may be determined either:
(a) by plotting the periodic functions graphically (see
worked Problems 13 and 16), or

Section 2

(c) When t = 8 ms,

214 Electrical and Electronic Principles and Technology
(b) by resolution of phasors by drawing or calculation
(see worked Problems 14 and 15)
Problem 13
...
By
plotting i1 and i2 on the same axes, using the same
scale, over one cycle, and adding ordinates at
intervals, obtain a sinusoidal expression for i1 + i2
...
14
...
Ordinates of i1 and i2 are added at,
say, 15◦ intervals (a pair of dividers are useful for this)
...
Two alternating voltages are
represented by v1 = 50 sin ωt volts and
v2 = 100 sin(ωt − π/6) V
...

Phasors are usually drawn at the instant when time t = 0
...
e
...

This is shown in Fig
...
11(a) where 0 is the point of
rotation of the phasors
...
3 + 8
...


Figure 14
...
e
...
14
...
e
...
14
...
10

The resultant waveform for i1 + i2 is shown by the broken line in Fig
...
10
...
The amplitude or peak value is
26
...
e
...
332 rads
Hence the sinusoidal expression for the resultant i1 + i2
is given by:
iR = i1 + i2 = 26
...
332) A

Alternatively, when two phasors are being added the
resultant is always the diagonal of the parallelogram, as
shown in Fig
...
11(c)
...

A more accurate solution is obtained by calculation,
using the cosine and sine rules
...
14
...
5 V

Alternating voltages and currents 215
Using the sine rule,

An alternative method of calculation is to use complex numbers (see ‘Engineering Mathematics’)
...
5
=
sin φ
sin 150◦
sin φ =

100 sin 150◦
145
...
3436
and φ = sin−1 0
...
096◦ = 0
...
Hence

π
3

= (20 + j0) + (5 + j8
...
66)
= 26
...
106◦ or 26
...
333 rad

vR = v1 + v2 = 145
...
35) V
Problem 15
...

(a) The relative positions of i1 and i2 at time t = 0
are shown as phasors in Fig
...
12(a)
...
14
...
33 rads leading i1
...
33) A

≡ 26
...
333) A
Problem 16
...
Obtain sinusoidal expressions for v1 − v2
(a) by plotting waveforms, and (b) by resolution of
phasors
...
14
...
For
example
at 30◦ , v1 − v2 = 60 − (−52) = 112 V
at 60◦ , v1 − v2 = 104 − 52 = 52 V
at 150◦ , v1 − v2 = 60 − 193 = −133 V and so on
...
12

(b) From Fig
...
12(b), by the cosine rule:
2
iR = 202 + 102 − 2(20)(10)(cos 120◦ )

from which iR = 26
...
46
=
sin φ
sin 120◦
from which φ = 19
...
e
...
333 rads)

Hence, by calculation,
iR = 26
...
333) A

Figure 14
...
14
...
e
...
73 radians)

Hence, by resolution of phasors,
vR = v1 − v2 = 143
...
721) volts

Hence, by drawing,
vR = v1 − v2 = 143 sin (ωt + 1
...
14
...
Since the
resultant of v1 − v2 is required, −v2 is drawn in
the opposite direction to +v2 and is shown by the
broken line in Fig
...
14(a)
...
14
...


Now try the following exercise
Exercise 79 Further problems on the
combination of periodic
functions
1
...
By plotting v1 and v2 on
the same axes, using the same scale, over one
cycle, obtain expressions for (a) v1 + v2 and
(b) v1 − v2
[(a) v1 + v2 = 12
...
32) V
(b) v1 − v2 = 4
...
Repeat Problem 1 using calculation
[(a) 12
...
324)
(b) 4
...
02)]

Section 2

3
...

or by calculation, find a sinusoidal expression
to represent i1 + i2
[23
...
588)]
Determine, either by plotting graphs and
adding ordinates at intervals, or by calculation, the following periodic functions in the
form v = Vm sin(ωt ± φ)

Figure 14
...
42
Sum of vertical components of v1 and v2
= 120 sin 0◦ + 200 sin 45◦ = 141
...
14
...
42)2 + (141
...
0
and

from which,

141
...
42
= tan 6
...
6013
= 81
...
61◦ or 1
...
10 sinωt + 4 sin(ωt + π/4)
[13
...
217)]
5
...
34 sin(ωt + 0
...
100 sin ωt − 70 sin(ωt − π/3)
[88
...
751)]
7
...
c
...
2t and
v2 = 90 sin (314
...

Determine (a) the voltage of the supply, in
trigonometric form, (b) the r
...
s
...

[(a) 229 sin(314
...
233) V
(b) 161
...
The voltages across three components in a
series circuit when connected across an a
...

supply are:
π
v1 = 30 sin 300πt −
volts,
6
π
v2 = 40 sin 300πt +
volts and
4
π
v3 = 50 sin 300πt +
volts
...
m
...
value of the
supply
...
39 sin(300πt + 0
...
67 ms (d) 68
...
7

diode D is switched off
...

Thus, an alternating, sinusoidal waveform applied to
the transformer primary is rectified into a unidirectional
waveform
...
14
...
e
...
c
...
It would, however, be satisfactory as
a battery charger
...
8, methods of smoothing
the output waveform are discussed
...
14
...
When P is sufficiently positive
with respect to Q, diode D1 conducts and current flows
(shown by the broken line in Fig
...
16)
...
14
...


Section 2

8
...
3t volts
and the voltage drop across one of the components is 18 sin (628
...
52) volts, calculate
(a) the voltage drop across the remainder of
the circuit, (b) the supply frequency, and (c)
the periodic time of the supply
...
96 sin(628
...
762) V
(b) 100 Hz (c) 10 ms]

Rectification

The process of obtaining unidirectional currents and
voltages from alternating currents and voltages is called
rectification
...


Half-wave rectification
Using a single diode, D, as shown in Fig
...
15, halfwave rectification is obtained
...
When P is negative with respect to Q,

Figure 14
...
The output
waveform is thus as shown in Fig
...
16
...
Section 14
...

A disadvantage of this type of rectifier is that centretapped transformers are expensive
...
15

Four diodes may be used in a bridge rectifier circuit,
as shown in Fig
...
17 to obtain full-wave rectification
...
r
...
Q
Current flow when Q
is positive w
...
t
...
18

F
OUTPUT
R

G

+
v
0
−

Voltage across capacitor
t

Q

D
...
voltage
output

P

X

Z
Ripple
Y

+

Section 2

taken from the remaining two corners)
...
14
...

Following the broken line in Fig
...
17:
When P is positive with respect to Q, current flows from
the transformer to point E, through diode D4 to point F,
then through load R to point H, through D2 to point G,
and back to the transformer
...
14
...
The output waveform is not
steady and needs improving; a method of smoothing is
explained in the next section
...
8

Smoothing of the rectified
output waveform

The pulsating outputs obtained from the half- and fullwave rectifier circuits are not suitable for the operation
of equipment that requires a steady d
...
output, such
as would be obtained from batteries
...
Smoothing is the process of removing the
worst of the output waveform variations
...
14
...
The improved waveforms for halfwave and full-wave rectifiers are shown in more detail
in Fig
...
19
...
C
...
17

X

Z
Ripple
Y

+
0
_

Time

(b) Full-wave rectifier

Figure 14
...
14
...
As the
waveform dies away, the capacitor discharges across
the load, as shown by XY
...
This process continues as shown
in Fig
...
19
...

The variation in potential between points X and Y is
called ripple, as shown in Fig
...
19; the object is to
reduce ripple to a minimum
...
14
...

The output voltage from the rectifier is applied to
capacitor C1 and the voltage across points AA is shown
in Fig
...
20, similar to the waveforms of Fig
...
19
...
m
...
’s are induced, as explained
in Chapter 9
...


Alternating voltages and currents 219
A

L

B

V

V

V

V

+

+

+

+

0
_

0
_

0
_

C1

A

C2

0
_

B

Figure 14
...
Briefly explain the principle of operation of
the simple alternator
2
...
What is the difference between an alternating
and a unidirectional waveform?
4
...

5
...
The mains supply voltage has a special shape
of waveform called a
...
What is a phasor quantity?
13
...
, and for a sine
wave, form factor =
...
Complete the statement:
Peak factor =
...

15
...
What do the symbols Im ,
ω and α represent?
16
...
c
...
c
...
Draw an appropriate circuit diagram suitable
for half-wave rectifications and explain its
operation
18
...
Explain, with a diagram, how full-wave rectification is obtained using a bridge rectifier
circuit
20
...
Define peak value
8
...
m
...
value?
9
...

10
...
m
...
which has a maximum value of
100 V?
11
...
× maximum value

Exercise 81

Multi-choice questions on
alternating voltages and
currents
(Answers on page 398)

1
...
m
...
in the inductor will oppose
the increase
...
m
...
will try to maintain the
current flow
...
14
...
A
further capacitor, C2 , completes the process
...
m
...
value
2
...
1 s
...
002 Hz
(d) 1 kHz

7
...
m
...
value
(c) average value
(d) peak value

3
...
14
...
m
...
will be:
(a) zero
(b) an r
...
s
...
State which of the following is false
...
414
(b) the r
...
s
...
707 × peak value
(c) the average value is 0
...
m
...
value
(d) the form factor is 1
...
An a
...
supply is 70
...
Which of the
following statements is false?
(a) The periodic time is 20 ms
(b) The peak value of the voltage is 70
...
m
...
value of the voltage is 70
...
21

4
...
c
...
Which of the following statements is false?
(a) It is cheaper to use a
...
than d
...

(b) Distribution of a
...
is more convenient
than with d
...
since voltages may be
readily altered using transformers
(c) An alternator is an a
...
generator
(d) A rectifier changes d
...
to a
...

6
...
Which of the following direct voltages, if applied to the lamp,
would cause the lamp to light with the same
brilliance?
(a) 100 V
(b) 63
...
7 V
(d) 141
...
An alternating voltage is given by
v = 100 sin(50πt − 0
...

Which of the following statements is true?
(a) The r
...
s
...
30 radians
11
...
m
...
or effective value
12
...
The marks for each question are shown in
brackets at the end of each question
...
1

Find also the current flowing in each of the other
two branches of the circuit
...
A d
...
voltage source has an internal resistance
of 2 and an open circuit voltage of 24 V
...

(5)
3
...
0 A
...
m
...
values
...
The instantaneous value of current in an a
...

circuit at any time t seconds is given by:
i = 50 sin(100πt − 0
...
Determine
(a) the peak to peak current, the frequency, the
periodic time and the phase angle (in degrees)
(b) the current when t = 0
(c) the current when t = 8 ms
(d) the first time when the voltage is a maximum
...

(14)

Section 2

1
...
RT4
...

Demonstrate that the same answer results from
each method
...
c
...
c
...
c
...
c
...
1

Purely resistive a
...
circuit

In a purely resistive a
...
circuit, the current IR and
applied voltage VR are in phase
...
15
...
2

VL
= 2πfL
IL

Purely inductive a
...
circuit

In a purely inductive a
...
circuit, the current IL lags the
applied voltage VL by 90◦ (i
...
π/2 rads)
...
15
...
XL is proportional to f as shown
in Fig
...
3

Single-phase series a
...
circuits 223
(a) Inductive reactance,
XL = 2πfL
= 2π(50)(40 × 10−3 ) = 12
...
1

V
240
=
= 19
...
57

(b) Inductive reactance,
XL = 2π(1000)(40 × 10−3 ) = 251
...
3

V
100
=
= 0
...
3

Purely capacitive a
...
circuit

Figure 15
...
c
...
e
...
See
Fig
...
4

Figure 15
...
(a) Calculate the reactance of a coil
of inductance 0
...
(b) A coil has a reactance of 124 in
a circuit with a supply of frequency 5 kHz
...


Figure 15
...
32) = 100
...

XC varies with frequency f as shown in Fig
...
5

XL
124
=
H = 3
...
A coil has an inductance of 40 mH
and negligible resistance
...


VC
1
=
IC
2πfC

Figure 15
...
Determine the capacitive reactance of
a capacitor of 10 μF when connected to a circuit of
frequency (a) 50 Hz (b) 20 kHz
(a) Capacitive reactance

1
2π(50)(10 × 10−6 )

=

106
= 318
...
73 A

Now try the following exercise
Exercise 82 Further problems on purely
inductive and capacitive
a
...
circuits

1
XC =
2πfC
=

=

V
XC

= 2π(50)(23 × 10−6 )(240)

=

(b)

I=

= 2πfCV

1
2πfC

XC =

Current

× 103 )(10

× 10−6 )

106
2π(20 × 103 )(10)

= 0
...
3 to 0
...
15
...
A capacitor has a reactance of 40
when operated on a 50 Hz supply
...

Since
XC =

1
2πfC

capacitance
C=

1
2πf XC

=

1
F
2π(50)(40)

=

106
μF
2π(50)(40)

= 79
...
Calculate the current taken by a 23 μF
capacitor when connected to a 240 V, 50 Hz supply
...
Calculate the reactance of a coil of inductance
0
...

[(a) 62
...
27 k ]
2
...
Calculate
the inductance of the coil
...
77 mH]
3
...
2 A
...

[0
...
An e
...
f
...

Determine (a) the reactance of the coil, and
(b) the current flowing in the coil
...
318 A]
5
...
Find the
p
...
across the inductor
...
7 V]
6
...
c
...
9 (b) 15
...
989 ]
7
...
Calculate the value
of its capacitance
...
79 μF]
8
...

[1
...
c
...
Two similar capacitors are connected in parallel to a 200 V, 1 kHz supply
...
628 A
...
25 μF]

15
...
c
...
c
...
15
...
In any a
...
series circuit the current is common
to each component and is thus taken as the reference
phasor
...
In a series R–L circuit the p
...
across
the resistance R is 12 V and the p
...
across the
inductance L is 5 V
...

From the voltage triangle of Fig
...
6, supply voltage
V=

122 + 52

V = 13 V

i
...


(Note that in a
...
circuits, the supply voltage is not the
arithmetic sum of the p
...
It is, in
fact, the phasor sum)
tan φ =

VL
5
=
VR
12

from which, circuit phase angle
Figure 15
...
15
...

For the R–L circuit:
V=

2
2
VR + V L

(by Pythagoras’ theorem)

and
tan φ =

VL
VR

(by trigonometric ratios)

In an a
...
circuit, the ratio applied voltage V to current
I is called the impedance, Z, i
...

Z=

V
I

If each side of the voltage triangle in Fig
...
6 is
divided by current I then the ‘impedance triangle’ is
derived
...
62◦ lagging

(‘Lagging’ infers that the current is ‘behind’ the voltage,
since phasors revolve anticlockwise)
Problem 7
...
55 mH
...
Determine also the phase
angle between the supply voltage and current
...
55 mH = 9
...
55 × 10−3 )
=3

Section 2

9
...
Determine its capacitance and the current
taken from the supply
...
92 μF, 0
...
15
...
87◦ lagging

Section 2

φ = tan−1

Problem 8
...
c
...
When connected to a 240 V, 50 Hz
supply the current is 20 A
...

Resistance

Problem 9
...
3 mH and
negligible resistance is connected in series with a
200 resistor to a 240 V, 50 Hz supply
...
d
...

L = 318
...
3183 H, R = 200 ,
V = 240 V and f = 50 Hz
...
15
...
3183) = 100
(b) Impedance
Z=

2
R 2 + XL

=

2002 + 1002 = 223
...
073 A
Z
223
...
d
...
c
...
c
...
073 × 100 = 107
...
d
...
c
...
c
...
073 × 200 = 214
...
39

Since XL = 2πfL, inductance,
L=

√
214
...
32 = 240 V,

XL
10
...
1 mH
2πf
2π(50)

This problem indicates a simple method for finding the
inductance of a coil, i
...
firstly to measure the current
when the coil is connected to a d
...
supply of known
voltage, and then to repeat the process with an a
...

supply
...
57◦ lagging
...
A coil consists of a resistance of
100 and an inductance of 200 mH
...
d
...
d
...


Single-phase series a
...
circuits 227
Since v = 200 sin 500t volts then Vm = 200 V and
ω = 2πf = 500 rad/s
Hence r
...
s
...
707 × 200 = 141
...
7

= ωL = 500 × 200 × 10−3 = 100
Inductive reactance

(a) Impedance
Z=
=

XL = 2πfL

2
R 2 + XL

= 2π(5 × 103 )(1
...
4

= 40

(b) Current
V
141
...
4
(c) P
...
across the resistance

Impedance,

I=

VR = IR = 1 × 100 = 100 V

(d) Phase angle between voltage and current is
given by:
XL
R

from which,
φ = tan−1

100
100

φ = 45◦ or

π
rads
4

Problem 11
...
273 mH is
connected in series with a pure resistance of 30
...
d
...
273 mH inductance
...

The circuit is shown in Fig
...
7(a)
Supply voltage, V = IZ
Current I =

VR
6
=
= 0
...
20)(50) = 10 V

P
...
across the inductance

tan φ =

Z=

Voltage across the 1
...
2)(40) = 8 V
The phasor diagram is shown in Fig
...
7(b)
(Note that in a
...
circuits, the supply voltage is not
the arithmetic sum of the p
...
’s across components but
the phasor sum)
Problem 12
...
2 mH and
resistance 20 is connected in series with a 60
resistor to a 240 V, 50 Hz supply
...
d
...
d
...
(f) Draw the circuit phasor diagram showing
all voltages
...
15
...
When
impedance’s are connected in series the individual resistance’s may be added to give the total circuit resistance
...
15
...

Inductive reactance XL = 2πfL
= 2π(50)(159
...


228 Electrical and Electronic Principles and Technology
2
...
Calculate the circuit impedance and the
current taken from the supply
...

[78
...
555 A, 39
...
8

(a) Circuit impedance,
Z=

2
R 2 + XL =

(b) Circuit current, I =

√
802 + 502 = 94
...
544 A
...
34

(c) Circuit phase angle φ = tan−1 XL /R
= tan−1 (50/80)
= 32◦ lagging
From Fig
...
8(a):

Section 2

(d) VR = IR = (2
...
6 V
(e) VCOIL = IZCOIL , where
√
2
2
ZCOIL = RC + XL = 202 + 502 = 53
...

Hence VCOIL = (2
...
85) = 137
...
15
...
544)(50) = 127
...

VRCOIL = IRC = (2
...
88 V
The 240 V supply voltage is the phasor sum of VCOIL
and VR as shown in the phasor diagram in Fig
...
9

3
...

Determine (a) the circuit impedance, (b) the
current flowing, (c) the p
...
across the resistance, and (d) the p
...
across the inductance
...
77 A
(c) 56
...
48 V]
4
...
c
...
When connected to a 200 V,
50 Hz a
...
supply the current is 25 A
...

[(a) 4 (b) 8 (c) 22
...
A resistor and an inductor of negligible resistance are connected in series to an a
...
supply
...
d
...
d
...
Calculate the
supply voltage and the phase angle between
voltage and current
...
13◦ lagging]
6
...
6 mH and negligible
resistance is connected in series with a 100
resistor to a 250 V, 50 Hz supply
...
d
...

[(a) 200 (b) 223
...
118 A
(d) 223
...
8 V (e) 63
...
9

15
...
c
...
c
...
Determine the impedance of a coil which has
a resistance of 12 and a reactance of 16
[20 ]

In an a
...
series circuit containing capacitance C and
resistance R, the applied voltage V is the phasor sum of
VR and VC (see Fig
...
10) and thus the current I leads
the applied voltage V by an angle lying between 0◦ and
90◦ (depending on the values of VR and VC ), shown as
angle α
...
15
...


Single-phase series a
...
circuits 229
(b) Current I = V /Z = 240/75
...
20 A
Phase angle between the supply voltage and current,
α = tan−1 (XC /R) hence
α = tan−1

70
...
54◦ leading

(‘Leading’ infers that the current is ‘ahead’ of the
voltage, since phasors revolve anticlockwise)
Figure 15
...
4, in an a
...
circuit, the ratio
applied voltage V to current I is called the impedance
Z, i
...
Z = V /I
If each side of the voltage triangle in Fig
...
10 is
divided by current I then the ‘impedance triangle’ is
derived
...
A resistor of 25 is connected in
series with a capacitor of 45 μF
...
Find also the phase angle between the
supply voltage and the current
...
The circuit diagram is as
shown in Fig
...
10
Capacitive reactance,
XC =
=

2
(a) Impedance Z = R2 + XC

Hence XC =
XC =
C=

1
hence,
2πfC

1
1
=
F = 88
...
87◦ leading
...
d
...
d
...
15
...


1
2πfC
1
= 70
...
742
= 75
...
11

Section 2

V=

Problem 14
...
A current of 3 A flows and the circuit
impedance is 50
...
d
...
d
...
Draw the phasor diagram
...
c
...
A voltage of 35 V is applied across a C–R series
circuit
...

[28 V]

Section 2

2
...
If a supply of 200 V,
100 Hz is connected across the arrangement
find (a) the circuit impedance, (b) the current flowing, and (c) the phase angle between
voltage and current
...
98 (b) 2
...
86◦ leading]
3
...
87 μF capacitor and a 30 resistor are
connected in series across a 150 V supply
...
d
...
d
...

[(a) 160 Hz (b) 90 V (c) 120 V]
4
...
12

30 resistor and 50 μF capacitor
...
d
...
d
...
05 (b) 4
...
8 V
(d) 113
...
81◦ leading]
5
...
c
...

Determine the supply frequency if the current
flowing in the circuit is 24 mA
[225 kHz]

15
...
c
...
c
...
15
...
VL and
VC are anti-phase, i
...
displaced by 180◦ , and there are
three phasor diagrams possible — each depending on
the relative values of VL and VC
...
c
...
15
...
15
...
87
5

= 49
...
15
...
This effect is called series
resonance (see Section 15
...

and

ZCOIL =

tan α =

Problem 15
...

Calculate (a) the current flowing, (b) the phase
difference between the supply voltage and current,
(c) the voltage across the coil and (d) the voltage
across the capacitor
...
15
...
72 = 38
...
91)(38
...
7
5

= 82
...
91)(31
...
15
...
The supply
voltage V is the phasor sum of VCOIL and VC
...
13

XL = 2πfL = 2π(50)(120 × 10−3 ) = 37
...
83
2πfC
2π(50)(100 × 10−6 )

Since XL is greater than XC the circuit is inductive
...
70 − 31
...
87
Impedance
Z=
=
(a) Current I =

R2 + (XL − XC )2
52 + 5
...
71

V
300
=
= 38
...
71

Figure 15
...
15
...
15
...


Figure 15
...
15
...

XC − XL = 31
...
34 = 15
...
492 = 27
...
73 = 1
...
15
...
e
...
15

Problem 16
...
25 μF capacitor
...

The circuit diagram is shown in Fig
...
16(a)
...
e
...
15
...

Inductive reactance,
XL = 2πfL = 2π(20 × 103 )(130 × 10−6 ) = 16
...
25 × 10−6 )
2πfC
2π(20 × 10
= 31
...
49
23

= 33
...
15
...
442)(8) = 11
...
342
= (1
...
09) = 24
...
832
= (1
...
36) = 48
...
Determine the p
...
’s V1 and V2 for
the circuit shown in Fig
...
17 if the frequency of
the supply is 5 kHz
...


Single-phase series a
...
circuits 233
The phasor sum of V1 and V2 gives the supply voltage
V of 100 V at a phase angle of 53
...
These
values may be determined by drawing or by calculation — either by resolving into horizontal and vertical
components or by the cosine and sine rules
...
17

Exercise 85 Further problems on R–L–C
a
...
circuits

and

XL = 2πfL
= 2π(5 × 103 )(0
...
985
2
V1 = IZ1 = I R2 + XL

= 5 42 + 8
...
18 V
Phase angle φ1 = tan−1

XL
= tan−1
R

8
...
0 lagging
For impedance Z2 : R2 = 8 and
1
1
=
XC =
2πfC
2π(5 × 103 )(1
...
A 40 μF capacitor in series with a coil of
resistance 8 and inductance 80 mH is connected to a 200 V, 100 Hz supply
...

[(a) 13
...
17 A (c) 52
...
1 V (e) 603
...
Find the values of resistance R and inductance
L in the circuit of Fig
...
19
...
545 H]
R

V2 = IZ2 = I

40 μF

I = 1
...
0
R2

L

240 V, 50 Hz
2
+ XC

=5

82

+ 25
...
2 V
...
0
8

= 72
...
15
...
19

3
...
The impedances
comprise:
(i) an inductance of 0
...
Draw the phasor
diagram
...
12 (b) 8
...
92◦ lagging
(d) 53
...
53 V, 76
...
For the circuit shown in Fig
...
20 determine
the voltages V1 and V2 if the supply frequency

Figure 15
...
Draw the phasor diagram and hence
determine the supply voltage V and the circuit
phase angle
...
0 V, V2 = 67
...
14◦ leading]

Figure 15
...
20

Problem 18
...
At
what frequency does resonance occur? Find the
current flowing at the resonant frequency
...
7 Series resonance
As stated in Section 15
...
15
...
This effect is called series
resonance
...
e
...
e
...
12 Hz
√
2π (125)(6)

At resonance, XL = XC and impedance Z = R
...
The current at resonance in a series
L–C–R circuit is 100 μA
...


where fr is the resonant frequency
...

(vi) Typical graphs of current I and impedance Z
against frequency are shown in Fig
...
21

(a) I = 100 μA = 100 × 10−6 A and
V = 2 mV = 2 × 10−3 V
...
Hence
R=

V
2 × 10−3
2 × 106
=
=
= 20
I
100 × 10−6 100 × 103

Single-phase series a
...
circuits 235
(b) At resonance XL = XC i
...

1
2πfC

Hence capacitance
C=

1
(2πf )2 L

=

1
F
(2π × 200 × 103 )2 (50 × 10−6 )

=

(106 )(106 )
μF
(4π)2 (1010 )(50)

(a) Resonant frequency
1

fr =

2π

15
...
15
...
5 = 8 A
Voltage across inductance, at resonance,

= 4525
...
4)(0
...
5 V)
Voltage magnification at resonance = VL /V or VC /V =
4525
...
255 i
...
at resonance, the voltage
across the reactance’s are 45
...
Hence the Q-factor of the circuit is
45
...
e
...
4 Hz or 1
...
25)
108

104
√
2π 2

= (8)(2π)(1125
...

Hence

Q-factor =

=

VL = IXL = (I)(2πfL)

voltage across L (or C)
supply voltage V

Q-factor =

0
...
0127 μF or 12
...
A series circuit comprises a coil of
resistance 2 and inductance 60 mH, and a 30 μF
capacitor
...


Section 2

2πfL =

Problem 20
...
25 μF and a resistor of resistance
12
...

Determine (a) the resonant frequency, and (b) the
current at resonance
...
36
2

Section 2

(a) Resonant frequency,
1
√

2π LC

1

=

100
103

2π
1

=
2π
=

20
108

=

2
106
1
√

2π 20
104

104
√ = 355
...
9 × 100 × 10−3 ) = 1118 V
Voltage across capacitance at resonance,
I
2πfr C
5
= 1118 V
=
2π(355
...
e
...
A coil of negligible resistance and
inductance 100 mH is connected in series with a
capacitance of 2 μF and a resistance of 10 across
a 50 V, variable frequency supply
...


fr =

2πfr L
R

VC 1118
VL
or
=
= 22
...
Find the resonant frequency of a series a
...
circuit consisting of a coil of resistance 10 and
inductance 50 mH and capacitance 0
...

Find also the current flowing at resonance if
the supply voltage is 100 V
...
183 kHz, 10 A]
2
...
2 mA
...
04 μF, find the circuit
resistance and inductance
...
25 k , 63
...
A coil of resistance 25
and inductance
100 mH is connected in series with a capacitance of 0
...
Calculate (a) the resonant
frequency, (b) the current at resonance and
(c) the factor by which the voltage across the
reactance is greater than the supply voltage
...
453 kHz (b) 8 A (c) 36
...
A coil of 0
...
If the current is in phase
with the supply voltage, determine the capacitance of the capacitor and the p
...
across its
terminals
...
26 μF, 3
...
Calculate the inductance which must be connected in series with a 1000 pF capacitor to
give a resonant frequency of 400 kHz
...
158 mH]
6
...
Determine the Q-factor of the circuit at
resonance
...
5 V, what
is the voltage across the capacitor?
[100, 150 V]

Single-phase series a
...
circuits 237
15
...
22 shows how current I varies with frequency
in an R–L–C series circuit
...
Also shown
are the points A and B where the current is 0
...
The power
delivered to the circuit is I 2 R
...
707 Ir , the power
is (0
...
5 Ir2 R, i
...
half the power that occurs
at frequency fr
...
The distance between
these points, i
...
( f2 − f1 ), is called the bandwidth
...
A filter in the form of a series
L–R–C circuit is designed to operate at a resonant
frequency of 5 kHz
...
Determine
the bandwidth of the filter
...
83

Since Qr = fr /( f2 − f1 ), bandwidth,
fr
5000
=
= 79
...
83

Selectivity is the ability of a circuit to respond more
readily to signals of a particular frequency to which it is
tuned than to signals of other frequencies
...
The higher the Q-factor,
the narrower the bandwidth and the more selective is
the circuit
...
A high Q-factor in a series power
circuit has disadvantages in that it can lead to dangerously high voltages across the insulation and may result
in electrical breakdown
...
22

It may be shown that

15
...
23

( f2 − f1 ) =

fr
Q

Power in a
...
circuits

In Figs
...
23(a)–(c), the value of power at any instant
is given by the product of the voltage and current at that
instant, i
...
the instantaneous power, p = vi, as shown
by the broken lines
...
c
...
15
...
27

(b) For a purely inductive a
...
circuit, the average
power is zero
...
15
...
c
...
See Fig
...
23(c)
Figure 15
...
The waveform for power (where p = vi)
is shown by the broken line, and its shape, and hence
average power, depends on the value of angle φ
...
272

= 66
...
33 = 1
...

To calculate power dissipation in an a
...
circuit two
formulae may be used:
(i) P = I 2 R = (1
...
33
= 0
...
658)(0
...
11

Figure 15
...
c
...
25(a) shows a phasor diagram in which the
current I lags the applied voltage V by angle φ
...
If each of the voltage phasors
is multiplied by I, Fig
...
25(b) is obtained and is known
as the ‘power triangle’
...
m
...
values)
Problem 24
...
Find the power dissipated in the resistor
...
m
...
value
of current
...
250 A
and r
...
s
...
707 × 0
...
Hence
power P = (0
...
250)2 (5000) = 156
...

Problem 25
...
Calculate the power dissipated
...
25

Apparent power,
S = VI voltamperes (VA)
True or active power,
P = VI cos φ watts (W)
Reactive power,
Q = VI sin φ reactive
voltamperes (var)

Single-phase series a
...
circuits 239
True power P
Apparent power S

For sinusoidal voltages and currents,
power factor =

P VI cos φ
=
S
VI

R
(from Fig
...
6)
Z
The relationships stated above are also true when current
I leads voltage V
...
e
...
f
...
A pure inductance is connected to a
150 V, 50 Hz supply, and the apparent power of the
circuit is 300 VA
...

Apparent power S = VI
...

Inductive reactance XL = V /I = 150/2 = 75
...
239 H
2πf
2π(50)

Problem 27
...
8
...

VI = 200 kVA = 200 × 103 and
p
...
= 0
...

Power output, P = VI cos φ = (200 × 103 )(0
...

Reactive power, Q = VI sin φ
...
8, then
φ = cos−1 0
...
87◦
...
87◦ = 0
...

Hence reactive power, Q = (200 × 103 )(0
...

Problem 28
...
5 lagging
...

True power P = 90 kW = VI cos φ and
power factor = 0
...

Apparent power, S = VI =

P
90
=
= 180 kVA
cos φ
0
...
5 = 60◦
hence sin φ = sin 60◦ = 0
...


Hence reactive power,
Q = VI sin φ = 180 × 103 × 0
...

Problem 29
...
Calculate (a) the
resistance, (b) the impedance, (c) the reactance,
(d) the power factor, and (e) the phase angle
between voltage and current
...
25
...

I
8

(b) Impedance Z =

2
(c) Since Z = R2 + XL , then
√
√
XL = Z 2 − R2 = 152 − 6
...
64
...
4167
(120)(8)

=

(e) p
...
= cos φ = 0
...
4167 = 65
...


Problem 30
...
5 from a 100 V, 60 Hz supply
...


(a) Power factor =
hence current,

true power
100
, i
...
0
...
5)(100)

(b) Power factor = 0
...
5 = 60◦ leading
(c) Power P = I 2 R hence resistance,
R=
(d) Impedance Z =

P
100
= 2 = 25
2
I
2

V 100
=
= 50
I
2

Section 2

Power factor =

240 Electrical and Electronic Principles and Technology
√
Z 2 − R2
√
= 502 −252 =43
...


(e) Capacitive reactance, XC =

XC = 1/2πfC
...
30)
= 61
...
c
...
A voltage v = 200 sin ωt volts is applied
across a pure resistance of 1
...
Find the
power dissipated in the resistor
...
33 W]

Section 2

2
...
Determine the true power and
the apparent power
...
3 VA]
3
...
c
...
Assuming a
power factor of 0
...
Find also the cost of running the
motor for 1 week continuously if 1 kWh of
electricity costs 12
...
43]
4
...
c
...
Assuming
a power factor of 0
...

[2
...
A transformer has a rated output of 100 kVA
at a power factor of 0
...
Determine the rated
power output and the corresponding reactive
power
...
A substation is supplying 200 kVA and
150 kvar
...

[132 kW, 0
...
A load takes 50 kW at a power factor of 0
...
Calculate the apparent power and the
reactive power
...
5 kVA, 37
...
A coil of resistance 400 and inductance
0
...

Calculate the power dissipated in the coil
...
452 W]

9
...

Calculate (a) the circuit impedance, (b) the
current flowing and (c) the power dissipated
in the circuit
...
9 (b) 0
...
The power taken by a series circuit containing
resistance and inductance is 240 W when connected to a 200 V, 50 Hz supply
...

[60 , 255 mH]
11
...
5 kHz supply,
is 0
...
Calculate (a) the resistance, (b) the impedance,
(c) the reactance, (d) the capacitance, (e) the
power factor, and (f) the phase angle between
voltage and current
...
745 (d) 11
...
571 (f) 55
...
A circuit consisting of a resistor in series with
an inductance takes 210 W at a power factor of 0
...
Find
(a) the current flowing, (b) the circuit phase
angle, (c) the resistance, (d) the impedance
and (e) the inductance
...
13◦ lagging (c) 4
...
143 (e) 9
...
A 200 V, 60 Hz supply is applied to a capacitive circuit
...
Calculate the
values of the resistance and capacitance
...
5 , 28
...
c
...
Complete the following statements:
(a) In a purely resistive a
...
circuit the
current is
...
c
...
the voltage by
...
c
...
Draw phasor diagrams to represent (a) a
purely resistive a
...
circuit (b) a purely inductive a
...
circuit (c) a purely capacitive a
...

circuit
3
...
What is capacitive reactance? State the symbol and formula for determining capacitive
reactance
5
...
What does ‘impedance’ mean when referring
to an a
...
circuit?
7
...
Derive from the triangle an expression
for (a) impedance, and (b) phase angle
8
...
From the triangle derive an expression
for (a) impedance, and (b) phase angle
9
...
Derive a formula for resonant frequency fr in
terms of L and C
11
...
State three formulae used to calculate the Qfactor of a series circuit at resonance
13
...
State a disadvantage of a high Q-factor in a
series power circuit
15
...
c
...
Show graphically that for a purely inductive
or purely capacitive a
...
circuit the average
power is zero
17
...
Define (a) apparent power (b) reactive power
19
...
c
...
An inductance of 10 mH connected across a
100 V, 50 Hz supply has an inductive reactance of
(a) 10π
(b) 1000π
(c) π
(d) π H
2
...
c
...
In question 2, the phase angle of the circuit
(a) decreases
(b) increases
(c) stays the same
4
...
c
...
In question 4, the phase angle of the circuit
(a) decreases
(b) increases
(c) stays the same
6
...
The capacitive reactance is
10
(a) 50 M
(b)
k
π
π
10
(c)
(d)
104
π
7
...
c
...
The supply voltage is
(a) 13 V
(b) 17 V
(c) 7 V
(d) 2
...
Inductive reactance results in a current that
(a) leads the voltage by 90◦
(b) is in phase with the voltage
(c) leads the voltage by π rad
(d) lags the voltage by π/2 rad

Section 2

(c) In a purely capacitive a
...
circuit the
current
...

degrees

242 Electrical and Electronic Principles and Technology
9
...
c
...
m
...
current and voltage gives the apparent power in an a
...

circuit
(c) Current is at a maximum at resonance
in an a
...
circuit
Apparent power
(d)
gives power factor
True power

supply
...
d
...
The amplitude of the current I flowing in the
circuit of Fig
...
26 is:
(a) 21 A
(b) 16
...
The impedance of a coil, which has a resistance of X ohms and an inductance of Y henrys, connected across a supply of frequency
K Hz, is
(a) 2π KY
(b) X + Y
√
(c)
X2 + Y 2
(d)
X 2 + (2πKY )2
11
...
When a capacitor is connected to an a
...

supply the current
(a) leads the voltage by 180◦
(b) is in phase with the voltage
(c) leads the voltage by π/2 rad
(d) lags the voltage by 90◦
13
...
c
...
In an R–L–C series a
...
circuit a current of
5 A flows when the supply voltage is 100 V
...
Which of the following
statements is false?
(a) The circuit is effectively inductive
(b) The apparent power is 500 VA
(c) The equivalent circuit reactance is 20
(d) The true power is 250 W
15
...
c
...
26

17
...
For the circuit shown in Fig
...
27, the value
of Q-factor is:
(a) 50
(b) 100
(c) 5 × 10−4
(d) 40

4Ω

400 mH
10 μF

V = 10 V

Figure 15
...
A series R–L–C circuit has a resistance of
8 , an inductance of 100 mH and a capacitance of 5 μF
...
c
...
c
...
c
...
1

Introduction

In parallel circuits, such as those shown in Figs
...
1
and 16
...

For any parallel a
...
circuit:
True or active power,

P = VI cos φ watts (W)

or

2
P = IR R watts

Apparent power,

S = VI voltamperes (VA)

Reactive power,

16
...
c
...
16
...
The supply current I is the
phasor sum of IR and IL and thus the current I lags the
applied voltage V by an angle lying between 0◦ and 90◦
(depending on the values of IR and IL ), shown as angle
φ in the phasor diagram
...
c
...

Figure 16
...
A 20 resistor is connected in
parallel with an inductance of 2
...
Calculate (a) the current in each
branch, (b) the supply current, (c) the circuit phase
angle, (d) the circuit impedance, and (e) the power
consumed
...
16
...
c
...
A 30 resistor is connected in parallel with
a pure inductance of 3 mH across a 110 V,
2 kHz supply
...

[(a) IR = 3
...
92 A (b) 4
...
51◦ lagging (d) 23
...
782 lagging]
2
...
Sketch the phasor
diagram and determine the inductance of the
coil
...
387 × 10−3 )
(b) From the phasor diagram, supply current,
2
2
IR + IL =

I=

32 + 4 2 = 5 A

16
...
c
...
16
...
The supply current I is
the phasor sum of IR and IC and thus the current I leads
the applied voltage V by an angle lying between 0◦ and
90◦ (depending on the values of IR and IC ), shown as
angle α in the phasor diagram
...
13◦ lagging
IR
3

(d) Circuit impedance,
Z=

V
60
=
= 12
I
5

(e) Power consumed
P = VI cos φ = (60)(5)(cos 53
...
2

From the phasor diagram:
Pythagoras’ theorem) where
IR =
tan α =

V
R

and

2
2
I = IR + IC (by

IC =

V
XC

IC
IR
IC
and cos α =
sin α =
IR
I
I

Single-phase parallel a
...
circuits 245

V
Circuit impedance, Z =
I
Problem 2
...
Calculate (a) the current in each branch,
(b) the supply current, (c) the circuit phase angle,
(d) the circuit impedance, (e) the power dissipated,
and (f ) the apparent power

Problem 3
...
The
supply current is 2 A at a power factor of 0
...
Determine the values of C and R

The circuit diagram is shown in Fig
...
3(a)
...
13°
V = 120 V
200 Hz

V
240
=
= 3A
R
80

Current in capacitor,
IC =

R

V
1
2πfC

= 2πfCV

Power factor = cos φ = 0
...
6 = 53
...

From the phasor diagram shown in Fig
...
3(b),
= 1
...
2622

= 3
...
13 = (2)(0
...
6 A

(Alternatively, IR and IC can be measured from the
scaled phasor diagram)
...
13◦ = (2)(0
...
3

= 2π(50)(30 × 106 )(240) = 2
...
16
...
262

V
IR

3
◦

= 37
...
88
I
3
...
757) cos 37
...
757) = 901
...
2

C=
=

IC
2πf V
1
...
61 μF

Section 2

(by trigonometric ratios)

246 Electrical and Electronic Principles and Technology
Now try the following exercise
Exercise 91 Further problems on R–C
parallel a
...
circuits
1
...
Calculate (a) the current in each
branch, (b) the supply current, (c) the circuit
phase angle, (d) the circuit impedance, (e) the
power consumed, (f ) the apparent power,
and (g) the circuit power factor
...

[(a) IR = 0
...
943 A (b) 1
...
46◦ leading (d) 8
...
25 W
(f) 11
...
553 leading]

Section 2

2
...

The supply current is 0
...
8 leading
...
55 μF]

16
...
16
...

The latter condition is not possible in practice due to circuit resistance inevitably being present (as in the circuit
described in Section 16
...

For the L–C parallel circuit,
IL =

V
XL

IC =

V
XC

I = phasor difference between IL and IC , and
V
Z=
I
Problem 4
...

Determine (a) the branch currents, (b) the supply
current and its phase angle, (c) the circuit
impedance, and (d) the power consumed
...
16
...
70
Capacitive reactance,
XC =

1
1
=
2πfC
2π(50)(25 × 10−6 )
= 127
...
653 A
XL
37
...
4

Theoretically there are three phasor diagrams
possible — each depending on the relative values of
IL and IC :
(i) IL > IC (giving a supply current, I = IL − IC lagging V by 90◦ )

IC =

V
100
=
= 0
...
3

(b) IL and IC are anti-phase, hence supply current,
I = IL − IC = 2
...
786 = 1
...
16
...
c
...
56
I
1
...
867)cos 90◦ = 0 W
Problem 5
...
1
Capacitive reactance,

Exercise 92 Further problems on L–C
parallel a
...
circuits
1
...
Determine (a) the branch currents, (b) the supply current, (c) the circuit
phase angle, (d) the circuit impedance and
(e) the power consumed
[(a) IC = 0
...
194 A (b) 0
...
44 (e) 0 W]
2
...
754 A, IL = 0
...
157 A
(c) 90◦ leading (d) 382
...
44
2π(150)(25 × 10−6 )

Current flowing in inductance,
IL =

V
100
=
= 0
...
1

Current flowing in capacitor,
IC =

V
100
=
= 2
...
44

(b) Supply current,

16
...
c
...
16
...
16
...
16
...
Rotating each and superimposing on
one another gives the complete phasor diagram shown
in Fig
...
5(d)

I = IC − IL = 2
...
884 = 1
...
16
...
93
I
1
...
5

The current ILR of Fig
...
5(d) may be resolved
into horizontal and vertical components
...
There are three
possible conditions for this circuit:

The circuit diagram is shown in Fig
...
6(a)
...
2 mH

(ii) ILR sin φ > IC (giving I lagging V by angle φ—as
shown in Fig
...
5(f))

ILR

(iii) IC = ILR sin φ1 (this is called parallel resonance,
see Section 16
...
16
...
These
are: (i) by a scaled phasor diagram, or (ii) by resolving
each current into their ‘in-phase’ (i
...
horizontal) and
‘quadrature’ (i
...
vertical) components, as demonstrated in problems 6 and 7
...
16
...
262 A

Z1

(i) IC > ILR sin φ1 (giving a supply current I leading
V by angle φ—as shown in Fig
...
5(e))

51
...
748 A

(b)

Figure 16
...
2 × 10−3 ) = 50
...
03

V
IC =
XC

Current in coil,

Supply current

ILR =

I = phasor sum of ILR and IC (by drawing)

V
240
=
= 3
...
03

Branch phase angle
=

(ILR cos φ1 )2 + (ILR sin φ1 ∼ IC )2
(by calculation)

φ1 = tan−1

XL
50
= tan−1
R
40
= tan−1 1
...
34◦ lagging

where ∼ means ‘the difference between’
...
16
...
A coil of inductance 159
...

Calculate (a) the current in the coil and its phase
angle, (b) the current in the capacitor and its phase
angle, (c) the supply current and its phase angle,
(d) the circuit impedance, (e) the power consumed,
(f ) the apparent power, and (g) the reactive power
...


1
1
=
2π fC
2π(50)(30 × 10−6 )
= 106
...
1
= 2
...
16
...

(c) The supply current I is the phasor sum of ILR and
IC
...
(Current I will always
be the diagonal of the parallelogram formed as in
Fig
...
6(b))
...
c
...
434) = 584
...
434)(sin 15
...
341 A
The vertical component of ILR
= −ILR sin 51
...
748 sin 51
...
927 A
The vertical component of IC
= IC sin 90◦ = 2
...
262 A

= 159
...
A coil of inductance 0
...
02 μF capacitor and is supplied at 40 V at a
frequency of 5 kHz
...
(c) Draw to
scale the phasor diagram and measure the supply
current and its phase angle; check the answer by
calculation
...

The circuit diagram is shown in Fig
...
8(a)
...
927 + 2
...
665 A
IH and IV are shown in Fig
...
7, from which,
I=

2
...
665)2 = 2
...
665
= 15
...
341

Hence the supply current I = 2
...
86◦
I H = 2
...
665 A
I

IC = 25
...
12 H

Section 2

Alternatively the current ILR and IC may be
resolved into their horizontal (or ‘in-phase’) and
vertical (or ‘quadrature’) components
...
34◦ = 3
...
34◦ = 2
...

The horizontal component of IC is

I

ILR C = 0
...
49°
I LR = 8
...
8

(a) Inductive reactance,
XL = 2π fL = 2π(5000)(0
...
7

Z1 =

R 2 + XL =

(d) Circuit impedance,
Z=

30002 + 37702

= 4818

240
V
= 98
...
434
I

(e) Power consumed,

Current in coil,
ILR =

P = VI cos φ = (240)(2
...
86◦
= 562 W
2
2
(Alternatively, P = IR R = ILR R (in this case)

= (3
...
30 mA
=
Z1
4818

Branch phase angle
φ = tan−1

XL
3770
= tan−1
R
3000
◦
= 51
...
Calculate (a) the current in the coil, (b) the current
in the capacitor, (c) the supply current and
its phase angle, (d) the circuit impedance,
(e) the power consumed, (f ) the apparent
power and (g) the reactive power
...

[(a) 1
...
943 A (c) 1
...
88◦
lagging (d) 194
...
5 W
(f ) 205
...
5 var]

1
1
=
XC =
2π fC
2π(5000)(0
...
13 mA leading V by 90◦

(c) Currents ILR and IC are shown in the phasor
diagram of Fig
...
8(b)
...
The current I is
measured as 19
...
5◦
...
49◦ )2 + (IC − ILR sin 51
...
34 mA

Section 2

and
φ = tan−1

IC − ILR sin 51
...
5◦

2
...
20 H
across a 100 V, 4 kHz supply
...
48 mA (b) 62
...
17 mA at 81
...
166 k (e) 0
...
50◦

16
...
068 k
I
19
...
34 × 10−3 ) cos 74
...
16
...
e
...

At this condition the supply current I is in-phase with
the supply voltage V
...
7 mW
2
(Alternatively, P = IR R
2
= ILR R

= (8
...
7 mW)
Now try the following exercise

Resonant frequency
When the quadrature component of ILR is equal to IC
then: IC = ILR sin φ1 (see Fig
...
9)
...
5)

from which,
2
ZLR = XL XC = (2πfr L)

Exercise 93 Further problems on LR–C
parallel a
...
circuit
1
...
4 mH is connected in parallel with a 15 μF

1
2πfr C

=

L
C

Hence
2
2
R 2 + XL

=

L
C

and

2
R 2 + XL =

L
C

(1)

Single-phase parallel a
...
circuits 251
Dynamic resistance
Since the current at resonance is in-phase with the voltage the impedance of the circuit acts as a resistance
...

From equation (2), impedance at resonance
=

=

Figure 16
...
e
...


L
− R2
C
L
R2
− 2
2C
L
L

1
2π

Applications of resonance

1
R2
− 2
LC L

(When R is negligible, then fr =
same as for series resonance)

1
√

2π LC

, which is the

Current at resonance
Current at resonance,
Ir = ILR cos φ1
=
=

V
ZLR

(from Fig
...
9)
R
ZLR

L
ohms
RC

Rejector circuit

i
...
parallel resonant frequency,
fr =

L
RC

V
VRC
L

(from Section 16
...


One use for resonance is to establish a condition of
stable frequency in circuits designed to produce a
...

signals
...
Just as
a pendulum can be used to stabilise the frequency of a
clock mechanism’s oscillations, so can a parallel circuit
be used to stabilise the electrical frequency of an a
...

oscillator circuit
...
A resonant circuit can
be used to ‘block’ (i
...
present high impedance toward)
a frequency or range of frequencies, thus acting as a sort
of frequency ‘filter’ to strain certain frequencies out of
a mix of others
...
In essence, this is how analogue radio receiver tuner
circuits work to filter, or select, one station frequency
out of the mix of different radio station frequency signals
intercepted by the antenna
...
The Q-factor
of a parallel resonant circuit is the ratio of the current
circulating in the parallel branches of the circuit to the
supply current, i
...
the current magnification
...
e
...

Note that in a parallel circuit the Q-factor is a measure of current magnification, whereas in a series
circuit it is a measure of voltage magnification
...

Problem 8
...
Determine (a) the
resonant frequency of the circuit and (b) the current
circulating in the capacitor and inductance at
resonance
...
16
...
97)(40 × 10−6 )(50)
= 0
...
97)(0
...
817 A)

Problem 9
...
20 H and
resistance 60 is connected in parallel with a
20 μF capacitor across a 20 V, variable frequency
supply
...

(a) Parallel resonant frequency,
fr =

1
2π

1
R2
− 2
LC
L

1
(60)2
−
(0
...
20)2
1 √
1 √
=
2 50 000 − 90 000 =
1 60 000
2π
2π
1
=
(400) = 63
...
10

1
6

= 64
...
c
...
20
L
=
= 166
...
12 A
RD
166
...
66)(0
...
33
R
60

Alternatively, Q-factor at resonance

Ic =

0
...
64)(108 )}

106
μF
0
...
51 × 108 )

(b) Dynamic resistance,
RD =

IC
Ir
V
=
XC

8002
(100 × 10−3 )2

= 0
...
515 nF

= current magnification (for a parallel circuit)
=

1

C=

100 × 10−3
L
=
CR
(9
...
14 k

V
1
2πfr C

= 2πfr CV

Ir =

= 2π(63
...
16 A
Hence Q-factor = IC /Ir = 0
...
12 = 1
...

Problem 10
...

Determine for the condition when the supply
current is a minimum: (a) the capacitance of the
capacitor, (b) the dynamic resistance, (c) the supply
current, and (d) the Q-factor
(a) The supply current is a minimum when the parallel
circuit is at resonance and resonant frequency,
1
fr =
2π

1
R2
− 2
LC
L

Transposing for C gives:
1
R2
− 2
LC
L
R2
1
(2πfr )2 + 2 =
L
LC
(2πfr )2 =

and C =

(c) Supply current at resonance,

1
L (2πfr )2 +

R2
L2

V
12
=
= 0
...
14 × 103

(d) Q-factor at resonance
=

2πfr L
2π(5000)(100 × 10−3 )
=
= 3
...
515 × 10−9 )(12)
= 3
...
913 × 10−3

Now try the following exercise
Exercise 94 Further problems on parallel
resonance and Q-factor
1
...
15 μF capacitor and a pure inductance
of 0
...
Determine
(a) the resonant frequency of the circuit, and
(b) the current circulating in the capacitor and
inductance
...
11 kHz (b) 38
...
A 30 μF capacitor is connected in parallel with
a coil of inductance 50 mH and unknown resistance R across a 120 V, 50 Hz supply
...

[(a) 37
...
94 A (c) 2
...
A coil of resistance 25
and inductance
150 mH is connected in parallel with a 10 μF
capacitor across a 60 V, variable frequency
supply
...

[(a) 127
...
10 A (d) 4
...
A coil having resistance R and inductance
80 mH is connected in parallel with a 5 nF
capacitor across a 25 V, 3 kHz supply
...

[(a) 3
...
318 k
(c) 5
...
41]
5
...
5 k and 0
...

Calculate (a) the capacitance of the capacitor when the supply current is a minimum,
(b) the dynamic resistance, and (c) the supply
current
...
8 k (c) 93
...
A parallel circuit as shown in Fig
...
11 is
tuned to resonance by varying capacitance C
...


Q-factor, (d) the bandwidth, (e) the current in
each branch, (f ) the supply current, and (g) the
power dissipated at resonance
...
533 pF (b) 5
...
9
(d) 11
...
915∠90◦ mA,
ILR = 15
...
863◦ mA (f ) 38 μA
(g) 7
...
7 Power factor improvement
For a particular power supplied, a high power factor
reduces the current flowing in a supply system, which
consequently lowers losses (i
...
I 2 R losses) and hence
results in cheaper running costs
...
Industrial loads
such as a
...
motors are essentially inductive (R–L) and
may have a low power factor
...
16
...
The supply current is reduced
from ILR to I, the phasor sum of ILR and IC , and the circuit power factor improves from cos φ1 to cos φ2 (see
Fig
...
12(b))
...
12
IC
C

I
V

Figure 16
...
A single-phase motor takes 50 A at a
power factor of 0
...
Determine (a) the current taken by a
capacitor connected in parallel with the motor to
correct the power factor to unity, and (b) the value
of the supply current after power factor correction
...
16
...


Single-phase parallel a
...
circuits 255
V = 240 V

M

I M = 50 A

IC

53
...


IM = 50 A
(b)

53
...
13

(a) A power factor of 0
...
6 i
...

−1

φ = cos

0
...
13

◦

Hence IM lags V by 53
...
16
...

If the power factor is to be improved to unity then
the phase difference between supply current I and
voltage V needs to be 0◦ , i
...
I is in phase with
V as shown in Fig
...
13(c)
...

ab = IM sin 53
...
8) = 40 A
Hence the capacitor current IC must be 40 A for
the power factor to be unity
...
A motor has an output of 4
...
625 lagging when operated from a 240 V, 50 Hz
supply
...
95 lagging by connecting a capacitor in parallel
with the motor
...

(a) Efficiency =
hence

power output
power input

80
4800
=
100 power input

4800
= 6000 W
0
...
625),
since cos φ = p
...
= 0
...
Thus current taken by
the motor,

and power input =

IM =

6000
= 40 A
(240)(0
...
16
...

The phase angle between IM and V is given by:
φ = cos−1 0
...
32◦ , hence the phasor diagram is as shown in Fig
...
14(b)
...
13◦ = 50(0
...

Problem 12
...
7 lagging
...
(c) If the power factor is
now raised to unity find the new kVA loading
...
f
...
7

(b) VI = 60 000 VA
hence I =

60 000
60 000
=
= 150 A
V
400

Figure 16
...
The phasor sum of IM and IC gives the supply current I, and has to be such as to change the
circuit power factor to 0
...
e
...
95 or 18
...
16
...
The horizontal component of IM
(shown as oa)
= IM cos 51
...
32◦ = 25 A
The horizontal component of I (also given by oa)

Problem 14
...
7 lagging, (iii) a 3 kVA motor operating at full
load and at a power factor of 0
...
Determine, for the lamps and
motor, (a) the total current, (b) the overall power
factor and (c) the total power
...
975 lagging
...
19◦
= 0
...
95 I
...
f
...
16
...
7, i
...
45
...
e
...
8, i
...
36
...
32 A
0
...
32◦
= 40 sin 51
...
22 A
Figure 16
...
19◦
= 26
...
19◦ = 8
...
e
...
22 − 8
...
52 kvar
1000
1000

In this problem the supply current has been reduced
from 40 A to 26
...
This means that the I 2 R losses
are reduced, and results in a saving of costs
...
87◦ + 8 cos 45
...
6 + 5
...
2 A
The vertical component of the currents
= 10 sin 0◦ + 12 sin 36
...
57◦
= 0 + 7
...
713 = 12
...
15(b), total current,
Fig
...
22 + 12
...
31 A at a phase angle
of φ = tan−1 (12
...
2) i
...
27
...

(b) Power factor
= cos φ = cos 27
...
890 lagging
(c) Total power,
P = VIL cos φ = (250)(28
...
890)
= 6
...
The capacitor
takes a current IC such that the supply current falls

Single-phase parallel a
...
circuits 257

oa = 28
...
13◦ = I cos 12
...
31 cos 27
...
84 A
cos 12
...
31 sin 27
...
84 sin 12
...
91 − 5
...
168 A
IC =

V
=
XC

V
1
2πfc

= 2πfCV

Figure 16
...
168
=
F = 91
...
890 to
0
...
27 μF capacitor is connected
in parallel with the loads
...
A 415 V alternator is supplying a load of 55 kW
at a power factor of 0
...
Calculate (a) the kVA loading and (b) the current
taken from the alternator
...

[(a) 84
...
9 A (c) 84
...
A single phase motor takes 30 A at a power factor of 0
...

Determine (a) the current taken by the capacitor connected in parallel to correct the power

factor to unity, and (b) the value of the supply
current after power factor correction
...
80 A (b) 19
...
A 20 non-reactive resistor is connected in
series with a coil of inductance 80 mH and
negligible resistance
...
Calculate (a) the reactance of the coil, (b) the
impedance of the circuit, (c) the current in
the circuit, (d) the power factor of the circuit,
(e) the power absorbed by the circuit, (f ) the
value of a power factor correction capacitor to
produce a power factor of unity, and (g) the
value of a power factor correction capacitor to
produce a power factor of 0
...

[(a) 25
...
12∠51
...
227∠−51
...
623
(e) 775
...
56 μF (g) 47
...
A motor has an output of 6 kW, an efficiency
of 75% and a power factor of 0
...

It is required to raise the power factor to
0
...
Determine (a) the current taken by the motor, (b) the supply current
after power factor correction, (c) the current
taken by the capacitor, (d) the capacitance of
the capacitor and (e) the kvar rating of the
capacitor
...
59 A (c) 25
...
2 μF (e) 6
...
A supply of 250 V, 80 Hz is connected across
an inductive load and the power consumed
is 2 kW, when the supply current is 10 A
...
What value of capacitance connected
in parallel with the load is needed to improve
the overall power factor to unity?
[R = 20 , L = 29
...
75 μF]
6
...
9 leading,
(ii) incandescent lamps taking a current of 6 A
at unity power factor, (iii) a motor taking a
current of 12 A at a power factor of 0
...
Determine the total current taken from
the supply and the overall power factor
...
31 A to I, lagging V by cos−1 0
...
e
...
84◦
...
16
...
98 lagging
...
74 A, 0
...
74 μF]

Exercise 96 Short answer questions on
single-phase parallel
a
...
circuits
1
...
Draw a phasor diagram for a two-branch
parallel circuit containing inductance L and
resistance R in one branch and capacitance
C in the other, connected across a supply
voltage V
3
...
c
...
Determine the quantities stated in
questions 1 to 8, selecting the correct answer from
the following list:
(a) 24 A
(b) 6
(c) 7
...
8 leading
(g) 7
...
44 kW
(m) 12
...
8 lagging
(s) 20 A

5
(j) tan−1 3 lagging
(l) 0
...
4 kW
(p) 0
...
92 kW

1
...
The capacitive reactance of the capacitor

4
...
The current flowing in the capacitor

5
...
The supply phase angle

6
...
Develop a formula for the resonant frequency
in an LR–C parallel circuit, in terms of
resistance R, inductance L and capacitance C
8
...
Develop a formula for the current at resonance in an LR–C parallel circuit in terms
of resistance R, inductance L, capacitance C
and supply voltage V
10
...
Explain a simple method of improving the
power factor of an inductive circuit
12
...
The supply current
6
...
The power consumed by the circuit
8
...
A two-branch parallel circuit consists of a
15 mH inductance in one branch and a 50 μF
capacitor in the other across a 120 V, 1/π kHz
supply
...
The following statements, taken correct to 2
significant figures, refer to the circuit shown
in Fig
...
17
...
80 H

Single-phase parallel a
...
circuits 259

XL = 4 Ω

R = 3Ω
ILR

IC

XC = 12
...
The magnitude of the impedance of the circuit
shown in Fig
...
18 is:
(a) 7
(b) 5
(c) 2
...
71

I
V = 250V,

5
kHz
2π

Figure 16
...
Which of the following statements is false?
(a) The supply current is a minimum at
resonance in a parallel circuit
(b) The Q-factor at resonance in a parallel
circuit is the voltage magnification
(c) Improving power factor reduces the current flowing through a system
(d) The circuit impedance is a maximum at
resonance in a parallel circuit
12
...
Which of the
following statements is false?
(a) The resonant frequency fr is 1
...
18

14
...
16
...
9
(k) Circuit power factor = 0
...
0 kW

Figure 16
...
e
...
1

Introduction

Attenuation is a reduction or loss in the magnitude of
a voltage or current due to its transmission over a line
...

A filter is frequency sensitive and is thus composed
of reactive elements
...

Between the pass band of a filter, where ideally the attenuation is zero, and the attenuation band, where ideally
the attenuation is infinite, is the cut-off frequency, this
being the frequency at which the attenuation changes
from zero to some finite value
...

Filters are used for a variety of purposes in nearly
every type of electronic communications and control

equipment
...

There are four basic types of filter sections:
(a)
(b)
(c)
(d)

low-pass
high-pass
band-pass
band-stop

17
...
The network between the
input port and the output port is a transmission network
for which a known relationship exists between the input
and output currents and voltages
...
1(a) shows a T-network, which is termed
symmetrical if ZA = ZB , and Fig
...
1(b) shows a
π-network which is symmetrical if ZE = ZF
...
17
...
17
...
Both networks

Filter networks 261
ZA

ZB

impedance for an asymmetrical network and its value
depends on which pair of terminals is taken to be the
input and which the output (there are thus two values of
iterative impedance, one for each direction)
...


ZC

(a)
ZD

17
...
3 shows simple unbalanced T- and π-section
filters using series inductors and shunt capacitors
...
17
...
This is an ideal characteristic and assumes
pure reactive elements
...
It is
for this reason that the networks shown in Fig
...
3(a)
and (b) are known as low-pass filters
...
1
Z B/ 2

Z A/ 2

ZC

Z A/ 2

Z B/ 2
(a)
Z D/ 2
(a)
ZF

ZE

(b)

Figure 17
...
2

shown have one common terminal, which may be
earthed, and are therefore said to be unbalanced
...
17
...
17
...

The input impedance of a network is the ratio of voltage to current at the input terminals
...
For
any passive two-port network it is found that a particular value of load impedance can always be found
which will produce an input impedance having the same
value as the load impedance
...
4

The electrical circuit diagram symbol for a low-pass
filter is shown in Fig
...
5
...
5

L
2

R0

R0

C

Summarising, a low-pass filter is one designed to
pass signals at frequencies below a specified cut-off
frequency
...
17
...
The characteristic may be improved somewhat closer to the ideal by connecting two or more
identical sections in cascade
...


(a)
L

R0

C
2

C
2

R0

Section 2

Attenuation

(b)

Figure 17
...
6

When rectifiers are used to produce the d
...
supplies
of electronic systems, a large ripple introduces undesirable noise and may even mask the effect of the signal
voltage
...

Filters are employed to isolate various sections of a
complete system and thus to prevent undesired interactions
...


(1)

When the frequency is very low, the characteristic impedance is purely resistive
...
Determine the cut-off frequency and
the nominal impedance for the low-pass
T-connected section shown in Fig
...
8
...
2 μF

Cut-off frequency and nominal
impedance calculations
A low-pass symmetrical T-network and a low-pass
symmetrical π-network are shown in Fig
...
7
...
8

Filter networks 263
Comparing Fig
...
8 with the low-pass section of
Fig
...
7(a), shows that:

From equation (1), cut-off frequency,
fc =

i
...
inductance,

L
= 100 mH,
2
L = 200 mH = 0
...
2 μF = 0
...


=
i
...


From equation (1), cut-off frequency,

=
i
...


f c = 1592 Hz

or

=

103
π(0
...
592 kHz

L
=
C

L
=
C

0
...
2 × 10−6

= 1000

To determine values of L and C
given R0 and f c

or 1 k

inductance L =

and
Problem 2
...
17
...

0
...
4
= 31
...
It may be
shown that:
1
capacitance C =
(3)
πR0 f c

From equation (2), nominal impedance,
R0 =

106
√
π 160

f c = 25
...
2 × 0
...
4 × 400 × 10−12 )

=

From equation (2), nominal impedance,

1
√
π LC
1

1

R0
πf c

Problem 3
...
Design
(a) a low-pass T-section filter, and (b) a low-pass
π-section filter to meet these requirements
...
e
...


200 pF

From equation (3), capacitance,

Figure 17
...
17
...
17
...
4 H
...
06 × 10−10 F = 106 pF

From equation (4), inductance,

C
= 200 pF,
2
i
...
capacitance,

(4)

−12

F,

L=

R0
600
=
H
πfc π(5 × 106 )
= 3
...
2 μH

Section 2

fc =

1
√
π LC

264 Electrical and Electronic Principles and Technology
19
...
1 μH

38
...
10

(a) A low-pass T-section filter is shown in Fig
...
10(a),
L
where the series arm inductances are each
2
38
...
17
...
e
...
1 μH
2
(b) A low-pass π-section filter is shown in Fig
...
10(b),
C
where the shunt arm capacitances are each
2
106
(see Fig
...
7(b)), i
...

= 53 pF
2

Section 2

Now try the following exercise
Exercise 98 Further problems on low-pass
filter sections
1
...
17
...

[(a) 1592 Hz; 5 k (b) 9545 Hz; 600 ]

a cut-off frequency of 1 kHz
...

[(a) Each series arm 79
...
6366 μF
(b) Series arm 159
...
3183 μF]
3
...
5 kHz
...
203 μF]
4
...
If the capacitance in
each of the shunt arms is 0
...

[72 mH]

17
...
5 H

0
...
12 shows simple unbalanced T- and π-section
filters using series capacitors and shunt inductors
...
17
...


0
...
8 nF

27
...
11

2
...
12

Once again this is an ideal characteristic assuming
pure reactive elements
...


Attenuation

Attenuation

Filter networks 265

Attenuation
band

Pass-band

0

Frequency

0

fC

fC

Frequency

Attenuation
band

Figure 17
...
15

Cut-off frequency and nominal
impedance calculations
A high-pass symmetrical T-network and a high-pass
symmetrical π-network are shown in Fig
...
16
...
14

Summarising, a high-pass filter is one designed to
pass signals at frequencies above a specified cut-off
frequency
...
17
...
Both of these conditions are impossible to achieve
in practice
...
In addition
to the resistive loss there is often an added loss due to
mismatching
...
However
the characteristic impedance of a filter section will vary
with frequency and the termination of the section may
be an impedance that does not vary with frequency in
the same way
...
13 showed an ideal high-pass filter section characteristic of attenuation against frequency
...
17
...


fc =

1
√

2C

R0

(5)

4π LC
2C

L

R0

(a)
C

R0

2L

2L

R0

(b)

Figure 17
...
This value of

Section 2

It is for this reason that the networks shown in
Figs
...
12(a) and (b) are known as high-pass filters
...
17
...


266 Electrical and Electronic Principles and Technology
characteristic impedance is then the nominal impedance of the section and is given by:
R0 =

L
C

(6)
200 μH

Problem 4
...
17
...

0
...
18

Comparing Fig
...
18 with the high-pass section of
Fig
...
16(b), shows that:

0
...
e
...


From equation (5), cut-off frequency,

Figure 17
...
17
...
17
...
2 μF,
i
...
capacitance,

C = 0
...
1 × 10−6 ,

and inductance,

L = 100 mH = 0
...


i
...


1
√

4π LC
1
4π

(10−4

× 4 × 10−9 )

= 1
...
1 × 0
...
1)

i
...
f c = 796 Hz

L
=
C
=

10−4
4 × 10−9
105
= 158
4

From equation (6), nominal impedance,
R0 =

L
0
...
1 × 10−6
= 1000
or 1 k

Problem 5
...
17
...


To determine values of L and C
given R0 and f c
If the values of the nominal impedance R0 and the cut-off
frequency fc are known for a high-pass T- or π-section,
it is possible to determine the values of inductance and
capacitance required to form the section
...
A filter section is required to pass all
frequencies above 25 kHz and to have a nominal
impedance of 600
...


are each 2C (see Fig
...
16(a)), i
...
2 × 5
...
61 nF
(b) A high-pass π-section filter is shown in
Fig
...
19(b), where the shunt arm inductances are each 2L (see Fig
...
6(b)), i
...

2 × 1
...
82 mH
...

From equation (7), capacitance,
C=

and
Exercise 99 Further problems on
high-pass filter sections

1
1
=
F
4πR0 fc 4π(600)(25 × 103 )
=

1012
pF
4π(600)(25 × 103 )

= 5305 pF

or

1
...
17
...

[(a) 22
...
14 k
(b) 281
...
305 nF
500 pF

500 pF

L=

R0
600
=
4πfc 4π(25 × 103 )

Section 2

From equation (8), inductance,
50 mH

= 0
...
91 mH
(a)

(a) A high-pass T-section filter is shown in
Fig
...
19(a), where the series arm capacitances
10
...
2 μF

10
...
91 mH
(b)

Figure 17
...
305 nF

3
...
82 mH

(b)

Figure 17
...
A filter section is required to pass all frequencies above 4 kHz and to have a nominal
impedance 750
...

[(a) Each series arm = 53
...
92 mH
(b) Series arm = 26
...
84 mH]

3
...
If the
nominal impedance of the section is 600 ,
determine the value of the capacitance in the
series arm
...
44 nF]
4
...
5 μF capacitor
...

[11
...
The characteristic of an ideal band-pass filter is
shown in Fig
...
21
...
23

17
...

Problem 7
...
The terminating impedance of the filter is
600
...


For the low-pass T-section filter:
0

fcH

f cL Frequency

Figure 17
...
fCH is the cut-off frequency of the
high-pass filter and fCL is the cut-off frequency of the
low-pass filter
...

The electrical circuit diagram symbol for a band-pass
filter is shown in Fig
...
22
...
22

A typical practical characteristic for a band-pass filter
is shown in Fig
...
23
...
They are common in the intermediatefrequency amplifiers of v
...
f
...
4 × 10−9 = 35
...
01273 H = 12
...
17
...
e
...
73
= 6
...
4 nF
...
37 mH

6
...
6 nF

35
...
6 nF

600 Ω

4
...
24

From equation (7), capacitance,

= 1
...
3 nF
From equation (8), inductance,
L=

600
R0
=
4πfc 4π(10 000)
= 4
...
77 mH
...
A band-pass filter is comprised of a low-pass
π-section filter having a cut-off frequency of
50 kHz, connected in series with a high-pass
π-section filter having a cut-off frequency of
40 kHz
...
Determine the values of the
components comprising the composite filter
...
95 mH,
each shunt arm 5
...
21 nF,
each shunt arm 2
...
17
...
3 = 26
...
77 mH
...
17
...

The attenuation against frequency characteristic will
be similar to Fig
...
23 where fCH = 10 kHz and
fCL = 15 kHz
...
6 Band-stop filters
A band-stop filter is one designed to pass signals with
all frequencies except those between two specified
cut-off frequencies
...
17
...


Attenuation

i
...


Pass-band

Stop-band

Pass-band

Exercise 100 Further problems on
band-pass filters
1
...
The terminating impedance of the filter is 600
...

[Low-pass T-section: each
series arm 4
...
53 nF
High-pass T-section: each
series arm 33
...
97 mH]

0

f cL

f cH

Frequency

Figure 17
...
As can be seen, for
a band-stop filter fCH > fCL , the stop-band being given
by the difference between these values
...
17
...


270 Electrical and Electronic Principles and Technology
5
...
filter
...
26

A typical practical characteristic for a band-stop filter
is shown in Fig
...
27
...
A network designed to pass signals with all
frequencies except those between two specified cut-off frequencies is called a
...


Attenuation

7
...
filter
...
A network designed to pass signals at frequencies above a specified cut-off frequency
is called a
...

9
...


0

fcL

Section 2

Pass-band

10
...


f cH
Stop-band

Pass-band

Figure 17
...
Usually such interference is from an odd harmonic of 50 Hz, for example,
250 Hz
...
A high-pass filter with cutoff frequency greater than 250 Hz would also remove
the interference, but some of the lower frequency
components of the audio signal would be lost as well
...
For more,
see Electrical Circuit Theory and Technology
...
Sketch (a) an ideal, and (b) a practical
attenuation/frequency characteristic for a
high-pass filter
...
Sketch (a) an ideal, and (b) a practical
attenuation/frequency characteristic for a
band-pass filter
...
State one application of a band-pass filter
...
Sketch (a) an ideal, and (b) a practical
attenuation/frequency characteristic for a
band-stop filter
...
State one application of a band-stop filter
...
Define a filter
...
Define the cut-off frequency for a filter
...
Define a two-port network
...
Define characteristic impedance for a
two-port network
...
A network designed to pass signals with
all frequencies except those between two
specified cut-off frequencies is called a:
(a) low-pass filter
(b) high-pass filter
(c) band-pass filter (d) band-stop filter
2
...
A network designed to pass signals with
frequencies between two specified cut-off
frequencies is called a:
(a) low-pass filter
(b) high-pass filter
(c) band-pass filter (d) band-stop filter
5
...
5 μF
in its shunt arm
...
6 Hz
(c) 711
...
9 Hz
6
...
The cut-off frequency of the
filter is:
(a) 25
...
29 kHz
(c) 17
...
59 kHz
The following refers to questions 7 and 8
...

7
...
68 μH in each series arm, 128
...
34 μH in each series arm, 256
...
68 μH in each series arm, 256
...
34 μH in each series arm, 128
...
A low-pass π-connected symmetrical filter
section is comprised of:
(a) 98
...
4 pF
in shunt arm
(b) 49
...
7 pF
in shunt arm

(c) 98
...
7 pF
in shunt arm
(d) 49
...
4 pF
in shunt arm
9
...
The cut-off frequency of the
filter is:
(a) 1592 Hz
(b) 1125 Hz
(c) 281 Hz
(d) 398 Hz

10
...
The cut-off frequency of the
filter is:
(a) 201
...
18 kHz
(c) 50
...
7 kHz
The following refers to questions 11 and 12
...

11
...
45 nF, shunt arm
1
...
90 nF, shunt arm
2
...
45 nF, shunt arm
2
...
90 nF, shunt arm
1
...
A high-pass π-connected symmetrical filter
section is comprised of:
(a) Series arm 4
...
04 mH
(b) Series arm 4
...
07 mH
(c) Series arm 2
...
07 mH
(d) Series arm 2
...
04 mH

Section 2

3
...
C
...
c
...
c
...
1

Introduction

When a d
...
voltage is applied to a capacitor C and
resistor R connected in series, there is a short period of
time immediately after the voltage is connected, during
which the current flowing in the circuit and voltages
across C and R are changing
...
c
...

These changing values are called transients
...
2 Charging a capacitor
(a) The circuit diagram for a series connected C − R
circuit is shown in Fig
...
1 When switch S is
closed then by Kirchhoff’s voltage law:
V = vC + vR

(1)

(b) The battery voltage V is constant
...
The voltage drop across R is given
by iR, where i is the current flowing in the circuit
...
C
...
1

At the instant of closing S, (initial circuit condition), assuming there is no initial charge on the
capacitor, q0 is zero, hence vCo is zero
...
e
...

This shows that the resistance to current is
solely due to R, and the initial current flowing,
i0 = I = V /R
...

Since V = vC + vR and V is a constant, then vR
decreases to i2 R, Thus vC is increasing and i and
vR are decreasing as time increases
...
e
...
e
...
It
follows from Equation (1) that vC = V
...
18
...
The curves showing the variation of
vR and i with time are called exponential decay
curves, and the graphs are called ‘resistor voltage/time’and ‘current/time’characteristics respectively
...
4)
...
3 Time constant for a C − R circuit
(a) If a constant d
...
voltage is applied to a series connected C − R circuit, a transient curve of capacitor
voltage vC is as shown in Fig
...
2(a)
...
18
...
Let the voltage be varied so that the current flowing in the circuit is
constant
...
3

Section 2

Figure 18
...
The
voltage vC1 is now (q1 /C) volts
...
Thus, Equation (2) is now
V = (q1 /C) + i1 R
...

(d) Let the capacitor voltage vC reach its final value
of V at time t2 seconds
...
The value of the time
constant is CR seconds, i
...
for a series connected
C − R circuit,
time constant τ = CR seconds

Section 2

Since the variable voltage mentioned in paragraph
(b) above can be applied at any instant during the
transient change, it may be applied at t = 0, i
...
at
the instant of connecting the circuit to the supply
...


18
...
632
of its steady state value (usually taken as 63
per cent of the steady state value), at a time
equal to two and a half time constants is 0
...
368 of its
initial value (usually taken as 37 per cent of
its initial value), at a time equal to two and a
half time constants is 0
...


The transient curves shown in Fig
...
2 have mathematical equations, obtained by solving the differential
equations representing the circuit
...
A 15 μF uncharged capacitor is
connected in series with a 47 k resistor across a
120 V, d
...
supply
...
From the characteristic,
determine the capacitor voltage at a time equal to
one time constant after being connected to the
supply, and also two seconds after being connected
to the supply
...


To construct an exponential curve, the time constant of the circuit and steady state value need to be
determined
...
705 s
Steady state value of vC = V , i
...
vC = 120 V
...
18
...
e
...
705 or about 3
...
The scale of the
vertical axis spans the change in the capacitor voltage,
that is, from 0 to 120 V
...

Point C is measured along AB so that AC is equal
to 1τ, i
...
AC = 0
...
Straight line OC is drawn
...
DE
is drawn vertically
...
C
...
From the characteristics determine the value
of capacitor voltage, resistor voltage and current
1
...


B

J
I

80
60

H

To draw the transient curves, the time constant of the
circuit and steady state values are needed
...
88 s

Figure 18
...
e
...
705 s
...

This procedure of

is repeated for vC values of 40, 60, 80 and 100 V, giving
points G, H, I and J
...
Drawing a smooth curve
through points 0, D, G, H, I, J and K gives the
exponential growth curve of capacitor voltage
...
It is a characteristic of all exponential growth curves, that after a
time equal to one time constant, the value of the transient is 0
...
In this problem,
0
...
84 V
...
[This value may be
checked using the equation vC = V (1 − e−t/τ ), where
V = 120 V, τ = 0
...
This calculation
gives vC = 112
...

The time for vC to rise to one half of its final
value, i
...
60 V, can be determined from the graph
and is about 0
...
[This value may be checked using
vC = V (1 − e−t/τ ) where V = 120 V, vC = 60 V and
τ = 0
...
489 s]
...
109 mA

Finally, vC = vR = i = 0
...
e
...
88 = 4
...
With reference to Fig
...
5,
to construct the decay curve:

Figure 18
...
A 4 μF capacitor is charged to 24 V
and then discharged through a 220 k resistor
...
e
...
4 s,

Section 2

(a) drawing a vertical line through point selected,
(b) at the steady-state value, drawing a horizontal line
corresponding to 1τ, and
(c) joining the first and last points,

i=

276 Electrical and Electronic Principles and Technology
(ii) the vertical scale is made to span the change
in capacitor voltage, i
...
0 to 24 V,
(iii) point A corresponds to the initial capacitor
voltage, i
...
24 V,
(iv) OB is made equal to one time constant and
line AB is drawn; this gives the initial slope
of the transient,
(v) the value of the transient after a time equal
to one time constant is 0
...
e
...
368 × 24 = 8
...
83 V,

Section 2

(vi) the value of the transient after a time equal
to two and a half time constants is 0
...
e
...
082 × 24 = 1
...
18
...
e
...
4 s, giving point E
...
At 1
...
4 V
...
5 and τ = 0
...
36 V]
(b) The voltage drop across the resistor is equal to the
capacitor voltage when a capacitor is discharging
through a resistor, thus the resistor voltage/time
characteristic is identical to that shown in Fig
...
5
Since vR = vC , then at 1
...
4 V (see (vii) above)
...
18
...
The values are:
point A: initial value of current = 0
...
368 × 0
...
040 mA
point D: at 2
...
082 × 0
...
009 mA
point E: at 5 τ, i = 0
Hence the current transient is as shown
...
5 s, the value of current, from the
characteristic is 0
...
109,
t = 1
...
88,
giving
i = 0
...
8 μA]
Problem 3
...
c
...
Determine: (a) the
initial value of the current flowing, (b) the time
constant of the circuit, (c) the value of the current
one second after connection, (d) the value of the
capacitor voltage two seconds after connection, and
(e) the time after connection when the resistor
voltage is 15 V
...

V = 20 V, C = 20 μF = 20 × 10−6 F,
R = 50 k = 50 × 103 V
(a) The initial value of the current flowing is
I=

V
20
= 0
...
3 the time constant,
τ = CR = (20 × 10−6 )(50 × 103 ) = 1 s
(c) Current, i = Ie−t/τ and working in mA units,
i = 0
...
4 × 0
...
147 mA
(d) Capacitor voltage,
vC = V (1 − e−t/τ ) = 20(1 − e−2/1 )
= 20(1 − 0
...
865
= 18
...
6

4
= ln 1
...
e
...
288 s
3

D
...
transients 277
Problem 4
...
5 μF capacitor and has
a time constant of 12 ms
...

(a) The time constantτ = CR, hence
τ
C
12 × 10−3
=
0
...
558) = 4
...

Problem 5
...
c
...
When the
supply is connected, calculate (a) the time constant,
(b) the maximum current, (c) the voltage across the
capacitor after 0
...
1 s, (f) the time for the capacitor voltage to
reach 45 V, and (g) the initial rate of voltage rise
...
25 s
(b) Current is a maximum when the circuit is first connected and is only limited by the value of resistance
in the circuit, i
...

Im =

45 = 100(1 − e−t/0
...
25
(i
...
gradient of the tangent at t = 0)

(g) Initial rate of voltage rise =

18
...
e
...
18
...
Initially, at the instant
of moving from A to B, the current flow is such that the
capacitor voltage vC is balanced by an equal and opposite voltage vR = iR
...
During the transient decay, by applying
Kirchhoff’s voltage law to Fig
...
7, vC = vR
...
5 s, then
vC = 100(1 − e−0
...
25 ) = 100(0
...
47 V

45
= 1 − e−t/0
...
25 = 1 −
= 0
...
55
0
...
25 ln 0
...
149 s

Figure 18
...
583

Alternatively, after one time constant the capacitor
voltage will have risen to 63
...
2% of its
final value, i
...
36
...
Hence, i = 36
...
368 × 4 = 1
...
1 s,
resistor voltage, vR = 100 e−0
...
25 = 67
...
472 mA

278 Electrical and Electronic Principles and Technology
Finally the transients decay exponentially to zero,
i
...
vC = vR = 0
...
18
...


Parts (b), (c) and (d) of this problem may be solved
graphically as shown in Problems 1 and 2 or by
calculation as shown below
...
8 s, R = 50 k = 50 × 103
(a) Since time constant, τ = CR, capacitance,
C=

0
...
8 from
which 1/5 = e−t/0
...
8 = 5 and taking natural logarithms
of each side, gives t/0
...
8 ln 5 = 1
...

(c) i = Ie−t/τ where the initial current flowing,
I=

100
V
=
= 2 mA
R
50 × 103

Section 2

Working in mA units,
i = Ie−t/τ = 2e(−0
...
8)
= 2e−0
...
535 = 1
...
8

The equations representing the transient curves during the discharge period of a series connected C − R
circuit are:
decay of voltage,
vC = vR = Ve

(−t/CR)

= Ve

(−t/τ)

decay of current, i = Ie(−t/CR) = Ie(−t/τ)
When a capacitor has been disconnected from the
supply it may still be charged and it may retain this
charge for some considerable time
...
This
is done by connecting a high value resistor across the
capacitor terminals
...
A capacitor is charged to 100 V and
then discharged through a 50 k resistor
...
8 s
...
5 s, and
(d) the voltage drop across the resistor when the
capacitor has been discharging for one second
...
8
= 100e−1
...
287 = 28
...
A 0
...
Determine (a) the initial discharge current,
(b) the time constant of the circuit, and (c) the
minimum time required for the voltage across the
capacitor to fall to less than 2 V
...
05 A or
R
4 × 103

50 mA

(b) Time constant τ = CR = 0
...
0004 s or

0
...
e
...

5τ = 5 × 0
...
C
...
An uncharged capacitor of 0
...
c
...
Determine, either graphically or by calculation the capacitor voltage
10 ms after the voltage has been applied
[39
...
A circuit consists of an uncharged capacitor connected in series with a 50 k resistor
and has a time constant of 15 ms
...
c
...

[(a) 0
...
33 V]
3
...
5 M resistor
...
Also find how long it takes for
the voltage to fall to 25 V
[105
...
53 s]
4
...

When the voltmeter reading is steady the battery is replaced with a shorting link
...

[55
...
When a 3 μF charged capacitor is connected
to a resistor, the voltage falls by 70 per cent
in 3
...
Determine the value of the resistor
...
08 M ]
6
...
c
...
Determine:
(a) the initial current flowing in the circuit,
(b) the time constant,

(c) the value of current when t is 50 ms and
(d) the voltage across the resistor 60 ms
after closing the switch
...
1 A (b) 50 ms
(c) 36
...
1 V]
7
...
c
...
Determine the time constant of the
circuit, the initial charging current, the
current flowing 120 ms after connecting to
the supply
...
67 mA, 1
...
An uncharged 80 μF capacitor is connected
in series with a 1 k resistor and is switched
across a 110 V supply
...
Determine also the value of
current flowing after (a) 40 ms and (b) 80 ms
[80 ms, 0
...
7 mA (b) 40
...
A resistor of 0
...
The battery is replaced
instantaneously by a conducting link
...
Determine from the graph the
approximate time for the capacitor voltage to
fall to 75 V
[9
...
A 60 μF capacitor is connected in series with
a 10 k resistor and connected to a 120 V
d
...
supply
...

[(a) 0
...
323 s ]
11
...
c
...
5 M resistor and a 2 μF capacitor in series
...

[(a) 35
...
87 V
(c) 12
...
(a) In the circuit shown in Fig
...
9,
with the switch in position 1, the
capacitor is uncharged
...
c
...


280 Electrical and Electronic Principles and Technology
is moved to position 2 at time t = 0 s,
calculate the (i) initial current through
the 0
...
5 s,
and (iii) the time taken for the voltage
across the capacitor to reach 12 V
...
5 s, the switch
is moved to position 3, calculate
(i) the initial current through the 1 M
resistor, (ii) the energy stored in the
capacitor 3
...
e
...

(c) Sketch a graph of the voltage across
the capacitor against time from t = 0 to
t = 5 s, showing the main points
...
05 V (iii) 0
...
30 μJ ]

Section 2

5 μF

2
1
3

1 MΩ

Figure 18
...
6

closed, then by Kirchhoff’s voltage law:
V = vL + vR

(3)

(b) The battery voltage V is constant
...
e
...

Hence, at all times:

40 V

0
...
10

Camera flash

The internal workings of a camera flash are an example
of the application of C − R circuits
...
When the capacitor
is fully charged, an indicator (red light) typically lets the
photographer know that the flash is ready for use
...
e
...
The current from the
capacitor is responsible for the bright light that is emitted
...
The capacitor must then be discharged
before the flash can be used again
...
7 Current growth in an
L − R circuit
(a) The circuit diagram for a series connected L − R
circuit is shown in Fig
...
10
...
m
...

in the inductance which is equal and opposite to V ,
hence V = vL + 0, i
...
vL = V
...

(d) A short time later at time t1 seconds after closing
S, current i1 is flowing, since there is a rate of
change of current initially, resulting in a voltage
drop of i1 R across the resistor
...
m
...
is reduced,
and Equation (4) becomes:
V =L

di1
+ i1 R
dt1

(e) A short time later still, say at time t2 seconds after
closing the switch, the current flowing is i2 , and
the voltage drop across the resistor increases to
i2 R
...

(f) Ultimately, a few seconds after closing S, the current flow is entirely limited by R, the rate of change
of current is zero and hence vL is zero
...
Under these conditions, steady state current flows, usually signified by I
...

(g) Curves showing the changes in vL , vR and i with
time are shown in Fig
...
11 and indicate that vL is
a maximum value initially (i
...
equal to V), decaying exponentially to zero, whereas vR and i grow

D
...
transients 281
growth of current flow,
i = I(1 − e−Rt/L ) = I(1 − e−t/τ )
The application of these equations is shown in
Problem 10
...
A relay has an inductance of 100 mH
and a resistance of 20
...
c
...
Use the ‘initial slope and three point’
method to draw the current/time characteristic and
hence determine the value of current flowing at a
time equal to two time constants and the time for
the current to grow to 1
...

Before the current/time characteristic can be drawn, the
time constant and steady-state value of the current have
to be calculated
...


18
...
3, the time constant of a
series connected L − R circuit is defined in the same
way as the time constant for a series connected C − R
circuit
...
e
...
18
...
e
...


18
...
4
...

Each of the transient curves shown in Fig
...
11 have
mathematical equations, and these are:
decay of induced voltage,
vL = Ve(−Rt/L) = Ve(−t/τ)
growth of resistor voltage,
vR = V(1 − e−Rt/L ) = V(1 − e−t/τ )

Figure 18
...
632 × I = 0
...
896 A
...
11

282 Electrical and Electronic Principles and Technology
At a time of 2
...
918 × I = 0
...
754 A
...

(d) A smooth curve is drawn through points 0, D, F and
H and this curve is the current/time characteristic
...
6 A
...
59 A]
...
5 A, the corresponding time is about
3
...
[Again, this may be checked by calculation,
using i = I(1 − e−t/τ ) where i = 1
...
466 ms]
...
2 s
R
15

Parts (c), (d) and (e) of this problem may be determined by drawing the transients graphically, as
shown in Problem 8 or by calculation as shown
below
...
m
...
, vL is given by vL = V e−t/τ
...
c
...
1 s and τ is 0
...
1/0
...
5
= 120 × 0
...
78 V

Section 2

Problem 9
...
04 H and
resistance 10 is connected to a 120 V, d
...
supply
...

V 120
(a) Final steady current, I = =
= 12 A
R
10
(b) Time constant of the circuit,
τ=

0
...
004 s or 4 ms
R
10

(d) When the current is 85 per cent of its final value,
i = 0
...
Also, i = I(1 − e−t/τ ), thus
0
...
85 = 1 − e−t/τ
τ = 0
...
85 = 1 − e−t/0
...
2 = 1 − 0
...
15
1
˙
et/0
...
6
0
...
2 = ln 6
...
2 per cent of
its final value of 12 A, i
...
in 4 ms the current rises
to 0
...
58 A
...
e
...


t
˙
ln e = ln 6
...
2
˙
ln e = 1, hence time t = 0
...
6 = 0
...
The winding of an electromagnet
has an inductance of 3 H and a resistance of 15
...
c
...
m
...
after
0
...
3 s
...
When I = 8, t = 0
...
2,
then
i = 8(1 − e−0
...
2 ) = 8(1 − e−1
...
2231) = 8 × 0
...
215 A

18
...
c
...
18
...
When

D
...
transients 283

Figure 18
...
The field winding of a 110 V, d
...

motor has a resistance of 15 and a time constant
of 2 s
...
From the characteristic
determine (a) the current flowing in the winding 3 s
after being shorted-out and (b) the time for the
current to decay to 5 A
...
e
...
e
...
3 A vertically
(ii) With reference to Fig
...
14, the initial slope is
obtained by making OB equal to 1 time constant,
(i
...
2 s), and joining AB
(iii) At, say, i = 6 A, let C be the point on AB corresponding to a current of 6 A
...
e
...
14

(iv) Repeat the procedure given in (iii) for current values of, say, 4 A, 2 A and 1 A, giving points F, G
and H
(v) Point J is at five time constants, when the value
of current is zero
...

This curve is the current/time characteristic
...
3 A [This may be checked by
˙
calculation using i = Ie−t/τ , where I = 7
...
64 A] The discrepancy between the two results is due to
relatively few values, such as C, F, G and H,
being taken
...
70 s [This may be checked by calcula˙
tion using i = Ie−t/τ , where i = 5, I = 7
...
766 s]
...

Problem 12
...
c
...
The time
constant of the circuit is 300 ms
...
Determine:
(a) the resistance of the coil, (b) the current flowing
in the circuit one second after the shorting link has

Section 2

S is moved to position B, the current value decreases,
causing a decrease in the strength of the magnetic field
...
By Lenz’s law, this voltage keeps current i
flowing in the circuit, its value being limited by R
...
The current decays exponentially to zero
and since vR is proportional to the current flowing, vR
decays exponentially to zero
...
The curves representing
these transients are similar to those shown in Fig
...
9
...

(a) The time constant,
circuit inductance
L
τ=
=
total circuit resistance
R + 10
L
6
− 10 =
− 10 = 10
τ
0
...


The time constant,
τ=

0
...
2 ms
R
1000

The steady-state current
I=

V
24
=
= 24 mA
R
1000

Thus R =

V
120
I=
=
= 6A
R
10 + 10
The transient current after 1 second,
i = Ie

Section 2

Thus

i = I(1 − e−t/τ )

and

t = 1τ
...
3

= 6e

˙

i = 6e−3
...
03567
= 0
...
e
...
6 A Using the equation
i = Ie−t/τ gives
0
...
3
0
...
3
6
6
or
et/0
...
6
Taking natural logarithms of each side of this
equation gives:
t
= ln 10
0
...
3 ln 10 = 0
...
368) = 15
...
248 V
(c) The voltage drop across the resistor,
vR = V (1 − e−t/τ )
When t = 3τ, vR = 24(1 − e−3τ/τ )
= 24(1 − e−3 )
= 22
...
e
...
An inductor has a negligible
resistance and an inductance of 200 mH and is
connected in series with a 1 k resistor to a 24 V,
d
...
supply
...
Find (a) the current flowing
in the circuit at a time equal to one time constant,
(b) the voltage drop across the inductor at a time
equal to two time constants and (c) the voltage drop
across the resistor after a time equal to three time
constants
...
A coil has an inductance of 1
...
c
...
Either by drawing the current/time
characteristic or by calculation determine the
value of the current flowing 60 ms after connecting the coil to the supply
...
32 A]
2
...
c
...
Either
by using a graphical method to draw the
exponential growth curve of current or by calculation determine the value of the current
flowing 100 ms after being connected to the
supply
...
97 A]

D
...
transients 285

4
...
c
...
Calculate:
(a) the time constant of the field winding,
(b) the value of current flow one time constant after being connected to the supply,
and
(c) the current flowing 50 ms after the supply
has been switched on
[(a) 25 ms (b) 6
...
65 A]
5
...
c
...
Calculate (a) the time constant, (b) the current after
1 time constant, (c) the time to develop maximum current, (d) the time for the current to
reach 2
...
[(a) 0
...
528 A (c) 0
...
147 s (e) 26
...
In the inductive circuit shown in Fig
...
15, the
switch is moved from position A to position
B until maximum current is flowing
...

[(a) 64
...
20 s
(c) 0
...
67 ms]
A
B

10 Ω

18
...
When the
d
...
supply is switched off the current falls rapidly, the
magnetic field collapses causing a large induced e
...
f
...
The high induced e
...
f
...
e
...
The energy from the
magnetic field will thus be aided by the supply voltage in
maintaining an arc, which could cause severe damage to
the switch
...
m
...
when the supply
switch is opened, a discharge resistor RD is connected
in parallel with the inductor as shown in Fig
...
16
...


Section 2

3
...
It is connected to a 50 V d
...

supply
...
1 s and (b) the time for the current
to grow to 1
...
984 A
(b) 0
...
16

18
...
If a rectangular waveform varying
from +E to −E is applied to a C − R circuit as shown in
Fig
...
17, output waveforms of the capacitor voltage
have various shapes, depending on the value of R
...
15
Figure 18
...
18
...
As the value
of R is increased, the waveform changes to that shown
in Fig
...
18(b)
...
18
...


Figure 18
...
c
...
18

Differentiator circuit
If a rectangular waveform varying from +E to −E
is applied to a series connected C − R circuit and the
waveform of the voltage drop across the resistor is
observed, as shown in Fig
...
19, the output waveform
alters as R is varied due to the time constant, (τ = CR),
altering
...
c
...

After a time of t seconds, the current flowing
is i amperes
...

1
...

2
...

3
...

4
...

5
...

6
...

7
...

8
...

9
...


Figure 18
...
18
...
Since
the change in capacitor voltage is from +E to −E,
the change in discharge current is 2E/R, resulting in
a change in voltage across the resistor of 2E
...
When R is large,
the waveform is as shown in Fig
...
20(b)
...
The initial value of the voltage drop across
the resistor is
...
Use this data to
answer questions 11 to 15
11
...


D
...
transients 287
12
...
seconds

26
...
per cent of its
final value in a time equal to the time constant

13
...

when compared with the time in question 12
above

27
...


14
...
per cent

29
...
seconds

An inductor of inductance L henrys and negligible resistance is connected in series with a
resistor of resistance R ohms and is switched
across a constant voltage d
...
supply of V
volts
...
Use
this data to answer questions 16 to 25
16
...
m
...
, vL , opposing the current
flow when t = 0 is
...
The voltage drop across the resistor when
t = 0 is vR =
...
The current flowing when t = 0 is
...
V , vR and vL are related by the equation
V =
...
The time constant of the circuit in terms of L
and R is
...
The steady-state value of the current is
reached in practise in a time equal to
...
The steady-state voltage across the inductor
is
...
The final value of the current flowing is
...
The steady-state resistor voltage is
...
The e
...
f
...
volts
A series-connected L − R circuit carrying
a current of I amperes is suddenly shortcircuited to allow the current to decay exponentially
...
If the value of R is halved, the time for the current to fall to zero is
...
With the aid of a circuit diagram, explain
briefly the effects on the waveform of the
capacitor voltage of altering the value of
resistance in a series connected C − R circuit, when a rectangular wave is applied to
the circuit
...
What do you understand by the term ‘integrator circuit’ ?
33
...
c
...
c
...
In questions 1 to 7, use this data to
select the correct answer from those given below:
(a) 10 ms
(b) 100 V
(c) 10 s
(d) 10 V
(e) 20 μA
(f) 1 s
(g) 0 V
(h) 50 V
(i) 1 ms
(k) 20 mA
(l) 0 A
(j) 50 μA
1
...
Determine the final voltage across the
capacitor
3
...
Determine the final voltage across the resistor

Section 2

15
...
seconds

28
...


288 Electrical and Electronic Principles and Technology
5
...
The value of the time constant of the circuit

6
...
The approximate value of the voltage across
the resistor after a time equal to the time
constant

7
...
The final value of the current flowing in the
circuit

In questions 8 and 9, a series connected
C − R circuit is suddenly connected to a d
...

source of V volts
...
The initial value of the voltage across the
inductor

Section 2

8
...
(a) The capacitor voltage is equal to the
voltage drop across the resistor
(b) The voltage drop across the resistor
decays exponentially
(c) The initial capacitor voltage is zero
(d) The initial voltage drop across the resistor is IR, where I is the steady-state
current
10
...
Which
of the following statements is false?
(a) The initial current flowing is V /R
amperes
(b) The voltage drop across the resistor is
equal to the capacitor voltage
(c) The time constant of the circuit is CR
seconds
(d) The current grows exponentially to a
final value of V /R amperes
An inductor of inductance 0
...
c
...
In questions 11 to 15, use this data to determine the
value required, selecting your answer from
those given below:
(a) 5 ms
(b) 12
...
4 A
(d) 500 ms
(e) 7
...
5 A
(g) 2 ms
(h) 0 V
(i) 0 A
(j) 20 V

15
...
The time constant for a circuit containing a
capacitance of 100 nF in series with a 5
resistance is:
(a) 0
...
The time constant for a circuit containing
an inductance of 100 mH in series with a
resistance of 4 is:
(a) 25 ms (b) 400 s (c) 0
...
The graph shown in Fig
...
21 represents
the growth of current in an L − R series circuit connected to a d
...
voltage V volts
...
21

Time t

Chapter 19

Operational amplifiers
At the end of this chapter you should be able to:
• recognize the main properties of an operational amplifier
• understand op amp parameters input bias current and offset current and voltage
• define and calculate common-mode rejection ratio
• appreciate slew rate
• explain the principle of operation, draw the circuit diagram symbol and calculate gain for the following
operational amplifiers:
inverter
non-inverter
voltage follower (or buffer)
summing
voltage comparator
integrator
differentiator
• understand digital to analogue conversion
• understand analogue to digital conversion

19
...
Now produced in
integrated-circuit (IC) form, op amps have many uses,
with one of the most important being as a high-gain d
...

and a
...
voltage amplifier
...
The circuit diagram symbol for an op amp is shown in
Fig
...
1 It has one output, V o , and two inputs;
the inverting input, V1 is marked −, and the
non-inverting input, V2 , is marked +

(i) a very high open-loop voltage gain Ao of around
105 for d
...
and low frequency a
...
, which
decreases with frequency increase

The operation of an op amp is most convenient from
a dual balanced d
...
power supply ±Vs (i
...
+Vs ,
0, −Vs ); the centre point of the supply, i
...
0 V, is

290 Electrical and Electronic Principles and Technology
+ Vs (Supply +)
Inverting
Input
V1

−

− Vs (Supply −)

Saturation

+ Vs

Output

+
V2

Vo

Vo

V2 > V1
P

Non-inverting
Input

0V (on power supply)

0

Q

(V2 − V1) μV

V2 < V1

Figure 19
...
The power supply
connections are not usually shown in a circuit diagram
...
e
...
Three situations are possible:
(i) if V2 > V1 , Vo is positive

− Vs

Figure 19
...
The limited linear behaviour is due to the very
high open-loop gain Ao , and the higher it is the greater
is the limitation
...

Problem 1
...
The input signals
are 2
...
35 V
...


Operational amplifiers nearly always use negative feedback, obtained by feeding back some, or all, of the
output to the inverting (−) input (as shown in Fig
...
5
in the next section)
...
This
reduces the new output of the amplifier and the resulting closed-loop gain A is then less than the open-loop
gain Ao
...
As long as
Ao
A, negative feedback gives:
(i) a constant and predictable voltage gain A,
(ii) reduced distortion of the output, and

From equation (1), output voltage,
Vo = Ao (V2 − V1 ) = 120(2
...
35)
= (120)(0
...

The advantages of using negative feedback outweigh
the accompanying loss of gain which is easily increased
by using two or more op amp stages
...
19
...

It is seen from Fig
...
2 that only within the very small
input range P0Q is the output directly proportional to the
input; it is in this range that the op amp behaves linearly
and there is minimum distortion of the amplifier output
...
e
...
Figure 19
...
At frequencies below
10 Hz the gain is constant, but at higher frequencies the
gain falls at a constant rate of 6 dB/octave (equivalent to
a rate of 20 dB per decade) to 0 dB
...
The value of

Operational amplifiers 291
Common-mode rejection ratio
106

The output voltage of an op amp is proportional to the
difference between the voltages applied to its two input
terminals
...
A signal applied to both
input terminals is called a common-mode signal and it
is usually an unwanted noise voltage
...
The
common-mode gain, Acom , is defined as:

Figure 19
...

f T = closed-loop voltage gain × bandwidth

(2)

In Fig
...
3, f T = 106 Hz or 1 MHz; a gain of 20 dB
(i
...
20 log10 10) gives a 100 kHz bandwidth, whilst
a gain of 80 dB (i
...
20 log10 104 ) restricts the bandwidth
to 100 Hz
...
2

Some op amp parameters

Acom =

Problem 2
...

From equation (3),
⎛

⎞
differential
voltage gain ⎟
⎜
CMRR = 20 log10 ⎝
⎠ dB
common mode
gain

The input bias current, IB , is the average of the currents into the two input terminals with the output at zero
volts, which is typically around 80 nA (i
...
80×10−9 A)
for a 741 op amp
...


⎛
Hence

⎜
90 = 20 log10 ⎝

from which

The input offset current, Ios , of an op amp is the difference between the two input currents with the output at
zero volts
...


and

⎞
× 103

150
⎟
⎠
common mode
gain

⎛
⎜
4
...
Due to imbalances within the amplifier
this is not always the case and a small output voltage
results
...
In a 741 op amp,
Vos is typically 1 mV
...


Input bias current

Input offset current

Vo
Vcom

104
...
74
104
...
A differential amplifier has an
open-loop voltage gain of 120 and a common input
signal of 3
...
An output signal
of 24 mV results
...

From equation (4), the common-mode gain,
Acom =

Vo
24 × 10−3
=
= 8 × 10−3 = 0
...
0

19
...
19
...
c
...
c
...
The non-inverting (+) terminal is held
at 0 V
...

Rf

I2

From equation (3), the
⎛

Ri

⎞

differential
⎜ voltage gain ⎟
⎟
CMRR = 20 log10 ⎜
⎝ common mode ⎠ dB
gain

−

I1
Vi

X
VA

+
VB

Vo
0V

Section 2

= 20 log10

120
0
...
5

= 20 log10 15 000 = 83
...
Figure 19
...
0
...


In an ideal op amp two assumptions are made, these
being that:
(i) each input draws zero current from the signal
source, i
...
their input impedance’s are infinite, and
(ii) the inputs are both at the same potential if the op
amp is not saturated, i
...
VA = VB in Fig
...
5
In Fig
...
5, VB = 0, hence VA = 0 and point X is
called a virtual earth
...
Hence
0

Time
Actual
output

− Vs

Figure 19
...

The closed-loop gain A is given by:
A=

Vo −Rf
=
Vi
Ri

(5)

Operational amplifiers 293
This shows that the gain of the amplifier depends only
on the two resistors, which can be made with precise
values, and not on the characteristics of the op amp,
which may vary from sample to sample
...
e
...
The input impedance of the
circuit is therefore Ri in parallel with the much greater
input impedance of the op amp, i
...
effectively Ri
...


Problem 4
...
19
...
Determine the output
voltage when the input voltage is: (a) +0
...
2 V
...
4 V,
−2000
(+0
...
8 V
1000

(b) When Vi = −1
...
6

Comparing Fig
...
6 with Fig
...
5, gives Ri = 10 k
and Rf = 1 M
(a) From equation (5), voltage gain,

A=

−2000
(−1
...
4 V
1000

Problem 5
...
19
...
Calculate
(a) the voltage gain, and (b) the output offset
voltage due to the input bias current
...
Hence, the offset voltage, Vos , at the
input due to the 100 nA input bias current, IB , is
given by:

Vos = IB

From equation (5),
Vo =

−

= −10

Thus an input of 100 mV will cause an output change
of 1 V
...
9 × 103 ) = 9
...
99 mV
(c) The effect of input bias current can be minimised
by ensuring that both inputs ‘see’ the same driving resistance
...
9 k (from part (b)) should be placed
between the non-inverting (+) terminal and
earth in Fig
...
6

Problem 6
...


Section 2

A=

× 103

R2 = 1 M Ω

294 Electrical and Electronic Principles and Technology
The voltage gain of an op amp, in decibels, is given by:
gain in decibels = 20 log10 (voltage gain)
from Chapter 10
...
The op amp shown in Fig
...
7 has an input
bias current of 90 nA at 20◦ C
...

[(a) −80 (b) 1
...
2 MΩ

With reference to Fig
...
5, and from equation (5),

R1 = 15 k Ω

Rf
A=
Ri
i
...

Hence

100 =

+

Vi

Vo

Rf
10 × 103

Rf = 100 × 10 × 103 = 1 M

From equation (2), Section 19
...
7

6
...

[(a) 3
...
78 MHz]

= 0
...
A differential amplifier has an open-loop voltage gain of 150 when the input signals are
3
...
40 V
...

[22
...
Calculate the differential voltage gain of an op
amp that has a common-mode gain of 6
...

[6 × 104 ]
3
...
0 V to both terminals
...
Determine the common-mode
gain and the CMRR
...
75 × 10−3 , 92
...
In the inverting amplifier of Fig
...
5 (on
page 292), Ri = 1
...
5 k
...
6 V (b) −0
...
0 V (b) +1
...
4

Op amp non-inverting amplifier

The basic circuit for a non-inverting amplifier is shown
in Fig
...
8 where the input voltage Vi (a
...
or d
...
) is
applied to the non-inverting (+) terminal of the op amp
...
Negative feedback is obtained by feeding back to
the inverting (−) terminal, the fraction of Vo developed
across Ri in the voltage divider formed by Rf and Ri
across Vo
...
19
...
9
Figure 19
...
19
...
19
...
7 × 103
= 1 + 2
...
13

Hence

A=

Vo Ri + Rf
Rf
=
=1+
Vi
Ri
Ri

For example, if Ri = 10 k

Vo = 1 +

(7)

Vi = (3
...
4) = −1
...
5
A=1+

R2
R1

× 103

100
= 1 + 10 = 11
10 × 103

Again, the gain depends only on the values of Ri and Rf
and is independent of the open-loop gain Ao
...
19
...
Thus Rf in
Fig
...
8 is zero and Ri is infinite
...
Also,
it is unaffected if the gain is altered by changing Rf
and/or Ri
...


Problem 7
...
19
...
7 k and R2 = 10 k
...
4 V, determine (a) the voltage gain (b) the
output voltage
...
10

From equation (6), A = 1/β (when Ao is very large)
...

Thus the voltage gain is nearly 1 and Vo = Vi to within
a few millivolts
...
19
...
It has an extremely high input impedance

Section 2

(b) Also from equation (7), output voltage,

296 Electrical and Electronic Principles and Technology
and a low output impedance
...
For example, it is used as the input stage of an analogue voltmeter
where the highest possible input impedance is required
so as not to disturb the circuit under test; the output voltage is measured by a relatively low impedance
moving-coil meter
...
6

Because of the existence of the virtual earth point, an
op amp can be used to add a number of voltages (d
...
or
a
...
) when connected as a multi-input inverting amplifier
...
Such circuits may
be used as ‘mixers’ in audio systems to combine the
outputs of microphones, electric guitars, pick-ups, etc
...

The circuit of an op amp summing amplifier having
three input voltages V1 , V2 and V3 applied via input
resistors R1 , R2 and R3 is shown in Fig
...
11
...

The virtual earth is also called the summing point of
the amplifier
...


50 k Ω

R2

I2

and

Rf
=3
R2

Problem 8
...
19
...


Rf

I

Rf
=4
R1
Rf
=1
R3

and V1 = V2 = V3 = +1 V, then

Op amp summing amplifier

I1

summation is said to have occurred
...

For example, if

R3

X

−

0
...
8 V

10 kΩ

Vo

V3

−

20 kΩ

+

1
...
11
Figure 19
...
e
...
5
0
...
2
+
+
10 × 103
20 × 103
30 × 103

= −(50 × 103 ) (5 × 10−5 + 4 × 10−5 + 4 × 10−5 )
= −(50 × 103 ) (13 × 10−5 )
= −6
...
e
...
It does this in an electronic digital voltmeter
...
19
...
Devise a light-operated alarm circuit
using an op amp, a LDR, a LED and a ±15 V
supply
...
19
...

Resistor R and the light dependent resistor (LDR)
form a voltage divider across the +15/0/−15 V supply
...
e
...
e
...
In the dark the resistance of the LDR is much
greater than that of R, so more of the 30 V across the
voltage divider is dropped across the LDR, causing V1
to fall below 0 V
...


Vo

V2

0V

Figure 19
...
The op amp is then saturated
...
8 Op amp integrator

i
...
when V2 exceeds V1 by 90 μV and Vo ≈ 9 V
...
e
...


The circuit for the op amp integrator shown in Fig
...
15
is the same as for the op amp inverting amplifier shown
in Fig
...
5, but feedback occurs via a capacitor C, rather
than via a resistor
...
14

Section 2

19
...
15

Section 2

1
CR

Vi dt

(9)

Since the inverting (−) input is used in Fig
...
15, Vo
is negative if Vi is positive, and vice versa, hence the
negative sign in equation (9)
...
19
...
e
...
Assuming
again that none of the input current I enters the op amp
inverting (−) input, then all of current I flows through
C and charges it up
...
Capacitor C therefore charges
at a constant rate and the potential of the output side of
C (= Vo , since its input side is zero) charges so that the
feedback path absorbs I
...
d
...
e
...
e
...
e
...
16

The output voltage is given by:
Vo = −

1

Vi
− Vo C = t
R
1
V =−
Vi t
CR

m = 2 and vertical axis intercept c = 0)
...
19
...
5 s when
the op amp saturates
...
A steady voltage of −0
...
5 μF
...

From equation (9), output voltage,
Vo = −
=−
=−

1
CR

Vi dt
1

(2
...
5

× 103 )

(− 0
...
75) dt = −2[−0
...
5t

When time t = 100 ms,
output voltage, Vo = (1
...
15 V
...

For example, if the input voltage Vi = −2 V and, say,
CR = 1 s, then

19
...
19
...


A graph of Vo /t will be ramp function as shown in
Fig
...
16 (Vo = 2t is of the straight line form
y = mx + c; in this case y = Vo and x = t, gradient,

(i) Let V1 volts be applied to terminal 1 and 0 V be
applied to terminal 2
...


+
R2
R3

If V1 > V2 , then:

Vo

Vo = (V1 − V2 ) −
0V

If V2 > V1 , then:
Vo = (V2 − V1 )

Figure 19
...
Then
I1 = V1 /R1
...
The volt drop across Rf , which is the
output voltage
Vo =

R3
R2 + R3

Vo
=
V2

V1
Rf
R1

Rf
R1

(12)

R3
R2 + R3

1+

Rf
R1

(13)

Problem 11
...
19
...
Determine the
output voltage Vo if:
(a) V1 = 5 mV and V2 = 0
(b) V1 = 0 and V2 = 5 mV
(c) V1 = 50 mV and V2 = 25 mV
(d) V1 = 25 mV and V2 = 50 mV
...


R3
R2 + R 3

Rf
100 × 103
V1 = −
R1
10 × 103

(b) From equation (11),

=

R3
R2 + R 3

1+

100
110

100
(5) mV = +50 mV
10

1+

Vo = (V1 − V2 ) −
= (50 − 25) −

V2
V2

−Rf
R1

Vo = (V2 − V1 )
= (50 − 25)

Vo
A=
V2
R3
+ −
R2 + R 3

Rf
−
R1

Rf
R1
100
mV = −250 mV
10

(d) V2 > V1 hence from equation (13),

and the voltage gain,

R3
=
R2 + R 3

V2

(c) V1 > V2 hence from equation (12),

V2 volts
...
e
...
If the input voltage for the op amp shown
in Fig
...
18, is −0
...
21 (b) −1
...
A steady voltage of −1
...
0 μF
...

[0
...
In the differential amplifier shown in
Fig
...
21, determine the output voltage, Vo ,
if: (a) V1 = 4 mV and V2 = 0 (b) V1 = 0 and
V2 = 6 mV (c) V1 = 40 mV and V2 = 30 mV
(d) V1 = 25 mV and V2 = 40 mV
[(a) −60 mV (b) +90 mV
(c) −150 mV (d) +225 mV]
120 kΩ

Output
voltage

6
...
18

1

8 kΩ

2

−
+

8 kΩ

Section 2

120 kΩ

2
...
19
...
19

3
...
19
...
9 V]
60 kΩ

0
...
5V

−
+

0
...
20

0V

Figure 19
...
10 Digital to analogue (D/A)
conversion
There are a number of situations when digital signals
have to be converted to analogue ones
...

A binary weighted resistor D/A converter is shown
in Fig
...
22 for a four-bit input
...
The
circuit uses an op amp as a summing amplifier (see
Section 19
...
Digitally controlled electronic switches are shown as S1 to S4
...
The input
voltages V1 to V4 applied to the op amp by the four-bit

Operational amplifiers 301

Vref

S1

1

V1

R

0

m
...
b
...
s
...


0V

Figure 19
...
e
...

From equation (8), page 296, the analogue output
voltage Vo is given by:
Vo = −

Rf
Rf
Rf
Rf
V1 +
V2 +
V3 +
V4
R
2R
4R
8R

Let Rf = R = 1 k , then:
1
1
1
Vo = − V1 + V2 + V3 + V4
2
4
8
With a four-bit input of 0001 (i
...
decimal 1), S4 connects 8R to VREF , i
...
V4 = VREF , and S1 , S2 and S3
connect R, 2R and 4R to 0 V, making V1 = V2 = V3 =
0
...
e
...
e
...


Again, if VREF = −8 V, then output voltage,
1
1
Vo = − 0 + (−8) + 0 + (−8) = +5 V
2
8
If the input is 0111 (i
...
decimal 7), the output voltage will be 7 V, and so on
...
Vo has a ‘stepped’
waveform, the waveform shape depending on the binary
input
...
19
...


19
...
This is an example
where an analogue to digital converter is needed
...
19
...
An op amp is
again used, in this case as a voltage comparator (see
Section 19
...
The analogue input voltage V2 , shown in

Section 2

0

302 Electrical and Electronic Principles and Technology
V2

Vo 14

Analogue
output
voltage

12
10

V1

(a)

t

1

8
(b) Comparator
output

6

0
t

4
2
(c) Pulse
generator
0000
0001
0010
0011
0100
0111
1000
1011
1100
1110
1010
1001
0111
0100
0010
0000

l
...
b
...
s
...


Section 2

Figure 19
...
19
...
c
...

The output from the comparator is applied to one
input of an AND gate and is a 1 (i
...
‘high’) until V1
equals or exceeds V2 , when it then goes to 0 (i
...
‘low’)
as shown in Fig
...
25(b)
...
19
...
When both inputs to
the AND gate are ‘high’, the gate ‘opens’ and gives
a ‘high’ output, i
...
a pulse, as shown in Fig
...
25(d)
...
The output pulses
from the AND gate are recorded by a binary counter

(e) Binary
output

0001
0010
0011
0100
0101
0110
0111
1000

Binary
input
Decimal

0
1
2
3
4
7
8
11
12
14
10
9
7
4
2
0

0

t

Figure 19
...
19
...
In practise, the ramp
generator is a D/A converter which takes its digital input
from the binary counter, shown by the broken lines in
Fig
...
24 As the counter advances through its normal
binary sequence, a staircase waveform with equal steps
(i
...
a ramp) is built up at the output of the D/A converter
(as shown by the first few steps in Fig
...
23
...
24

Binary
counter

Reset

m
...
b
l
...
b

4-bit
digital
output

Operational amplifiers 303
Now try the following exercises
19
...
List three main properties of an op amp

3
...
Sketch a typical gain/bandwidth characteristic for an op amp
5
...
Define common-mode rejection ratio
7
...
In an inverting amplifier, the closed-loop gain
A is given by: A =
...
Explain the principle of operation of an
op amp non-inverting amplifier
10
...

11
...
Explain the principle of operation of an
op amp summing amplifier
13
...

14
...
Explain the principle of operation of an
op amp integrator
16
...

17
...
Explain the principle of operation of a binary
weighted resistor digital to analogue converter using a four-bit input

Exercise 110

Multi-choice questions on
operational amplifiers
(Answers on page 399)

1
...
The input signals are
2
...
4 V
...
Which of the following statements relating to
operational amplifiers is true?
(a) It has a high open-loop voltage gain
at low frequency, a low input impedance
and low output impedance
(b) It has a high open-loop voltage gain at
low frequency, a high input impedance
and low output impedance
(c) It has a low open-loop voltage gain at
low frequency, a high input impedance
and low output impedance
(d) It has a high open-loop voltage gain
at low frequency, a low input impedance
and high output impedance
3
...
The common-mode gain of
the operational amplifier is:
(a) 1
...
2
(c) 1
...
2 × 10−5
4
...
19
...
2 V
(b) +1
...
2 V
(d) −1
...
The 3 k resistor in Fig
...
26 is replaced by
one of value 0
...
If the op amp has an
input bias current of 80 nA, the output offset
voltage is:
(a) 79
...
2 nV

Section 2

2
...
6 V

V

o

Figure 19
...
In the op amp shown in Fig
...
27, the voltage
gain is:
(a) −3
(b) +4
(c) +3
(d) −4

8
...
0 V is applied to an
op amp integrator having component values
of R = 100 k and C = 10 μF
...
In the differential amplifier shown in
Fig
...
29, the output voltage, Vo , is:
(a) +1
...
92 mV
(c) −1
...
1 V

+

0
...
29
Output
voltage

5 kΩ

Figure 19
...
For the op amp shown in Fig
...
28, the
output voltage, Vo , is:
(a) −1
...
8 V

−
Vo

0
...
28

+

10
...
The marks for each question are shown in
brackets at the end of each question
...
A coil of resistance 20 and inductance 200 mH is
connected in parallel with a 4 μF capacitor across a
50 V, variable frequency supply
...

(10)

6
...
1, determine the value of the output
voltage, V0
(3)
7
...
2,
determine the output voltage, V0 when:
(a) V1 = 4 mV and V2 = 0
(b) V1 = 0 and V2 = 5 mV
(c) V1 = 20 mV and V2 = 10 mV
(6)
120 kΩ

V1

3
...

Determine the Q-factor of the circuit at resonance
...
The winding of an electromagnet has an inductance
of 110 mH and a resistance of 5
...
When it is
connected to a 110 V, d
...
supply, calculate (a) the
steady state value of current flowing in the winding,
(b) the time constant of the circuit, (c) the value of
the induced e
...
f
...
01 s, (d) the time for the
current to rise to 75 per cent of it’s final value, and
(e) the value of the current after 0
...

(11)
5
...
65 lagging from a 300 V, 50 Hz supply
...

(7)

V2

1

20 k Ω

2

−
+

20 k Ω
120 k Ω

OV

Figure RT5
...
A filter section is to have a characteristic impedance
at zero frequency of 600 and a cut-off frequency
of 2
...
Design (a) a low-pass T-section filter,
and (b) a low-pass π-section filter to meet these
requirements
...
Determine the cut-off frequency and the nominal
impedance for a high-pass π-connected section having a 5 nF capacitor in its series arm and inductances
of 1 mH in each of its shunt arms
...
5 V
1
...
1

Vo

Vo

Section 2

1
...
Calculate (a) the resistance,
(b) the impedance, (c) the reactance, (d) the power
factor, and (e) the phase angle between voltage and
current
...
C
...
637 Im
π

For a sine wave: IAV =

1
2π

Ir =

2
2
2
2
i1 + i 2 + i 2 + · · · + in
n

2πfr L
IC
=
R
Ir

L
CR

RD =

P = VI cos φ or I 2 R

1
I = √ Im or 0
...
F2

L
2

L

L
2

R0
C

Figure F1

L=

High-pass T or π:

1
√

VC
2πfr L
1
1
=
=
=
V
R
2πfr CR
R
or

L
C

See Fig
...
C
...
1

Introduction

Generation, transmission and distribution of electricity
via the National Grid system is accomplished by threephase alternating currents
...
20
...
Most consumers
ω

are fed by means of a single-phase a
...
supply
...
The neutral is usually
connected via protective gear to earth, the earth wire
being coloured green
...
c
...
The majority of single-phase
supplies are obtained by connection to a three-phase
supply (see Fig
...
5, page 313)
...
2 Three-phase supply

R
N

S
R1

Induced
EMF
e
0

Figure 20
...
20
...
The result is
three independent supplies of equal voltages which are
each displaced by 120◦ from each other as shown in
Fig
...
2(b)
...
20
...
2

(ii) The phase-sequence is given by the sequence in
which the conductors pass the point initially taken
by the red conductor
...

A three-phase a
...
supply is carried by three conductors, called ‘lines’ which are coloured red, yellow and
blue
...
d
...
A fourth conductor, called the neutral (coloured black, and connected through protective
devices to earth) is often used with a three-phase supply
...
20
...
To reduce the number of wires it is usual to
interconnect the three phases
...
Sources of three-phase supplies, i
...

alternators, are usually connected in star, whereas
three-phase transformer windings, motors and
other loads may be connected either in star or
delta
...
3 Star connection

Figure 20
...

(iv) From Fig
...
3 it can be seen that the phase currents (generally denoted by Ip ) are equal to their
respective line currents IR , IY and IB , i
...
for a star
connection:
I L = Ip
(v) For a balanced system:
IR = IY = IB ,

VRY = VYB = VBR ,

ZR = ZY = ZB

and the current in the neutral conductor, IN = 0
When a star-connected system is balanced, then
the neutral conductor is unnecessary and is often
omitted
...
20
...
In the phasor
diagram of Fig
...
4(b), phasor VY is reversed
(shown by the broken line) and then added
phasorially to VR (i
...
VRY = VR + (−VY ))
...
e
...
20
...

(ii) The voltages, VR , VY and VB are called phase
voltages or line to neutral voltages
...


VR = VY = VB

VY

VRY

VR
30°

−VY
120°

(a)
120°

VB
(b)

Figure 20
...
5

(vii) The star connection of the three phases of a supply,
together with a neutral conductor, allows the use
of two voltages — the phase voltage and the line
voltage
...
The standard electricity supply to consumers in Great Britain is 415/240 V,
50 Hz, 3-phase, 4-wire alternating current, and a
diagram of connections is shown in Fig
...
5
For most of the 20th century, the supply voltage
in the UK in domestic premises has been 240 V a
...

(r
...
s
...
In 1988, a European-wide agreement
was reached to change the various national voltages,
which ranged at the time from 220 V to 240 V, to a
common European standard of 230 V
...
However, as an interim measure, electricity suppliers can work with an asymmetric
voltage tolerance of 230 V +10%/−6% (i
...
216
...
The old standard was 240 V ± 6% (i
...
225
...
4 V), which is mostly contained within the new
range, and so in practice suppliers have had no reason
to actually change voltages
...
e
...

European harmonisation required this to be changed to
400 V +10%/−6% (i
...
376 V to 440 V)
...

Many of the calculations following are based on the
240 V/415 V supply voltages which have applied for
many years and are likely to continue to do so
...
Three loads, each of resistance 30 ,
are connected in star to a 415 V, 3-phase supply
...

A ‘415 V, 3-phase supply’ means that 415 V is the line
voltage, VL
√
(a) For a star connection, VL = 3 Vp
...
6 V
or 240 V, correct to 3 significant figures
...
A star-connected load consists of
three identical coils each of resistance 30 and
inductance 127
...
If the line current is 5
...

Inductive reactance
XL = 2πfL = 2π(50)(127
...


314 Electrical and Electronic Principles and Technology
lagging
...


Impedance of each phase
Zp =

2
R 2 + XL =

302 + 402 = 50

For a star connection
IL = Ip =

Vp
Zp

Hence phase voltage,
Vp = Ip Zp = (5
...
Hence VY is
reversed and added phasorially to VR
...
e
...
Similarly, VYB = VY − VB and
VBR = VB − VR
Problem 4
...
20
...
Determine (a) the current in
each line and (b) the current in the neutral
conductor
...
A balanced, three-wire,
star-connected, 3-phase load has a phase voltage of
240 V, a line current of 5 A and a lagging power
factor of 0
...
Draw the complete phasor diagram
...
20
...


Figure 20
...
6

Section 3

Procedure to construct the phasor diagram:
(i) Draw VR = VY = VB = 240 V and spaced 120◦
apart
...

(ii) Power factor = cos φ = 0
...
Hence the
load phase angle is given by cos−1 0
...
e
...
20
...
Since each load is resistive
the currents are in phase with the phase voltages and are hence mutually displaced by 120◦
...

Figure 20
...
oa represents IR in magnitude and direction
...
Three identical capacitors are connected in star
to a 400 V, 50 Hz 3-phase supply
...

[165
...
8

4
...
If the line current is
30 A, find the value of L
...
78 mH]
5
...
Determine the current
flowing in each of the four conductors
...
95 A, IY = 86
...
25 A, IN = 37
...
9

magnitude and direction
...
oc
represents the resultant, IN By measurement, IN = 43 A
...
65
...
50
...
652 + 37
...
3 A
...
4

Delta connection

(i) A delta (or mesh) connected load is shown in
Fig
...
10 where the end of one load is connected
to the start of the next load
...
10

1
...

Determine (a) the phase voltage, (b) the phase
current and (c) the line current
...
62 A (c) 4
...
A star-connected load consists of three identical coils, each of inductance 159
...
If the supply frequency is
50 Hz and the line current is 3 A determine
(a) the phase voltage and (b) the line voltage
...
20
...
e
...
20
...
20
...
e
...
Three coils each having resistance
3 and inductive reactance 4 are connected (i) in
star and (ii) in delta to a 415 V, 3-phase supply
...

(i) For a star connection: IL = Ip and VL =

√
3 Vp
...
11

Problem 5
...
3 mH are connected in
delta to a 440 V, 50 Hz, 3-phase supply
...


Phase voltage,
VL
415
Vp = √ = √ = 240 V
3
3
(b) Impedance per phase,

Phase impedance, Zp = 50 (from Problem 2) and for
a delta connection, Vp = VL
(a) Phase current,

Thus when the load is connected in delta, three times
the line current is taken from the supply than is taken if
connected in star
...
Three identical capacitors are
connected in delta to a 415 V, 50 Hz, 3-phase
supply
...

√
For a delta connection IL = 3 Ip
...
66 A
3
3
Capacitive reactance per phase,
Vp
VL
=
Ip
Ip

(since for a delta connection VL = Vp )
...
92
8
...
8 A
Ip =
Zp
Zp
50
(b) For a delta connection,
√
√
IL = 3 Ip = 3(8
...
24 A

XC =

Zp =

1
2
F = 66
...
92)

Ip = Vp /Zp = 240/5 = 48 A
Line current,
IL = Ip = 48 A
(ii) For a delta connection: VL = Vp and IL =

√
3 Ip
...
Three loads, each of resistance 50 are connected in delta to a 400 V, 3-phase supply
...

[(a) 400 V (b) 8 A (c) 13
...
Three inductive loads each of resistance 75
and inductance 318
...
Determine (a) the phase voltage, (b) the phase
current, and (c) the line current
[(a) 415 V (b) 3
...
75 A]
3
...
If the
line current is 12 A determine the capacitance
of each of the capacitors
...
13 μF]
4
...
If the line
current is 30 A, find the value of L
[73
...
A 3-phase, star-connected alternator delivers
a line current of 65 A to a balanced deltaconnected load at a line voltage of 380 V
...

[(a) 219
...
53 A]
6
...
What
value of capacitance must be connected in
delta in order to take the same line current?
[8 μF]

Vp = VL

and

IL
Ip = √
3

hence
√
IL
P = 3VL √ cos φ = 3 VL IL cos φ
3
Hence for either a star or a delta balanced connection
the total power P is given by:
√
P = 3 VL IL cos φ watts
2
or P = 3Ip Rp watts

Total volt-amperes
√
S = 3 VL IL volt-amperes
Problem 8
...
Determine the total
power dissipated by the resistors
...
Phase current,
Ip =

Vp
Vp
240
= 20 A
=
=
Zp
12
Rp

20
...
4 kW
or power

hence
√
VL
P = 3 √ IL cos φ = 3 VL IL cos φ
3

2
P = 3Ip Rp = 3(20)2 (12) = 14
...
If a
load is balanced then the total power P is given by:
P = 3 × power consumed by one phase
...

For a star connection,

318 Electrical and Electronic Principles and Technology
Problem 9
...
c
...
If the voltage and
current to the motor are 400 V and 8
...


Power dissipated,
√
√
P = 3 VL IL cos φ = 3(415)(14
...
6042)
= 6
...
50)2 (10) = 6
...
6 A and
√
power, P = 3 VL IL cos φ
Hence

VL = Vp = 415 V,
Zp = 16
...
6042
lagging (from above)
...
55 = 25
...


P
3 VL IL

5000
3(400)(8
...
839

Problem 10
...
Determine the total power
dissipated in each case
...
08) = 43
...

Power dissipated,
√
P = 3 VL IL cos φ
√
= 3(415)(43
...
6042) = 18
...
08)2 (10) = 18
...


XL = 2πf L = 2π(50)(42 × 10−3 ) = 13
...

Phase impedance,
2
Zp = R2 + XL = 102 + 13
...
55
...
A 415 V, 3-phase a
...
motor has a
power output of 12
...
77 lagging and with an efficiency of 85
per cent
...


VL = 415 V
and phase voltage,
√
√
VP = VL / 3 = 415/ 3 = 240 V
...
Hence
85/100 = 12 750/power input from which,
12 750 × 100
85
= 15 000 W or 15 kW

power input =

Phase current,

Section 3

Ip = Vp /Zp = 240/16
...
50 A
...
50 A
...
55
= 0
...


√
3 VL IL cos φ, hence line current,

IL = √
=√

P
3(415)(0
...
77)

= 27
...
10
Ip = √ = √ = 15
...
6

Measurement of power in
three-phase systems

Power in three-phase loads may be measured by the
following methods:
(i) One-wattmeter method for a balanced load
Wattmeter connections for both star and delta are
shown in Fig
...
12

1
...

[(a) 9
...
04 kW]
2
...

[1
...
A balanced delta-connected load has a line
voltage of 400 V, a line current of 8 A and a
lagging power factor of 0
...
Draw a complete
phasor diagram of the load
...
21 kW]
4
...

When connected to a 3-phase supply the loads
consume 1
...
Calculate (a) the power factor of the load, (b) the phase current, (c) the
line current and (d) the supply voltage
...
406 (b) 10 A (c) 17
...
53 V]

Figure 20
...
20
...
Similar
connections are made for a delta-connected load
...
The input voltage, current and power to a
motor is measured as 415 V, 16
...
Determine the power factor of the
system
...
509]

Figure 20
...


Section 3

6
...
c
...
25 kW and operates at a power factor of
0
...
If the motor is delta connected determine
(a) the power input, (b) the line current and
(c) the phase current
...
39 kW (b) 21
...
68 A]

320 Electrical and Electronic Principles and Technology
It is possible, depending on the load power factor,
for one wattmeter to have to be ‘reversed’to obtain
a reading
...

(iii) Three-wattmeter method for a three-phase, 4wire system for balanced and unbalanced loads
(see Fig
...
14)
...
15

wattmeter 2 Hence total instantaneous power,
p = (wattmeter 1 reading)
+ (wattmeter 2 reading)
= p1 + p2

Figure 20
...
(a) Show that the total power in a
3-phase, 3-wire system using the two-wattmeter
method of measurement is given by the sum of the
wattmeter readings
...

(b) Draw a phasor diagram for the two-wattmeter
method for a balanced load
...


The moving systems of the wattmeters are unable
to follow the variations which take place at
normal frequencies and they indicate the mean
power taken over a cycle
...

(b) The phasor diagram for the two-wattmeter method
for a balanced load having a lagging current is
shown in Fig
...
16, where VRB = VR − VB and
VYB = VY − VB (phasorially)
...
20
...


Section 3

Total instantaneous power, p = eR iR + eY iY + eB iB
and in any 3-phase system iR + iY + iB = 0; hence
iB = −iR − iY Thus,
p = eR iR + eY iY + eB (−iR − iY )
= (eR − eB )iR + (eY − eB )iY
However, (eR − eB ) is the p
...
across wattmeter 1
in Fig
...
15 and (eY − eB ) is the p
...
across

Figure 20
...

Hence
P1
cos 30◦ cos φ + sin 30◦ sin φ
=
P2
cos 30◦ cos φ − sin 30◦ sin φ
(from compound angle formulae, see ‘Engineering Mathematics’)
...
17

tan φ

Hence Ip = Vp /Zp = 400/50 = 8 A
...
86 A
...
86 A is the current supplied by the
alternator
...
e
...
6
...
86)(0
...
76 kW
...

Problem 13
...
Calculate (a) the current
supplied by the alternator and (b) the output power
and the kVA of the alternator, neglecting losses in
the line between the alternator and load
...
20
...
86)
= 9
...

Problem 14
...
The load is connected to a
400 V, 50 Hz, 3-phase supply
...

Draw the complete phasor diagram for the load
...
79
2πfC
2π(50)(80 × 10−6 )

Phase impedance,

hence Vp = 400 V
...
792 = 49
...


Power factor = cos φ = Rp /Zp
= 302 + 402 = 50


...
83 = 0
...
602 = 52
...


√ 8−4
√ P1 − P2
= 3
3
P1 + P 2
8+4
√
√ 1
4
1
= 3
= 3
=√
12
3
3

(b) tan φ =

Phase current,
Ip = Vp /Zp

and

Vp = VL

for a delta connection
...
83 = 8
...

√
Hence IL = 3(8
...
90 A
(c) Total power dissipated,
√
P = 3 VL IL cos φ
√
= 3(400)(13
...
602) = 5
...
90) = 9
...
866
Problem 16
...
The power factor is 0
...
Determine the
readings of each wattmeter
...
20
...
6 = cos φ
...
6 =
53
...
13◦ = 1
...
Hence
1
...
3333)
√
3

i
...

P1 − P2 = 9
...
237
i
...


21
...
62 kW

P1 =

Section 3

Figure 20
...
Two wattmeters are connected to
measure the input power to a balanced 3-phase load
by the two-wattmeter method
...

(a) Total input power,
P = P1 + P2 = 8 + 4 = 12 kW

Hence wattmeter 1 reads 10
...
62) = 1
...
Two wattmeters indicate 10 kW and
3 kW respectively when connected to measure the
input power to a 3-phase balanced load, the reverse
switch being operated on the meter indicating the
3 kW reading
...


Three-phase systems 323
Since the reversing switch on the wattmeter had to be
operated the 3 kW reading is taken as −3 kW

P 1 − P2 =

(a) Total input power,
P = P1 + P2 = 10 + (−3) = 7 kW

tan−1 3
...
73◦

Power factor = cos φ = cos 72
...
297
Problem 18
...
Calculate for each
connection the readings on each of two wattmeters
connected to measure the power by the
two-wattmeter method
...

When the coils are star-connected the wattmeter
readings are thus 8
...
275 kW
√
(b) Delta connection: VL = Vp and IL = 3 Ip
Phase current, Ip =

Vp
415
=
= 36
...

ZP 11
...
69)2 (8) = 32 310 W

Hence P1 + P2 = 32 310 W
tan φ =

√ P1 − P2
3
P1 + P2

√
3(P1 − P2 )
thus 1 =
32 310

(3)

from which,

VL
415
Phase voltage, Vp = √ = √
3
3
and phase impedance,
2
2
Rp + XL

(10 766)(1)
= 6216 W
√
3

Adding Equations (1) and (2) gives:

√ 10 − (−3)
= 3
10 + (−3)
√ 13
= 3
= 3
...


= 82 + 82 = 11
...
18 A
=
Zp
11
...
48 kW and
6
...

√ P1 − P2
3
P1 + P 2
√
3(P1 − P2 )
hence tan 45◦ =
10 766
tan φ =

Exercise 114 Further problems on the
measurement of power in
3-phase circuits
1
...
If
the wattmeter readings are 9
...
4 kW
determine (a) the total output power, and
(b) the load power factor
[(a) 14
...
909]

Section 3

2
P = 3Ip Rp = 3(21
...
8 kW is found by the two-wattmeter method
to be the power input to a 3-phase motor
...
85
[5
...
569 kW]
3
...
5 kW and
2
...
5 kW having to be
reversed
...
277]
4
...
0 and an inductive reactance of 3
...
Calculate for
each connection the readings on each of two
wattmeters connected to measure the power by
the two-wattmeter method
...
15 kW, 5
...
46 kW, 17
...
A 3-phase, star-connected alternator supplies a
delta connected load, each phase of which has
a resistance of 15 and inductive reactance
20
...

[(a) 27
...
52 kW, 19
...
Each phase of a delta-connected load comprises a resistance of 40 and a 40 μF capacitor in series
...
66 A (b) 8
...
605 kW (d) 5
...
7 Comparison of star and delta
connections
(i) Loads connected in delta dissipate three times
more power than when connected in star to the
same supply
...
To
achieve the same phase current in a star-connected
system as in a delta-connected system, the line
√
voltage in the star system is 3 times the line voltage in the delta system
...


20
...
This
means a saving of copper (or aluminium) and thus
the original installation costs are less
...
3 (vii))
(iii) Three-phase motors are very robust, relatively
cheap, generally smaller, have self-starting properties, provide a steadier output and require little
maintenance compared with single-phase motors
...
Explain briefly how a three-phase supply is
generated
2
...
State the two ways in which phases of a threephase supply can be interconnected to reduce
the number of conductors used compared
with three single-phase systems
4
...
When may the neutral conductor of a starconnected system be omitted?
6
...
What is the standard electricity supply to
domestic consumers in Great Britain?
8
...
By what methods may power be measured in
a three-phase system?
10
...
Loads connected in star dissipate
...
Name three advantages of three-phase systems over single-phase systems

a smaller line current than loads connected in star
(b) When using the two-wattmeter method
of power measurement the power factor
is unity when the wattmeter readings are
the same
(c) A
...
may be distributed using a singlephase system with two wires, a threephase system with three wires or a
three-phase system with four wires
(d) The national standard phase sequence
for a three-phase supply is R, Y, B
Three loads, each of resistance 16 and inductive
reactance 12 are connected in delta to a 400 V,
3-phase supply
...
4) kW
√
(d) 20 A
(e) 6
...
2 kW (k) 100 A
(l) 400 V
(m) 28
7
...
Line voltage

Exercise 116

Multi-choice questions on
three-phase systems
(Answers on page 399)

Three loads, each of 10 resistance, are connected in star to a 400 V, 3-phase supply
...
Line voltage
2
...
Phase current

9
...
Phase current
11
...
Total power dissipated in the load
13
...

The line voltage is:
(a) 720 V
(b) 440 V
(c) 340 V
(d) 240 V
14
...
The phase
current is:
(a) 40 A
(b) 10 A
(c) 20 A
(d) 30 A

5
...
Which of the following statements is false?
(a) For the same power, loads connected
in delta have a higher line voltage and

15
...
The phase
voltage is:
(a) 19
...
35 kV
(d) 7
...
Line current

326 Electrical and Electronic Principles and Technology
16
...
The
power factor may be determined from:
√ P1 + P2
√ P1 − P2
(a) 3
(b) 3
P1 − P2
P1 + P2

Section 3

(P1 − P2 )
(c) √
3(P1 + P2 )

(P1 + P2 )
(d) √
3(P1 − P2 )

17
...
The line voltage is:
(a) 440 V
(b) 330 V
(c) 191 V
(d) 110 V

Chapter 21

Transformers
At the end of this chapter you should be able to:
• understand the principle of operation of a transformer
• understand the term ‘rating’ of a transformer
• use V1 /V2 = N1 /N2 = I2 /I1 in calculations on transformers
• construct a transformer no-load phasor diagram and calculate magnetising and core loss components of the
no-load current
• state the e
...
f
...
44 f m N and use it in calculations
• construct a transformer on-load phasor diagram for an inductive circuit assuming the volt drop in the windings
is negligible
• describe transformer construction
• derive the equivalent resistance, reactance and impedance referred to the primary of a transformer
• understand voltage regulation
• describe losses in transformers and calculate efficiency
• appreciate the concept of resistance matching and how it may be achieved
• perform calculations using R1 = (N1 /N2 )2 RL
• describe an auto transformer, its advantages/disadvantages and uses
• describe an isolating transformer, stating uses
• describe a three-phase transformer
• describe current and voltage transformers

21
...
In fact, one of the
main advantages of a
...
transmission and distribution
is the ease with which an alternating voltage can be
increased or decreased by transformers
...
Being static they have a long life and
are very stable
...

A transformer is represented in Fig
...
1(a) as consisting of two electrical circuits linked by a common
ferromagnetic core
...
A circuit diagram symbol for a
transformer is shown in Fig
...
1(b)
...
c
...
1

Combining equations (1) and (2) gives:

21
...
This alternating
flux links with both primary and secondary coils and
induces in them e
...
f
...

The induced e
...
f
...
In an ideal transformer, the rate of change of flux
is the same for both primary and secondary and thus
E1 /N1 = E2 /N2 i
...
the induced e
...
f
...

Assuming no losses, E1 = V1 and E2 = V2

Section 3

Hence

V1
V2
=
N1
N2

or

V1
N1
=
V2
N2

(1)

(V1 /V2 ) is called the voltage ratio and (N1 /N2 ) the turns
ratio, or the ‘transformation ratio’ of the transformer
...
If N2 is
greater then N1 then V2 is greater than V1 and the device
is termed a step-up transformer
...
In an ideal transformer losses are
neglected and a transformer is considered to be 100 per
cent efficient
...
e
...
With
reference to Fig
...
1(a), the transformer rating is either
V1 I1 or V2 I2 , where I2 is the full-load secondary current
...
A transformer has 500 primary turns
and 3000 secondary turns
...

For an ideal transformer, voltage ratio = turns ratio i
...

N1
240
V1
500
=
hence
=
V2
N2
V2
3000
Thus secondary voltage
V2 =

(240)(3000)
= 1440 V or 1
...
An ideal transformer with a turns
ratio of 2:7 is fed from a 240 V supply
...

A turns ratio of 2:7 means that the transformer has
2 turns on the primary for every 7 turns on the secondary
(i
...
a step-up transformer); thus (N1 /N2 ) = (2/7)
...
Thus the secondary voltage
V2 =

(2)

(3)

(240)(7)
= 840 V
2

Transformers 329
Problem 3
...
Calculate the secondary voltage
and current
...
e
...

=

V1
V2
N1
N2

V2 = V1
Also,

N1
N2

=

or secondary voltage

I2
I1

I2 = I1

1
= 240
8

N1
N2

8
=3
1

Turns ratio =

I2
, from which,
I1
I1 = I2

V2
V1

12
...
625 A
20

Problem 5
...
Determine the
primary voltage if the supply current is 4 A
...

(V1 /V2 ) = (I2 /I1 ), from which the primary voltage
V1 = V2

I2
I1

= 120

N2
N1

= 2500

1
10

= 250 V

(b) Minimum value of load resistance,
RL =
(c)

N1
N2

=

I2
I1

I1 = I2

V2
V1

=

250
20

= 12
...


from which primary current
N1
N2

= 20

1
10

= 2A

Now try the following exercise

12
= 12
...
e
...


= 24 A

V1
240
N1
=
= 20
=
N2
V2
12

N1
N2

V2 = V1

hence secondary current

V1 = 240 V, V2 = 12 V, I2 = (P/V2 ) = (150/12) = 12
...


=

Since

= 30 volts

Problem 4
...

Calculate the transformer turns ratio and the current
taken from the supply
...
5 kV = 2500 V
...
A transformer has 600 primary turns connected to a 1
...
Determine the
number of secondary turns for a 240 V output
voltage, assuming no losses
...
An ideal transformer with a turns ratio of 2:9
is fed from a 220 V supply
...

[990 V]
3
...
If the primary voltage is 160 V, determine the secondary voltage
assuming an ideal transformer
...
A 5 kVA single-phase transformer has
a turns ratio of 10:1 and is fed from a 2
...

Neglecting losses, determine (a) the full-load
secondary current, (b) the minimum load resistance
which can be connected across the secondary
winding to give full load kVA, (c) the primary
current at full load kVA
...
An ideal transformer with a turns ratio of 3:8
has an output voltage of 640 V
...

[240 V]
5
...
Calculate the
secondary voltage
...
A transformer primary winding connected
across a 415 V supply has 750 turns
...
66 kV is
required
...
An ideal transformer has a turns ratio of
15:1 and is supplied at 180 V when the primary current is 4 A
...

[12 V, 60 A]
8
...
Neglecting losses, calculate
the value of the secondary current
...
A transformer has a primary to secondary
turns ratio of 1:15
...
If
the load current is 3 A determine the primary
current
...

[16 V, 45 A]
10
...
4 kV supply
...

[(a) 50 A (b) 4 (c) 4
...
A 20 resistance is connected across the
secondary winding of a single-phase power
transformer whose secondary voltage is
150 V
...

[225 V, 3:2]

21
...
On no-load

the primary winding takes a small no-load current I0
and since, with losses neglected, the primary winding
is a pure inductor, this current lags the applied voltage
V1 by 90◦
...
21
...
The primary induced
e
...
f
...
The secondary induced e
...
f
...

A no-load phasor diagram for a practical transformer
is shown in Fig
...
2(b)
...
When losses are considered then the no-load
current I0 is the phasor sum of two components — (i) IM ,
the magnetising component, in phase with the flux,
and (ii) IC , the core loss component (supplying the
hysteresis and eddy current losses)
...
21
...

Power factor on no-load = cos φ0 = (IC /I0 )
...
e
...
A 2400 V/400 V single-phase
transformer takes a no-load current of 0
...
Determine the values of the
magnetising and core loss components of the
no-load current
...

V1 = 2400 V, V2 = 400 V and I0 = 0
...
e
...

i
...

Hence

400 = (2400)(0
...
3333
(2400)(0
...
3333 = 70
...
21
...
5 sin 70
...
471 A
...
5 cos 70
...
167 A
Problem 8
...
8 A
when its primary is connected to a 240 volt, 50 Hz
supply, the secondary being on open circuit
...


Transformers 331

Figure 21
...
A 500 V/100 V, single-phase transformer takes
a full load primary current of 4 A
...

[(a) 20 A (b) 2 kVA]
2
...
8 A and the iron
loss is 500 W
...

[0
...
152 A]
Figure 21
...
8 A and V = 240 V
(a) Power absorbed = total core loss = 72 = V1 I0 cos φ0
...
30 A

3
...
If
the power absorbed is 120 watts, calculate
(a) the iron loss current, (b) the power factor
on no-load, and (c) the magnetising current
...
40 A (b) 0
...
917 A]

(b) Power factor at no load,
IC
0
...
375
I0
0
...
21
...
82 − 0
...
74 A

21
...
m
...
equation of a transformer

The magnetic flux set up in the core of a transformer
when an alternating voltage is applied to its primary
winding is also alternating and is sinusoidal
...
The time for 1 cycle of the

Section 3

cos φ0 =

332 Electrical and Electronic Principles and Technology
alternating flux is the periodic time T , where T = (1/f )
seconds
The flux rises sinusoidally from zero to its maximum value in (1/4) cycle, and the time for (1/4)
cycle is (1/4f ) seconds
...
m
...
induced in each
turn = 4 f m volts
...
m
...
will be induced in each turn of
both primary and secondary windings
...
m
...
value = form factor × average value =
1
...
m
...
e
...
f
...
44 f

m

volts

m N1

volts

m N2

(4)

volts

(5)

Dividing equation (4) by equation (5) gives:
E1
E2

=

E
4
...
44)(50)(100)

= 9
...
01 mWb
[Alternatively, equation (4) could have been used,
where

m

and r
...
s
...
m
...
induced in secondary,
E2 = 4
...
2
Problem 9
...

Determine (a) the primary and secondary current,
(b) the number of primary turns, and (c) the
maximum value of the flux
...
44)(50)(2000)

Problem 10
...
The
cross-sectional area of the core is 300 cm2
...


(a) From equation (4),
e
...
f
...
44 f
i
...
250 = 4
...
01 mWb as above]

V1 = 4000 V, V2 = 200 V, f = 50 Hz, N2 = 100 turns

I1 =

from which,

=

E1 = 4
...
m
...
value of e
...
f
...
44 f

m N2

=

m

r
...
s
...
11 (see Chapter 14)

m

4000
(100) = 2000 turns
200

(N2 ) =

(c) From equation (5), E2 = 4
...
11 × 4 f

V1
V2

N1 =

V 1 N1
=
from which, primary
V 2 N2

=

m N1 volts
m (25) from which, maximum

250
Wb = 0
...
44)(50)(25)

However, m = Bm × A, where Bm = maximum
flux density in the core and A = cross-sectional
area of the core (see Chapter 7)
...
04505 from which,
0
...
50 T

maximum flux density, Bm =

Transformers 333
V1 N1
N2
=
from which, V2 = V1
V2 N2
N1
induced in the secondary winding,
V2 = (250)

300
25

i
...
voltage

= 3000 V or 3 kV

Problem 11
...
5 T and an effective core cross-sectional area of
50 cm2
...

The e
...
f
...
44 f m N
and maximum flux, m = B × A = (1
...
44 f m N1 then primary turns,
N1 =

E1
4
...
44)(50)(75 × 10−4 )
= 300 turns

=

Since E2 = 4
...
44 f

=

N2 =

m

then secondary turns,

100
(4
...
A 4500 V/225 V, 50 Hz single-phase
transformer is to have an approximate e
...
f
...
4 T
...

E1 E2
=
= 15
N1 N2
E1 4500
Hence primary turns, N1 =
=
= 300
15
15
E2 255
=
= 15
and secondary turns, N2 =
15
15
(b) E
...
f
...
44 f m N1 from which,
(a) E
...
f
...
0676 Wb
4
...
44)(50)(300)

Now flux, m = Bm × A, where A is the crosssectional area of the core,
hence area,

A=

m

Bm

=

0
...
4

= 0
...
m
...
equation
1
...
Calculate (a) the primary and secondary current,
(b) the number of primary turns, and (c) the
maximum value of the flux
...
5 A, 600 A (b) 800 (c) 9
...
A single-phase, 50 Hz transformer has 40 primary turns and 520 secondary turns
...

When the primary winding is connected to a
300 volt supply, determine (a) the maximum
value of flux density in the core, and (b) the
voltage induced in the secondary winding
...
25 T (b) 3
...
A single-phase 800 V/100 V, 50 Hz transformer has a maximum core flux density of
1
...
Calculate the number of turns on
the primary and secondary windings
...
A 3
...
m
...
per
turn of 22 V and operate with a maximum
flux of 1
...
Calculate (a) the number of
primary and secondary turns, and (b) the
cross-sectional area of the core
...
8 cm2 ]

21
...
m
...
E2 in the secondary
...
Assuming an equal number of turns
on primary and secondary windings, then E1 = E2 , and
let the load have a lagging phase angle φ2
...
21
...
When a load is connected across the
secondary winding a current I2 flows in the secondary

Section 3

(b)

334 Electrical and Electronic Principles and Technology

Figure 21
...
The resulting secondary e
...
f
...
However this does not happen since reduction of the core flux reduces E1 , hence
a reflected increase in primary current I1 occurs which
provides a restoring m
...
f
...
m
...
’s are equal, but in opposition, and
the core flux remains constant
...
21
...
I0 , shown
at a phase angle φ0 to V1 , is the no-load current of the
transformer (see Section 21
...


Figure 21
...
20, then cos φ0 = 0
...
2 = 78
...

In the phasor diagram shown in Fig
...
5, I2 = 100 A
is shown at an angle of φ = 31
...

The no-load current I0 = 5 A is shown at an angle of
φ0 = 78
...
Current I1 is the phasor sum of I1
and I0 , and by drawing to scale, I1 = 44 A and angle
φ1 = 37◦
...
A single-phase transformer has
2000 turns on the primary and 800 turns on the
secondary
...
20 lagging
...
85 lagging
...
m
...
Then
I1 N1 = I2 N2
i
...


Section 3

from which,

I1 (2000) = (100)(800)
(100)(800)
2000
= 40 A

I1 =

If the power factor of the secondary is 0
...
85, from which, φ2 = cos−1 0
...
8◦
...
2) + (40)(0
...
0 A
and

I1 sin φ1 = 0c + 0d
= I0 sin φ0 + I1 sin φ2
= (5) sin 78
...
8◦
= 25
...
02 + 25
...
59 A and tan φ1 = (25
...
0) from which,
φ1 = tan−1 (25
...
0) = 36
...
Hence the power
factor of the primary = cos φ1 = cos 36
...
80
...
A single-phase transformer has 2400 turns on
the primary and 600 turns on the secondary
...
25 lagging
...
8 lagging
...
26 A, 0
...
6 Transformer construction
(i) There are broadly two types of single-phase
double-wound transformer constructions — the
core type and the shell type, as shown in
Fig
...
6
...


laminated silicon steel or stalloy, the laminations reducing eddy currents and the silicon steel
keeping hysteresis loss to a minimum
...
Small power transformers have many
applications, examples including welding and
rectifier supplies, domestic bell circuits, imported
washing machines, and so on
...
f
...
A typical application of a
...
transformers is in an audio
amplifier system
...
f
...
Ferrite is a
ceramic material having magnetic properties similar to silicon steel, but having a high resistivity
...
e
...

Applications of r
...
transformers are found in radio
and television receivers
...

(vi) Cooling is achieved by air in small transformers
and oil in large transformers
...
6

(ii) For power transformers, rated possibly at several MVA and operating at a frequency of 50 Hz
in Great Britain, the core material used is usually

Figure 21
...

R1 and R2 represent the resistances of the primary and
secondary windings and X1 and X2 represent the reactances of the primary and secondary windings, due to
leakage flux
...
Reactance
X takes the magnetising component Im
...
21
...

It is often convenient to assume that all of the
resistance and reactance as being on one side of the
transformer
...
21
...
7 Equivalent circuit of a
transformer

336 Electrical and Electronic Principles and Technology

Figure 21
...
8

by inserting an additional resistance R2 in the primary
circuit such that the power absorbed in R2 when carrying the primary current is equal to that in R2 due to the
secondary current, i
...

2
2
I1 R2 = I2 R2

from which,

R 2 = R2

I2
I1

2

V1
V2

= R2

Section 3

Re = R1 + R2

V1
V2

2

Xe = X1 + X2

V1
V2

2
R2 + X e
e

cos φe =

2

(6)

By similar reasoning, the equivalent reactance in the
primary circuit is given by Xe = X1 + X2

i
...


Ze =

(8)

If φe is the phase angle between I1 and the volt drop
I1 Ze then

Then the total equivalent resistance in the primary circuit Re is equal to the primary and secondary resistances
of the actual transformer
...
e
...
21
...

Problem 14
...
The primary and
secondary resistances are 0
...
01
respectively and the corresponding leakage
reactances are 1
...
04 respectively
...


Transformers 337

Figure 21
...
e
...
25 + 0
...
41

(b) From equation (7), equivalent reactance,
Xe = X1 + X2
i
...


2

V1
V2

Xe = 1
...
04

600
150

2

= 1
...
412 + 1
...
69

(d) From equation (9),
Re
0
...
69
0
...
96◦
1
...
A transformer has 1200 primary turns and 200
secondary turns
...
2 and 0
...
2 and 0
...
Calculate

(a) the equivalent resistance, reactance and
impedance referred to the primary winding,
and (b) the phase angle of the impedance
...
92 , 3
...
14 (b) 72
...
8

Regulation of a transformer

When the secondary of a transformer is loaded, the
secondary terminal voltage, V2 , falls
...
This is
called the regulation of the transformer and it is usually expressed as a percentage of the secondary no-load
voltage, E2
...
Typical values of voltage
regulation are about 3% in small transformers and about
1% in large transformers
...
A 5 kVA, 200 V/400 V, single-phase
transformer has a secondary terminal voltage of
387
...
Determine the regulation
of the transformer
...
6
× 100%
400
12
...
1%
=

Section 3

V1
V2

338 Electrical and Electronic Principles and Technology
Problem 16
...
A tap changing device is set to
operate when the percentage regulation drops
below 2
...
Determine the load voltage at which
the mechanism operates
...
If R1 and R2 are the primary and secondary winding resistances then the total copper
2
2
loss is I1 R1 + I2 R2
(b) Iron losses are constant for a given value of frequency and flux density and are of two types —
hysteresis loss and eddy current loss
...
5 =

240 − V2
240

(i) Hysteresis loss is the heating of the core
as a result of the internal molecular structure reversals which occur as the magnetic
flux alternates
...
(See
Chapter 7)

× 100%

(ii) Eddy current loss is the heating of the core
due to e
...
f
...

These induced e
...
f
...
Owing to the
low resistance of the core, eddy currents
can be quite considerable and can cause a
large power loss and excessive heating of the
core
...
e
...

This increases the resistance of the eddy current path, and reduces the value of the eddy
current
...
5)(240)
= 240 − V2
100

∴
i
...


6 = 240 − V2

from which, load voltage, V2 = 240 − 6 = 234 volts
Now try the following exercise
Exercise 122 Further problems on
regulation
1
...
5 volts when loaded
...

[2
...
A transformer has an open circuit voltage
of 110 volts
...
Calculate the load voltage at which the tap-changer
operates
...
7 volts]

η=

i
...


Section 3

21
...

(a) Copper losses are variable and result in a heating
of the conductors, due to the fact that they possess

output power
input power − losses
=
input power
input power

η=1−

losses
input power

(11)

and is usually expressed as a percentage
...
A 200 kVA rated transformer has a
full-load copper loss of 1
...
Determine the transformer efficiency at full
load and 0
...


=

output power
input power
input power − losses
input power

=1−

losses
input power

Full-load output power = VI cos φ = (200)(0
...

Total losses = 1
...
0 = 2
...
5 = 172
...

Hence efficiency = 1 −

2
...
5

= 1 − 0
...
9855 or 98
...
5 and R2 = 0
...
5) + (1250)2 (0
...
5 = 4762
...
Determine the efficiency of the
transformer in Problem 17 at half full-load and 0
...

Half full-load power output = (1/2)(200)(0
...

Copper loss (or I 2 R loss) is proportional to current
squared
...
375 kW
...
375 = 86
...
Hence
efficiency = 1 −

(a) Rating = 400 kVA = V1 I1 = V2 I2
...
375
= 1−
86
...
01592
= 0
...
41%
Problem 19
...
5 and a
secondary winding resistance of 0
...
The iron
loss is 2
...
If the

= 4762
...
5 W = 7
...
85) = 340 kW
Input power = output power + losses
= 340 kW + 7
...
2625 kW
Efficiency, η = 1 −
= 1−

losses
input power
7
...
2625

× 100%
× 100%

= 97
...
5 × 2 = 1190
...
Hence total
loss on half load = 1190
...
625 W
or 3
...

1
Output power on half full load = 2 (340)
= 170 kW
...
691 kW
= 173
...
85, determine the
efficiency of the transformer (a) on full load, and
(b) on half load
...
691
173
...
87%

Maximum efficiency
It may be shown that the efficiency of a transformer
is a maximum when the variable copper loss (i
...

2
2
I1 R1 + I2 R2 ) is equal to the constant iron losses
...
A 500 kVA transformer has a full
load copper loss of 4 kW and an iron loss of 2
...

Determine (a) the output kVA at which the
efficiency of the transformer is a maximum, and
(b) the maximum efficiency, assuming the power
factor of the load is 0
...
The corresponding
total copper loss = (4 kW)(x 2 )
...
√
Hence 4x 2 = 2
...
5/4 and x = 2
...
791
...
791 × 500 = 395
...

(b) Total loss at maximum efficiency
= 2 × 2
...
5 kVA × p
...

= 395
...
75 = 296
...
625 + 5 = 301
...
625

× 100%

× 100% = 98
...
A single-phase transformer has a voltage ratio
of 6:1 and the h
...
winding is supplied at 540 V
...
8 lagging
...

[(a) 2
...
16 kW (c) 5 A]
2
...

The transformer has full-load copper losses of
800 W and iron losses of 500 W
...
8
power factor
...
10%]
3
...
8 power
factor
...
81%]
4
...
Calculate its efficiency for a load of 60 kW at 0
...

[97
...
Determine the efficiency of a 15 kVA transformer for the following conditions:
(i) full-load, unity power factor
(ii) 0
...
8 power factor
Assume that iron losses are 200 W and
the full-load copper loss is 300 W
...
77% (ii) 96
...
62%]
6
...
4 and a secondary winding
resistance of 0
...
The iron loss is 2 kW
and the primary and secondary voltages are
4 kV and 200 V respectively
...
78, determine the efficiency
of the transformer (a) on full load, and (b) on
half load
...
84% (b) 97
...
A 250 kVA transformer has a full load copper
loss of 3 kW and an iron loss of 2 kW
...
80
...
1 kVA (b) 97
...
10 Resistance matching

i
...


Varying a load resistance to be equal, or almost equal, to
the source internal resistance is called matching
...

With d
...
generators or secondary cells, the internal
resistance is usually very small
...

A method of achieving maximum power transfer between a source and a load (see Section 13
...
A transformer
may be used as a resistance matching device by
connecting it between the load and the source
...
With reference to Fig
...
10:
RL =

V2
I2

and R1 =

V1
I1

R1 =

2

N1
N2

RL

Hence by varying the value of the turns ratio, the
equivalent input resistance of a transformer can be
‘matched’ to the internal resistance of a load to achieve
maximum power transfer
...
A transformer having a turns ratio of
4:1 supplies a load of resistance 100
...

From above, the equivalent input resistance,
R1 =
=

2

N1
N2
4
1

RL
2

(100) = 1600

Problem 22
...
Calculate the
optimum turns ratio of a transformer which would
match a load resistance of 7 to the output
resistance of the amplifier
...
21
...


Figure 21
...
11

V2

N2
I1 =
N1

and

I2

The equivalent input resistance, R1 of the transformer
needs to be 112 for maximum power transfer
...
e
...
Determine the optimum value of
load resistance for maximum power transfer if the
load is connected to an amplifier of output
resistance 150 through a transformer with a turns
ratio of 5:1
The equivalent input resistance R1 of the transformer
needs to be 150 for maximum power transfer
...
25 A

2
P = I2 RL = (1
...
28 × 103 )

= 2000 watts or 2 kW

N2
N1

2

1 2
5

=6

= 150

= 10

Power dissipated in load resistor RL ,

2

N1
N2

(b) For an ideal transformer

Problem 24
...
If the load across
the secondary winding is 1
...


Problem 25
...
c
...
Determine (a) the value of the load
resistance and (b) the power dissipated in the load
...
21
...
21
...
13

(a) For maximum power transfer R1 needs to be equal
to 15 k
...
12

(a) Turns ratio
N1
N2

R1 =
=

V1
V2

220
1760

=

=

1
8

Equivalent input resistance of the transformer
...
e
...

Primary current,
I1 =

Primary current,

2

from which, load resistance,

2

(1
...
8 mA
30 000
30 000

N1 /N2 = I2 /I1 from which, I2 = I1 (N1 /N2 ) =
(0
...


Transformers 343
Power dissipated in the load RL ,
2
P = I2 RL = (20 × 10−3 )2 (24)

= 9600 × 10−6 W = 9
...
A transformer having a turns ratio of 8:1 supplies a load of resistance 50
...

[3
...
What ratio of transformer turns is required to
make a load of resistance 30 appear to have
a resistance of 270 ?
[3:1]
3
...

[12 ]
4
...
If the load across the
secondary winding is 720 determine (a) the
primary current flowing and (b) the power
dissipated in the load resistance
...
5 kW]
5
...
Determine the turns ratio
of the coupling transformer
...
An a
...
source of 20 V and internal resistance
20 k is matched to a load by a 16:1 singlephase transformer
...

[(a) 78
...
11

Auto transformers

An auto transformer is a transformer which has part
of its winding common to the primary and secondary
circuits
...
21
...
21
...
14

transformer
...
Since the current is less in this section,
the cross-sectional area of the winding can be reduced,
which reduces the amount of material necessary
...
15 shows the circuit diagram symbol for
an auto transformer
...
15

Problem 26
...
Assuming an ideal transformer,
determine the current in each section of the
winding
...
5 A
V1
320

and secondary current,
I2 =

20 × 103
20 × 103
=
= 80 A
V2
250

Hence current in common part of the winding
= 80 − 62
...
5 A
The current flowing in each section of the transformer
is shown in Fig
...
16
...
This is explained below
...
The closer N2 is to N1 , the greater the saving in
copper
...
Determine the saving in the volume
of copper used in an auto transformer compared
with a double-wound transformer for (a) a
200 V:150 V transformer, and (b) a 500 V:100 V
transformer
...
16

The volume, and hence weight, of copper required
in a winding is proportional to the number of turns and
to the cross-sectional area of the wire
...
e
...

Volume of copper in an auto transformer

see Fig
...
14(b)
∝ N1 I1 − N2 I1 + N2 I2 − N2 I1
∝ N1 I1 + N2 I2 − 2N2 I1
(since N2 I2 = N1 I1 )

Volume of copper in a double-wound transformer
∝ N1 I1 + N2 I2 ∝ 2N1 I1

volume of copper in a
double-wound transformer

=

2N1 I1 − 2N2 I1
2N1 I1

If (N2 /N1 ) = x then
(volume of copper in an auto transformer)
= (1 − x) (volume of copper in a doublewound transformer)
(12)

Section 3

If, say, x = (4/5) then (volume of copper in auto
transformer)

=

1
5

(volume of copper in a
double-wound transformer)
(volume in double-wound transformer)

i
...
a saving of 80%
...
25) (volume of copper in
double-wound transformer)
= 25% (of copper in a
double-wound transformer)

x=

100
V2
=
= 0
...
75) (volume of copper in
double-wound transformer)

(b) For a 500 V:100 V transformer,

=

= 1−

V2
150
= 0
...
Hence
volume of copper in
an auto transformer

x=

Hence from equation (12), (volume of copper in
auto transformer)

∝ (N1 − N2 )I1 + N2 (I2 − I1 )

∝ 2N1 I1 − 2N2 I1

(a) For a 200 V:150 V transformer,

= (1 − 0
...
8) (volume in double-wound transformer)
= 80% of copper in a double-wound transformer
Hence the saving is 20%
...
A single-phase auto transformer has a voltage ratio of 480 V:300 V and supplies a
load of 30 kVA at 300 V
...

[I1 = 62
...
5 A]
2
...
a saving in cost since less copper is needed (see
above)

21
...
An isolating transformer is a 1:1 ratio transformer with several important applications, including bathroom shaver-sockets,
portable electric tools, model railways, and so on
...
13 Three-phase transformers

2
...
a higher efficiency, resulting from lower

Auto transformers are used for reducing the voltage when starting induction motors (see Chapter 23)
and for interconnecting systems that are operating at
approximately the same voltage
...
a continuously variable output voltage is achievable
if a sliding contact is used
5
...


Disadvantages of auto transformers

Section 3

The primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the
secondary winding the full primary voltage appears
across the secondary
...
They basically consist of three pairs of singlephase windings mounted on one core, as shown in
Fig
...
17, which gives a considerable saving in the
amount of iron used
...
21
...
21
...
The windings may be with the primary
delta-connected and the secondary star-connected, or
star-delta, star-star or delta-delta, depending on its use
...
17

346 Electrical and Electronic Principles and Technology

Figure 21
...
21
...
21
...

Problem 28
...
If the supply
voltage is 2
...

√
(a) For a star-connection, VL = 3 Vp (see Chapter 20)
...
64 volts
...
64)

50
500

Section 3

= 138
...
4 kV = 2400 volts
...

Now try the following exercise
Exercise 126 A further problem on the
three-phase transformer
1
...
If the supply
voltage is 1
...

[(a) 649
...
5 V]

21
...
With a d
...
movingcoil ammeter the current required to give full scale
deflection is very small — typically a few milliamperes
...
However, even
with shunt resistors added it is not possible to measure
very large currents
...
c
...
20

Figure 21
...

In a double-wound transformer:

Problem 29
...
The secondary winding is
connected to an ammeter with a resistance of
0
...
The resistance of the secondary winding is
0
...
If the current in the primary winding is
300 A, determine (a) the reading on the ammeter,
(b) the potential difference across the ammeter and
(c) the total load (in VA) on the secondary
...
A typical arrangement is shown in
Fig
...
19
...


(b) P
...
across the ammeter = I2 RA , (where RA is the
ammeter resistance) = (5)(0
...
75 volts
...
15 + 0
...
40
...
m
...
in secondary = (5)(0
...
0 V
...
0)(5) = 10 VA
...
A current transformer has two turns on the primary winding and a secondary winding of 260
turns
...
2
...
3
...

[(a) 5 A (b) 1 V (c) 7
...

For very large currents the transformer core can be
mounted around the conductor or bus-bar
...

It is very important to short-circuit the secondary winding before removing the ammeter
...

Current transformer circuit diagram symbols are shown
in Fig
...
20
...
15 Voltage transformers

5
...
These are normal double-wound transformers with a large number
of turns on the primary, which is connected to a high
voltage supply, and a small number of turns on the
secondary
...
21
...


6
...
Draw a phasor diagram for an ideal transformer on no-load
8
...
m
...
equation for a transformer
9
...
Name two types of transformer construction
11
...
Name three core materials used in r
...
transformers
13
...
f
...
f
...
21

14
...
State the expressions for equivalent resistance
and reactance of a transformer, referred to the
primary

V1
N1
=
V2
N2
the secondary voltage,

16
...
Name two sources of loss in a transformer

Thus if the arrangement in Fig
...
21 has 4000 primary turns and 20 secondary turns then for a voltage of
22 kV on the primary, the voltage on the secondary,
V2 = V1

N2
N1

20
= (22 000)
4000

= 110 volts

Now try the following exercises

18
...
What are eddy currents? How may they be
reduced in transformers?
20
...
What is the condition for maximum efficiency
of a transformer?
22
...
What is a transformer?

23
...
Explain briefly how a voltage is induced in
the secondary winding of a transformer

24
...
Draw the circuit diagram symbol for a transformer

25
...
State the relationship between turns and voltage ratios for a transformer

26
...
In what applications are auto transformers
used?
28
...
Describe briefly the construction of a threephase transformer
30
...
Describe how a current transformer operates
32
...
Describe how a voltage transformer operates

Multi-choice questions on
transformers
(Answers on page 399)

1
...
m
...
equation of a transformer of secondary turns N2 , magnetic flux density Bm ,
magnetic area of core a, and operating at
frequency f is given by:
(a) E2 = 4
...
44
volts
a
N2 Bm f
(c) E2 =
volts
a
(d) E2 = 1
...
7A

3
...
22

5
...
A 1 kV/250 V transformer has 500 turns on
the secondary winding
...
The core of a transformer is laminated to:
(a) limit hysteresis loss
(b) reduce the inductance of the windings
(c) reduce the effects of eddy current loss
(d) prevent eddy currents from occurring

2
...
21
...
3 A (b) 1
...
6 A

V1

4
...
The number of turns on
the secondary is:
(a) 550
(b) 250
(c) 4000
(d) 25

8
...
If the primary current is 2
...
A transformer has 800 primary turns and 100
secondary turns
...
5 V
(d) 20 V
A 100 kVA, 250 V/10 kV, single-phase transformer has a full-load copper loss of 800 W
and an iron loss of 500 W
...
For the statements in

Section 3

Exercise 129

3
...
If the output current is 5 A, the input
current is:
(a) 50 A
(b) 5 A
(c) 2
...
5 A

350 Electrical and Electronic Principles and Technology
questions 10 to 16, select the correct answer
from the following list:
(a) 81
...
32%
(d) 80 kW
(e) 3
(f) 4800
(g) 1
...
40% (i) 100 kW
(j) 98
...
3 kW
(m) 96
...
The total full-load losses
11
...
8 power
factor
12
...
8 power factor
13
...
8 power factor
14
...
The transformer efficiency at half full-load,
0
...
The number of secondary winding turns

Section 3

17
...
An ideal transformer has a turns ratio of
1:5 and is supplied at 200 V when the primary current is 3 A
...
6 kVA
(e) The secondary voltage is 1 kV
(f) The secondary current is 0
...
Iron losses in a transformer are due to:
(a) eddy currents only
(b) flux leakage
(c) both eddy current and hysteresis losses
(d) the resistance of the primary and secondary windings
20
...
The value of the load resistance for
maximum power transfer is:
(a) 100
(b) 1 k
(c) 100 m
(d) 1 m

Revision Test 6
This revision test covers the material contained in Chapters 20 to 21
...


2
...
If the instrument readings are
10 kW and 6 kW, determine (a) the total power
input, and (b) the load power factor
...
An ideal transformer connected to a 250 V mains,
supplies a 25 V, 200 W lamp
...

(5)
4
...
Determine
(a) the primary and secondary currents, (b) the number of primary turns, and (c) the maximum value of
flux
...
Determine the percentage regulation of an 8 kVA,
100 V/200 V, single phase transformer when its
secondary terminal voltage is 194 V when loaded
...
A 500 kVA rated transformer has a full-load copper
loss of 4 kW and an iron loss of 3 kW
...
80
power factor, and (b) at half full load and 0
...

(10)
7
...
A single-phase auto transformer has a voltage ratio
of 250 V:200 V and supplies a load of 15 kVA at
200 V
...

(3)

Section 3

1
...
Calculate for each connection (a) the line and phase
voltages, (b) the phase and line currents, and (c) the
total power dissipated
...
C
...
c
...
c
...
m
...
in an armature winding using E = 2p nZ/c
• describe types of d
...
generator and their characteristics
• calculate generated e
...
f
...
c
...
c
...
m
...
for a d
...
motor using E = V − Ia Ra
• calculate the torque of a d
...
motor using T = EIa /2πn and T = p ZIa /πc
• describe types of d
...
motor and their characteristics
• state typical applications of d
...
motors
• describe a d
...
motor starter
• describe methods of speed control of d
...
motors
• list types of enclosure for d
...
motors

22
...
Thus an electric motor
converts electrical energy into mechanical energy
...
4, page 91
...
Thus, a generator converts
mechanical energy to electrical energy
...
C
...
2, page 98
...
2 The action of a commutator
In an electric motor, conductors rotate in a uniform magnetic field
...
22
...
A voltage is
applied at points A and B in Fig
...
1(a)

to the current flowing from E to F (from Fleming’s left
hand rule)
...
22
...
22
...
e
...
This apparent reversal in the direction of current flow is achieved
by a process called commutation
...
22
...
Thus the forces act to give continuous rotation
in an anti-clockwise direction
...
22
...

In practice, there are many conductors on the rotating
part of a d
...
machine and these are attached to many
commutator segments
...
22
...

Poor commutation results in sparking at the trailing
edge of the brushes
...


Figure 22
...
3 D
...
machine construction
The basic parts of any d
...
machine are shown in
Fig
...
3, and comprise:
(a) a stationary part called the stator having,
(i) a steel ring called the yoke, to which are
attached
(ii) the magnetic poles, around which are the

Section 3

A force, F, acts on the loop due to the interaction
of the magnetic field of the permanent magnets and the
magnetic field created by the current flowing in the loop
...
e
...
The force is made up of two parts, one
acting vertically downwards due to the current flowing
from C to D and the other acting vertically upwards due

Figure 22
...
4

Figure 22
...
e
...
22
...
2),
(b) a rotating part called the armature mounted in
bearings housed in the stator and having,
(iv) a laminated cylinder of iron or steel called
the core, on which teeth are cut to house the
(v) armature winding, i
...
a single or multiloop conductor system, and
(vi) the commutator, (see Section 22
...
These are called wave windings and lap
windings
...
Wave wound
generators produce high voltage, low current
outputs
...
The total current output
divides equally between them
...


Section 3

22
...
c
...
22
...
If the field winding is connected in series with the armature, as shown in
Fig
...
4(b), then the machine is said to be series wound
...


Depending on whether the electrical machine is series
wound, shunt wound or compound wound, it behaves
differently when a load is applied
...
c
...
The characteristics shown in the following sections are theoretical, since they neglect the
effects of armature reaction
...
In a generator,
armature reaction results in a reduced output voltage,
and in a motor, armature reaction results in increased
speed
...


22
...
m
...
generated in an armature
winding
Z = number of armature conductors,
= useful flux per pole, in webers,
p = number of pairs of poles
n = armature speed in rev/s

The e
...
f
...
m
...
generated by one of the parallel paths
...
Hence flux cut
by one conductor per second = 2p n Wb and so the
average e
...
f
...
C
...
m
...
between
brushes = (average e
...
f
...
m
...
,

2p nZ
i
...

generated e
...
f
...
However 2πn is the angular velocity ω
in radians per second, hence the generated e
...
f
...
m
...
E ∝

ω

(2)

Problem 1
...
If
the flux per pole is 20 mWb, determine the
generated e
...
f
...

Generated e
...
f
...
A 4-pole generator has a lap-wound
armature with 50 slots with 16 conductors per slot
...
Determine the
speed at which the machine must be driven to
generate an e
...
f
...

E = 240 V, c = 2 p (for a lap winding), Z = 50 × 16 = 800
and = 30 × 10−3 Wb
...
m
...

2p nZ
2p nZ
=
=
c
2p

i
...


E=

nZ

= (0
...
Determine the generated e
...
f
...

Generated e
...
f
...
A d
...
shunt-wound generator
running at constant speed generates a voltage of
150 V at a certain value of field current
...

The generated e
...
f
...
e
...
It follows that E = k n, where
k is a constant
...
e
...
An 8-pole, lap-wound armature has
1200 conductors and a flux per pole of 0
...

Determine the e
...
f
...


356 Electrical and Electronic Principles and Technology
The initial conditions are E1 = 150 V,
= 1 and
n = n1
...
8 of the initial value, i
...

2 = 0
...
Since the generator is running at constant
speed, n2 = n1
Thus

E1
=
E2

that is,

E2 = 150 × 0
...
8

1 n2

=

1
0
...

Problem 6
...
c
...
m
...
of 200 V
...

From Equation (2), generated e
...
f
...
If the useful flux per pole is
15 mWb, determine the generated e
...
f
...
A 6-pole generator has a lap-wound armature
with 40 slots with 20 conductors per slot
...
Calculate the speed at
which the machine must be driven to generate
an e
...
f
...
A 4-pole armature of a d
...
machine has 1000
conductors and a flux per pole of 20 mWb
...
m
...
generated when running at 600 rev/min when the armature is
(a) wave-wound (b) lap-wound
...
A d
...
generator running at 25 rev/s generates
an e
...
f
...
Determine the percentage
increase in the flux per pole required to
generate 180 V at 20 rev/s
[50%]

Let E1 = 200 V, n1 = 30 rev/s
and flux per pole at this speed be

22
...
C
...
These groupings are:

Let E2 = 250 V, n2 = 20 rev/s
and flux per pole at this speed be
Since
Hence
from which,

E∝

2

E1
=
E2
200
=
250

n then

2

=

(i) Separately-excited generators, where the field
winding is connected to a source of supply other
than the armature of its own machine
...
875

(ii) Self-excited generators, where the field winding
receives its supply from the armature of its own
machine, and which are sub-divided into (a) shunt,
(b) series, and (c) compound wound generators
...
5
per cent

Now try the following exercise

Section 3

D
...
generators

Exercise 130 Further problems on
generator e
...
f
...
A 4-pole, wave-connected armature of a d
...

machine has 750 conductors and is driven

22
...
c
...
22
...

When a load is connected across the armature terminals, a load current Ia will flow
...
m
...
E due to a volt
drop caused by current flowing through the armature
resistance, shown as Ra

D
...
machines 357
i
...


terminal voltage, V = E − Ia Ra

or

generated e
...
f
...
5, generated e
...
f
...
10)(20)
Hence
=
E2
(0
...
10)(25)
from which, E2 =
(0
...
5 volts
(b)

150 (0
...
08)(20)
from which, e
...
f
...
5

(150)(0
...
10)(20)

= 120 volts

Problem 7
...
m
...
of 200 V and
has an armature current of 30 A on load
...
30
...
10)(20)
=
E4
(0
...
07)(24)
(0
...
m
...
, E4 =

With reference to Fig
...
5, terminal voltage,
V = E − I a Ra
= 200 − (30)(0
...
A generator is connected to a 60
load and a current of 8 A flows
...
m
...


Characteristics
The two principal generator characteristics are the generated voltage/field current characteristics, called the
open-circuit characteristic and the terminal voltage/
load current characteristic, called the load characteristic
...
22
...
22
...


(a) Terminal voltage, V = Ia RL = (8)(60) = 480 volts
(b) Generated e
...
f
...
A separately-excited generator
develops a no-load e
...
f
...
10 Wb
...
m
...
when (a) the speed
increases to 25 rev/s and the pole flux remains
unchanged, (b) the speed remains at 20 rev/s and
the pole flux is decreased to 0
...
07 Wb
...
6

A separately-excited generator is used only in special
cases, such as when a wide variation in terminal p
...
is
required, or when exact control of the field current is
necessary
...


(b) Shunt wound generator
In a shunt wound generator the field winding is connected in parallel with the armature as shown in

Section 3

E = V + Ia Ra

358 Electrical and Electronic Principles and Technology
Fig
...
7
...


(b) Armature current Ia = If + I
V
210
= 4
...
2 + 100 = 104
...
m
...
E = V + Ia Ra
= 210 + (104
...
168
= 214
...
7

For the circuit shown in Fig
...
7
terminal voltage, V = E − Ia Ra
or

The generated e
...
f
...
5), hence at constant speed, since ω = 2πn,
E ∝
...
Hence the open circuit characteristic
is as shown in Fig
...
9(a)
...
m
...
, E = V + Ia Ra

I a = I f + I from Kirchhoff’s current law, where
Ia = armature current, If = field current (= V /Rf ) and
I = load current
...
A shunt generator supplies a 20 kW
load at 200 V through cables of resistance, R =
100 m
...
m
...

generated in the armature
...
22
...

Hence terminal voltage, V = 200 + 10 = 210 volts
...
9

As the load current on a generator having constant
field current and running at constant speed increases, the
value of armature current increases, hence the armature
volt drop, Ia Ra increases
...
Since E
is constant, V decreases with increasing load
...
22
...
In practice,
the fall in voltage is about 10 per cent between no-load
and full-load for many d
...
shunt-wound generators
...
This then
avoids excessive variation of the terminal voltage
...


(c) Series-wound generator
Figure 22
...
22
...


D
...
machines 359
series windings, designed to combine the advantages
of each
...
22
...
22
...
The latter is the most
generally used form of d
...
generator
...
10

Characteristic
The load characteristic is the terminal voltage/current
characteristic
...
m
...
E, is proportional
to ω and at constant speed ω(=2πn) is a constant
...
For values of current below
magnetic saturation of the yoke, poles, air gaps and
armature core, the flux
is proportional to the current, hence E ∝ I
...
m
...
is
approximately constant
...
A typical load characteristic for a series generator
is shown in Fig
...
11
...
12

Problem 11
...
If the field resistance,
Rf = 40 , the series resistance, RSe = 0
...
04 , determine the
e
...
f
...

The circuit is shown in Fig
...
13
...
02)
= 1
...


Figure 22
...
13

In a series-wound generator, the field winding is in
series with the armature and it is not possible to have a
value of field current when the terminals are open circuited, thus it is not possible to obtain an open-circuit
characteristic
...
c
...


P
...
across the field winding = p
...
across
armature = V1 = 200 + 1
...
6 V
Field current If =

V1
201
...
04 A
Rf
40

Armature current, Ia = I + If = 80 + 5
...
04 A
= 201
...
04)(0
...
6 + 3
...
m
...
, E = V1 + Ia Ra

360 Electrical and Electronic Principles and Technology
Characteristics
In cumulative-compound machines the magnetic flux
produced by the series and shunt fields are additive
...

A large number of series winding turns results
in an over-compounded characteristic, as shown in
Fig
...
14, in which the full-load terminal voltage exceeds the no-load voltage
...
22
...


3
...
m
...
of 180 V at an armature speed
of 15 rev/s and a flux per pole of 0
...

Calculate the generated e
...
f
...
125 Wb
(c) the speed increases to 25 rev/s and the
pole flux is decreased to 0
...
5 volts (c) 270 volts]
4
...
2
...
05 , determine (a) the
terminal voltage, (b) the e
...
f
...
68 volts]
5
...
If the field resistance is 30 ,
the series resistance 0
...
05 , determine the e
...
f
...
5 volts]

Figure 22
...
22
...
However even this latter characteristic is a little better than that for a shunt generator alone
...


Now try the following exercise
Exercise 131 Further problems on the
d
...
generator

Section 3

1
...
m
...
of 240 V and has an
armature current of 50 A on load
...
A generator is connected to a 50 load and a
current of 10 A flows
...
5 , determine (a) the terminal
voltage, and (b) the generated e
...
f
...
A d
...
generator has a generated e
...
f
...
Determine the generated
e
...
f
...
2 V]
7
...
c
...
1
...
m
...
when the generator is supplying 50 kW, neglecting the field current of the
generator
...
8

D
...
machine losses

As stated in Section 22
...
When such conversions take place,
certain losses occur which are dissipated in the form
of heat
...
C
...

(ii) Iron (or core) loss, due to hysteresis and eddycurrent losses in the armature
...
At constant speed, the iron loss is
assumed constant
...
At
constant speed, these losses are assumed to be
constant
...
This loss is approximately proportional
to the load current
...
e
...
A 10 kW shunt generator having an
armature circuit resistance of 0
...
Determine the efficiency of the
generator at full load, assuming the iron, friction
and windage losses amount to 600 W
...
22
...


22
...
c
...

(Ia a
f
If the output current is I, then the output power is
2
VI
...
Hence
efficiency, η =

output
, i
...

input

Figure 22
...

Field current, If = V /Rf = 250/125 = 2 A
...
75)
+(2)(250) + 600

10 000
12 423

= 80
...
The Greek letter, ‘η’ (eta) is used to
signify efficiency and since the units are, power/power,
then efficiency has no units
...
c
...
A 15 kW shunt generator having an armature
circuit resistance of 0
...
Determine the efficiency
of the generator at full load, assuming the iron,
friction and windage losses amount to 1 kW
[82
...
The armature of a d
...
machine has
a resistance of 0
...
Calculate the e
...
f
...


(a) As a generator, generated e
...
f
...
25)
= 300 + 25
= 325volts

22
...
C
...
c
...
c
...
The only difference is that in a generator the
generated e
...
f
...
m
...
is less than the
terminal voltage
...
C
...
c
...


(b) As a motor, generated e
...
f
...
m
...
),
E = V − Ia Ra , from Equation (5),
= 300 − (80)(0
...
m
...

When a d
...
motor rotates, an e
...
f
...
By Lenz’s law this
induced e
...
f
...
m
...
, and the supply voltage, V is
given by:
V = E + Ia Ra

or

E = V − Ia R a

(5)

Problem 13
...
c
...
The armature resistance is 0
...
Determine
the back e
...
f
...


Section 3

For a motor, V = E + Ia Ra hence back e
...
f
...
2)
= 240 − 10 = 230 volts

Exercise 133 Further problems on
back e
...
f
...
A d
...
motor operates from a 350 V supply
...
4 determine the
back e
...
f
...
The armature of a d
...
machine has a resistance
of 0
...

Calculate the e
...
f
...
Determine the generated e
...
f
...
c
...
1 and
it (a) is running as a motor connected to a 230 V
supply, the armature current being 60 A, and
(b) is running as a generator with a terminal
voltage of 230 V, the armature current being
80 A
[(a) 224 V (b) 238 V]

D
...
machines 363
From Equation (7),

22
...
c
...
c
...
7 Nm

Multiplying each term by current Ia gives:
2
VIa = EIa + Ia Ra

The term VIa is the total electrical power supplied to
the armature, the term I2 Ra is the loss due to armaa
ture resistance, and the term EIa is the mechanical
power developed by the armature
...
e
...
m
...
E = V − Ia Ra = 250 − (40)(1) = 210 V
(7)

E
...
f
...
e
...
An 8-pole d
...
motor has a
wave-wound armature with 900 conductors
...
Determine the
torque exerted when a current of 30 A flows in each
armature conductor
...


Problem 17
...
c
...
The armature has
500 conductors and a resistance of 1
...
Calculate (a) the speed and (b) the
torque developed when the armature current is 40 A
...
7 Nm
2πn
2π(15)

(6)

From Section 22
...
m
...
E generated
is given by

Hence

V = 350 V, Ra = 0
...

Back e
...
f
...
5) = 320 V
...
66 Nm
2πn
2π(21)

Problem 18
...
c
...
Determine the torque developed by
a 350 V d
...
motor having an armature resistance of
0
...
The armature current
is 60 A
...
The armature current of the generator is
16 A at this value of torque
...

Determine the armature current at this new value of
torque
...

The torque at flux 1 and armature current Ia1 is
T1 = k 1 Ia1 Similarly, T2 = k 2 Ia2
By division

T1
k
=
T2
k

Hence

25
1 × 16
=
35
0
...
e
...
35 A
0
...
35 A
Problem 19
...
c
...
If
the torque on the shaft driving the generator is
12 Nm, determine (a) the efficiency of the generator
and (b) the power loss in the generator
...
9, the efficiency of a generator =
output power/input power × 100 per cent
...
e
...

The input power to a generator is the mechanical power in the shaft driving the generator, i
...

T ω or T (2πn) watts, where T is the torque in Nm
and n is speed of rotation in rev/s
...
e
...
6%

(b) The input power = output power + losses
Hence, T (2πn) = VI + losses
i
...
losses = T (2πn) − VI
= (12)(2π)

1500
60
− [(100)(15)]

i
...
power loss = 1885 − 1500 = 385 W

Now try the following exercise
Exercise 134 Further problems on losses,
efficiency, and torque
1
...
c
...
7 Nm when it is running at
1250 rev/min
...
3 A, determine the voltage at the
terminals of the generator
[123
...
A 220 V, d
...
generator supplies a load of
37
...
Determine
the shaft torque of the diesel motor driving
the generator, if the generator efficiency is
78 per cent
[65
...
A 4-pole d
...
motor has a wave-wound armature with 800 conductors
...
Calculate the torque exerted
when a current of 40 A flows in each armature
conductor
[203
...
Calculate the torque developed by a 240 V
d
...
motor whose armature current is 50 A,
armature resistance is 0
...
1 Nm]
5
...
c
...
The armature has 800 conductors and
a resistance of 0
...
If the useful flux per
pole is 40 mWb and the armature current is
30 A, calculate (a) the speed and (b) the torque
developed
[(a) 5
...
8 Nm]
6
...
c
...
If

D
...
machines 365
the torque on the shaft driving the generator
is 35
...
4 per cent (b) 748
...
6
= 28
...
m
...

E = V − Ia Ra = 240 − (28
...
4) = 228
...
12 Types of d
...
motor and their
characteristics

The two principal characteristics are the torque/armature
current and speed/armature current relationships
...


(a) Shunt wound motor
In the shunt wound motor the field winding is in parallel with the armature across the supply as shown in
Fig
...
16
...
11)
...
e
...
Since
is constant, it follows
that T ∝ Ia , and the characteristic is as shown
in Fig
...
17

Figure 22
...
22
...
m
...
, E =V − I a Ra
Supply current, I = I a + If
from Kirchhoff’s current law

Problem 20
...
If the field winding resistance
Rf = 150 and the armature resistance Ra = 0
...
m
...


(a) Field current If =

V
240
= 1
...
17

(ii) The armature circuit of a d
...
motor has resistance due to the armature winding and brushes,
Ra ohms, and when armature current Ia is flowing through it, there is a voltage drop of Ia Ra
volts
...
22
...
Also, even though
the machine is a motor, because conductors are
rotating in a magnetic field, a voltage, E ∝ ω,
is generated by the armature conductors
...
5, E ∝ n, hence
n ∝ E/ i
...


speed of rotation, n ∝

E

∝

V − I a Ra

(9)

Section 3

Supply voltage, V= E + I a Ra

366 Electrical and Electronic Principles and Technology
For a shunt motor, V ,
and Ra are constants,
hence as armature current Ia increases, Ia Ra
increases and V − Ia Ra decreases, and the speed
is proportional to a quantity which is decreasing and is as shown in Fig
...
18 As the load on
the shaft of the motor increases, Ia increases and
the speed drops slightly
...
c
...
Due
to this relatively small drop in speed, the d
...

shunt-wound motor is taken as basically being a
constant-speed machine and may be used for driving lathes, lines of shafts, fans, conveyor belts,
pumps, compressors, drilling machines and so on
...
4 = 188 V

and

E2 = 200 − 45 × 0
...
Since the flux is
constant, 1 = 2
...
e
...
5 × 182
= 21
...
78 × 60 rev/min i
...
1307 rev/min
...
18

Figure 22
...
22
...

Problem 21
...
c
...
4 and at a certain
load has an armature current of 30 A and runs at
1350 rev/min
...

The relationship E ∝ n applies to both generators and
motors
...
A 220 V, d
...
shunt-wound motor
runs at 800 rev/min and the armature current is
30 A
...
4
...


(a) For a d
...
shunt-wound motor, E = V − Ia Ra
...
m
...
, E1 = 220 = 30 ×
0
...
The generated e
...
f
...
2 V Hence, the voltage
drop due to the armature resistance is 220 − 187
...
e
...
8 V
...
8/0
...
This increase in current is about three times the initial value and causes
an increase in torque, (T ∝ Ia )
...

(b) T ∝ Ia and, since the torque is constant,
is reduced by 10 per
1 Ia1 = 2 Ia2
...
9 1
...
9 1 ×
Ia2 i
...
the steady state value of armature current,
Ia2 = 30/0
...
33 A
...
C
...
22
...


Figure 22
...
Thus n ∝ (V − IR)/I
where R is the combined resistance of the series
field and armature circuit
...

Hence the theoretical speed/current characteristic is as shown in Fig
...
22
...


For the series motor shown in Fig
...
20,
Supply voltage V = E + I(Ra + Rf )
or generated e
...
f
...

(i) The torque/current characteristic
It is shown in Section 22
...

Since the armature and field currents are the same
current, I, in a series machine, then T ∝ I over
a limited range, before magnetic saturation of
the magnetic circuit of the motor is reached, (i
...

the linear portion of the B–H curve for the yoke,
poles, air gap, brushes and armature in series)
...
After magnetic saturation,
almost becomes a constant and T ∝ I
...
22
...


Figure 22
...
A typical speed/torque characteristic is
shown in Fig
...
23
...
21

A d
...
series motor takes a large current on starting and the characteristic shown in Fig
...
21
shows that the series-wound motor has a large
torque when the current is large
...
23

368 Electrical and Electronic Principles and Technology
motors are used for traction (such as trains, milk
delivery vehicles, etc
...


Problem 23
...
2 and a series field resistance of
0
...
It is connected to a 240 V supply and at a
particular load runs at 24 rev/s when drawing 15 A
from the supply
...
m
...

at this load (b) Calculate the speed of the motor
when the load is changed such that the current is
increased to 30 A
...


(c) Compound wound motor
There are two types of compound wound motor:
(i) Cumulative compound, in which the series winding is so connected that the field due to it assists
that due to the shunt winding
...

Figure 22
...
22
...


(a) With reference to Fig
...
20, generated e
...
f
...
24

= 240 − (15)(0
...
3)
= 240 − 7
...
5 volts
(b) When the current is increased to 30 A, the generated
e
...
f
...
2 + 0
...
m
...
E ∝ n thus
E1
=
E2
i
...


232
...
5
(2 1 )n2

1 n1
2 n2
2

=2

1

Characteristics
A compound-wound motor has both a series and a shunt
field winding, (i
...
one winding in series and one in
parallel with the armature), and is usually wound to
have a characteristic similar in shape to a series wound
motor (see Figs
...
21–22
...
A limited amount of
shunt winding is present to restrict the no-load speed
to a safe value
...
Generally,
compound-wound motors are used for heavy duties, particularly in applications where sudden heavy load may
occur such as for driving plunger pumps, presses, geared
lifts, conveyors, hoists and so on
...
22
...


Hence

Section 3

speed of motor, n2 =

(24)(225)
= 11
...
5)(2)

As the current has been increased from 15 A to
30 A, the speed has decreased from 24 rev/s to
11
...
Its speed/current characteristic is similar
to Fig
...
22
...
25

D
...
machines 369
Efficiency,

22
...
c
...
9, that the efficiency of a d
...

machine is given by:
efficiency, η =

output power
× 100%
input power

2
Also, the total losses = Ia Ra + If V + C (for a shunt
motor) where C is the sum of the iron, friction and
windage losses
...
8 − 2560 − 1500
25 600
20 503
...
1%

− If V − C

Hence efficiency,
η=

=
=

the input power = VI
= VI

2
VI − Ia Ra − If V − C
× 100%
VI
⎛
⎞
(320) (80) − (72)2 (0
...
A 250 V series motor draws a
current of 40 A
...
15
and the field resistance is 0
...
Determine the
maximum efficiency of the motor
...
22
...
A 320 V shunt motor takes a total
current of 80 A and runs at 1000 rev/min
...
5 kW,
the shunt field resistance is 40 and the armature
resistance is 0
...


The circuit is shown in Fig
...
26
...

Armature current Ia = I − If = 80 − 8 = 72 A
...


2
However for a series motor, If = 0 and the Ia Ra loss
2 (R + R ) Hence efficiency,
needs to be I a
f

η=
Figure 22
...
27

370 Electrical and Electronic Principles and Technology
For maximum efficiency I 2 (Ra + Rf ) = C
...
6%

(400)(10) − (10)2 (2) − 300
(400)(10)
4000 − 200 − 300
4000

=

(250)(40)

× 100%

=
=

(250)(40) − 2(40)2 (0
...
05)

3500
4000

× 100%

× 100%

× 100% = 87
...
A 200 V d
...
motor develops a shaft
torque of 15 Nm at 1200 rev/min
...

output power
× 100%
...
The input power is the electrical power
in watts supplied to the motor, i
...
VI watts
...
e
...
8 A

Problem 27
...
c
...
If the total resistance of the motor
is 2 and the iron, friction and windage losses
amount to 300 W, determine the efficiency of the
motor
...
c
...
A 240 V shunt motor takes a total current of
80 A
...
4 , determine
(a) the current in the armature, and (b) the
back e
...
f
...
8 V]
2
...
c
...
Find the approximate value of the speed of the motor when
connected to a 200 V supply, assuming the flux
decreases by 30 per cent and neglecting the
armature volt drop
...
A series motor having a series field resistance
of 0
...
15 ,
is connected to a 220 V supply and at a particular load runs at 20 rev/s when drawing 20 A
from the supply
...
m
...
generated at this load
...
Assume the flux
increases by 25 per cent
[212 V, 15
...
A 500 V shunt motor takes a total current of
100 A and runs at 1200 rev/min
...
25 and the iron, friction and windage
losses amount to 2 kW, determine the overall
efficiency of the motor
...
95 per cent]
5
...
If

D
...
machines 371
its efficiency at this load is 88 per cent, find
the current taken from the supply
...
94 A]
6
...
c
...
Supply voltage: 500 V, current taken
from the supply: 42
...
Determine the efficiency of the motor correct to the nearest 0
...

[78
...
A series motor drives a load at 1500 rev/min
and takes a current of 20 A when the supply voltage is 250 V
...
5 and the iron, friction and
windage losses amount to 400 W, determine
the efficiency of the motor
...
A series-wound motor is connected to a d
...

supply and develops full-load torque when the
current is 30 A and speed is 1000 rev/min
...

[21
...
14 D
...
motor starter
If a d
...
motor whose armature is stationary is switched
directly to its supply voltage, it is likely that the fuses
protecting the motor will burn out
...
Thus, additional resistance must be added to
the armature circuit at the instant of closing the switch
to start the motor
...
Thus the value of
the additional armature resistance can then be reduced
...
m
...

is such that no additional resistance is required in the
armature circuit
...
c
...
22
...


Figure 22
...
For a shunt-wound motor,
the field winding is connected to stud 1 or to L via a
sliding contact on the starting handle, to give maximum
field current, hence maximum flux, hence maximum
torque on starting, since T ∝ Ia
...


22
...
c
...
c
...
The speed is varied either by varying
the value of flux, , or by varying the value of Ra
...
22
...


Figure 22
...


Section 3

7
...

The field resistance is 40 m and the armature
resistance is 0
...
Determine the maximum
efficiency of the motor
...
Thus only
speeds above that given without a shunt field regulator
can be obtained by this method
...
A 500 V shunt motor runs at its
normal speed of 10 rev/s when the armature current
is 120 A
...
2
...
5 is connected in series with
the armature, the shunt field remaining constant
(b) Determine the speed when the current is 60 A
and the shunt field is reduced to 80 per cent of its
normal value by increasing resistance in the field
circuit
...
22
...
m
...
at
120 A, E1 = V − Ia Ra = 500 − (120)(0
...

When Ia = 60 A,
E2 = 500 − (60)(0
...
5)
= 500 − (60)(0
...
e
...
e
...
62 rev/s
476

3

= 0
...
82 rev/s
(0
...

(a) The speed of a d
...
series-wound motor is given by:
n=k

V − IR

where k is a constant, V is the terminal voltage,
R is the combined resistance of the armature and
series field and is the flux
...
This is achieved
by putting a variable resistance in parallel with the
field winding and reducing the field current, and
hence flux, for a given value of supply current
...
22
...
A variable resistor connected in parallel with the series-wound field to control speed is
called a diverter
...
Problem
29 below demonstrates this method
...
8 1 n3

1 n1

1 (10)

1 n1

from which,

Since resistor R is in series with the armature, it carries
the full armature current and results in a large power loss
in large motors where a considerable speed reduction is
required for long periods
...


Now

E3 = 500 − (60)(0
...
22
...
m
...
when Ia = 60 A,

Figure 22
...
C
...
22
...

This effectively increases the value of R in the
equation
n=k

E2 = V − Ia2 (Ra + R)

Hence e
...
f
...
62)(0
...
04)
= 300 − (100
...
14)
= 300 − 14
...
9 volts

V − IR
Now e
...
f
...
Since the additional
resistor carries the full supply current, a large
power loss is associated with large motors in
which a considerable speed reduction is required
for long periods
...


E1
=
E2

At 300 V, e
...
f
...
8Ia2 n2

286
...
9
(0
...
62)n2

Hence
and

Problem 29
...
The armature
resistance is 0
...
Determine the speed when developing
full load torque but with a 0
...
(Assume that the flux is
proportional to the field current)
...
9)(90)(15)
(286
...
8)(100
...
74 rev/s
Thus the speed of the motor has increased from 15 rev/s
(i
...
900 rev/min) to 16
...
e
...
2 diverter resistance in parallel with the
series winding
...
A series motor runs at 800 rev/min
when the voltage is 400 V and the current is 25 A
...
4 and the series field
resistance is 0
...
Determine the resistance to be
connected in series to reduce the speed to
600 rev/min with the same current
...
1 + 0
...
15)
= 300 − 13
...
5 volts
With the 0
...
22
...
m
...
,

= 400 − (25)(0
...
2)

(0
...
05)
(0
...
05)
=
= 0
...
2 + 0
...
25

= 400 − (25)(0
...
22
...
2
I = 0
...
2 + 0
...
8 Ia2 then (90)(90) = (Ia2 )(0
...

∝ Ia1 and
1

902
0
...
62 A
0
...

Thus E ∝ n or E ∝ n and
E1
n1
=
E2
n2

from which,

2
Ia2 =

Hence

and

Ia2

from which,

385
800
=
E2
600
E2 =

(385)(600)
= 288
...
22
...
75 = 400 − 25(0
...
2 + R)

Rearranging gives:
0
...
75
= 4
...
45 − 0
...
e
...
85
...
85 has
reduced the speed from 800 rev/min to 600 rev/min
...
c
...
A 350 V shunt motor runs at its normal speed
of 12 rev/s when the armature current is 90 A
...
3
...
4 is connected in
series with the armature, the shunt field
remaining constant
(b) Find the speed when the current is 45 A
and the shunt field is reduced to 75 per
cent of its normal value by increasing
resistance in the field circuit
...
83 rev/s (b) 16
...

The most common type of protection is the screenprotected type, where ventilation is achieved by fitting
a fan internally, with the openings at the end of the motor
fitted with wire mesh
...

A flame-proof type is usually cooled by the conduction of heat through the motor casing
...


Now try the following exercises
Exercise 137 Short answer questions
on d
...
machines
1
...
converts mechanical energy into
electrical energy
2
...
converts electrical energy into
mechanical energy
3
...
Poor commutation may cause sparking
...
State any five basic parts of a d
...
machine

3
...
The armature resistance is 0
...
05
...
15
diverter in parallel with the field winding
...
A series motor runs at 900 rev/min when the
voltage is 420 V and the current is 40 A
...
3 and the series field
resistance is 0
...
Calculate the resistance to
be connected in series to reduce the speed to
720 rev/min with the same current
...
The e
...
f
...
State what
p, , n, Z and c represent
...
16 Motor cooling
Motors are often classified according to the type of
enclosure used, the type depending on the conditions

6
...
What is armature reaction? How can it be
overcome?

9
...
c
...
with the armature circuit
10
...
c
...

11
...
c
...
It is called a
...
C
...
Sketch a typical open-circuit characteristic
for (a) a separately excited generator (b) a
shunt generator (c) a series generator
13
...
State one application for (a) a shunt generator (b) a series generator (c) a compound
generator
15
...
c
...
The efficiency of a d
...
machine is given by
the ratio (
...
The torque produced by a d
...
motor is
proportional to
...
A starter is necessary for a d
...
motor because
the generated e
...
f
...
at low speeds
30
...
c
...
if the value of resistance of the shunt
field regulator is increased
31
...
c
...
if the
value of resistance in the armature circuit is
increased
32
...
c
...
as the value of the armature
current increases

17
...
m
...
, E,
terminal voltage, armature current and armature resistance for a d
...
motor is E =
...
At a large value of torque, the speed of a d
...

series-wound motor is
...
The torque T of a d
...
motor is given by
T = p ZIa /πc newton metres
...
At a large value of field current, the generated e
...
f
...
c
...


19
...
In a d
...
machine
(a) generated e
...
f
...


35
...


21
...
State two applications for each of the following motors:
(a) shunt
(b) series
(c) compound
In questions 23 to 26, an electrical machine
runs at n rev/s, has a shaft torque of T , and
takes a current of I from a supply voltage V
23
...
watts
24
...
watts
25
...

watts
26
...
watts
27
...
m
...
of a d
...
volts

36
...
c
...
Explain briefly why these two distinct methods are used and why the field
current plays a significant part in controlling
the speed of a d
...
motor
...
Name three types of motor enclosure

Exercise 138

Multi-choice questions on
d
...
machines
(Answers on page 399)

1
...
c
...
c
...
c
...
c
...
Sketch typical characteristics of torque/armature current for
(a) a shunt motor
(b) a series motor
(c) a compound motor

376 Electrical and Electronic Principles and Technology
A shunt-wound d
...
machine is running at
n rev/s and has a shaft torque of T Nm
...
c
...
The armature
resistance of the machine is Ra ohms, the
armature current is Ia A and the generated
voltage is E volts
...
The input power when running as a generator
3
...
The input power when running as a motor
5
...
The generated voltage when running as a
motor
7
...
The generated voltage when running as a
generator
9
...
Which of the following statements is false?
(a) A commutator is necessary as part of a
d
...
motor to keep the armature rotating
in the same direction
(b) A commutator is necessary as part of a
d
...
generator to produce unidirectional
voltage at the terminals of the generator
(c) The field winding of a d
...
machine is
housed in slots on the armature
(d) The brushes of a d
...
machine are usually made of carbon and do not rotate
with the armature
11
...
c
...
m
...

(a) remains the same (b) is doubled (c) is
halved
12
...
c
...
If the flux per pole of a shunt-wound d
...

generator is halved, the generated e
...
f
...
In a series-wound generator running at constant speed, as the load current increases, the
terminal voltage
(a) increases
(b) decreases
(c) stays the same
15
...
c
...
Which of the following statements is false?
(a) A series-wound motor has a large starting torque
(b) A shunt-wound motor must be permanently connected to its load
(c) The speed of a series-wound motor
drops considerably when load is applied
(d) A shunt-wound motor is essentially a
constant-speed machine
17
...
c
...
The armature resistance of a d
...
motor is
0
...
m
...
is 196 V at full speed
...
In d
...
generators iron losses are made up of:
(a) hysteresis and friction losses
(b) hysteresis, eddy current and brush contact losses
(c) hysteresis and eddy current losses

D
...
machines 377
(d) hysteresis, eddy current and copper
losses
20
...
The supply voltage to a d
...
motor is 240 V
...
m
...
is 230 V and the armature
resistance is 0
...
With a d
...
motor, the starter resistor:
(a) limits the armature current to a safe
starting value
(b) controls the speed of the machine
(c) prevents the field current flowing
through and damaging the armature
(d) limits the field current to a safe starting
value

S
R

Terminal
voltage

Q
P

0

Load current

Figure 22
...
A commutator is a device fitted to a generator
...
c
...
c
...
From Fig
...
31, the expected characteristic
for a shunt-wound d
...
generator is:
(a) P
(b) Q
(c) R
(d) S

Chapter 23

Three-phase induction
motors
At the end of this chapter you should be able to:
• appreciate the merits of three-phase induction motors
• understand how a rotating magnetic field is produced
• state the synchronous speed, ns = ( f /p) and use in calculations
• describe the principle of operation of a three-phase induction motor
• distinguish between squirrel-cage and wound-rotor types of motor
• understand how a torque is produced causing rotor movement
• understand and calculate slip
• derive expressions for rotor e
...
f
...
1

Introduction

In d
...
motors, introduced in Chapter 22, conductors on
a rotating armature pass through a stationary magnetic
field
...

Its name is derived from the fact that the current in the
rotor is induced by the magnetic field instead of being
supplied through electrical connections to the supply
...
The principal disadvantage of a three-phase
induction motor is that its speed cannot be readily
adjusted
...
2

Production of a rotating
magnetic field

When a three-phase supply is connected to symmetrical
three-phase windings, the currents flowing in the windings produce a magnetic field
...

With reference to Fig
...
1, the windings are represented by three single-loop conductors, one for each
phase, marked RS RF , YS YF and BS BF , the S and F
signifying start and finish
...


Figure 23
...
23
...
If the value of
current in a winding is positive, the assumption is made
that it flows from start to finish of the winding, i
...
if it
is the red phase, current flows from RS to RF , i
...
away
from the viewer in RS and towards the viewer in RF
...
e
...
At time, say t1 , shown in Fig
...
1(a), the
current flowing in the red phase is a maximum positive
value
...
5 times the maximum
value and are negative
...
23
...
The resulting magnetic field is as shown, due to the ‘solenoid’ action and
application of the corkscrew rule
...
87 times its maximum
value and is positive, the current in the yellow phase is
zero and the current in the blue phase is about 0
...
Hence the currents
and resultant magnetic field are as shown in Fig
...
1(c)
...
5 of their maximum values and the current in the blue
phase is a maximum negative value
...
23
...

Similar diagrams to Fig
...
1(b), (c) and (d) can be
produced for all time values and these would show that
the magnetic field travels through one revolution for
each cycle of the supply voltage applied to the stator
windings
...
The three coils shown in
Fig
...
2(a), are connected in star to a three-phase supply
...
The directions of φA , φB and φC do not alter, but
their magnitudes are proportional to the currents flowing in the coils at any particular time
...
23
...
e
...
e
...

These currents give rise to the magnetic fluxes φA , φB
and φC , whose magnitudes and directions are as shown

Section 3

Three-phase induction motors 379

380 Electrical and Electronic Principles and Technology
The magnetic fluxes and the resultant magnetic flux
are as shown in Fig
...
2(e)
Inspection of Fig
...
2(c), (d) and (e) shows that the
magnitude of the resultant magnetic flux, , in each
1
case is constant and is 1 2 × the maximum value of φA ,
φB or φC , but that its direction is changing
...


23
...
23
...
For this reason, it is called a
2-pole system and an induction motor using three phase
windings only is called a 2-pole induction motor
...
23
...
2

in Fig
...
2(c)
...
23
...
At time t2 ,
the currents flowing are:
iB , 0
...
866 × maximum negative value
...
23
...

At time t3 ,
iB is 0
...
5 × maximum value and is positive
...
3

Three-phase induction motors 381

f
ns = rev/s
p
Problem 1
...
Determine
the synchronous speed of the motor in rev/min
...
Since
the motor is connected to a 50 hertz supply, f = 50
...
Thus, synchronous
speed, ns = (50/1) = 50 rev/s = 50 × 60 rev/min
= 3000 rev/min
...
A stator winding supplied from a
three-phase 60 Hz system is required to produce a
magnetic flux rotating at 900 rev/min
...

Synchronous speed,
ns = 900 rev/min =

900
rev/s = 15 rev/s
60

Since
ns =

f
p

then p =

f
ns

=

60
15

=4

Hence the number of pole pairs is 4 and thus the
number of poles is 8

Problem 3
...
Calculate
the frequency of the supply voltage
...
The synchronous speed of a 3-phase, 4-pole
induction motor is 60 rev/s
...

[120 Hz]
2
...
Calculate the equivalent number of pairs of poles of the motor
...
A 6-pole, 3-phase induction motor is connected to a 300 Hz supply
...

[100 rev/s]

23
...
c
...

It is wound to give a 2-pole, 4-pole, 6-pole,
...
The rotor, corresponding to the armature of a
d
...
machine, is built up of laminated iron, to reduce
eddy currents
...
23
...
A cross-sectional view of a three-phase
induction motor is shown in Fig
...
4(b)
...
The current distribution in the
stator windings are shown in Fig
...
3(a), for the time
t shown in Fig
...
3(b)
...

This is called a 4-pole system and an induction motor
using six phase windings is called a 4-pole induction
motor
...

In general, if f is the frequency of the currents in the
stator windings and the stator is wound to be equivalent to p pairs of poles, the speed of revolution of the
rotating magnetic field, i
...
the synchronous speed, ns
is given by:

382 Electrical and Electronic Principles and Technology

Figure 23
...
6 Slip

Figure 23
...
If the slots are skewed, better starting
and quieter running is achieved
...
The squirrel-cage motor
is cheap, reliable and efficient
...
With this type there are phase windings in slots, similar to those in the stator
...
The slip rings are used to add
external resistance to the rotor circuit, particularly for
starting (see Section 23
...

The principle of operation is the same for both the
squirrel cage and the wound rotor machines
...
5

Principle of operation of a
three-phase induction motor

When a three-phase supply is connected to the stator
windings, a rotating magnetic field is produced
...
m
...
is induced
in it and since it is joined, via the end conducting rings,
to another bar one pole pitch away, a current flows in
the bars
...
23
...
Similar forces are applied to all the conductors on the rotor, so that a torque is produced causing
the rotor to rotate
...
As
the rotor speed increases, the rate at which the rotating
magnetic field cuts the rotor bars is less and the frequency of the induced e
...
f
...

If the rotor runs at the same speed as the rotating magnetic field, no e
...
f
...

Thus the rotor slows down
...

When there is no load on the rotor, the resistive forces
due to windage and bearing friction are small and the
rotor runs very nearly at synchronous speed
...
m
...
’s in the rotor bars
and hence the rotor current, force and torque increase
...
e
...
Thus
slip, s =

ns − n r
ns

× 100%

Typical values of slip between no load and full load
are about 4 to 5 per cent for small motors and 1
...

Problem 4
...
The
rotor runs at 1455 rev/min at full load
...

(a) The number of pairs of poles, p = (4/2) = 2
...
The synchronous
speed, ns = ( f /p) = (50/2) = 25 rev/s
...
25
25

× 100%

= 3%
Problem 5
...
If the slip is 2 per cent at a certain load,
determine (a) the synchronous speed, (b) the speed
of the rotor, and (c) the frequency of the induced
e
...
f
...

(a) f = 60 Hz and p = (2/2) = 1
...

(b) Since slip,
s=

ns − nr
ns

× 100%

2=

60 − nr
60

× 100

Hence
2 × 60
= 60 − nr
100
i
...

nr = 60 −

2 × 60
= 58
...
e
...
8 × 60 = 3528 rev/min
(c) Since the synchronous speed is 60 rev/s and that
of the rotor is 58
...
8) = 1
...

Thus the frequency of the e
...
f
...
2)( 2 ) = 1
...

2
Problem 6
...
Determine
the synchronous speed
...
04 =

ns − 20
ns

from which, ns (0
...
04 ns = ns (1 − 0
...

Hence synchronous speed,
20
˙
= 20
...
04
˙
= (20
...
A 6-pole, 3-phase induction motor runs at
970 rev/min at a certain load
...

[3%]
2
...

If the full load slip is 2
...
m
...
’s
[(a) 750 rev/min (b) 731 rev/min (c) 1
...
A three-phase induction motor is supplied
from a 60 Hz supply and runs at 1710 rev/min
when the slip is 5 per cent
...

[1800 rev/min]
4
...
Calculate
(a) the synchronous speed,
(b) the slip and
(c) the frequency of the rotor induced e
...
f
...
7 Rotor e
...
f
...
m
...


× 100%

Rotor speed, nr = (1200/60) = 20 rev/s and s = 4
...
23
...
25 rev/s
...
5 rev/s
or (12
...
06 =

ns − nr
ns
12
...
5

(0
...
5) = 12
...
6

and rotor speed,
nr = 12
...
06)(12
...
m
...
at standstill is given by
E2 =

N2
N1

= 11
...

When an induction motor is running, the induced
e
...
f
...
The
induced e
...
f
...
Hence when
running, rotor e
...
f
...
e
...
m
...
per phase = s

N2
N1

E1

(2)

Rotor frequency
The rotor e
...
f
...
Thus the frequency of the rotor e
...
f
...
A 12-pole, 3-phase, 50 Hz induction motor
runs at 475 rev/min
...
5 Hz]
2
...
Determine
(a) the slip, and
(b) the rotor speed, in rev/min
[(a) 0
...
8 Rotor impedance and current

However (ns − nr )/ns is the slip s and (ns p) is the
supply frequency f , hence
fr = sf

(3)

Problem 7
...
Determine (a) the slip, and
(b) the rotor speed
...


Rotor reactance
Rotor reactance varies with the frequency of the rotor
current
...

When running, reactance per phase,
Xr = 2πfr L

(a) From Equation (3), fr = sf
...
06 or 6%
50

= 2π(s f )L

from equation (3)

= s(2π f L)
i
...


X r = sX 2

(4)

Three-phase induction motors 385
Figure 23
...


23
...
If P2 is the power
input to the rotor from the rotating field, and Pm is the
mechanical power output (including friction losses)

from which,
Figure 23
...
e
...
Hence

At standstill, slip s = 1, then
Z2 =

2
2
R2 + X2

(6)

slip, s =

rotor copper loss
I 2 R2
= r
rotor input
P2

(9)

or power input to the rotor,

Rotor current
From Fig
...
6 and 23
...
10 Induction motor losses and
efficiency

and when running, current,
Er
=
Zr

Figure 23
...

Motor efficiency,

E1

R2 + (sX2 )2
2

(8)

η=

output power
Pm
=
× 100%
input power
P1

Section 3

Ir =

N2
s
N1

Figure 23
...
The power supplied to a three-phase
induction motor is 32 kW and the stator losses are
1200 W
...

(a)

Input power to rotor = stator input power
− stator losses

(a)

n s − nr
ns

Slip, s =

× 100%

ns − 0
...
65)(100) = 65%
Input power to rotor = 30
...
2 kW

=

= 30
...
8

i
...


from which, rotor copper loss = (0
...
8)
= 1
...
8 − 1
...
26 kW
(c) Output power of motor

(b) Power developed by rotor
= input power to rotor
−rotor copper loss
= 30
...
02 = 10
...
78 − 0
...
03 kW
Efficiency,

= power developed by the rotor
− friction and windage losses

η=

= 29
...
75 = 28
...
51
32

× 100%

× 100%

= 89
...
8)
100

= 20
...
The speed of the induction motor of
Problem 8 is reduced to 35 per cent of its
synchronous speed by using external rotor
resistance
...


output power
input power
10
...
34%
Now try the following exercise
Exercise 142 Further problems on losses
and efficiency
1
...

If the slip is 4 per cent, determine
(a) the rotor copper loss,
(b) the total mechanical power developed by
the rotor,

Three-phase induction motors 387
(c) the output power of the motor if friction
and windage losses are 1 kW, and
(d) the efficiency of the motor, neglecting
rotor iron losses
...
92 kW (b) 46
...
08 kW
(d) 90
...
By using external rotor resistance, the speed of
the induction motor in Problem 1 is reduced
to 40 per cent of its synchronous speed
...

[(a) 28
...
40%]

If there are m phases then torque,
⎞
⎛
N2 2 2
E1 R2 ⎟
⎜s N
m
⎟
⎜
1
T=
⎟
⎜ 2
2πns ⎝ R2 + (sX2 )2 ⎠
i
...

⎛

2

N2
⎜m N
⎜
1
T =⎜
⎝ 2πns

=k

⎞
⎟
sE2 R2
⎟
1
⎟
⎠ R2 + (sX 2 )2
2

(11)

2
sE1 R2
2
R2 + (sX2 )2

where k is a constant for a particular machine, i
...

2
sE1 R2

torque, T∝

23
...
e
...
e
...
Thus maximum torque occurs when
rotor resistance and rotor reactance are equal, i
...
when
R2 = Xr
Problems 10 to 13 following illustrate some of the
characteristics of three-phase induction motors
...
A 415 V, three-phase, 50 Hz, 4 pole,
star-connected induction motor runs at 24 rev/s on
full load
...
35 and 3
...
85:1
...


(a) Synchronous speed,
ns = ( f /p) = (50/2) =
25 rev/s or (25 × 60) = 1500 rev/min
(b) Slip, s =

ns − nr
ns

=

25 − 24
= 0
...
1(239
...
35
0
...
352

= (0
...
29
0
...
18 Nm

(f) For maximum torque, slip s = 0
...
e
...
85)2
2π(25)

from part (c)

= (0
...
6 volts
3

=

2
sE1 R2
2
R2 + (sX2 )2

Tm = (0
...
1 =

(c) Phase voltage,

Full load torque,
⎛
N2
⎜m N
⎜
1
T =⎜
⎝ 2πns

Hence maximum torque,

(0
...
6)2 (0
...
35)2 + (0
...
5)2

803
...
01380)
0
...
1)(25) = 25 − nr and
nr = 25 − (0
...
5 = 22
...
e
...
Hence,
⎛

N2
⎜m N
⎜
1
starting torque = ⎜
⎝ 2πns

2

⎞
⎟
2
E1 R2
⎟
⎟
2
2
⎠ R2 + X2

= 78
...
05)
= 11 770 watts
Hence, power output = Pm − mechanical losses
= 11 770 − 770
= 11 000 W

from Equation (11) with s = 1
= (0
...
6)2 0
...
352 + 3
...
01380)

20 092
...
3725

i
...
starting torque = 22
...
05 Nm but the starting torque is only 22
...
35
Slip,

s=

R2
0
...
1
X2
3
...
Determine for the induction motor
in Problem 10 at full load, (a) the rotor current,
(b) the rotor copper loss, and (c) the starting current
...
04)(0
...
6)

=
=

√
(c) Power input, P1 = 3 VL IL cos φ (see Chapter 20) and cos φ = p
...
= 0
...
91 × 1000
= √
= 20
...
87
P1

0
...
04 × 3
...
1464
= 21
...
37696

(b) Rotor copper
loss per phase = Ir2 R2

Problem 13
...


= (21
...
35)

= 490
...
85)(239
...
90 A
2
2
0
...
52
R2 + X2

(Note that the starting current of 57
...
61 A)
Problem 12
...
87 lagging
...
770 kW from part (d),
Problem 10
...
35 W =
0
...

Stator input power,
P1 = Pm + rotor copper loss + rotor stator loss
= 11
...
49035 + 0
...
91 kW
(b) Net power output = 11 kW from part (d), Problem 10
...
91

= 85
...
At the
moment of starting, slip, s = 1
...
5
...
5 − 0
...
15 per phase could be added to
the rotor resistance to give maximum torque at starting
(see Section 23
...

Now try the following exercise
Exercise 143 Further problems on the
torque equation
1
...
5 rev/s
on full load
...
4 and 4
...
8:1
...

[(a) 50 rev/s or 3000 rev/min (b) 0
...
43 Nm (d) 6
...
74 Nm
(f) 45 rev/s or 2700 rev/min (g) 8
...
45 W
Total copper loss (for 3 phases)
= 3 × 163
...
For the induction motor in Problem 1, calculate
at full load
(a) the rotor current,
(b) the rotor copper loss, and
(c) the starting current
...
27 A (b) 211
...
96 A]
3
...
84
[(a) 7
...
75% (c) 13
...
For the induction motor in Problem 1, determine the resistance of the rotor winding
required for maximum starting torque
[4
...
35 and X2 = 3
...

Curve P in Fig
...
9 is a typical characteristic for an
induction motor
...
The normal operating conditions are
between 0 and X, thus it can be seen that for normal operation the speed variation with load is quite small — the
induction motor is an almost constant-speed machine
...
23
...
c
...


23
...
Also, from Problem 10, parts (e) and (f), it is
seen that the speed at which maximum torque occurs
is determined by the value of the rotor resistance
...
From
these observations, the torque-speed and torque-slip
characteristics of an induction motor are as shown in
Fig
...
9
...
9

Figure 23
...
23
...
However, as can be seen, the motor
has a full load slip of over 30 per cent, which results
in a drop in efficiency
...
13 Starting methods for
induction motors
Squirrel-cage rotor
(i) Direct-on-line starting
With this method, starting current is high and
may cause interference with supplies to other
consumers
...
However, the starting
torque is seriously reduced (see Equation (12)), so
the voltage is reduced only sufficiently to give the
required reduction of the starting current
...
23
...
A doublethrow switch connects the auto transformer in
circuit for starting, and when the motor is up

Figure 23
...

(iii) Star-delta starting
With this method, for starting, the connections
to the stator phase winding are star-connected,
so that the voltage across each phase winding
√
is (1/ 3) (i
...
0
...
For
running, the windings are switched to deltaconnection
...
23
...


Wound rotor
When starting on load is necessary, a wound rotor induction motor must be used
...
A face-plate type starter is used, and as the
resistance is gradually reduced, the machine characteristics at each stage will be similar to Q, S, R and P of
Fig
...
13
...
23
...
For very large induction motors,
very gradual and smooth starting is achieved by a liquid
type resistance
...
14

Advantages of squirrel-cage
induction motors

The advantages of squirrel-cage motors compared with
the wound rotor type are that they:
(i) are cheaper and more robust
(ii) have slightly higher efficiency and power factor
(iii) are explosion-proof, since the risk of sparking
is eliminated by the absence of slip rings and
brushes
...
Curves R and S
of Fig
...
9 are characteristics for values of rotor resistance’s between those of P and Q
...

A squirrel-cage induction motor would normally
follow characteristic P
...
However it has a poor starting torque
and must be started off-load or very lightly loaded (see
Section 23
...
Also, on starting, the current can
be four or five times the normal full load current, due
to the motor acting like a transformer with secondary
short circuited
...

A wound-rotor induction motor would follow characteristic P when the slip-rings are short-circuited,
which is the normal running condition
...
23
...

In general, for three-phase induction motors, the
power factor is usually between about 0
...
9 lagging, and the full load efficiency is usually about 80–90
per cent
...
Any voltage
variations therefore would seriously affect the induction
motor performance
...
12

Figure 23
...
15 Advantages of wound rotor
induction motors

Section 3

The advantages of the wound rotor motor compared with
the cage type are that they:
(i) have a much higher starting torque
(ii) have a much lower starting current
(iii) have a means of varying speed by use of external
rotor resistance
...
16

Double cage induction motor

The advantages of squirrel-cage and wound rotor induction motors are combined in the double cage induction
motor
...
The outer cage has high resistance conductors so
that maximum torque is achieved at or near starting
...
The torque-speed characteristic of the

Three-phase induction motors 393
inner cage is that of a normal induction motor, as shown
in Fig
...
14
...
The combined characteristic of inner and outer
cages is shown in Fig
...
14 The double cage induction
motor is highly efficient when running
...
Explain briefly how slip-frequency currents
are set up in the rotor bars of a 3-phase induction motor and why this frequency varies with
load
5
...
Define the slip of an
induction motor and explain why its value
depends on the load on the rotor
...
Write down the two properties of the magnetic field produced by the stator of a threephase induction motor
7
...
speed
8
...
proportional to
supply frequency

Figure 23
...
17 Uses of three-phase induction
motors
Three-phase induction motors are widely used in industry and constitute almost all industrial drives where a
nearly constant speed is required, from small workshops
to the largest industrial enterprises
...
The squirrel cage rotor type is the most
widely used of all a
...
motors
...
The synchronous speed of a three-phase
induction motor is
...
The type of rotor most widely used in a threephase induction motor is called a
...
The slip of a three-phase induction motor is

...

12
...
%
13
...

14
...

Rotor input power

15
...
Name three advantages that a three-phase
induction motor has when compared with a
d
...
motor
2
...
c
...
Explain briefly, with the aid of sketches, the
principle of operation of a 3-phase induction
motor

16
...


occurs

when

17
...
State two methods of starting squirrel-cage
induction motors
19
...
Describe briefly a double cage induction
motor

Section 3

Exercise 144 Short answer questions on
three-phase induction motors

394 Electrical and Electronic Principles and Technology
21
...
State two advantages of wound rotor
machines compared with cage rotor machines

4
...
The frequency
of the supply to the stator is:
(a) 50 Hz
(b) 100 Hz
(c) 25 Hz
(d) 12
...
Name any three applications of three-phase
induction motors

Questions 5 and 6 refer to a three-phase
induction motor
...
Which of the following statements about a
three-phase squirrel-cage induction motor is
false?
(a) It has no external electrical connections
to its rotor
(b) A three-phase supply is connected to its
stator
(c) A magnetic flux which alternates is produced
(d) It is cheap, robust and requires little or
no skilled maintenance

Section 3

2
...
Which of the following statements is false
when referring to a three-phase induction
motor?
(a) The synchronous speed is half the supply frequency when it has four poles
(b) In a 2-pole machine, the synchronous
speed is equal to the supply frequency
(c) If the number of poles is increased, the
synchronous speed is reduced
(d) The synchronous speed is inversely proportional to the number of poles

5
...
m
...
’s
increases with load on the rotor
(d) The torque on the rotor is due to the
interaction of magnetic fields
6
...
In questions 7 to 10,
determine the correct answers for the quantities stated, selecting your answer from the
list given below:
(a) 12
...
The synchronous speed
8
...
The percentage slip
10
...
m
...
’s in the rotor
11
...
The slip speed of an induction motor depends
upon:
(a) armature current (b) supply voltage
(c) mechanical load (d) eddy currents
13
...
The slip speed of an induction motor:
(a) is zero until the rotor moves and then
rises slightly
(b) is 100 per cent until the rotor moves and
then decreases slightly
(c) is 100 per cent until the rotor moves and
then falls to a low value
(d) is zero until the rotor moves and then
rises to 100 per cent

16
...
The number of poles is:
(a) 2
(b) 8
(c) 6
(d) 4
17
...
The
rotor runs at 2880 rev/min at full load
...
17%
(b) 92%
(c) 4%
(d) 96%
18
...

The rotor speed is:
(a) 427
...
A four-pole induction motor when supplied
from a 50 Hz supply experiences a 5 per cent
slip
...
75 rev/s
(c) 26
...
875 rev/s

Revision Test 7
This revision test covers the material contained in Chapters 22 and 23
...

1
...
Determine the e
...
f
...

(6)
2
...
c
...
3 and is connected to a 200 V supply
...
m
...
generated when it is running (a) as
a generator giving 80 A (b) as a motor taking
80 A
...
A 15 kW shunt generator having an armature circuit
resistance of 1 and a field resistance of 160
generates a terminal voltage of 240 V at full-load
...

(6)
4
...
c
...
The useful flux per
pole is 40 mWb
...

(4)

Section 3

5
...
The
armature resistance is 0
...
Calculate the speed,

in rev/min when the current is 50 A and a resistance
of 0
...

(7)
6
...
The rotor
runs at 1155 rev/min at full load
...

(6)
7
...
If
the slip is 4 per cent determine (a) the rotor copper loss, (b) the total mechanical power developed
by the rotor, (c) the output power of the motor
if frictional and windage losses are 1
...

(9)
8
...
25 rev/s on full load
...
2
and 2 respectively and the effective rotor-stator
turns ratio is 0
...
Calculate (a) the synchronous speed, (b) the slip, and (c) the full load
torque
...
C
...
m
...
E =

√
3 Ip

Generator: E = V + Ia Ra

Two-wattmeter method
P = P1 + P2

E = 4
...
Exercise 4 (page 7)
1 (c)
2 (d)
3 (c)

4 (a)
5 (c)
6 (b)

7 (b)
8 (c)

Chapter 9
...
Exercise 15 (page 27)
1 (c)
2 (d)

3 (b)
4 (d)

5 (d)
6 (c)

7 (b)
8 (c)

4 (c)
5 (b)
6 (d)

7 (d)
8 (b)
9 (c)

4 (c)
5 (a)

12 (a)
13 (c)

8 (b)
9 (c)

10 (d)
11 (d)

11 (a)
12 (b)

7 (c)
8 (a)
9 (i)
10 ( j)
11 (g)
12 (c)

13 (b)
14 (p)
15 (d)
16 (o)
17 (n)
18 (b)

19 (d)
20 (a)
21 (d)
22 (c)
23 (a)

3 (d)
4 (c)

5 (b)
6 (b)

7 (c)
8 (d)

9 (a)
10 (b)

Chapter 12
...
Exercise 75 (page 202)

Chapter 6
...
Exercise 63 (page 152)

Chapter 5
...
Exercise 18 (page 38)
1 (d)
2 (a)
3 (b)

4 (b)
5 (c)
6 (a)

Chapter 10
...
Exercise 10 (page 19)
1 (b)
2 (b)
3 (c)

1 (c)
2 (b)
3 (c)

4 (c)
5 (a)

6 (b)
7 (b)

8 (a)
9 (c)

10 (c)
11 (d)

1 (d)
2 (c)
3 (b)
4 (c)

5 (a)
6 (d)
7 (c)

8 (a)
9 (c)
10 (c)

11 (b)
12 (d)
13 (d)

14 (b)
15 (c)
16 (a)

Chapter 14
...
Exercise 37 (page 82)
1 (d)
2 (b)
3 (b)
4 (c)

5 (c)
6 (d)
7 (a)

8 (c)
9 (c)
10 (c)

11 (a) and (d),
(b) and (f ),
(c) and (e)

12 (a)
13 (a)

4 (a)
5 (d)
6 (c)

7 (b)
8 (c)

9 (b)
10 (c)

11 (b)
12 (d)

Chapter 15
...
Exercise 41 (page 94)
1 (d)
2 (c)

1 (c)
2 (d)
3 (d)

3 (d)
4 (a)

5 (b)
6 (c)

7 (d)
8 (a)

9 (a)
10 (b)

1 (c)
2 (a)
3 (b)
4 (b)

5 (a)
6 (b)
7 (a)
8 (d)

9 (d)
10 (d)
11 (b)
12 (c)

13 (b)
14 (c)
15 (b)
16 (b)

17 (c)
18 (a)
19 (d)

Answers to multiple choice questions 399
Chapter 16
...
Exercise 129 (page 349)
12 (d)
13 (c)
14 (b)

Chapter 17
...
Exercise 110 (page 303)
1 (c)
2 (b)

3 (b)
4 (d)

5 (a)
6 (b)

7 (d)
8 (a)

5 (c)
6 (a)
7 (b)
8 (a)

9 (b)
10 (g)
11 (d)
12 (a)

13 (h)
14 (k)
15 ( j)
16 (f )

17 (c)
18 (b) and (c)
19 (c)
20 (b)

Chapter 22
...
Exercise 106 (page 287)
1 (c)
2 (b)
3 (b)
4 (g)

1 (a)
2 (d)
3 (a)
4 (b)

1 (b)
2 (e)
3 (e)
4 (c)
5 (c)

6 (a)
7 (d)
8 (f )
9 (b)
10 (c)

11 (b)
12 (a)
13 (b)
14 (a)
15 (d)

16 (b)
17 (b)
18 (b)
19 (c)
20 (b)

21 (b)
22 (a)
23 (c)
24 (d)

Chapter 23
...
Exercise 116 (page 325)
5 (f )
6 (a)
7 (g)
8 (l)

9 (l)
10 (d)
11 (f )

12 ( j)
13 (d)
14 (b)

15 (c)
16 (b)
17 (c)

Section 3

1 (g)
2 (c)
3 (a)
4 (a)

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Index
Absolute permeability, 74
permittivity, 58
potential, 51
A
...
bridges, 132
generator, 205
values, 207
Acceptor circuit, 234
Active power, 238, 243
Advantages of:
squirrel cage induction motor, 391
three-phase systems, 324
wound rotor induction motor, 392
Air capacitors, 67
Alkaline cell, 29
Alternating voltages and currents, 205
Alternative energy, 37
Aluminium, 142
Ammeter, 12, 112
Amplifier gain, 292, 294
Amplifier, transistor, 165
Amplitude, 119, 207
Analogue instruments, 111
to digital conversion, 301
Angular velocity, 211
Anode, 29, 147
Antimony, 142
Apparent power, 238, 243
Armature, 354
reaction, 354
Arsenic, 142
Asymmetrical network, 260
Atoms, 11, 141
Attenuation, 260
bands, 260
Attraction-type m
...
instrument, 111
Audio frequency transformer, 335
Auto transformer, 343
Avalanche breakdown, 148
effect, 145
Average value, 207
Back e
...
f
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
m
...
, 5, 30
equation of transformer, 331
in armature winding, 354
induced in conductors, 98
of a cell, 30
Emitter, 155
Energy, 6, 16
stored in:
capacitor, 66
inductor, 103
Equivalent circuit of transformer, 335
Farad, 57
Faraday’s laws, 97
Ferrite, 81
Field effect transistor, 163
amplifiers, 165
characteristics, 163
Filter, 218, 251
networks, 260
Fleming’s left hand rule, 89
Fleming’s right hand rule, 97
Fluke, 115
Force, 4
on a charge, 93
on a current-carrying conductor, 88
Form factor, 208
Formulae, lists of, 176, 306, 397
Forward bias, 144
characteristics, 145
Forward transconductance, 164
Frequency, 119, 206
Friction and windage losses, 361
Fuel cells, 36
Full wave rectification, 217
Fundamental, 126
Fuses, 17
Gallium arsenide, 142
Galvanometer, 130
Generator: 352
a
...
, 205
d
...
, 352
Geothermal energy, 37
hotspots, 37
Germanium, 140, 142
Grip rule, 87
Half-power points, 237
Half-wave rectification, 217
Harmonics, 126
Heating effects of current, 17
Henry, 101
Hertz, 206

High-pass filter, 264
Hole, 141, 142
Hydroelectricity, 37
Hydrogen cell, 37
Hysteresis, 81
loop, 81
loss, 81, 338
Impedance, 225, 229
triangle, 225, 230
Indium, 142
Indium arsenide, 142
Induced e
...
f
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
c
...
i
...
c
...
c
...
c
...
m
...
value, 119, 207
Rotation of loop in magnetic field, 100
Rotor copper loss, 385
Saturation flux density, 81
Scale, 111
Schottky diodes, 150
Screw rule, 86
Secondary cells, 29, 34
Selectivity, 237
Self-excited generators, 356
Self inductance, 101
Semiconductor diodes, 140, 147
Semiconductors, 140
Semiconductor materials, 141
Separately-excited generators, 356
Series:
a
...
circuits, 272
d
...
circuits, 41

406 Index
Series: (continued)
circuit, 41
connected capacitors, 62
lamps, 52
resonance, 234
winding, 354
wound generator, 358
motor, 367, 372
Shells, 11, 141
Shell type transformer, 335
Shunt, 112
field regulator, 371
winding, 354
wound generator, 357
motor, 365, 371
Siemen, 6
Silicon, 140, 142
Silicon controlled rectifiers, 149
Silver oxide battery, 29
Simple cell, 29
Sine wave, 206
Single-phase:
parallel a
...
circuit, 243
series a
...
circuit, 222
voltage, 311
Sinusoidal waveform equation, 211
S
...
units, 3
Slew rate, 292
Slip, 382
Smoothing of rectified waveform, 218
Solar energy, 37
Solar panels, 37
Solenoid, 86
Spectrum analysis, 126
Speed control of d
...
motors, 371
Squirrel-cage rotor induction motors, 381, 291
advantages of, 391
Star connection, 312
point, 312
Star/delta comparison, 314
Stator, 353
Steady state, 273
Stopbands, 260
Sub-multiples of units, 13
Sub-system, 10
Summing amplifier, 296, 300
Superposition theorem, 183
Switched-mode power supplies, 151
Switching inductive circuits, 285
Symbols, electrical, 10
Symmetrical network, 260
Synchronous speed, 379, 380
System, electrical, 9
T-network, 260
Tangent method, 274

Telephone receiver, 88
Temperature coefficient of resistance, 22
Tesla, 73
Thermal generation of electron-hole pairs, 143
Thévenin’s theorem, 188
Thévenin and Norton equivalent circuits, 197
Three-phase:
induction motor, 378
supply, 311
systems, 311
advantages of, 324
power, 317
transformers, 345
Thyristors, 149
Tidal power, 37
Time constant:
CR circuit, 273
LR circuit, 281
Titanium oxide capacitor, 68
Torque equation:
for induction motor, 387
of a d
...
machine, 363
Torque-speed characteristics of induction motor, 390
Transformation ratio, 328
Transformers, 327
auto, 343
construction, 335
current, 346
e
...
f
...
I

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