39
Electromagnetic induction
At the end of this chapter you should be able to: Figure 39.1(a) shows a coil of wire connected to a centre-
zero galvanometer, which is a sensitive ammeter with the
ž understand how an e.m.f. may be induced in a zero-current position in the centre of the scale.
conductor
(a) When the magnet is moved at constant speed towards
ž state Faraday’s laws of electromagnetic induction the coil (Fig. 39.1(a)), a deflection is noted on the gal-
vanometer showing that a current has been produced in
ž state Lenz’s law the coil.
ž use Fleming’s right-hand rule for relative directions
(b) When the magnet is moved at the same speed as in
ž appreciate that the induced e.m.f., E D Blv or (a) but away from the coil the same deflection is noted
E D Blv sin q but is in the opposite direction (see Fig. 39.1(b)).
ž calculate induced e.m.f. given B, l, v and q and (c) When the magnet is held stationary, even within the coil,
determine relative directions no deflection is recorded.
ž define inductance L and state its unit (d) When the coil is moved at the same speed as in (a) and
the magnet held stationary the same galvanometer deflec-
ž define mutual inductance M dI1 tion is noted.
dt
ž calculate mutual inductance using E2 D (e) When the relative speed is, say, doubled, the galvanome-
ter deflection is doubled.
ž describe the principle of operation of a transformer
(f) When a stronger magnet is used, a greater galvanometer
ž perform calculations using V1 D N1 D I2 for an deflection is noted.
V2 N2 I1
(g) When the number of turns of wire of the coil is increased,
ideal transformer a greater galvanometer deflection is noted.
ž define the ‘rating’ of a transformer
39.1 Introduction to electromagnetic SN SN
induction
Direction of Galvanometer
When a conductor is moved across a magnetic field so as movement (a)
to cut through the lines of force (or flux), an electromotive
force (e.m.f.) is produced in the conductor. If the conductor (b)
forms part of a closed circuit then the e.m.f. produced causes
an electric current to flow round the circuit. Hence an e.m.f. SN (c)
(and thus current) is ‘induced’ in the conductor as a result of
its movement across the magnetic field. This effect is known Fig. 39.1
as ‘electromagnetic induction’.
288 Science for Engineering
Figure 39.1(c) shows the magnetic field associated with the Summarising:
magnet. As the magnet is moved towards the coil, the
magnetic flux of the magnet moves across, or cuts, the coil. First finger Field
It is the relative movement of the magnetic flux and
the coil that causes an e.m.f. and thus current, to be Thumb Motion
induced in the coil. This effect is known as electromagnetic
induction. The laws of electromagnetic induction stated SEcond finger E.m.f.
in Section 39.2 evolved from experiments such as those
described above. In a generator, conductors forming an electric circuit are
made to move through a magnetic field. By Faraday’s law
39.2 Laws of electromagnetic induction an e.m.f. is induced in the conductors and thus a source of
e.m.f. is created. A generator converts mechanical energy
Faraday’s laws of electromagnetic induction state: into electrical energy. (The action of a simple a.c. generator
is described in Chapter 40)
(i) An induced e.m.f. is set up whenever the magnetic field
linking that circuit changes. The induced e.m.f. E set up between the ends of the
conductor shown in Fig. 39.3 is given by:
(ii) The magnitude of the induced e.m.f. in any circuit is
proportional to the rate of change of the magnetic flux E = Bl v volts
linking the circuit.
Magnetic N Conductor
Lenz’s law states: flux density l
v
The direction of an induced e.m.f. is always such that it B S
tends to set up a current opposing the motion or the change
of flux responsible for inducing that e.m.f. Fig. 39.3
An alternative method to Lenz’s law of determining relative where B, the flux density, is measured in teslas, l, the length
directions is given by Fleming’s right-hand rule (often of conductor in the magnetic field, is measured in metres,
called the generator rule) which states: and v, the conductor velocity, is measured in metres per
second.
Let the thumb, first finger and second finger of the right
hand be extended such that they are all at right angles If the conductor moves at an angle q° to the magnetic field
to each other (as shown in Fig. 39.2). If the first finger (instead of at 90° as assumed above) then
points in the direction of the magnetic field and the thumb
points in the direction of motion of the conductor relative E = Bl v sin q volts
to the magnetic field, then the second finger will point in
the direction of the induced e.m.f. Problem 1. A conductor 300 mm long moves at a
uniform speed of 4 m/s at right-angles to a uniform
Motion magnetic field of flux density 1.25 T. Determine the
current flowing in the conductor when (a) its ends are
Magnetic field open-circuited, (b) its ends are connected to a load of
20 resistance.
Induced e.m.f When a conductor moves in a magnetic field it will have an
e.m.f. induced in it but this e.m.f. can only produce a current
Fig. 39.2 if there is a closed circuit.
300
Induced e.m.f. E D Blv D ⊲1.25⊳ 1000 ⊲4⊳ D 1.5 V
Electromagnetic induction 289
(a) If the ends of the conductor are open circuited no current km m 1 h
will flow even though 1.5 V has been induced. and v D 400 h ð 1000 km ð 60 ð 60 s
(b) From Ohm’s law, current, ⊲400⊳⊲1000⊳ 4000
D 3600 D 36 m/s
E 1.5
I D R D 20 D 0.075 A or 75 mA Hence
Problem 2. At what velocity must a conductor 75 mm E D Blv D ⊲40 ð 10 6⊳⊲36⊳ 4000 D 0.16 V
long cut a magnetic field of flux density 0.6 T if an e.m.f. 36
of 9 V is to be induced in it? Assume the conductor,
the field and the direction of motion are mutually Problem 5. The diagram shown in Fig. 39.4 repre-
perpendicular. sents the generation of e.m.f’s. Determine (i) the direc-
tion in which the conductor has to be moved in
Induced e.m.f. E D Blv, hence velocity v D E Fig. 39.4(a), (ii) the direction of the induced e.m.f. in
Bl Fig. 39.4(b), (iii) the polarity of the magnetic system in
Fig. 39.4(c).
Hence velocity,
N
9 SN
v D ⊲0.6⊳⊲75 ð 10 3⊳
9 ð 103
D 0.6 ð 75 D 200 m=s
Problem 3. A conductor moves with a velocity of S (b)
15 m/s at an angle of (a) 90° (b) 60° and (c) 30° to
a magnetic field produced between two square-faced (a) (c)
poles of side length 2 cm. If the flux leaving a pole face
is 5 µWb, find the magnitude of the induced e.m.f. in Fig. 39.4
each case.
v D 15 m/s, length of conductor in magnetic field, The direction of the e.m.f., and thus the current due to the
l D 2 cm D 0.02 m, A D ⊲2 ð 2⊳ cm2 D 4 ð 10 4 m2 and e.m.f. may be obtained by either Lenz’s law or Fleming’s
D 5 ð 10 6 Wb Right-hand rule (i.e. GeneRator rule).
(i) Using Lenz’s law: The field due to the magnet and the
(a) E90 D Blv sin 90° D A lv sin 90° field due to the current-carrying conductor are shown
in Fig. 39.5(a) and are seen to reinforce to the left of
D 5 ð 10 6 ⊲0.02⊳⊲15⊳⊲1⊳ D 3.75 mV the conductor. Hence the force on the conductor is to
4 ð 10 4 the right. However Lenz’s law states that the direction
of the induced e.m.f. is always such as to oppose the
(b) E60 D Blv sin 60° D E90 sin 60° effect producing it. Thus the conductor will have to
be moved to the left.
D 3.75 sin 60° D 3.25 mV
(ii) Using Fleming’s right-hand rule:
(c) E30 D Blv sin 30° D E90 sin 30°
D 3.75 sin 30° D 1.875 mV First finger Field, i.e. N ! S, or right to left;
Problem 4. The wing span of a metal aeroplane is ThuMb Motion, i.e. upwards;
36 m. If the aeroplane is flying at 400 km/h, determine
the e.m.f. induced between its wing tips. Assume the SEcond finger E.m.f.
vertical component of the earth’s magnetic field is
40 µT. i.e. towards the viewer or out of the paper, as shown
in Fig. 39.5(b).
Induced e.m.f. across wing tips, E D Blv
B D 40 µT D 40 ð 10 6 T, l D 36 m (iii) The polarity of the magnetic system of Fig. 39.4(c) is
shown in Fig. 39.5(c) and is obtained using Fleming’s
right-hand rule.
290 Science for Engineering
Force on N S
conductor N
Motion of
conductor
SN
(b)
S
(a) (c)
Fig. 39.5
Now try the following exercise 39.3 Self inductance
Exercise 172 Further problems on induced e.m.f. Inductance is the name given to the property of a circuit
(Answers on page 357) whereby there is an e.m.f. induced into the circuit by the
change of flux linkages produced by a current change, as
1. A conductor of length 15 cm is moved at 750 mm/s stated in Chapter 41, page 310.
at right-angles to a uniform flux density of 1.2 T.
Determine the e.m.f. induced in the conductor. When the e.m.f. is induced in the same circuit as that in
which the current is changing, the property is called self
2. Find the speed that a conductor of length 120 mm inductance, L.
must be moved at right angles to a magnetic field
of flux density 0.6 T to induce in it an e.m.f. of The unit of inductance is the henry, H.
1.8 V
39.4 Mutual inductance
3. A 25 cm long conductor moves at a uniform speed of
8 m/s through a uniform magnetic field of flux density When an e.m.f. is induced in a circuit by a change of flux
1.2 T. Determine the current flowing in the conductor due to current changing in an adjacent circuit, the property
when (a) its ends are open-circuited, (b) its ends are is called mutual inductance, M.
connected to a load of 15 ohms resistance.
Mutually induced e.m.f. in the second coil,
4. A straight conductor 500 mm long is moved with
constant velocity at right angles both to its length E2 = −M dI1 volts
and to a uniform magnetic field. Given that the e.m.f. dt
induced in the conductor is 2.5 V and the velocity
is 5 m/s, calculate the flux density of the magnetic where M is the mutual inductance between two coils, in
field. If the conductor forms part of a closed circuit henrys, and dI1 is the rate of change of current in the first
of total resistance 5 ohms, calculate the force on the
conductor. dt
coil.
5. A car is travelling at 80 km/h. Assuming the back axle
of the car is 1.76 m in length and the vertical compo- Problem 6. Calculate the mutual inductance between
nent of the earth’s magnetic field is 40 µT, find the two coils when a current changing at 200 A/s in one
e.m.f. generated in the axle due to motion. coil induces an e.m.f. of 1.5 V in the other.
6. A conductor moves with a velocity of 20 m/s at an Induced e.m.f. jE2j D M dI1 , i.e. 1.5 D M⊲200⊳.
angle of (a) 90° (b) 45° (c) 30°, to a magnetic field dt
produced between two square-faced poles of side Thus mutual inductance,
length 2.5 cm. If the flux on the pole face is 60 mWb,
find the magnitude of the induced e.m.f. in each M D 1.5 D 0.0075 H or 7.5 mH
case. 200
Electromagnetic induction 291
Problem 7. The mutual inductance between two coils 39.5 The transformer
is 18 mH. Calculate the steady rate of change of cur-
rent in one coil to induce an e.m.f. of 0.72 V in the A transformer is a device which uses the phenomenon of
other. mutual induction to change the values of alternating voltages
and currents. In fact, one of the main advantages of a.c. trans-
Induced e.m.f. jE2j D M dI1 mission and distribution is the ease with which an alternating
dt voltage can be increased or decreased by transformers.
Hence rate of change of current,
Losses in transformers are generally low and thus effi-
dI1 D jE2j D 0.72 D 40 A=s ciency is high. Being static they have a long life and are
dt M 0.018 very stable.
Problem 8. Two coils have a mutual inductance of Transformers range in size from the miniature units used in
0.2 H. If the current in one coil is changed from 10 A to electronic applications to the large power transformers used
4 A in 10 ms, calculate (a) the average induced e.m.f. in in power stations. The principle of operation is the same for
the second coil, (b) the change of flux linked with the each.
second coil if it is wound with 500 turns.
A transformer is represented in Fig. 39.6(a) as consisting
(a) Induced e.m.f. E2 D M dI1 D ⊲0.2⊳ 10 4 of two electrical circuits linked by a common ferromagnetic
dt 10 ð 10 3 core. One coil is termed the primary winding which is
connected to the supply of electricity, and the other the
secondary winding, which may be connected to a load.
A circuit diagram symbol for a transformer is shown in
Fig. 39.6(b).
D 120 V Flux Φ
d
I1 I2
(b) From Section 15.8, induced e.m.f. jE2j D N dt , hence V2
d D jE2j dt Primary Secondary
N A.C.
supply
V1
Load
Thus the change of flux, winding winding
N1 turns N2 turns
d D ⊲120⊳⊲10 ð 10 3⊳ D 2.4 mWb
500 Ferromagnetic core
(a)
Now try the following exercise
Exercise 173 Further problems on mutual induc- (b)
tance (Answers on page 357)
Fig. 39.6
1. The mutual inductance between two coils is 150 mH.
Find the magnitude of the e.m.f. induced in one coil Transformer principle of operation
when the current in the other is increasing at a rate of
30 A/s When the secondary is an open-circuit and an alternating
2. Determine the mutual inductance between two coils voltage V1 is applied to the primary winding, a small current -
when a current changing at 50 A/s in one coil induces called the no-load current I0 - flows, which sets up a magnetic
an e.m.f. of 80 mV in the other. flux in the core. This alternating flux links with both primary
3. Two coils have a mutual inductance of 0.75 H. Calcu- and secondary coils and induces in them e.m.f.’s of E1 and
late the magnitude of the e.m.f. induced in one coil
when a current of 2.5 A in the other coil is reversed E2 respectively by mutual induction.
in 15 ms.
The induced e.m.f. E in a coil of N turns is given by
4. The mutual inductance between two coils is 240 mH.
If the current in one coil changes from 15 A to 6 A in ED N d volts, where d is the rate of change of flux
12 ms, calculate (a) the average e.m.f. induced in the dt dt
other coil, (b) the change of flux linked with the other (see Chapter 41, page 311). In an ideal transformer, the rate
coil if it is wound with 400 turns.
of change of flux is the same for both primary and secondary
292 Science for Engineering
and thus E1 D E2 i.e. the induced e.m.f. per turn is For an ideal transformer,
N1 N2
N1 D V1 hence 2 240
constant. N2 V2 7 D V2
Assuming no losses, E1 D V1 and E2 D V2
V1 D V2 or V1 D N1 Thus the secondary voltage,
Hence N1 N2 V2 N2 ⊲1⊳ ⊲240⊳⊲7⊳
V2 D 2 D 840 V
V1 is called the voltage ratio and N1 the turns ratio, or
V2 N2 Problem 11. An ideal transformer has a turns ratio of
the ‘transformation ratio’ of the transformer. If N2 is less 8:1 and the primary current is 3 A when it is supplied
than N1 then V2 is less than V1 and the device is termed at 240 V. Calculate the secondary voltage and current.
a step-down transformer. If N2 is greater then N1 then
V2 is greater than V1 and the device is termed a step-up A turns ratio of 8:1 means N1 D 8 i.e. a step-down
transformer. N2 1
transformer.
When a load is connected across the secondary winding, a
N1 D V1
current I2 flows. In an ideal transformer losses are neglected N2 V2
and a transformer is considered to be 100% efficient.
Hence input power = output power, or V1I1 D V2I2 i.e. in
an ideal transformer, the primary and secondary ampere-
turns are equal
Thus V1 D I2 ⊲2⊳ hence, secondary voltage
V2 I1
Combining equations (1) and (2) gives: V2 D V1 N2 D 240 1 D 30 volts
N1 8
V1 = N1 = I2
V2 N2 I1 ⊲3⊳ Also, N1 D I2 hence, secondary current
N2 I1
The rating of a transformer is stated in terms of the volt-
amperes that it can transform without overheating. With I2 D I1 N1 8 D 24 A
reference to Fig. 39.6(a), the transformer rating is either V1I1 N2 D3 1
or V2I2, where I2 is the full-load secondary current.
Problem 9. A transformer has 500 primary turns and Problem 12. An ideal transformer, connected to a
3000 secondary turns. If the primary voltage is 240 V, 240 V mains, supplies a 12 V, 150 W lamp. Calculate
determine the secondary voltage, assuming an ideal the transformer turns ratio and the current taken from
transformer. the supply.
For an ideal transformer, voltage ratio D turns ratio V1 D 240 V, V2 D 12 V, and since power P D VI, then
i.e. V1 D N1 hence 240 500 P 150
D I2 D V2 D 12 D 12.5 A
V2 N2 V2 3000
Thus secondary voltage,
⊲240⊳⊲3000⊳ Turns ratio D N1 D V1 D 240 D 20
V2 D 500 D 1440 V or 1.44 kV N2 V2 12
Problem 10. An ideal transformer with a turns ratio V1 D I2 from which, I1 D I2 V2
of 2:7 is fed from a 240 V supply. Determine its output V2 I1 V1
voltage.
12
D 12.5 240
A turns ratio of 2:7 means that the transformer has 2 turns on Hence the current taken from the supply,
the primary for every 7 turns on the secondary (i.e. a step-up
transformer); thus N1 D 2 12.5
N2 7 I1 D 20 D 0.625 A
Electromagnetic induction 293
Problem 13. A 12 resistor is connected across the Exercise 175 Short answer questions on electromag-
secondary winding of an ideal transformer whose sec- netic induction
ondary voltage is 120 V. Determine the primary voltage
if the supply current is 4 A. 1. What is electromagnetic induction?
2. State Faraday’s laws of electromagnetic induction
Secondary current 3. State Lenz’s law
4. Explain briefly the principle of the generator
I2 D V2 D 120 D 10 A 5. The direction of an induced e.m.f. in a generator may
R2 12 be determined using Fleming’s . . . . . .. rule
V1 D I2 6. The e.m.f. E induced in a moving conductor may
V2 I1 be calculated using the formula E D Blv. Name the
quantities represented and their units
from which the primary voltage
7. What is self-inductance? State its symbol and unit
V1 D V2 I2 D 120 10 D 300 volts 8. What is mutual inductance? State its symbol and unit
I1 4
9. The mutual inductance between two coils is M. The
e.m.f. E2 induced in one coil by the current changing
dI1
Now try the following exercises at dt in the other is given by: E2 D ...... volts
Exercise 174 Further problems on the transformer 10. What is a transformer?
principle of operation (Answers on
page 357) 11. Explain briefly how a voltage is induced in the
secondary winding of a transformer
1. A transformer has 600 primary turns connected to a
1.5 kV supply. Determine the number of secondary 12. Draw the circuit diagram symbol for a transformer
turns for a 240 V output voltage, assuming no losses.
13. State the relationship between turns and voltage ratios
2. An ideal transformer with a turns ratio of 2:9 is fed for a transformer
from a 220 V supply. Determine its output voltage.
14. How is a transformer rated?
3. An ideal transformer has a turns ratio of 12:1 and is
supplied at 192 V. Calculate the secondary voltage. 15. Briefly describe the principle of operation of a trans-
former
4. A transformer primary winding connected across a
415 V supply has 750 turns. Determine how many Exercise 176 Multi-choice questions on electromag-
turns must be wound on the secondary side if an output netic induction (Answers on page 357)
of 1.66 kV is required.
1. A current changing at a rate of 5 A/s in a coil of
5. An ideal transformer has a turns ratio of 12:1 and is inductance 5 H induces an e.m.f. of:
supplied at 180 V when the primary current is 4 A.
Calculate the secondary voltage and current. (a) 25 V in the same direction as the applied voltage
6. A step-down transformer having a turns ratio of 20:1 (b) 1 V in the same direction as the applied voltage
has a primary voltage of 4 kV and a load of 10 kW.
Neglecting losses, calculate the value of the secondary (c) 25 V in the opposite direction to the applied
current. voltage
7. A transformer has a primary to secondary turns ratio of (d) 1 V in the opposite direction to the applied volt-
1:15. Calculate the primary voltage necessary to sup- age
ply a 240 V load. If the load current is 3 A determine
the primary current. Neglect any losses. 2. A bar magnet is moved at a steady speed of 1.0 m/s
towards a coil of wire which is connected to a centre-
8. A 20 resistance is connected across the secondary zero galvanometer. The magnet is now withdrawn
winding of a single-phase power transformer whose along the same path at 0.5 m/s. The deflection of the
secondary voltage is 150 V. Calculate the primary galvanometer is in the:
voltage and the turns ratio if the supply current is 5 A,
neglecting losses. (a) same direction as previously, with the magnitude
of the deflection doubled
(b) opposite direction as previously, with the mag-
nitude of the deflection halved
294 Science for Engineering
(c) same direction as previously, with the magnitude 8. The mutual inductance between two coils, when a
of the deflection halved current changing at 20 A/s in one coil induces an
e.m.f. of 10 mV in the other, is:
(d) opposite direction as previously, with the mag-
nitude of the deflection doubled (a) 0.5 H (b) 200 mH (c) 0.5 mH (d) 2 H
3. An e.m.f. of 1 V is induced in a conductor moving 9. A transformer has 800 primary turns and 100 second-
at 10 cm/s in a magnetic field of 0.5 T. The effective ary turns. To obtain 40 V from the secondary winding
length of the conductor in the magnetic field is: the voltage applied to the primary winding must be:
(a) 20 cm (b) 5 m (c) 20 m (d) 50 m (a) 5 V (b) 20 V (c) 2.5 V (d) 320 V
4. Which of the following is false? 10. A step-up transformer has a turns ratio of 10. If the
output current is 5 A, the input current is:
(a) Fleming’s left-hand rule or Lenz’s law may be
used to determine the direction of an induced (a) 50 A (b) 5 A (c) 2.5 A (d) 0.5 A
e.m.f.
11. A 440 V/110 V transformer has 1000 turns on the pri-
(b) An induced e.m.f. is set up whenever the mag- mary winding. The number of turns on the secondary
netic field linking that circuit changes is:
(c) The direction of an induced e.m.f. is always such (a) 550 (b) 250 (c) 4000 (d) 25
as to oppose the effect producing it
12. A 1 kV/250 V transformer has 500 turns on the sec-
(d) The induced e.m.f. in any circuit is proportional ondary winding. The number of turns on the primary
to the rate of change of the magnetic flux linking is:
the circuit
(a) 2000 (b) 125 (c) 1000 (d) 250
5. A strong permanent magnet is plunged into a coil
and left in the coil. What is the effect produced on 13. The power input to a mains transformer is 200 W. If
the coil after a short time? the primary current is 2.5 A, the secondary voltage is
2 V and assuming no losses in the transformer, the
(a) The coil winding becomes hot turns ratio is:
(a) 80:1 step up (b) 40:1 step up
(b) The insulation of the coil burns out
(c) A high voltage is induced (c) 80:1 step down (d) 40:1 step down
(d) There is no effect
14. An ideal transformer has a turns ratio of 1:5 and is
6. Self-inductance occurs when: supplied at 200 V when the primary current is 3 A.
Which of the following statements is false?
(a) the current is changing
(a) The turns ratio indicates a step-up transformer
(b) the circuit is changing
(b) The secondary voltage is 40 V
(c) the flux is changing
(c) The secondary current is 15 A
(d) the resistance is changing
(d) The transformer rating is 0.6 kVA
7. Faraday’s laws of electromagnetic induction are
related to: (e) The secondary voltage is 1 kV
(a) the e.m.f. of a chemical cell (f) The secondary current is 0.6 A
(b) the e.m.f. of a generator
(c) the current flowing in a conductor
(d) the strength of a magnetic field
Electromagnetic induction 295
Assignment 12 4. A conductor, 25 cm long is situtated at right angles to a
magnetic field. Determine the strength of the magnetic
This assignment covers the material contained field if a current of 12 A in the conductor produces a
Chapter 37 to 39. The marks for each question are
shown in brackets at the end of each question. force on it of 4.5 N. (5)
5. An electron in a television tube has a charge of
1.5 ð 10 19 C and travels at 3 ð 107 m/s perpendicular
1. Determine the value of currents I1 to I5 shown in to a field of flux density 20 µT. Calculate the force
Fig. A12.1. (5) exerted on the electron in the field. (5)
2. Find the current flowing in the 5 resistor of the
circuit shown in Fig. A12.2 using Kirchhoff’s laws. 10 V 3V
2Ω
Find also the current flowing in each of the other two 5Ω
1Ω
branches of the circuit. (11)
3. The maximum working flux density of a lifting elec-
tromagnet is 1.7 T and the effective area of a pole face
is circular in cross-section. If the total magnetic flux
produced is 214 mWb, determine the radius of the pole
face. (6)
Fig. A12.2
I2 6. A lorry is travelling at 100 km/h. Assuming the vertical
component of the earth’s magnetic field is 40 µT and
the back axle of the lorry is 1.98 m, find the e.m.f.
24 A 9A 6A generated in the axle due to motion. (6)
7. Two coils, P and Q, have a mutual inductance of
I1 100 mH. If a current 3 A in coil P is reversed in 20 ms,
10 A I3 I4 determine (a) the average e.m.f. induced in coil Q, and
I5
11 A (b) the flux change linked with coil Q if it wound with
200 turns. (6)
Fig. A12.1 8. An ideal transformer connected to a 250 V mains,
supplies a 25 V, 200 W lamp. Calculate the transformer
turns ratio and the current taken from the supply. (6)
40
Alternating voltages and currents
At the end of this chapter you should be able to: ;;yyy;y;N;;yyyy;;y;y;y;ω
ž appreciate why a.c. is used in preference to d.c. S
ž describe the principle of operation of an a.c. gener-
Fig. 40.1
ator
ž distinguish between unidirectional and alternating An e.m.f. is generated in the coil (from Faraday’s Laws)
which varies in magnitude and reverses its direction at regular
waveforms intervals. The reason for this is shown in Fig. 40.2. In posi-
ž define cycle, period or periodic time T and fre- tions (a), (e) and (i) the conductors of the loop are effectively
moving along the magnetic field, no flux is cut and hence no
quency f of a waveform e.m.f. is induced. In position (c) maximum flux is cut and
1 hence maximum e.m.f. is induced. In position (g), maximum
flux is cut and hence maximum e.m.f. is again induced. How-
ž perform calculations involving T D ever, using Fleming’s right-hand rule, the induced e.m.f. is in
f the opposite direction to that in position (c) and is thus shown
as E. In positions (b), (d), (f) and (h) some flux is cut and
ž define instantaneous, peak, mean and r.m.s. values, hence some e.m.f. is induced. If all such positions of the coil
and form and peak factors for a sine wave are considered, in one revolution of the coil, one cycle of
alternating e.m.f. is produced as shown. This is the principle
ž calculate mean and r.m.s. values and form and peak of operation of the a.c. generator (i.e. the alternator).
factors for given waveforms
40.1 Introduction
Electricity is produced by generators at power stations and
then distributed by a vast network of transmission lines
(called the National Grid system) to industry and for domestic
use. It is easier and cheaper to generate alternating current
(a.c.) than direct current (d.c.) and a.c. is more conveniently
distributed than d.c. since its voltage can be readily altered
using transformers. Whenever d.c. is needed in preference to
a.c., devices called rectifiers are used for conversion.
40.2 The a.c. generator 40.3 Waveforms
Let a single turn coil be free to rotate at constant angular If values of quantities which vary with time t are plotted to a
velocity symmetrically between the poles of a magnet system base of time, the resulting graph is called a waveform. Some
as shown in Fig. 40.1. typical waveforms are shown in Fig. 40.3. Waveforms (a)
Alternating voltages and currents 297
(a) (b) (c) (d) (e) (f) (g) (h) (i)
N N N N NNN NN
SS SS SSSSS
+E
Induced e.m.f. Revolutions of loop
0
1131 3 1
8482 4
–E
Fig. 40.2 + ++
Fig. 40.3
V IV
0 t0 t0 t
– – Rectangular – Square
pulse
(a) (b) wave (c)
+ + t I i 1 Im t2
I V 0 t1 i2 t
t0
0
– Triangular (f)
– wave
Sawtooth (e)
wave
(d)
+ Em Peak-to-peak P Time
e.m.f. e1 e.m.f.
t2
0 t1
e2
–
Periodic Sine wave
time, T (g)
298 Science for Engineering
and (b) are unidirectional waveforms, for, although they Now try the following exercise
vary considerably with time, they flow in one direction only
(i.e. they do not cross the time axis and become negative). Exercise 177 Further problems on frequency and
Waveforms (c) to (g) are called alternating waveforms since periodic time (Answers on page 357)
their quantities are continually changing in direction (i.e.
alternately positive and negative). 1. Determine the periodic time for the following frequen-
cies:
A waveform of the type shown in Fig. 40.3(g) is called a (a) 2.5 Hz (b) 100 Hz (c) 40 kHz
sine wave. It is the shape of the waveform of e.m.f. produced
by an alternator and thus the mains electricity supply is of 2. Calculate the frequency for the following periodic
‘sinusoidal’ form. times:
(a) 5 ms (b) 50 µs (c) 0.2 s
One complete series of values is called a cycle (i.e. from
O to P in Fig. 40.3(g)). 3. An alternating current completes 4 cycles in 5 ms.
What is its frequency?
The time taken for an alternating quantity to complete one
cycle is called the period or the periodic time, T , of the 40.4 A.C. values
waveform.
Instantaneous values are the values of the alternating quan-
The number of cycles completed in one second is called tities at any instant of time. They are represented by small
the frequency, f , of the supply and is measured in hertz, letters, i, v, e, etc., (see Figs. 40.3(f) and (g)).
Hz. The standard frequency of the electricity supply in Great
Britain is 50 Hz. The largest value reached in a half cycle is called the
peak value or the maximum value or the crest value or
11 the amplitude of the waveform. Such values are represented
T = or f = by Vm, Im, Em, etc. (see Figs. 40.3(f) and (g)). A peak-to-
peak value of e.m.f. is shown in Fig. 40.3(g) and is the
fT difference between the maximum and minimum values in
a cycle.
Problem 1. Determine the periodic time for frequen-
cies of (a) 50 Hz and (b) 20 kHz. The average or mean value of a symmetrical alternating
quantity, (such as a sine wave), is the average value measured
11 over a half cycle, (since over a complete cycle the average
(a) Periodic time T D f D 50 D 0.02 s or 20 ms value is zero).
11 area under the curve
(b) Periodic time T D f D 20 000 D 0.00005 s or 50 µs Average or mean value =
Problem 2. Determine the frequencies for periodic length of base
times of (a) 4 ms (b) 8 µs.
The area under the curve is found by approximate meth-
1 1 1000 ods such as the trapezoidal rule, the mid-ordinate rule or
(a) Frequency f D T D 4 ð 10 3 D 4 D 250 Hz Simpson’s rule. Average values are represented by VAV, IAV,
EAV, etc.
1 1 1 000 000
(b) Frequency f D T D 8 ð 10 6 D 8 For a sine wave:
D 125 000 Hz or 125 kHz or average value = 0.637 U maximum value
0.125 MHz 2
Problem 3. An alternating current completes 5 cycles i.e. U maximum value
in 8 ms. What is its frequency? p
Time for one cycle D 8 ms D 1.6 ms D periodic time T. The effective value of an alternating current is that current
5 which will produce the same heating effect as an equiva-
lent direct current. The effective value is called the root
1 1 1000 10 000 mean square (r.m.s.) value and whenever an alternating
Frequency f D T D 1.6 ð 10 3 D 1.6 D 16 quantity is given, it is assumed to be the rms value. For
D 625 Hz
Alternating voltages and currents 299
example, the domestic mains supply in Great Britain is 200 v4
240 V and is assumed to mean ‘240 V rms’. The sym- v3
bols used for r.m.s. values are I, V, E, etc. For a non- 150 v2
sinusoidal waveform as shown in Fig. 40.4 the r.m.s. value
is given by: 100 v1
I D i21 C i22 C Ð Ð Ð C in2 Voltage (V) 50
n
05 10 15 20 Time
where n is the number of intervals used. (ms)
–200
+ 10 (a)
Current 0 i1 i2 i3 0Current (A) 8 16 Time (ms)
– –10 (b)
Time
Fig. 40.5
Fig. 40.4 (a) Triangular waveform (Fig. 40.5(a))
For a sine wave: (i) Time for one complete cycle D 20 ms D periodic
rms value = 0.707 U maximum value time, T
1
i.e. p U maximum value 11
2 Hence frequency f D T D 20 ð 10 3
r.m.s. value 1000
Form factor = average value D 20 D 50 Hz
For a sine wave, form factor D 1.11 (ii) Area under the triangular waveform for a half
Peak factor = maximum value cycle D 1 ð base ð height D 1 ð ⊲10 ð 10 3⊳ ð 200
r.m.s. value 2 2
For a sine wave, peak factor D 1.41. D 1 volt second
The values of form and peak factors give an indication of
Average value of waveform D area under curve
the shape of waveforms. length of base
Problem 4. For the periodic waveforms shown in 1 volt second
Fig. 40.5 determine for each: (i) the frequency, D
(ii) the average value over half a cycle, (iii) the r.m.s.
value, (iv) the form factor, and (v) the peak factor. 10 ð 10 3 second
1000
D D 100 V
10
(iii) In Fig. 40.5(a), the first 1 cycle is divided into 4
4
intervals
Thus rms value D v12 C v22 C v23 C v42
4
252 C 752 C 1252 C 1752
D
4
D 114.6 V
300 Science for Engineering
(Note that the greater the number of intervals cho- 80
sen, the greater the accuracy of the result. For Maximum value = 76 A
example, if twice the number of ordinates as that
chosen above are used, the r.m.s. value is found to 70
be 115.6 V)
60
r.m.s. value 114.6
(iv) Form factor D average value D 100 D 1.15 Current i (A) 50
RMS value = 43.77 A
maximum value 200
(v) Peak factor D r.m.s. value D 114.6 D 1.75 40
Average value = 35.1 A
(b) Rectangular waveform (Fig. 40.5(b))
30
(i) Time for one complete cycle D 16 ms D periodic
time, T 20 30 49 63 73 72 30
1 1 1000 19
Hence frequency, f D T D 16 ð 10 3 D 16
D 62.5 Hz 10 19
10
(ii) Average value over half a cycle
3
area under curve 10 ð ⊲8 ð 10 3⊳
D length of base D 8 ð 10 3 D 10 A 0 1.0 2.0 3.0 4.0
1.25 3.8
(iii) The r.m.s. value D i21 C i22 C i32 C i24 D 10 A 2
4 Time t (ms)
5.0
however many intervals are chosen, since the wave-
form is rectangular.
r.m.s. value 10 Fig. 40.6
(iv) Form factor D average value D 10 D 1
maximum value 10 (b) Instantaneous value of current after 1.25 ms is 19 A, from
(v) Peak factor D r.m.s. value D 10 D 1 Fig. 40.6
Instantaneous value of current after 3.8 ms is 70 A, from
Problem 5. The following table gives the correspond- Fig. 40.6
ing values of current and time for a half cycle of alter-
nating current: (c) Peak or maximum value D 76 A
time t (ms) 0 0.5 1.0 1.5 2.0 2.5 area under curve
current i (A) 0 7 14 23 40 56 (d) Mean or average value D length of base
time t (ms) 3.0 3.5 4.0 4.5 5.0 Using the mid-ordinate rule with 10 intervals, each of
current i (A) 68 76 60 5 0 width 0.5 ms gives: area under curve
Assuming the negative half cycle is identical in shape D ⊲0.5 ð 10 3⊳[3 C 10 C 19 C 30 C 49 C 63
to the positive half cycle, plot the waveform and find
(a) the frequency of the supply, (b) the instantaneous + 73 + 72 + 30 + 2] (see Fig. 40.6)
values of current after 1.25 ms and 3.8 ms, (c) the peak
or maximum value, (d) the mean or average value, and D ⊲0.5 ð 10 3⊳⊲351⊳
(e) the r.m.s. value of the waveform.
Hence mean or average value D ⊲0.5 ð 10 3⊳⊲351⊳
5 ð 10 3
D 35.1 A
The half cycle of alternating current is shown plotted in 32 C 102 C 192 C 302 C 492 C 632
Fig. 40.6.
(e) R.m.s. value D C 732 C 722 C 302 C 22
(a) Time for a half cycle D 5 ms; hence the time for one 10
cycle, i.e. the periodic time, T D 10 ms or 0.01 s
11
Frequency, f D T D 0.01 D 100 Hz
19 157
D 10 D 43.8 A
Alternating voltages and currents 301
Problem 6. Calculate the r.m.s. value of a sinusoidal 5
current of maximum value 20 A
For a sine wave, r.m.s. value D 0.707 ð maximum value Current (A) 0 10 Time
D 0.707 ð 20 D 14.14 A 5 (ms)
Problem 7. Determine the peak and mean values for –5
a 240 V mains supply. (a)
For a sine wave, r.m.s. value of voltage V D 0.707 ð Vm Voltage (V) 20
A 240 V mains supply means that 240 V is the r.m.s. value, 0
2 4 Time
V 240
hence Vm D 0.707 D 0.707 D 339.5 V = peak value –20 (ms)
Mean value VAV D 0.637Vm D 0.637 ð 339.5 D 216.3 V (b)
Problem 8. A supply voltage has a mean value of 24
150 V. Determine its maximum value and its r.m.s.
value. Current (A)
For a sine wave, mean value D 0.637 ð maximum value 0
2 4 6 8 Time
Hence maximum value D mean value D 150 D 235.5 V (ms)
0.637 0.637
–24
R.m.s. value D 0.707 ð maximum value D 0.707 ð 235.5 (c)
D 166.5 V 100
Now try the following exercises Voltage (V) 0
1 2 3 4 Time
Exercise 178 Further problems on a.c. values of (ms)
waveforms (Answers on page 357)
–100 (d)
1. An alternating current varies with time over half a
cycle as follows: Fig. 40.7
Current (A) 0 0.7 2.0 4.2 8.4 8.2 3. An alternating voltage is triangular in shape, rising at
time (ms) 0 1 2 3 4 5 a constant rate to a maximum of 300 V in 8 ms and
then falling to zero at a constant rate in 4 ms. The
Current (A) 2.5 1.0 0.4 0.2 0 negative half cycle is identical in shape to the positive
time (ms) 6 7 8 9 10 half cycle. Calculate (a) the mean voltage over half a
cycle, and (b) the r.m.s. voltage.
The negative half cycle is similar. Plot the curve and
determine: 4. Calculate the r.m.s. value of a sinusoidal curve of
(a) the frequency maximum value 300 V.
(b) the instantaneous values at 3.4 ms and 5.8 ms
(c) its mean value and 5. Find the peak and mean values for a 200 V mains
(d) its r.m.s. value supply.
2. For the waveforms shown in Fig. 40.7 determine for 6. A sinusoidal voltage has a maximum value of 120 V.
each (i) the frequency (ii) the average value over half Calculate its r.m.s. and average values.
a cycle (iii) the r.m.s. value (iv) the form factor (v) the
peak factor.
302 Science for Engineering
7. A sinusoidal current has a mean value of 15.0 A.
Determine its maximum and r.m.s. values.
N
Exercise 179 Short answer questions on alternating S
voltages and currents
Fig. 40.8
1. Briefly explain the principle of operation of the
simple alternator 5. The supply of electrical energy for a consumer is
usually by a.c. because:
2. What is meant by (a) waveform (b) cycle (a) transmission and distribution are more easily
effected
3. What is the difference between an alternating and a (b) it is most suitable for variable speed motors
unidirectional waveform?
(c) the volt drop in cables is minimal
4. The time to complete one cycle of a waveform is
called the . . . . . . (d) cable power losses are negligible
5. What is frequency? Name its unit 6. Which of the following statements is false?
6. The mains supply voltage has a special shape of (a) It is cheaper to use a.c. than d.c.
waveform called a . . . . . .
(b) Distribution of a.c. is more convenient than with
7. Define peak value d.c. since voltages may be readily altered using
transformers
8. What is meant by the r.m.s. value?
(c) An alternator is an a.c. generator
9. What is the mean value of a sinusoidal alternating
e.m.f. which has a maximum value of 100 V? (d) A rectifier changes d.c. to a.c.
10. The effective value of a sinusoidal waveform is
. . . . . . ð maximum value.
Exercise 180 Multi-choice questions on alternating 7. An alternating voltage of maximum value 100 V is
voltages and currents (Answers on applied to a lamp. Which of the following direct
page 358) voltages, if applied to the lamp, would cause the lamp
to light with the same brilliance?
1. The number of complete cycles of an alternating
current occurring in one second is known as: (a) 100 V (b) 63.7 V (c) 70.7 V (d) 141.4 V
(a) the maximum value of the alternating current 8. The value normally stated when referring to alternat-
(b) the frequency of the alternating current ing currents and voltages is the:
(c) the peak value of the alternating current
(d) the r.m.s. or effective value (a) instantaneous value (b) r.m.s. value
(c) average value (d) peak value
2. The value of an alternating current at any given 9. State which of the following is false. For a sine wave:
instant is:
(a) the peak factor is 1.414
(a) a maximum value (b) a peak value (b) the r.m.s. value is 0.707 ð peak value
(c) the average value is 0.637 ð r.m.s. value
(c) an instantaneous value (d) an r.m.s. value (d) the form factor is 1.11
3. An alternating current completes 100 cycles in 0.1 s. 10. An a.c. supply is 70.7 V, 50 Hz. Which of the fol-
Its frequency is: lowing statements is false?
(a) 20 Hz (b) 100 Hz (c) 0.002 Hz (d) 1 kHz (a) The periodic time is 20 ms
(b) The peak value of the voltage is 70.7 V
4. In Fig. 40.8, at the instant shown, the generated e.m.f. (c) The r.m.s. value of the voltage is 70.7 V
will be: (d) The peak value of the voltage is 100 V
(a) zero (b) an r.m.s. value
(c) an average value (d) a maximum value
41
Capacitors and inductors
At the end of this chapter you should be able to: + X
d.c. supply Y
ž explain how a capacitor stores energy
VQ −
ž Appreciate Q D It, E D , D D , Q D CV Fig. 41.1
dA
metal plates X and Y separated by an insulator, which could
D be air. Since the plates are electrical conductors each will
and E D e0er and perform calculations using these contain a large number of mobile electrons. Because the
formulae plates are connected to a d.c. supply the electrons on plate
ž state the unit of capacitance X, which have a small negative charge, will be attracted
ž define a dielectric to the positive pole of the supply and will be repelled
ž state the value of e0 and appreciate typical values from the negative pole of the supply on to plate Y. X will
of er become positively charged due to its shortage of electrons
ž appreciate that the capacitance C of a parallel plate whereas Y will have a negative charge due to its surplus of
capacitor is given by C D e0erA⊲n 1⊳ electrons.
d The difference in charge between the plates results in
ž perform calculations involving the parallel plate a p.d. existing between them, the flow of electrons dying
away and ceasing when the p.d. between the plates equals
capacitor the supply voltage. The plates are then said to be charged
ž define dielectric strength and there exists an electric field between them. Figure 41.2
ž determine the energy stored in a capacitor and state shows a side view of the plates with the field represented
by ‘lines of electrical flux’. If the plates are disconnected
its unit from the supply and connected together through a resistor the
ž describes practical types of capacitor surplus of electrons on the negative plate will flow through
ž define inductance and state its unit the resistor to the positive plate. This is called discharging.
ž state the factors which affect the inductance of an The current flow decreases to zero as the charges on the
plates reduce. The current flowing in the resistor causes it to
inductor liberate heat showing that energy is stored in the electric
ž describe examples of practical inductors and their field.
circuit diagrams
ž calculate the energy stored in the magnetic field of
an inductor and state its unit
41.1 Capacitors and capacitance
A capacitor is a device capable of storing electrical energy.
Figure 41.1 shows a capacitor consisting of a pair of parallel
304 Science for Engineering
I is undesirable but has to be accepted, minimised or compen-
sated for. There are other situations, such as in capacitors,
+X where capacitance is a desirable property.
The ratio of electric flux density, D, to electric field
strength, E, is called absolute permittivity, e, of a dielectric.
+ Thus D
d.c. supply De
E
−
− Electric Permittivity of free space is a constant, given by
Y field
e0 D 8.85 ð 10 12 F/m.
Relative permittivity,
I flux density of the field in a dielectric
er D flux density of the field in a vacuum
Fig. 41.2
Summary of important formulae and definitions (er has no units). Examples of the values of er include:
From Chapter 33, charge Q is given by: air D 1, polythene D 2.3, mica D 3–7, glass D 5–10,
ceramics D 6–1000.
Absolute permittivity, e D e0er, thus
Q = I U t coulombs D
E = e0er
where I is the current in amperes, and t the time in seconds. Problem 1. (a) Determine the p.d. across a 4 µF capa-
A dielectric is an insulating medium separating charged citor when charged with 5 mC (b) Find the charge on
a 50 pF capacitor when the voltage applied to it is 2 kV
surfaces.
Electric field strength, electric force, or voltage gradient, (a) C D 4 µF D 4 ð 10 6 F and Q D 5 mC D 5 ð 10 3 C
p.d. across dielectric
E D thickness of dielectric
i.e. V Q
E = volts=m Since C D
d
V
Electric flux density, then Q 5 ð 10 3 5 ð 106 5000
V D C D 4 ð 10 6 D 4 ð 103 D 4
D = Q C=m2 Hence p.d. V = 1250 V or 1.25 kV
A
(b) C D 50 pF D 50 ð 10 12 F and V D 2 kV D 2000 V
Charge Q on a capacitor is proportional the applied voltage,
V, i.e. Q / V Q D CV D 50 ð 10 12 ð 2000
D 5ð2 D 0.1 ð 10 6
108
Q = CV Hence, charge Q = 0.1 µC
where the constant of proportionality, C, is the capacitance Problem 2. A direct current of 4 A flows into a pre-
viously uncharged 20 µF capacitor for 3 ms. Determine
Capacitance, Q the p.d. between the plates.
C=
V
The unit of capacitance is the farad, F (or more usually I D 4 A, C D 20 µF D 20ð10 6 F and t D 3 ms D 3ð10 3 s.
µF D 10 6 F or pF D 10 12 F), which is defined as the
capacitance of a capacitor when a p.d. of one volt appears Q D It D 4 ð 3 ð 10 3 C
across the plates when charged with one coulomb.
VD Q D 4 ð 3 ð 10 3 D 12 ð 106 D 0.6 ð 103
Every system of electrical conductors possesses capaci- C 20 ð 10 6 20 ð 103
tance. For example, there is capacitance between the con-
ductors of overhead transmission lines and also between the D 600 V
wires of a telephone cable. In these examples the capacitance
Hence, the p.d. between the plates is 600 V
Capacitors and inductors 305
Problem 3. A 5 µF capacitor is charged so that the 4. For how long must a charging current of 2 A be fed
p.d. between its plates is 800 V. Calculate how long to a 5 µF capacitor to raise the p.d. between its plates
the capacitor can provide an average discharge current by 500 V.
of 2 mA.
5. A direct current of 10 A flows into a previously
C D 5 µF D 5 ð 10 6 F, V D 800 V and uncharged 5 µF capacitor for 1 ms. Determine the p.d.
I D 2 mA D 2 ð 10 3 A. between the plates.
Q D CV D 5 ð 10 6 ð 800 D 4 ð 10 3 C
6. A capacitor uses a dielectric 0.04 mm thick and oper-
Also, Q D It. ates at 30 V. What is the electric field strength across
Q 4 ð 10 3 the dielectric at this voltage?
Thus, t D I D 2 ð 10 3 D 2 s 7. A charge of 1.5 µC is carried on two parallel rectangu-
Hence, the capacitor can provide an average discharge lar plates each measuring 60 mm by 80 mm. Calculate
current of 2 mA for 2 s the electric flux density. If the plates are spaced 10 mm
apart and the voltage between them is 0.5 kV deter-
Problem 4. Two parallel rectangular plates measuring mine the electric field strength.
20 cm by 40 cm carry an electric charge of 0.2 µC.
Calculate the electric flux density. If the plates are 41.2 The parallel plate capacitor
spaced 5 mm apart and the voltage between them is
0.25 kV determine the electric field strength. For a parallel-plate capacitor, experiments show that capac-
itance C is proportional to the area A of a plate, inversely
Area A D 20 cm ð 40 cm D 800 cm2 D 800 ð 10 4 m2 and proportional to the plate spacing d (i.e. the dielectric thick-
charge Q D 0.2 µC D 0.2 ð 10 6 C. ness) and depends on the nature of the dielectric:
Electric flux density, Capacitance, C = e0er A(n − 1) farads
d
Q 0.2 ð 10 6 0.2 ð 104
D D A D 800 ð 10 4 D 800 ð 106 where e0 D 8.85 ð 10 12 F/m (constant),
er D relative permittivity,
D 2000 6 D 2.5 µC/m2 A D area of one of the plates, in m2,
800 ð 10 d D thickness of dielectric in m
Voltage V D 0.25 kV D 250 V and plate spacing, and n D number of plates
d D 5 mm D 5 ð 10 3 m. Problem 5. (a) A ceramic capacitor has an effective
plate area of 4 cm2 separated by 0.1 mm of ceramic of
Electric field strength, relative permittivity 100. Calculate the capacitance of
the capacitor in picofarads. (b) If the capacitor in part
V 250 (a) is given a charge of 1.2 µC what will be the p.d.
E D d D 5 ð 10 3 D 50 kV/m between the plates?
Now try the following exercise (a) Area A D 4 cm2 D 4 ð 10 4 m2,
Exercise 181 Further problems on capacitors and d D 0.1 mm D 0.1 ð 10 3 m, e0 D 8.85 ð 10 12 F/m,
capacitance (Answers on page 358) er D 100 and n D 2
(Where appropriate take e0 as 8.85 ð 10 12 F/m.) Capacitance, C D e0erA farads
d
1. Find the charge on a 10 µF capacitor when the applied
voltage is 250 V. D 8.85 ð 10 12 ð 100 ð 4 ð 10 4
0.1 ð 10 3
2. Determine the voltage across a 1000 pF capacitor to F
charge it with 2 µC.
D 8.85 ð 4 F D 8.85 ð 4 ð 1012 pF
3. The charge on the plates of a capacitor is 6 mC when 1010 1010
the potential between them is 2.4 kV. Determine the
capacitance of the capacitor. D 3540 pF
(b) Q D CV thus V D Q D 1.2 ð 10 6 V D 339 V
C 3540 ð 10 12
306 Science for Engineering
Problem 6. A waxed paper capacitor has two parallel 4. A parallel plate capacitor is made from 25 plates, each
plates, each of effective area 800 cm2. If the capacitance 70 mm by 120 mm interleaved with mica of relative
of the capacitor is 4425 pF determine the effective permittivity 5. If the capacitance of the capacitor is
thickness of the paper if its relative permittivity is 2.5 3000 pF determine the thickness of the mica sheet.
A D 800 cm2 D 800 ð 10 4 m2 D 0.08 m2, 5. A capacitor is constructed with parallel plates and has
C D 4425 pF D 4425 ð 10 12 F, e0 D 8.85 ð 10 12 F/m and a value of 50 pF. What would be the capacitance of
er D 2.5 the capacitor if the plate area is doubled and the plate
spacing is halved?
Since C D e0erA then d D e0erA
dC 41.3 Capacitors connected in parallel and
series
8.85 ð 10 12 ð 2.5 ð 0.08
D 4425 ð 10 12 D 0.0004 m For n parallel-connected capacitors, the equivalent capac-
itance, C, is given by:
Hence, the thickness of the paper is 0.4 mm
C = C1 + C2 + C3 · · · + Cn
Problem 7. A parallel plate capacitor has nineteen
interleaved plates each 75 mm by 75 mm separated (similar to resistors connected in series.)
by mica sheets 0.2 mm thick. Assuming the relative Also, total charge, Q, is given by:
permittivity of the mica is 5, calculate the capacitance
of the capacitor. Q = Q1 + Q2 + Q3
n D 19 thus n 1 D 18, A D 75 ð 75 D 5625 mm2 D For n series-connected capacitors the equivalent capaci-
5625 ð 10 6 m2, er D 5, e0 D 8.85 ð 10 12 F/m and tance, C, is given by:
d D 0.2 mm D 0.2 ð 10 3 m.
1111 1
Capacitance, C D e0erA⊲n 1⊳ = + + +···+
d C C1 C2 C3 Cn
8.85 ð 10 12 ð 5 ð 5625 ð 10 6 ð 18 (similar to resistors connected in parallel.)
D 0.2 ð 10 3 F When connected in series the charge on each capacitor is
D 0.0224 µF or 22.4 nF
the same.
Now try the following exercise
Problem 8. Calculate the equivalent capacitance of
two capacitors of 6 µF and 4 µF connected (a) in parallel
and (b) in series.
Exercise 182 Further problems on parallel plate (a) In parallel, equivalent capacitance
capacitors (Answers on page 358)
C D C1 C C2 D 6 µF C 4 µF D 10 µF
(Where appropriate take e0 as 8.85 ð 10 12 F/m)
(b) For the special case of two capacitors in series:
1. A capacitor consists of two parallel plates each of
area 0.01 m2, spaced 0.1 mm in air. Calculate the 1 D 1 C 1 D C2 C C1
capacitance in picofarads. C C1 C2 C1 C C2
2. A waxed paper capacitor has two parallel plates, each Hence C = C1C2 product
of effective area 0.2 m2. If the capacitance is 4000 pF C1 + C2 i.e. sum
determine the effective thickness of the paper if its
relative permittivity is 2. Thus 6 ð 4 24
C D 6 C 4 D 10 D 2.4 µF
3. How many plates has a parallel plate capacitor having
a capacitance of 5 nF, if each plate is 40 mm by 40 mm Problem 9. What capacitance must be connected in
and each dielectric is 0.102 mm thick with a relative series with a 30 µF capacitor for the equivalent capaci-
permittivity of 6. tance to be 12 µF?
Capacitors and inductors 307
Let C D 12 µF (the equivalent capacitance), C1 D 30 µF and The voltage across the 6 µF capacitor,
C2 be the unknown capacitance.
Q 0.6 ð 10 3
For two capacitors in series V2 D C2 D 6 ð 10 6 D 100 V
The voltage across the 12 µF capacitor,
11 1
DC Q 0.6 ð 10 3
V3 D C3 D 12 ð 10 6 D 50 V
C C1 C2 [Check: In a series circuit V D V1 C V2 C V3
Hence 11 1 D C1 C V1 C V2 C V3 D 200 C 100 C 50
and D
C2 C C1 CC1 D 350 V D supply voltage]
C2 D CC1 D 12 ð 30 D 360 D 20 µF In practice, capacitors are rarely connected in series unless
C1 C 30 12 18 they are of the same capacitance. The reason for this can
be seen from the above problem where the lowest valued
Problem 10. Capacitance’s of 3 µF, 6 µF and 12 µF are capacitor (i.e. 3 µF) has the highest p.d. across it (i.e. 200 V)
connected in series across a 350 V supply. Calculate which means that if all the capacitors have an identical
(a) the equivalent circuit capacitance, (b) the charge on construction they must all be rated at the highest voltage.
each capacitor, and (c) the p.d. across each capacitor.
The circuit diagram is shown in Fig. 41.3.
C1 = 3 µF C2 = 6 µF C3 = 12 µF Now try the following exercise
V1 V2 V3 Exercise 183 Further problems on capacitors in par-
I V = 350 v allel and series (Answers on page 358)
Fig. 41.3 1. Capacitors of 2 µF and 6 µF are connected (a) in paral-
lel and (b) in series. Determine the equivalent capac-
(a) The equivalent circuit capacitance C for three capacitors itance in each case.
in series is given by:
2. Find the capacitance to be connected in series with a
11 1 1 10 µF capacitor for the equivalent capacitance to be
DCC 6 µF.
C C1 C2 C3 3. What value of capacitance would be obtained if capac-
itors of 0.15 µF and 0.10 µF are connected (a) in series
i.e. 1 1 1 1 4C2C1 7 and (b) in parallel.
C D 3 C 6 C 12 D 12 D 12
4. Two 6 µF capacitors are connected in series with one
Hence the equivalent circuit capacitance, having a capacitance of 12 µF. Find the total equivalent
circuit capacitance. What capacitance must be added
in series to obtain a capacitance of 1.2 µF?
5. For the arrangement shown in Fig. 41.4 find (a) the
equivalent circuit capacitance and (b) the voltage
across a 4.5 µF capacitor.
12 5 4.5 µF 4.5 µF 4.5 µF
C = 7 = 1 7 µF or 1.714 µF
(b) Total charge QT D CV, 3 µF
hence 12 6 ð 350 D 600 µC or 0.6 mC 1 µF 1 µF
QT D 7 ð 10
Since the capacitors are connected in series 0.6 mC is
the charge on each of them.
(c) The voltage across the 3 µF capacitor, 500 V
Q 0.6 ð 10 3 Fig. 41.4
V1 D C1 D 3 ð 10 6 D 200 V
308 Science for Engineering
6. In Fig. 41.5 capacitors P, Q and R are identical and 41.5 Energy stored in capacitors
the total equivalent capacitance of the circuit is 3 µF.
Determine the values of P, Q and R. The energy, W, stored by a capacitor is given by
2 µF 3.5 µF W = 1 CV 2 joules
4.5 µF 2
Problem 12. (a) Determine the energy stored in a 3 µF
capacitor when charged to 400 V (b) Find also the
average power developed if this energy is dissipated
in a time of 10 µs.
(a) Energy stored,
PQ R W D 1 CV2 joules
2
Fig. 41.5 D 1 6 ð 4002 D 3 ð 16 ð 10 2
2 ð 3 ð 10 2
D 0.24 J
41.4 Dielectric strength energy 0.24
(b) Power D time D 10 ð 10 6 W D 24 kW
The maximum amount of field strength that a dielec- Problem 13. A 12 µF capacitor is required to store 4 J
tric can withstand is called the dielectric strength of the of energy. Find the p.d. to which the capacitor must be
material. charged.
Dielectric strength, Em = Vm 1 CV2 2W
d 2 C
Energy stored W D hence V2 D
Problem 11. A capacitor is to be constructed so that and p.d. V D 2W 2 ð 4
its capacitance is 0.2 µF and to take a p.d. of 1.25 kV D
across its terminals. The dielectric is to be mica which C 12 ð 10 6
has a dielectric strength of 50 MV/m. Find (a) the thick-
ness of the mica needed, and (b) the area of a plate 2 ð 106
assuming a two-plate construction (Assume er for mica D D 816.5 V
to be 6).
3
V Now try the following exercise
(a) Dielectric strength, E D
Exercise 184 Further problems on energy stored in
d capacitors (Answers on page 358)
i.e. V 1.25 ð 103 m D 0.025 mm (Assume e0 D 8.85 ð 10 12 F/m.)
dD D
E 50 ð 106 1. When a capacitor is connected across a 200 V supply
the charge is 4 µC. Find (a) the capacitance and (b) the
(b) Capacitance, C D e0erA hence, energy stored.
d
2. Find the energy stored in a 10 µF capacitor when
area Cd 0.2 ð 10 6 ð 0.025 ð 10 3 charged to 2 kV.
AD D
e0er 8.85 ð 10 12 ð 6 m2 3. A 3300 pF capacitor is required to store 0.5 mJ of
energy. Find the p.d. to which the capacitor must be
D 0.09416 m2 charged.
D 941.6 cm2
Capacitors and inductors 309
4. A bakelite capacitor is to be constructed to have a Mica sheets
capacitance of 0.04 µF and to have a steady working
potential of 1 kV maximum. 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.
41.6 Practical types of capacitor Metal foil
(lead or aluminium)
Practical types of capacitor are characterised by the material
used for their dielectric. The main types include: variable air, Fig. 41.7
mica, paper, ceramic, plastic, titanium oxide and electrolytic.
working voltage rating and a long service life and are used
1. Variable air capacitors. These usually consist of two sets in high frequency circuits with fixed values of capacitance
of metal plates (such as aluminium), one fixed, the other up to about 1000 pF.
variable. The set of moving plates rotate on a spindle as
shown by the end view of Fig. 41.6. 3. Paper capacitors. A typical paper capacitor is shown in
Fig. 41.8 where the length of the roll corresponds to the
capacitance required.
Spindle
Moving Metal end cap for
plate connection to
metal foil
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
Fixed ;;;;;;;;;;;;;;;;;
plate ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
Fig. 41.6
As the moving plates are rotated through half a revo- Metal Paper
lution, the meshing, and therefore the capacitance, varies foil
from a minimum to a maximum value. Variable air capac-
itors are used in radio and electronic circuits where very Fig. 41.8
low losses are required, or where a variable capacitance
is needed. The maximum value of such capacitors is The whole is usually impregnated with oil or wax
between 500 pF and 1000 pF. to exclude moisture, and then placed in a plastic or
aluminium container for protection. Paper capacitors are
2. Mica capacitors. A typical older type construction is made in various working voltages up to about 150 kV and
shown in Fig. 41.7. are used where loss is not very important. The maximum
Usually the whole capacitor is impregnated with wax value of this type of capacitor is between 500 pF and
and placed in a bakelite case. Mica is easily obtained 10 µF. Disadvantages of paper capacitors include variation
in thin sheets and is a good insulator. However, mica in capacitance with temperature change and a shorter
is expensive and is not used in capacitors above about service life than most other types of capacitor.
0.2 µF.
A modified form of mica capacitor is the silvered mica 4. Ceramic capacitors. These are made in various forms,
type. The mica is coated on both sides with a thin layer of each type of construction depending on the value of
silver which forms the plates. Capacitance is stable and
less likely to change with age. Such capacitors have a
constant capacitance with change of temperature, a high
310 Science for Engineering
capacitance required. For high values, a tube of ceramic Ceramic
material is used as shown in the cross section of Fig. 41.9. disc
For smaller values the cup construction is used as shown
in Fig. 41.10, and for still smaller values the disc con- Conducting
struction shown in Fig. 41.11 is used. Certain ceramic coatings
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.
Ceramic capacitors are available in the range 1 pF to
0.1 µF and may be used in high frequency electronic cir-
cuits subject to a wide range of temperatures.
;;;;;;;;;;;;;;;;;;
Connection Connection Fig. 41.11
;;;;;;;;
Ceramic Conducting 7. Electrolytic capacitors. Construction is similar to the
tube coating paper capacitor with aluminium foil used for the plates
and with a thick absorbent material, such as paper,
Fig. 41.9 (eg. silver) impregnated with an electrolyte (ammonium borate), sep-
arating the plates. The finished capacitor is usually assem-
Connection Ceramic cup bled in an aluminium container and hermetically sealed.
Its operation depends on the formation of a thin alu-
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; minium oxide layer on the positive plate by electrolytic
action when a suitable direct potential is maintained
between the plates. This oxide layer is very thin and forms
the dielectric. (The absorbent paper between the plates
is a conductor and does not act as a dielectric). Such
capacitors must always be used on d.c. 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. Electrolytic capacitors are manufactured with
working voltages from 6 V to 600 V, although accuracy is
generally not very high. These capacitors possess a much
larger capacitance than other types of capacitors of sim-
ilar dimensions due to the oxide film being only a few
microns thick. The fact that they can be used only on d.c.
supplies limit their usefulness.
Conducting 41.7 Discharging capacitors
coating
When a capacitor has been disconnected from the supply it
Fig. 41.10 may still be charged and it may retain this charge for some
considerable time. Thus precautions must be taken to ensure
5. Plastic capacitors. Some plastic materials such as polyst- that the capacitor is automatically discharged after the supply
yrene and Teflon can be used as dielectrics. Construction is switched off. This is done by connecting a high value
is similar to the paper capacitor but using a plastic film resistor across the capacitor terminals.
instead of paper. Plastic capacitors operate well under
conditions of high temperature, provide a precise value of 41.8 Inductance
capacitance, a very long service life and high reliability.
Inductance is the property of a circuit whereby there is an
6. Titanium oxide capacitors have a very high capacitance e.m.f. induced into the circuit by the change of flux linkages
with a small physical size when used at a low tempe- produced by a current change. When the e.m.f. is induced in
rature.
Capacitors and inductors 311
the same circuit as that in which the current is changing, the Induced e.m.f. E D dI ⊲12⊳⊲4⊳ D 48 volts.
property is called self-inductance L. L D
dt
The unit of inductance is the henry, H:
A circuit has an inductance of one henry when an e.m.f. of Problem 17. An e.m.f. of 1.5 kV is induced in a coil
one volt is induced in it by a current changing at the rate when a current of 4 A collapses uniformly to zero in
of one ampere per second. 8 ms. Determine the inductance of the coil.
Induced e.m.f. in a coil of N turns, Change in current,
d8 dI D ⊲4 0⊳ D 4 A, dt D 8 ms D 8 ð 10 3 s,
E = −N volts
dI 4 4000
dt dt D 8 ð 10 3 D 8 D 500 A/s
and E D 1.5 kV D 1500 V
where d is the change in flux in Webers, and dt is the time
taken for the flux to change in seconds (i.e. d is the rate
dt
of change of flux).
dI
Induced e.m.f. in a coil of inductance L henrys, Since jEj D L
dt
dI inductance, jEj 1500
E = −L volts L D dI D 500 D 3 H
dt dt
where dI is the change in current in amperes and dt is the (Note that jEj means the ‘magnitude of E’ which disregards
time taken for the current to change in seconds (i.e. dI is the minus sign.)
dt Now try the following exercise
the rate of change of current).
Exercise 185 Further problems on inductance
The minus sign in each of the above two equations remind (Answers on page 358)
us of its direction (given by Lenz’s law). 1. Find the e.m.f. induced in a coil of 200 turns when
there is a change of flux of 30 mWb linking with it in
Problem 14. Determine the e.m.f. induced in a coil of 40 ms.
200 turns when there is a change of flux of 25 mWb
linking with it in 50 ms. 2. An e.m.f. of 25 V is induced in a coil of 300 turns
when the flux linking with it changes by 12 mWb. Find
d 25 ð 10 3 the time, in milliseconds, in which the flux makes the
Induced e.m.f. E D N D ⊲200⊳ change.
dt 50 ð 10 3
3. An ignition coil having 10 000 turns has an e.m.f. of
D −100 volts 8 kV induced in it. What rate of change of flux is
required for this to happen?
Problem 15. A flux of 400 µWb passing through a
150-turn coil is reversed in 40 ms. Find the average 4. A flux of 0.35 mWb passing through a 125-turn coil is
e.m.f. induced. reversed in 25 ms. Find the magnitude of the average
e.m.f. induced.
Since the flux reverses, the flux changes from C400 µWb to
400 µWb, a total change of flux of 800 µWb. 5. Calculate the e.m.f. induced in a coil of inductance
6 H by a current changing at a rate of 15 A/s.
Induced e.m.f. E D d ⊲150⊳ 800 ð 10 6
N D 40 ð 10 3
dt
150 ð 800 ð 103
D 40 ð 106
Hence, the average e.m.f. induced, E = −3 volts. 41.9 Inductors
Problem 16. Calculate the e.m.f. induced in a coil A component called an inductor is used when the property
of inductance 12 H by a current changing at the rate of inductance is required in a circuit. The basic form of an
of 4 A/s. inductor is simply a coil of wire. Factors which affect the
inductance of an inductor include:
312 Science for Engineering
(i) the number of turns of wire – the more turns the higher Insulator
the inductance. Wire
(ii) the cross-sectional area of the coil of wire – the greater Fig. 41.14
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.
41.10 Practical inductors 41.11 Energy stored by inductors
Two examples of practical inductors are shown in Fig. 41.12, An inductor possesses an ability to store energy. The energy
and the standard electrical circuit diagram symbols for air- stored, W, in the magnetic field of an inductor is given by:
cored and iron-cored inductors are shown in Fig. 41.13.
W = 1 LI 2 joules
Laminated 2
Iron iron core
core Problem 18. An 8 H inductor has a current of 3 A
flowing through it. How much energy is stored in the
magnetic field of the inductor?
Wire Coil of Energy stored, W D 1 LI2 D 1 ⊲8⊳⊲3⊳2 D 36 joules
(a) wire 22
Fig. 41.12 (b) Now try the following exercise
Air-cored inductor Exercise 186 Further problems on energy stored
Iron-cored inductor (Answers on page 358)
Fig. 41.13 1. An inductor of 20 H has a current of 2.5 A flowing in
it. Find the energy stored in the magnetic field of the
inductor.
2. Calculate the value of the energy stored when a current
of 30 mA is flowing in a coil of inductance 400 mH.
3. The energy stored in the magnetic field of an inductor
is 80 J when the current flowing in the inductor is 2 A.
Calculate the inductance of the coil.
An iron-cored inductor is often called a choke since, when 41.12 Inductance of a coil
used in a.c. circuits, it has a choking effect, limiting the
current flowing through it. Inductance is often undesirable in If a current changing from 0 to I amperes, produces a flux
a circuit. To reduce inductance to a minimum the wire may change from 0 to webers, then dI D I and d D . Then,
be bent back on itself, as shown in Fig. 41.14, so that the from Section 41.8,
magnetising effect of one conductor is neutralised by that
of the adjacent conductor. The wire may be coiled around N LI
an insulator, as shown, without increasing the inductance. induced e.m.f. E D D
Standard resistors may be non-inductively wound in this
manner. tt
from which, inductance of coil, N8
L = henrys
I
Capacitors and inductors 313
Problem 19. Calculate the coil inductance when a cur- 2. An e.m.f. of 2 kV is induced in a coil when a current
rent of 4 A in a coil of 800 turns produces a flux of of 5 A collapses uniformly to zero in 10 ms. Determine
5 mWb linking with the coil. the inductance of the coil.
For a coil, inductance L D N D ⊲800⊳⊲5 ð 10 3⊳ D 1H 3. An average e.m.f. of 60 V is induced in a coil of induc-
I 4 tance 160 mH when a current of 7.5 A is reversed.
Calculate the time taken for the current to reverse.
Problem 20. A flux of 25 mWb links with a 1500 turn
coil when a current of 3 A passes through the coil. 4. A coil of 2500 turns has a flux of 10 mWb linking
Calculate (a) the inductance of the coil, (b) the energy with it when carrying a current of 2 A. Calculate the
stored in the magnetic field, and (c) the average e.m.f. coil inductance and the e.m.f. induced in the coil when
induced if the current falls to zero in 150 ms. the current collapses to zero in 20 ms.
(a) Inductance, 5. When a current of 2 A flows in a coil, the flux linking
with the coil is 80 µWb. If the coil inductance is 0.5 H,
N ⊲1500⊳⊲25 ð 10 3⊳ calculate the number of turns of the coil.
LD D D 12.5 H
I3 Exercise 188 Short answer questions on capacitors
and inductors
(b) Energy stored,
1. How can an ‘electric field’ be established between
W D 1 LI2 D 1 ⊲12.5⊳⊲3⊳2 D 56.25 J two parallel metal plates?
2 2
2. What is capacitance?
(c) Induced emf, 3. State the unit of capacitance.
dI 3 0 ......
ED L D ⊲12.5⊳ D −250 V 4. Complete the statement: Capacitance D
dt 150 ð 10 3
ÐÐÐÐÐÐ
(Alternatively, 5. Complete the statements:
ED d ⊲1500⊳ 25 ð 10 3 250 V (a) 1 µF D . . . F (b) 1 pF D . . . F
N D 150 ð 10 3 D 6. Comp.l.e.te. . .the statement: Electric field strength
dt
ED
since if the current falls to zero so does the flux.) ÐÐÐÐÐÐ
Problem 21. When a current of 1.5 A flows in a coil 7. Comp.l.e.te. . . the statement: Electric flux density
the flux linking with the coil is 90 µWb. If the coil DD
inductance is 0.60 H, calculate the number of turns of ÐÐÐÐÐÐ
the coil.
8. Draw the electrical circuit diagram symbol for a
N capacitor.
For a coil, L D ,
9. Name two practical examples where capacitance is
I present, although undesirable.
thus LI ⊲0.6⊳⊲1.5⊳ D 10 000 turns 10. The insulating material separating the plates of a
ND D capacitor is called the . . . . . ..
90 ð 10 6
11. 10 volts applied to a capacitor results in a charge of
Now try the following exercises 5 coulombs. What is the capacitance of the capacitor?
Exercise 187 Further problems on the inductance of 12. Three 3 µF capacitors are connected in parallel. The
a coil (Answers on page 358) equivalent capacitance is . . ..
1. A flux of 30 mWb links with a 1200 turn coil when 13. Three 3 µF capacitors are connected in series. The
a current of 5 A is passing through the coil. Calculate equivalent capacitance is . . ..
(a) the inductance of the coil, (b) the energy stored in
the magnetic field, and (c) the average e.m.f. induced 14. State an advantage of series-connected capacitors.
if the current is reduced to zero in 0.20 s. 15. Name three factors upon which capacitance depends.
16. What does ‘relative permittivity’ mean?
17. Define ‘permittivity of free space’.
18. What is meant by the ‘dielectric strength’ of a mate-
rial?
19. State the formula used to determine the energy stored
by a capacitor.
314 Science for Engineering The capacitance of a capacitor
(a) is proportional to the cross-sectional area of the
20. Name five types of capacitor commonly used.
21. Sketch a typical rolled paper capacitor. plates
22. Explain briefly the construction of a variable air (b) is proportional to the distance between the plates
capacitor. (c) depends on the number of plates
23. State three advantages and one disadvantage of mica (d) is proportional to the relative permittivity of the
capacitors. dielectric
24. Name two disadvantages of paper capacitors. 7. Which of the following statement is false?
25. Between what values of capacitance are ceramic
(a) An air capacitor is normally a variable type
capacitors normally available. (b) A paper capacitor generally has a shorter service
26. What main advantages do plastic capacitors possess?
27. Explain briefly the construction of an electrolytic life than most other types of capacitor
(c) An electrolytic capacitor must be used only on
capacitor.
28. What is the main disadvantage of electrolytic capac- a.c. supplies
(d) Plastic capacitors generally operate satisfactorily
itors?
29. Name an important advantage of electrolytic capaci- under conditions of high temperature
tors. 8. The energy stored in a 10 µF capacitor when charged
30. What safety precautions should be taken when a to 500 V is
(a) 1.25 mJ (b) 0.025 µJ (c) 1.25 J (d) 1.25 C
capacitor is disconnected from a supply?
31. Define inductance and name its unit. 9. The capacitance of a variable air capacitor is at
32. What factors affect the inductance of an inductor? maximum when
33. What is an inductor? Sketch a typical practical induc- (a) the movable plates half overlap the fixed plates
(b) the movable plates are most widely separated
tor. from the fixed plates
34. Explain how a standard resistor may be non-induc- (c) both sets of plates are exactly meshed
(d) the movable plates are closer to one side of the
tively wound. fixed plate than to the other
35. Energy W stored in the magnetic field of an inductor
10. When a voltage of 1 kV is applied to a capacitor, the
is given by: W D . . . . . . joules. charge on the capacitor is 500 nC. The capacitance
of the capacitor is:
Exercise 189 Multi-choice questions on capacitors (a) 2 ð 109F (b) 0.5 pF (c) 0.5 mF (d) 0.5 nF
and inductance (Answers on page 358)
11. When a magnetic flux of 10 Wb links with a circuit
1. The capacitance of a capacitor is the ratio of 20 turns in 2 s, the induced e.m.f. is:
(a) charge to p.d. between plates (a) 1 V (b) 4 V (c) 100 V (d) 400 V
(b) p.d. between plates to plate spacing
(c) p.d. between plates to thickness of dielectric 12. A current of 10 A in a coil of 1000 turns produces
(d) p.d. between plates to charge a flux of 10 mWb linking with the coil. The coil
inductance is:
2. The p.d. across a 10 µF capacitor to charge it with (a) 106 H (b) 1 H (c) 1 µH (d) 1 mH
10 mC is
(a) 10 V (b) 1 kV (c) 1 V (d) 10 V 13. Which of the following statements is false?
The inductance of an inductor increases:
3. The charge on a 10 pF capacitor when the voltage (a) with a short, thick, coil
applied to it is 10 kV is (b) when wound on an iron core
(a) 100 µC (b) 0.1 C (c) 0.1 µC (d) 0.01 µC (c) as the number of turns increases
(d) as the cross-sectional area of the coil decreases
4. Four 2 µF capacitors are connected in parallel. The
equivalent capacitance is 14. Self-inductance occurs when:
(a) 8 µF (b) 0.5 µF (c) 2 µF (d) 6 µF (a) the current is changing
(b) the circuit is changing
5. Four 2 µF capacitors are connected in series. The (c) the flux is changing
equivalent capacitance is (d) the resistance is changing
(a) 8 µF (b) 0.5 µF (c) 2 µF (d) 6 µF
6. State which of the following is false.
42
Electrical measuring instruments and
measurements
At the end of this chapter you should be able to: 42.1 Introduction
ž recognise the importance of testing and measure- Tests and measurements are important in designing, evaluat-
ments in electric circuits ing, maintaining and servicing electrical circuits and equip-
ment. In order to detect electrical quantities such as current,
ž appreciate the essential devices comprising an ana- voltage, resistance or power, it is necessary to transform an
logue instrument electrical quantity or condition into a visible indication. This
is done with the aid of instruments (or meters) that indi-
ž explain the operation of attraction and repulsion cate the magnitude of quantities either by the position of a
type moving-iron instruments pointer moving over a graduated scale (called an analogue
instrument) or in the form of a decimal number (called a
ž explain the operation of a moving-coil rectifier digital instrument).
instrument
42.2 Analogue instruments
ž compare moving-coil, moving-iron and moving coil
rectifier instruments All analogue electrical indicating instruments require three
essential devices:
ž calculate values of shunts for ammeters and multi-
pliers for voltmeters (a) A deflecting or operating device. A mechanical force
is produced by the current or voltage which causes the
ž understand the advantages of electronic instruments pointer to deflect from its zero position.
ž understand the operation of an ohmmeter/megger (b) A controlling device. The controlling force acts in oppo-
sition to the deflecting force and ensures that the deflec-
ž appreciate the operation of multimeters/Avometers tion shown on the meter is always the same for a given
measured quantity. It also prevents the pointer always
ž understand the operation of a wattmeter going to the maximum deflection. There are two main
types of controlling device – spring control and gravity
ž appreciate instrument ‘loading’ effect control.
ž understand the operation of a C.R.O. for d.c. and (c) A damping device. The damping force ensures that the
a.c. measurements pointer comes to rest in its final position quickly and
without undue oscillation. There are three main types
ž calculate periodic time, frequency, peak to peak of damping used – eddy-current damping, air-friction
values from waveforms on a C.R.O. damping and fluid-friction damping.
ž understand null methods of measurement for a
Wheatstone bridge and d.c. potentiometer
ž appreciate the most likely source of errors in mea-
surements
ž appreciate calibration accuracy of instruments
316 Science for Engineering
There are basically two types of scale – linear and non- Pointer Scale
linear. I
;;;;;;;;;;;;;; Pivot and controlling
A linear scale is shown in Fig. 42.1(a), where the divi- ;;; spring
sions or graduations are evenly spaced. The voltmeter shown ;;;;;;;;;;;;;;
has a range 0–100 V, i.e. a full-scale deflection (f.s.d.) of Air-piston damping
100 V. A non-linear scale is shown in Fig. 42.1(b) where
the scale is cramped at the beginning and the graduations
are uneven throughout the range. The ammeter shown has a
f.s.d. of 10 A.
Solenoid Soft iron
disc
40 60 (a) ATTRACTION TYPE
20 80
Scale
0 100
V
(a) ;;; Solenoid
Pointer Pivot and controlling
spring
2 58
0 A Fixed piece of iron
(b) Movable piece of iron
Fig. 42.1 10
(b) REPULSION TYPE
Fig. 42.2
Id m.c.
milliammeter
42.3 Moving-iron instrument + Im Average
0
(a) An attraction type of moving-iron instrument is shown A value
diagrammatically in Fig. 42.2(a). When current flows in – t = 0.637 Im
the solenoid, a pivoted soft-iron disc is attracted towards
the solenoid and the movement causes a pointer to move Ia
across a scale.
i t
(b) In the repulsion type moving-iron instrument shown 0
diagrammatically in Fig. 42.2(b), two pieces of iron are
placed inside the solenoid, one being fixed, and the other Fig. 42.3
attached to the spindle carrying the pointer. When current
passes through the solenoid, the two pieces of iron are in Fig. 42.3 to provide an indication of alternating currents
magnetised in the same direction and therefore repel each and voltages (see Chapter 40). The average value of the full
other. The pointer thus moves across the scale. The force wave rectified current is 0.637 Im. However, a meter being
moving the pointer is, in each type, proportional to I2 and used to measure a.c. is usually calibrated in r.m.s. values.
because of this the direction of current does not matter. For sinusoidal quantities the indication is 0.707Im i.e. 1.11
The moving-iron instrument can be used on d.c. or a.c.;
the scale, however, is non-linear. 0.637Im
times the mean value.
42.4 The moving-coil rectifier instrument
Rectifier instruments have scales calibrated in r.m.s. quan-
A moving-coil instrument, which measures only d.c., may be tities and it is assumed by the manufacturer that the a.c. is
used in conjunction with a bridge rectifier circuit as shown sinusoidal.
Electrical measuring instruments and measurements 317
42.5 Comparison of moving-coil, moving-iron and moving-coil rectifier instruments
Type of Moving-coil Moving-iron Moving-coil rectifier
instrument Direct current and voltage
Direct and alternating currents Alternating current and voltage
Suitable for and voltage (reading in r.m.s. (reads average value but scale
measuring value) is adjusted to give r.m.s. value
for sinusoidal waveforms)
Scale Linear Non-linear Linear
Hairsprings Hairsprings
Method of control Eddy current Air Hairsprings
Method of – Eddy current
damping
Frequency limits 20–200 Hz 20–100 kHz
Advantages 1 Linear scale 1 Robust construction 1 Linear scale
Disadvantages 2 High sensitivity 2 Relatively cheap 2 High sensitivity
3 Well shielded from stray 3 Measures dc and ac 3 Well shielded from stray
4 In frequency range
magnetic fields magnetic fields
4 Lower power consumption 20–100 Hz reads rms 4 Low power consumption
correctly regardless of 5 Good frequency range
1 Only suitable for dc supply wave-form
2 More expensive than 1 More expensive than moving
1 Non-linear scale iron type
moving iron type 2 Affected by stray magnetic
3 Easily damaged 2 Errors caused when supply
fields is non-sinusoidal
3 Hysteresis errors in dc
circuits
4 Liable to temperature errors
5 Due to the inductance of the
solenoid, readings can be
affected by variation of
frequency
(For the principle of operation of a moving-coil instrument, see Chapter 38, page 283).
42.6 Shunts and multipliers Thus the value of the shunt,
An ammeter, which measures current, has a low resistance RS = Ia ra ohms
(ideally zero) and must be connected in series with the circuit. IS
A voltmeter, which measures p.d., has a high resistance The milliammeter is converted into a voltmeter by connecting
(ideally infinite) and must be connected in parallel with the a high value resistance (called a multiplier) in series with it
part of the circuit whose p.d. is required. as shown in Fig. 42.4(b).
There is no difference between the basic instrument used From Fig. 42.4(b),
to measure current and voltage since both use a milliammeter
as their basic part. This is a sensitive instrument which gives V D Va C VM D Ira C IRM
f.s.d. for currents of only a few milliamperes. When an
ammeter is required to measure currents of larger magnitude, Thus the value of the multiplier,
a proportion of the current is diverted through a low-value
resistance connected in parallel with the meter. Such a RM = V − Ira ohms
diverting resistor is called a shunt. I
From Fig. 42.4(a), VPQ D VRS. Hence Iara D ISRS
318 Science for Engineering
I P Ia A QI Hence Rs D Ia ra D ⊲0.04⊳⊲25⊳ D 0.02002 D 20.02 mZ
Is ra Is 49.96
Thus for the moving-coil instrument to be used as an ammeter
with a range 0–50 A, a resistance of value 20.02 m needs
to be connected in parallel with the instrument.
R Rs S Problem 2. A moving-coil instrument having a resis-
(shunt) tance of 10 , gives a f.s.d. when the current is 8 mA.
Calculate the value of the multiplier to be connected in
(a) series with the instrument so that it can be used as a
voltmeter for measuring p.d.s. up to 100 V.
(multiplier)
RM
A
ra
Va VM The circuit diagram is shown in Fig. 42.6,
I
where ra D resistance of instrument D 10 ,
V RM D resistance of multiplier,
I D total permissible instrument current D 8 mA
(b) D 0.008 A,
V D total p.d. required to give f.s.d. D 100 V
Fig. 42.4
Problem 1. A moving-coil instrument gives a f.s.d. I =8 mA RM
when the current is 40 mA and its resistance is 25 . A VM
Calculate the value of the shunt to be connected in ra
parallel with the meter to enable it to be used as an
ammeter for measuring currents up to 50 A. Va
V =100 V
Fig. 42.6
The circuit diagram is shown in Fig. 42.5, V D Va C VM D Ira C IRM
where ra D resistance of instrument D 25 , i.e. 100 D ⊲0.008⊳⊲10⊳ C ⊲0.008⊳RM
Rs D resistance of shunt,
Ia D maximum permissible current flowing in or 100 0.08 D 0.008RM
instrument D 40 mA D 0.04 A, thus 99.92
Is D current flowing in shunt and
I D total circuit current required to gives RM D 0.008 D 12 490 D 12.49 kZ
f.s.d. D 50 A
I =50 A Ia = 40 mA 50 A Hence for the moving-coil instrument to be used as a volt-
Is A meter with a range 0–100 V, a resistance of value 12.49 k
ra needs to be connected in series with the instrument.
V
Rs Now try the following exercise
Fig. 42.5 Exercise 190 Further problems on shunts and mul-
tipliers (Answers on page 358)
Since I D Ia C Is 0.04 D 49.96 A
then Is D I Ia D 50 1. A moving-coil instrument gives f.s.d. for a current
V D Iara D IsRs of 10 mA. Neglecting the resistance of the instru-
ment, calculate the approximate value of series resis-
tance needed to enable the instrument to measure
up to
(a) 20 V (b) 100 V (c) 250 V
Electrical measuring instruments and measurements 319
2. A meter of resistance 50 has a f.s.d. of 4 mA. a self-contained source of voltage, such as a battery. Initially,
Determine the value of shunt resistance required in terminals XX are short-circuited and R adjusted to give f.s.d.
order that f.s.d. should be on the milliammeter. If current I is at a maximum value and
(a) 15 mA (b) 20 A (c) 100 A voltage E is constant, then resistance R D E/I is at a mini-
mum value. Thus f.s.d. on the milliammeter is made zero on
3. A moving-coil instrument having a resistance of 20 , the resistance scale. When terminals XX are open circuited
gives a f.s.d. when the current is 5 mA. Calculate the no current flows and R⊲D E/O⊳ is infinity, 1.
value of the multiplier to be connected in series with
the instrument so that it can be used as a voltmeter for R
measuring p.d.’s up to 200 V. A
4. A moving-coil instrument has a f.s.d. of 20 mA and a E
resistance of 25 . Calculate the values of resistance
required to enable the instrument to be used (a) as (a) x Unknown x
a 0–10 A ammeter, and (b) as a 0–100 V voltmeter. resistance
State the mode of resistance connection in each case.
1000 500 200 100
42.7 Electronic instruments 1 F.S.D 3
2000 1 F.S.D 4 F.S.D
Electronic measuring instruments have advantages over 2
instruments such as the moving-iron or moving-coil meters,
in that they have a much higher input resistance (some as ∞ 4 0
high as 1000 M) and can handle a much wider range of 0 F.S.D
frequency (from d.c. up to MHz). Resistance(Ω)
Current(mA)
The digital voltmeter (DVM) is one which provides a
digital display of the voltage being measured. Advantages (b)
of a DVM over analogue instruments include higher accu-
racy and resolution, no observational or parallex errors (see Fig. 42.7
Section 42.16) and a very high input resistance, constant on
all ranges. A digital multimeter is a DVM with additional The milliammeter can thus be calibrated directly in ohms.
circuitry which makes it capable of measuring a.c. voltage, A cramped (non-linear) scale results and is ‘back to front’,
d.c. and a.c. current and resistance. as shown in Fig. 42.7(b). When calibrated, an unknown
resistance is placed between terminals XX and its value
Instruments for a.c. measurements are generally calibrated determined from the position of the pointer on the scale. An
with a sinusoidal alternating waveform to indicate r.m.s. ohmmeter designed for measuring low values of resistance
values when a sinusoidal signal is applied to the instrument. is called a continuity tester. An ohmmeter designed for
Some instruments, such as the moving-iron and electro- measuring high values of resistance (i.e. megohms) is called
dynamic instruments, give a true r.m.s. indication. With other an insulation resistance tester (e.g. ‘Megger’).
instruments the indication is either scaled up from the mean
value (such as with the rectified moving-coil instrument) or 42.9 Multimeters
scaled down from the peak value. Sometimes quantities to be
measured have complex waveforms, and whenever a quantity Instruments are manufactured that combine a moving-coil
is non-sinusoidal, errors in instrument readings can occur if meter with a number of shunts and series multipliers, to
the instrument has been calibrated for sine waves only. Such provide a range of readings on a single scale graduated
waveform errors can be largely eliminated by using electronic to read current and voltage. If a battery is incorporated
instruments. then resistance can also be measured. Such instruments are
called multimeters or universal instruments or multirange
42.8 The ohmmeter instruments. An ‘Avometer’ is a typical example. A par-
ticular range may be selected either by the use of separate
An ohmmeter is an instrument for measuring electrical resis- terminals or by a selector switch. Only one measurement can
tance. A simple ohmmeter circuit is shown in Fig. 42.7(a). be performed at a time. Often such instruments can be used
Unlike the ammeter or voltmeter, the ohmmeter circuit does in a.c. as well as d.c. circuits when a rectifier is incorporated
not receive the energy necessary for its operation from the in the instrument.
circuit under test. In the ohmmeter this energy is supplied by
320 Science for Engineering
(a) Resistance of voltmeter, Rv D sensitivity ð f.s.d.
42.10 Wattmeters Hence, Rv D ⊲10 k/V⊳ ð ⊲200 V⊳ D 2000 k D 2 M
A wattmeter is an instrument for measuring electrical Current flowing in voltmeter,
power in a circuit. Fig. 42.8 shows typical connections of
a wattmeter used for measuring power supplied to a load. Iv D V D 100 D 50 ð 10 6A
The instrument has two coils: Rv 2 ð 106
(i) a current coil, which is connected in series with the load, Power dissipated by
like an ammeter, and
voltmeter D VIv D ⊲100⊳⊲50 ð 10 6⊳ D 5 mW
(ii) a voltage coil, which is connected in parallel with the
load, like a voltmeter. When R D 250 , current in resistor,
V 100
ML
IR D R D 250 D 0.4 A
Current Power dissipated in load resistor R
coil
D VIR D ⊲100⊳⊲0.4⊳ D 40 W
V+ Thus the power dissipated in the voltmeter is insignificant
in comparison with the power dissipated in the load.
Supply Voltage Load (b) When R D 2 M, current in resistor,
coil
IR D V D 100 D 50 ð 10 6A
R 2 ð 106
Fig. 42.8 Power dissipated in load resistor R
42.11 Instrument ‘loading’ effect D VIR D 100 ð 50 ð 10 6 D 5 mW
Some measuring instruments depend for their operation on In this case the higher load resistance reduced the power
power taken from the circuit in which measurements are dissipated such that the voltmeter is using as much power
being made. Depending on the ‘loading’ effect of the instru- as the load.
ment (i.e. the current taken to enable it to operate), the
prevailing circuit conditions may change. Problem 4. An ammeter has a f.s.d. of 100 mA and
a resistance of 50 . The ammeter is used to mea-
The resistance of voltmeters may be calculated since each sure the current in a load of resistance 500 when the
have a stated sensitivity (or ‘figure of merit’), often stated supply voltage is 10 V. Calculate (a) the ammeter read-
in ‘k per volt’ of f.s.d. A voltmeter should have as high a ing expected (neglecting its resistance), (b) the actual
resistance as possible ( ideally infinite). In a.c. circuits the current in the circuit, (c) the power dissipated in the
impedance of the instrument varies with frequency and thus ammeter, and (d) the power dissipated in the load.
the loading effect of the instrument can change.
From Fig. 42.10,
Problem 3. Calculate the power dissipated by the volt- V 10
meter and by resistor R in Fig. 42.9 when (a) R D 250
(b) R D 2 M. Assume that the voltmeter sensitivity (or (a) expected ammeter reading D R D 500 D 20 mA
figure of merit) is 10 k/V.
R V 100 V R = 500 Ω 0 – 100 mA
0 – 200 V A
ra = 50 Ω
10 V
Fig. 42.9 Fig. 42.10
Electrical measuring instruments and measurements 321
V 10 (a) the approximate value of current (neglecting the
(b) Actual ammeter reading D R C ra D 500 C 50 ammeter resistance),
D 18.18 mA (b) the actual current in the circuit,
Thus the ammeter itself has caused the circuit conditions (c) the power dissipated in the ammeter,
to change from 20 mA to 18.18 mA.
(d) the power dissipated in the 1 k resistor.
(c) Power dissipated in the ammeter
2. (a) A current of 15 A flows through a load having
D I2ra D ⊲18.18 ð 10 3⊳2⊲50⊳ D 16.53 mW a resistance of 4 . Determine the power dissipated
in the load. (b) A wattmeter, whose current coil has
(d) Power dissipated in the load resistor a resistance of 0.02 is connected (as shown in
Fig. 42.11) to measure the power in the load. Deter-
D I2R D ⊲18.18 ð 10 3⊳2⊲500⊳ D 165.3 mW mine the wattmeter reading assuming the current in
the load is still 15 A.
Problem 5. (a) A current of 20 A flows through a
load having a resistance of 2 . Determine the power 42.12 The cathode ray oscilloscope
dissipated in the load. (b) A wattmeter, whose current
coil has a resistance of 0.01 is connected as shown The cathode ray oscilloscope (c.r.o.) may be used in
in Fig. 42.13. Determine the wattmeter reading. the observation of waveforms and for the measurement of
voltage, current, frequency, phase and periodic time. For
I = 20 A 0.01 Ω examining periodic waveforms the electron beam is deflected
horizontally (i.e. in the X direction) by a sawtooth generator
R=2Ω Supply acting as a timebase. The signal to be examined is applied
to the vertical deflection system (Y direction) usually after
Fig. 42.11 amplification.
(a) Power dissipated in the load, Oscilloscopes normally have a transparent grid of 10 mm
by 10 mm squares in front of the screen, called a graticule.
P D I2R D ⊲20⊳2⊲2⊳ D 800 W Among the timebase controls is a ‘variable’ switch which
gives the sweep speed as time per centimetre. This may be
(b) With the wattmeter connected in the circuit the total in s/cm, ms/cm or µs/cm, a large number of switch positions
resistance RT is 2 C 0.01 D 2.01 being available. Also on the front panel of a c.r.o. is a Y
The wattmeter reading is thus amplifier switch marked in volts per centimetre, with a large
number of available switch positions.
I2RT D ⊲20⊳2⊲2.01⊳ D 804 W
(i) With direct voltage measurements, only the Y amplifier
Now try the following exercise ‘volts/cm’ switch on the c.r.o. is used. With no voltage
applied to the Y plates the position of the spot trace on
Exercise 191 Further problems on instrument ‘load- the screen is noted. When a direct voltage is applied
ing’ effects (Answers on page 358) to the Y plates the new position of the spot trace is an
indication of the magnitude of the voltage. For example,
1. A 0–1 A ammeter having a resistance of 50 is used in Fig. 42.12(a), with no voltage applied to the Y plates,
to measure the current flowing in a 1 k resistor when the spot trace is in the centre of the screen (initial
the supply voltage is 250 V. Calculate: position) and then the spot trace moves 2.5 cm to the
final position shown, on application of a d.c. voltage.
With the ‘volts/cm’ switch on 10 volts/cm the magnitude
of the direct voltage is 2.5 cm ð 10 volts/cm, i.e. 25 volts
(ii) With alternating voltage measurements, let a sinu-
soidal waveform be displayed on a c.r.o. screen as
shown in Fig. 42.12(b). If the time/cm switch is on, say,
5 ms/cm then the periodic time T of the sinewave is
5 ms/cm ð 4 cm, i.e. 20 ms or 0.02 s
11
Since frequency f D , frequency = = 50 Hz
T 0.02
322 Science for Engineering
Final position (a) To obtain a spot trace on a typical c.r.o. screen:
Initial position
(i) Switch on the c.r.o.
52 1 0.5
10 0.2 (ii) Switch the timebase control to off. This control is
calibrated in time per centimetres – for example,
20 V/cm 5 ms/cm or 100 µs/cm. Turning it to zero ensures
30 no signal is applied to the X-plates. The Y-plate
(a) 4050 input is left open-circuited.
Peak-to-peak (iii) Set the intensity, X-shift and Y-shift controls to
value about the mid-range positions.
52 1 0.5 (iv) A spot trace should now be observed on the screen.
0.2 If not, adjust either or both of the X and Y-shift
10 controls. The X-shift control varies the position of
the spot trace in a horizontal direction whilst the
20 V/cm Y-shift control varies its vertical position.
30 (v) Use the X and Y-shift controls to bring the spot to
40 50 the centre of the screen and use the focus control
to focus the electron beam into a small circular
Periodic 5 ms 100 µs spot.
time
(b) 50 ms (b) To obtain a continuous horizontal trace on the screen
Fig. 42.12 S/cm the same procedure as in (a) is initially adopted. Then
the timebase control is switched to a suitable position,
2s initially the millisecond timebase range, to ensure that
the repetition rate of the sawtooth is sufficient for the
If the ‘volts/cm’ switch is on, say, 20 volts/cm then the persistence of the vision time of the screen phosphor to
amplitude or peak value of the sinewave shown is hold a given trace.
20 volts/cm ð 2 cm, i.e. 40 V
Problem 7. For the c.r.o. square voltage waveform
peak voltage shown in Fig. 42.13 determine (a) the periodic time,
Since r.m.s. voltage D p (b) the frequency and (c) the peak-to-peak voltage. The
‘time/cm’ (or timebase control) switch is on 100 µs/cm
2 and the ‘volts/cm’ (or signal amplitude control) switch
(see chapter 40), is on 20 V/cm.
40
r.m.s.voltage D p D 28.28 volts.
2
Double beam oscilloscopes are useful whenever two
signals are to be compared simultaneously.
The c.r.o. demands reasonable skill in adjustment and use.
However its greatest advantage is in observing the shape of
a waveform – a feature not possessed by other measuring
instruments.
Problem 6. Describe how a simple c.r.o. is adjusted to Fig. 42.13
give (a) a spot trace, (b) a continuous horizontal trace
on the screen, explaining the functions of the various (In Figures 42.13 to 42.19 assume that the squares shown are
controls. 1 cm by 1 cm)
Electrical measuring instruments and measurements 323
(a) The width of one complete cycle is 5.2 cm
Hence the periodic time,
T D 5.2 cm ð 100 ð 10 6 s/cm D 0.52 ms
11
(b) Frequency, f D T D 0.52 ð 10 3 D 1.92 Hz
(c) The peak-to-peak height of the display is 3.6 cm, hence
the
peak-to-peak voltage D 3.6 cm ð 20 V/cm D 72 V
Problem 8. For the c.r.o. display of a pulse wave- Fig. 42.15
form shown in Fig. 42.14 the ‘time/cm’ switch is on
50 ms/cm and the ‘volts/cm’ switch is on 0.2 V/cm. (a) The width of one complete cycle is 4 cm. Hence the
Determine (a) the periodic time, (b) the frequency, periodic time, T is 4 cm ð 500 µs/cm, i.e. 2 ms
(c) the magnitude of the pulse voltage.
11
Fig. 42.14 Frequency, f D T D 2 ð 10 3 D 500 Hz
(a) The width of one complete cycle is 3.5 cm
(b) The peak-to-peak height of the waveform is 5 cm. Hence
Hence the periodic time, the peak-to-peak
T D 3.5 cm ð 50 ms/cm D 175 ms
voltage D 5 cm ð 5 V/cm D 25 V
1
(c) Amplitude D 2 D 25 V D 12.5 V
(d) The peak value of voltage is the amplitude, i.e. 12.5 V,
and
r.m.s. voltage D peak voltage D 12.5 D 8.84 V
p p
22
Problem 10. For the double-beam oscilloscope dis-
plays shown in Fig. 42.16 determine (a) their frequency,
(b) their r.m.s. values, (c) their phase difference. The
‘time/cm’ switch is on 100 µs/cm and the ‘volts/cm’
switch on 2 V/cm.
11 B
(b) Frequency, f D T D 0.175 D 5.71 Hz A
(c) The height of a pulse is 3.4 cm hence the magnitude of
the pulse
voltage D 3.4 cm ð 0.2 V/cm D 0.68 V
Problem 9. A sinusoidal voltage trace displayed by a φ
c.r.o. is shown in Fig. 42.15. If the ‘time/cm’ switch is
on 500 µs/cm and the ‘volts/cm’ switch is on 5 V/cm, Fig. 42.16
find, for the waveform, (a) the frequency, (b) the
peak-to-peak voltage, (c) the amplitude, (d) the r.m.s.
value.
324 Science for Engineering
(a) The width of each complete cycle is 5 cm for both 2. For the pulse waveform shown in Fig. 42.18, find
waveforms. (a) its frequency, (b) the magnitude of the pulse
Hence the periodic time, T, of each waveform is voltage
5 cm ð 100 µs/cm, i.e. 0.5 ms
Frequency of each waveform,
11
f D T D 0.5 ð 10 3 D 2 kHz
(b) The peak value of waveform A is 2 cm ð 2 V/cm D 4 V,
4
hence the r.m.s. value of waveform A D p D 2.83 V
2
The peak value of waveform B is 2.5 cmð2 V/cm D 5 V,
hence the r.m.s. value of waveform B D 52 D 3.54 V
p
(c) Since 5 cm represents 1 cycle, then 5 cm represents 360°,
360 2 1 0.5 5 ms
i.e. 1 cm represents 5 D 72° 5 0.2
10
The phase angle f D 0.5 cm D 0.5 cm ð 72°/cm D 36° 20 v/cm 50 ms 100 µs
30
Hence waveform A leads waveform B by 36◦ 40 time/cm
50 500 ms
Now try the following exercise Fig. 42.18 2s
Exercise 192 Further problems on the cathode ray 3. For the sinusoidal waveform shown in Fig. 42.19,
oscilloscope (Answers on page 358) determine (a) its frequency, (b) the peak-to-peak volt-
age, (c) the r.m.s. voltage
1. For the square voltage waveform displayed on a c.r.o.
shown in Fig. 42.17, find (a) its frequency, (b) its
peak-to-peak voltage
2 1 0.5 5 ms 2 1 0.5 5 ms 100 µs
5 0.2 5 0.2 50 ms
10
20 v/cm 100 µs 10
30
40 50 ms 20 v/cm time/cm
50 time/cm 30
40
Fig. 42.17 50 2s
Fig. 42.19
Electrical measuring instruments and measurements 325
42.13 Null method of measurement Problem 11. In a Wheatstone bridge ABCD, a gal-
vanometer is connected between A and C, and a battery
A null method of measurement is a simple, accurate and between B and D. A resistor of unknown value is con-
widely used method which depends on an instrument read- nected between A and B. When the bridge is balanced,
ing being adjusted to read zero current only. The method the resistance between B and C is 100 , that between
assumes: C and D is 10 and that between D and A is 400 .
Calculate the value of the unknown resistance.
(i) if there is any deflection at all, then some current is
flowing The Wheatstone bridge is shown in Fig. 42.21 where Rx is
the unknown resistance. At balance, equating the products of
(ii) if there is no deflection, then no current flows (i.e. a null opposite ratio arms, gives:
condition)
⊲Rx⊳⊲10⊳ D ⊲100⊳⊲400⊳
Hence it is unnecessary for a meter sensing current flow to be ⊲100⊳⊲400⊳
calibrated when used in this way. A sensitive milliammeter
or microammeter with centre zero position setting is called and Rx D 10 D 4000
a galvanometer. Two examples where the method is used Hence, the unknown resistance, Rx = 4 kZ
are in the Wheatstone bridge (see Section 42.14), in the d.c.
potentiometer (see Section 42.15) A
42.14 Wheatstone bridge 400 Ω RX
Figure 42.20 shows a Wheatstone bridge circuit which DGB
compares an unknown resistance Rx with others of known
values, i.e. R1 and R2, which have fixed values, and R3, which 10 Ω 100 Ω
is variable. R3 is varied until zero deflection is obtained on
the galvanometer G. No current then flows through the meter, C
VA D VB, and the bridge is said to be ‘balanced’.
A
R1 R3
G
Fig. 42.21
42.15 D.C. potentiometer
R2 Rx The d.c. potentiometer is a null-balance instrument used for
B determining values of e.m.f.’s and p.d.s. by comparison with
a known e.m.f. or p.d. In Fig. 42.22(a), using a standard cell
Fig. 42.20
of known e.m.f. E1, the slider S is moved along the slide wire
until balance is obtained (i.e. the galvanometer deflection is
zero), shown as length l1
The standard cell is now replaced by a cell of unknown
e.m.f. E2 (see Fig. 42.22(b)) and again balance is obtained
(shown as l2).
Since E1 / l1 and E2 / l2 then
At balance, E1 D l1 and E2 = E1 l2 volts
E2 l2 l1
R1Rx D R2R3 i.e. Rx = R2R3 ohms. A potentiometer may be arranged as a resistive two-element
R1 potential divider in which the division ratio is adjustable
326 Science for Engineering
Supply source V 42.16 Measurement errors
V
Errors are always introduced when using instruments to
l1 l2 measure electrical quantities. The errors most likely to occur
S in measurements are those due to:
S
E2 G (i) the limitations of the instrument
Slide wire (b) (ii) the operator
of uniform (iii) the instrument disturbing the circuit
E1 G cross-section (i) Errors in the limitations of the instrument
Standard The calibration accuracy of an instrument depends on the
precision with which it is constructed. Every instrument
cell has a margin of error which is expressed as a percent-
(a) age of the instruments full scale deflection. For example,
industrial grade instruments have an accuracy of š2% of
Fig. 42.22 f.s.d. Thus if a voltmeter has a f.s.d. of 100 V and it indi-
cates 40 V say, then the actual voltage may be anywhere
to give a simple variable d.c. supply. Such devices may between 40 š (2% of 100), or 40 š 2, i.e. between 38 V
be constructed in the form of a resistive element carrying and 42 V.
a sliding contact which is adjusted by a rotary or linear
movement of the control knob. When an instrument is calibrated, it is compared against a
standard instrument and a graph is drawn of ‘error’ against
Problem 12. In a d.c. potentiometer, balance is ‘meter deflection’.
obtained at a length of 400 mm when using a standard
cell of 1.0186 volts. Determine the e.m.f. of a dry cell A typical graph is shown in Fig. 42.23 where it is seen that
if balance is obtained with a length of 650 mm. the accuracy varies over the scale length. Thus a meter with
a š2% f.s.d. accuracy would tend to have an accuracy which
E1 D 1.0186 V, l1 D 400 mm and l2 D 650 mm. is much better than š2% f.s.d. over much of the range.
With reference to Fig. 42.22,
(ii) Errors by the operator
E1 D l1
E2 l2 It is easy for an operator to misread an instrument. With
linear scales the values of the sub-divisions are reasonably
from which, E2 D E1 l2 D ⊲1.0186⊳ 650 easy to determine; non-linear scale graduations are more
l1 400 difficult to estimate. Also, scales differ from instrument to
instrument and some meters have more than one scale (as
D 1.655 volts. with multimeters) and mistakes in reading indications are
easily made. When reading a meter scale it should be viewed
Now try the following exercise from an angle perpendicular to the surface of the scale at the
location of the pointer; a meter scale should not be viewed
Exercise 193 Further problems on the Wheatstone ‘at an angle’.
bridge and d.c. potentiometer (Answers
on page 358) (iii) Errors due to the instrument disturbing the circuit
1. In a Wheatstone bridge PQRS, a galvanometer is Any instrument connected into a circuit will affect that cir-
connected between Q and S and a voltage source cuit to some extent. Meters require some power to operate,
between P and R. An unknown resistor Rx is connected but provided this power is small compared with the power
between P and Q. When the bridge is balanced, the in the measured circuit, then little error will result. Incor-
resistance between Q and R is 200 , that between rect positioning of instruments in a circuit can be a source
R and S is 10 and that between S and P is 150 . of errors. For example, let a resistance be measured by the
Calculate the value of Rx voltmeter-ammeter method as shown in Fig. 42.24. Assum-
ing ‘perfect’ instruments, the resistance should be given by
2. Balance is obtained in a d.c. potentiometer at a length the voltmeter reading divided by the ammeter reading (i.e.
of 31.2 cm when using a standard cell of 1.0186 volts. R D V/I). However, in Fig. 42.24(a), V/I D R C ra and in
Calculate the e.m.f. of a dry cell if balance is obtained Fig. 42.24(b) the current through the ammeter is that through
with a length of 46.7 cm.
Electrical measuring instruments and measurements 327
2% Upper limit
1% of error
Error 0 1 1 3 Deflection
–1% 4 2 4
– 2% F.S.D F.S.D F.S.D F.S.D
Lower limit
of error
Fig. 42.23
R R 6. State two advantages and two disadvantages of a
A moving coil instrument
I ra V
A 7. Name two advantages of electronic measuring instru-
V ments compared with moving coil or moving iron
instruments
(a) (b)
8. Briefly explain the principle of operation of an ohm-
Fig. 42.24 meter
the resistor plus that through the voltmeter. Hence the volt- 9. Name a type of ohmmeter used for measuring (a) low
meter reading divided by the ammeter reading will not give resistance values (b) high resistance values
the true value of the resistance R for either method of con-
nection. 10. What is a multimeter?
11. When may a rectifier instrument be used in prefer-
Now try the following exercises
ence to either a moving coil or moving iron instru-
Exercise 194 Short answer questions on electrical ment?
measuring instruments and measure- 12. Name five quantities that a c.r.o. is capable of mea-
ments suring
13. What is meant by a null method of measurement?
1. What is the main difference between an analogue and 14. Sketch a Wheatstone bridge circuit used for measur-
a digital type of measuring instrument? ing an unknown resistance in a d.c. circuit and state
the balance condition
2. Name the three essential devices for all analogue 15. How may a d.c. potentiometer be used to measure
electrical indicating instruments p.d.’s
16. Define ‘calibration accuracy’ as applied to a measur-
3. Complete the following statements: ing instrument
17. State three main areas where errors are most likely
(a) An ammeter has a . . . resistance and is connected to occur in measurements
. . . with the circuit
Exercise 195 Multi-choice questions on electrical
(b) A voltmeter has a . . . resistance and is connected measuring instruments and measure-
. . . with the circuit ments (Answers on page 359)
4. State two advantages and two disadvantages of a 1. Which of the following would apply to a moving coil
moving coil instrument instrument?
(a) An uneven scale, measuring d.c.
5. What effect does the connection of (a) a shunt (b) a
multiplier have on a milliammeter?
328 Science for Engineering
(b) An even scale, measuring a.c. Q
P
(c) An uneven scale, measuring a.c.
2 1 0.5 5 ms 100 µs
(d) An even scale, measuring d.c. 5 0.2
10 50 ms
2. In question 1, which would refer to a moving iron 20 v/cm
instrument? 30 time/cm
40
3. In question 1, which would refer to a moving coil 2s
rectifier instrument? 50
4. Which of the following is needed to extend the range Fig. 42.26
of a milliammeter to read voltages of the order of
100 V?
(a) a parallel high-value resistance
(b) a series high-value resistance
(c) a parallel low-value resistance
(d) a series low-value resistance
5. Figure 42.25 shows a scale of a multi-range ammeter.
What is the current indicated when switched to a 25 A
scale?
(a) 84 A (b) 5.6 A (c) 14 A (d) 8.4 A
40 50 60 Fig. 42.26 shows double-beam c.r.o. waveform traces.
For the quantities stated in questions 11 to 17, select the
30 70 correct answer from the following:
10 60 75 90 105 90 100 (a) 30 V (b) 0.2 s (c) 50 V
20 45 80 15 (e) 54° leading 250
(h) 100 µs
0 15 30 120 135 150 (d) p (f) p V
2 2
(g) 15 V 50
(i) p V
0
2
(j) 250 V (k) 10 kHz (l) 75 V
25
3p
(m) 40 µs (n) rads lagging (o) p V
10 2
Fig. 42.25 (p) 5 Hz (q) 30 V (r) 25 kHz
p
A sinusoidal waveform is displayed on a c.r.o. sreen. The 2
peak-to-peak distance is 5 cm and the distance between 75 3p
cycles is 4 cm. The ‘variable’ switch is on 100 µs/cm and (s) p V (t) rads leading
the ‘volts/cm’ switch is on 10 V/cm. In questions 6 to 10,
select the correct answer from the following: 2 10
(a) 25 V (b) 5 V (c) 0.4 ms (d) 35.4 V 11. Amplitude of waveform P
(e) 4 ms (f) 50 V (g) 250 Hz (h) 2.5 V 12. Peak-to-peak value of waveform Q
(i) 2.5 kHz (j) 17.7 V 13. Periodic time of both waveforms
14. Frequency of both waveforms
6. Determine the peak-to-peak voltage
7. Determine the periodic time of the waveform 15. R.m.s. value of waveform P
8. Determine the maximum value of the voltage 16. R.m.s. value of waveform Q
9. Determine the frequency of the waveform
10. Determine the r.m.s. value of the waveform 17. Phase displacement of waveform Q relative to wave-
form P
18. A potentiometer is used to:
(a) compare voltages (b) measure power factor
(c) compare currents (d) measure phase sequence
Electrical measuring instruments and measurements 329
Assignment 13 magnetic field, and (c) the average e.m.f. induced if
the current falls to zero in 300 ms. (6)
This assignment covers the material contained in 6. A moving coil instrument gives a f.s.d. when the cur-
chapters 40 to 42. The marks for each question are
shown in brackets at the end of each question. rent is 50 mA and has a resistance of 40 . Determine
the value of resistance required to enable the instru-
ment to be used (a) as a 0–5 A ammeter, and (b) as
1. An alternating current completes 3 cycles in 12 ms. a 0–200 V voltmeter. State the mode of connection in
What is its frequency? (5) each case. (8)
2. For the periodic waveform shown in Fig. A13.1 deter- 7. A sinusoidal voltage trace displayed on a c.r.o. is shown
in Fig. A13.2; the ‘time/cm’ switch is on 50 ms/cm
mine: (a) the frequency (b) the average value over and the ‘volts/cm’ switch is on 2 V/cm. Determine for
the waveform (a) the frequency (b) the peak-to-peak
half a cycle (use 8 intervals for a half cycle) (c) the voltage (c) the amplitude (d) the r.m.s. value. (8)
r.m.s. value (d) the form factor, and (e) the peak
factor (12)
Voltage 100 2.5 5 7.5 10 Time(ms)
(V) 50
0 Fig. A13.2
– 50
–100 8. An ammeter has a f.s.d. of 200 mA and a resistance
of 40 . The ammeter is used to measure the current
Fig. A13.1 in a load of resistance 400 when the supply voltage
is 20 V. Calculate (a) the ammeter reading expected
(neglecting its resistance), (b) the actual current in the
3. A sinusoidal voltage has a mean value of 30 V. Deter- circuit, (c) the power dissipated in the ammeter, and
mine it’s maximum and r.m.s. values. (4) (d) the power dissipated in the load. (8)
4. An e.m.f. of 2.5 kV is induced in a coil when a 9. In a Wheatstone bridge PQRS, a galvanometer is
connected between P and R, and a battery between
current of 2 A collapses to zero in 5 ms. Calculate the Q and S. A resistor of unknown value is connected
between P and Q. When the bridge is balanced, the
inductance of the coil. (4) resistance between Q and R is 150 , that between
R and S is 25 and that between S and P is 500 .
5. A flux of 15 mWb links with a 2000 turn coil when a Calculate the value of the unknown resistance. (5)
current of 2 A passes through the coil. Calculate (a) the
inductance of the coil, (b) the energy stored in the
Section 4 Engineering Systems
43
Introduction to engineering systems
At the end of this chapter you should be able to: 43.2 Systems
ž Define a system, sub-system, component and ele- Systems comprise a number of elements, components or sub-
ments systems that are connected together in a particular way to
perform a desired function. The individual elements of an
ž State and understand common types of sys- engineering system interact together to satisfy a particular
tems – electromechanical, communications, fluidic, functional requirement.
electrochemical, information, control, transport
A basic system has (i) a function or purpose, (ii) inputs
ž Represent a system block diagram (for example, raw materials, energy, and control values) and
ž Define a transducer as an energy conversion device outputs (for example, finished products, processed materials,
converted energy), (iii) a boundary, and (iv) a number of
and state examples smaller linked components or elements.
ž Explain system control – analogue, digital, sequen-
43.3 Types of systems
tial, combinational, open- and closed-loop, off/on,
hysteresis, proportional Some of the most common types of system used in engineer-
ž Discuss possible system responses, with examples ing are:
ž Define negative and positive feedback
ž Appreciate factors involved in evaluating system (i) Electromechanical systems – for example, a vehicle
response electrical system comprising battery, starter motor, igni-
tion coil, contact breaker and distributor.
43.1 Introduction
(ii) Communications systems – for example, a local area
The study of engineering systems is a considerable topic in network comprising file server, coaxial cable, network
itself and whole books have been written on the subject. This adaptors, several computers and a laser printer.
chapter offers a brief overview of the subject area, sufficient
to form a basis for further study.
Introduction to engineering systems 331
(iii) Fluidic systems – for example, a vehicle braking sys- A.C. Supply
tem comprising foot-operated lever, master cylinder,
slave cylinder, piping and fluid reservoir. Microphone Loudspeaker
(iv) Electrochemical systems – for example, a cell that Amplifier
uses gas as a fuel to produce electricity, pure water
and heat. Fig. 43.2
(v) Information systems – for example, a computerized output transducer converts the electrical energy supplied to
airport flight arrival system. it back into acoustic energy.
A sub-system is a part of a system that performs an iden-
(vi) Control systems – for example, a microcomputer- tified function within the whole system; the amplifier in
based controller that regulates the flow and temperature Figure 43.2 is an example of a sub-system.
of material used in a die casting process.
A component or element is usually the simplest part of a
(vii) Transport systems – for example, an overhead con- system that has a specific and well-defined function – for
veyor for transporting gravel from a quarry to a nearby example, the microphone in Figure 43.2. The illustrations
processing site. in Figures 43.1 and 43.2 are called block diagrams and
engineering systems, which can often be quite complicated,
43.4 Transducers can be better understood when broken down in this way. It is
not always necessary to know precisely what is inside each
A transducer is a device that responds to an input signal sub-system in order to know how the whole system functions.
in one form of energy and gives an output signal bearing
a relationship to the input signal but in a different form of As another example of an engineering system, Figure 43.3
energy. For example, a tyre pressure gauge is a transducer, illustrates a temperature control system containing a
converting tyre pressure into a reading on a scale. Similarly, a heat source – a gas boiler, a fuel controller – an electrical
photo-cell converts light into a voltage, and a motor converts solenoid valve, a thermostat and a source of electrical energy.
electric current into shaft velocity. The system of Figure 43.3 can be shown in block diagram
form as in Figure 43.4; the thermostat compares the actual
43.5 System diagrams room temperature with the desired temperature and switches
the heating on or off.
A simple practical processing system is shown in Fig-
ure 43.1 comprising an input, an output, and an energy 240 V
source, together with unwanted inputs and outputs that may,
or may not, be present.
Solenoid Gas
boiler
Fuel
supply
Fig. 43.1 Thermostat
Set temperature
A typical public address system is shown in Figure 43.2. A Radiators
microphone is used as an input transducer to collect acoustic Enclosed space
energy in the form of sound pressure waves and convert
this into electrical energy in the form of small voltages and Fig. 43.3
currents. The signal from the microphone is then amplified
by means of an electronic circuit containing transistors and/or There are many types of engineering systems and all such
integrated circuits before it is applied to a loudspeaker. This systems may be represented by block diagrams.
332 Science for Engineering
Thermostat Analogue control
+ Error Heating Enclosure Analogue control involves the use of signals and quantities
Temperature − system that are continuously variable. Within analogue control sys-
Temperature tems, signals are represented by voltages and currents that
command Actual of enclosure can take any value between two set limits. Figure 43.5(a)
shows how the output of a typical analogue system varies
temperature with time.
Fig. 43.4 Analogue control systems are invariably based on the use
of operational amplifiers (see Figure 43.6). These devices
43.6 System Control are capable of performing mathematical operations such as
addition, subtraction, multiplication, division, integration and
The behaviour of most systems is subject to variations differentiation.
in input (supply) and in output (demand). The behaviour
of a system may also change in response to variations Inputs − Output
in the characteristics of the components that make up the +
system. In practice it is desirable to include some means
of regulating the output of the system so that it remains Fig. 43.6
relatively immune to these three forms of variation. System
control involves maintaining the system output at the desired Digital control
level regardless of any disturbances that may affect (i) the
input quantities, (ii) any unwanted variations in systems Digital control involves the use of signals and quantities that
components, or (iii) the level of demand (or ‘loading’) on vary in discrete steps. Values that fall between two adjacent
the output.
Different control methods are appropriate to different types
of system. The overall control strategy can be based on
analogue or digital techniques and may be classed as either
sequential or combinational.
(a)
(b)
Fig. 43.5
Introduction to engineering systems 333
steps must take one or other value. Figure 43.5(b) shows how INPUT ELECTRIC OUTPUT
the output of a typical digital system varies with time. ELECTRICITY FAN HEATER WARM AIR
Digital control systems are usually based on logic devices COLD AIR
(such as AND-gates, OR-gates, NAND-gates and NOR-
gates), or microprocessor based computer systems. Fig. 43.7
Sequential control systems Closed-loop control
Many systems are required to perform a series of opera- From above, open-loop control has some significant disad-
tions in a set order. For example, the ignition system of vantages. What is required is some means of closing the loop
a gas boiler may require the following sequence of oper- in order to make a continuous automatic comparison of the
ations: (i) operator’s start button is pressed, (ii) fan motor actual value of the output compared with the setting of the
operates, (iii) delay 60 seconds, (iv) open gas supply valve, input control variable. In the above example, the chef actually
(v) ignition operates for 2 seconds, (vi) if ignition fails, close closes the loop on an intermittent basis. In effect, the cooker
gas supply valve, delay 60 seconds, then stop fan motor, relies on human intervention in order to ensure consistency
(vii) if ignition succeeds, boiler will continue to operate until of the food produced.
either the stop switch operates or flame fails.
All practical engineering systems make use of closed-loop
The components of simple sequential systems often control. In some cases, the loop might be closed by a human
include timers, relays, counters, and so on; however, digital operator who determines the deviation between the desired
logic and microprocessor-based controllers are used in more and actual output. In most cases, however, the action of the
complex systems. system is made fully automatic and no human intervention
is necessary other than initially setting the desired value of
Combinational control systems the output. The principle of a closed-loop control system is
illustrated in Figure 43.8.
Combinational control systems take several inputs and per-
form comparisons on a continuous basis. In effect, everything
happens at the same time – there are no delays or predeter-
mined sequences that would be associated with sequential
controllers. An aircraft instrument landing system (ILS)
for example, makes continuous comparisons of an aircraft’s
position relative to the ILS radio beam. Where any deviation
is detected, appropriate correction is applied to the aircraft’s
flight controls.
43.7 Control methods Fig. 43.8
Open-loop control Reasons for making a system fully automatic include:
In a system that employs open-loop control, the value of the (i) Some systems use a very large number of input variables
input variable is set to a given value in the expectation that and it may be difficult or impossible for a human
the output will reach the desired value. In such a system operator to keep track of them all.
there is no automatic comparison of the actual output value
with the desired output value in order to compensate for (ii) Some processes are extremely complex and there may
any differences. A simple example of an open-loop control be significant interaction between the input variables.
method is the manual adjustment of the regulator that controls
the flow of gas to a burner on the hob of a gas cooker. This (iii) Some systems may have to respond very quickly to
adjustment is carried out in the expectation that food will be changes in variables.
raised to the correct temperature in a given time and without
burning. Other than the occasional watchful eye of the chef, (iv) Some systems require a very high degree of precision.
there is no means of automatically regulating the gas flow in
response to the actual temperature of the food. The analogue closed-loop control system shown in Fig-
ure 43.9 provides speed control for the d.c. motor, M.
Another example of an open-loop control system is that The actual motor speed is sensed by means of a small d.c.
of an electric fan heater as shown in the block diagram of tachogenerator, G, coupled to the output shaft. The voltage
Figure 43.7. produced by the tachogenerator is compared with that pro-
duced at the slider of a potentiometer, R, which is used to set
334 Science for Engineering
Fig. 43.9
Fig. 43.10
the desired speed. The comparison of the two voltages, (i.e. operation, the process of switching on and off will continue
that of the tachogenerator with that corresponding to the set indefinitely.
point), is performed by an operational amplifier connected as
a comparator. The output of the comparator stage is applied Hysteresis
to a power amplifier that supplies current to the d.c. motor.
Energy is derived from a mains-powered d.c. power supply In practice, a small amount of hysteresis is built into most
comprising transformer, rectifier and smoothing circuits. on/off control systems. In the previous example, this hystere-
sis is designed to prevent the heater switching on and off too
On/off control rapidly which, in turn, would result in early failure of the
switch contacts fitted to the thermostat. Note, however, that
On/off control is the simplest form of control and it merely the presence of hysteresis means that, at any instant of time,
involves turning the output on and off repeatedly in order to the actual temperature will always be somewhere between
achieve the required level of output variable. Since the output the upper and lower threshold values (see Figure 43.10).
of the system is either fully on or fully off at any particular
instant of time (i.e. there is no half-way state), this type of Proportional control
control is sometimes referred to as ‘discontinuous’.
On/off control is crude but can be effective in many simple
The most common example of an on/off control system is applications. A better method is to vary the amount of
that of a simple domestic room heater. A variable thermostat correction applied to the system according to the size of the
is used to determine the set point (SP) temperature value. deviation from the set point (SP) value. As the difference
When the actual temperature is below the SP value, the heater between the actual value and the desired output becomes less,
is switched on (i.e. electrical energy is applied to the heating the amount of correction becomes correspondingly smaller.
element). Eventually, the room temperature will exceed the A simple example of this type of control is in the control of
SP value and, at this point, the heater is switched off. Later, the water level in a water tank.
the temperature falls due to heat loss, the room temperature
will once again fall below the SP value and the heater will
once again be switched on (see Figure 43.10). In normal
Introduction to engineering systems 335
43.8 System response Consider the case of the motor speed control system
shown in Figure 43.12 where the output shaft is connected
In a perfect system, the output value, C, will respond instan- to a substantial flywheel. The flywheel effectively limits
taneously to a change in the input, S. There will be no the acceleration of the motor speed when the set point
delay when changing from one value to another and no time (S) is increased. Furthermore, as the output speed reaches
required for the output to ‘settle’ to its final value. The ideal the desired value, the inertia present will keep the speed
state is shown in Figure 43.11(b). In practice, real-world sys- increasing despite the reduction in voltage (C) applied to the
tems take time to reach their final state. Indeed, a very sudden motor. Thus the output shaft speed overshoots the desired
change in output may, in some cases, be undesirable. Fur- value before eventually falling back to the required value.
thermore, inertia is present in many systems. Increasing the gain present in the system will have the
effect of increasing the acceleration but this, in turn, will
also produce a correspondingly greater value of overshoot.
Conversely, decreasing the gain will reduce the overshoot
but at the expense of slowing down the response. The actual
response of the system represents a compromise between
speed and an acceptable value of overshoot. Figure 43.11(c)
shows the typical response of a system to the step input
shown in Figure 43.11(a).
Second-order response
The graph shown in Figure 43.11(c) is known as a ‘second-
order’ response. This response has two basic components,
an exponential growth curve and a damped oscillation. The
oscillatory component can be reduced (or eliminated) by
artificially slowing down the response of the system. This
is known as ‘damping’. The optimum value of damping
is that which just prevents overshoot. When a system is
‘underdamped’, some overshoot is still present. Conversely,
an ‘overdamped’ system may take a significantly greater
time to respond to a sudden change in input. These damping
states are illustrated in Figure 43.13.
Fig. 43.11 43.9 Negative and positive feedback
Most systems use negative feedback in order to precisely
control the operational parameters of the system and to
maintain the output over a wide variation of the internal
parameters of the system. In the case of an amplifier, for
example, negative feedback can be used not only to stabilize
the gain but also to reduce distortion and improve band-
width.
Fig. 43.12
336 Science for Engineering
6. State three reasons for making a system fully auto-
matic.
7. Define (a) negative feedback (b) positive feedback
8. Explain the terms (a) underdamped (b) overdamped.
9. State four measurements that may be involved in
evaluating a system response.
10. State four possible sources of error that may exist
when evaluating system response.
Fig. 43.13 Exercise 197 Multi-choice questions on engineering
systems (Answers on page 359)
The amount of feedback determines the overall (or closed-
loop) gain. Because this form of feedback has the effect of 1. A networked computer database containing details of
reducing the overall gain of the circuit, this form of feedback railway timetables is an example of:
is known as ‘negative feedback’. An alternative form of
feedback, where the output is fed back in such a way as to (a) an electromechanical system
reinforce the input (rather than to subtract from it) is known (b) a transport system
as ‘positive feedback’. (c) a computer system
(d) an information system
43.10 Evaluation of system response 2. An anti-lock braking system fitted to an articulated
lorry is an example of:
Depending on the equipment available, the evaluation of sys-
tems can be carried out practically. These can involve mea- (a) an electromechanical system
surements of (i) accuracy, (ii) repeatability, (iii) stability, (b) a fluidic system
(iv) overshoot, (v) settling time after disturbances, (v) sensi- (c) a logic system
tivity, and (vi) rate of response. (d) an information system
When evaluating systems, it is important to recognize 3. Which one of the following system components is a
that (i) all variables are dependent on time, (ii) velocity is the device used to store electrical energy?
rate of change of distance, (iii) current is the rate of change (a) a flywheel (b) an actuator
of charge, and (iv) flow is the rate of change of quantity. (c) a battery (d) a transformer
Sources of error also exist, for example (i) dimensional 4. Which one of the following is NOT a system out-
tolerance, (ii) component tolerance, (iii) calibration toler- put device?
ance, (iv) instrument accuracy, (v) instrument resolution, and
(vi) human observation. (a) a linear actuator (b) a stepper motor
Now try the following exercise (c) a relay (d) a tachogenerator
Exercise 196 Short answer questions on engineer- 5. Which one of the following is NOT a system input
ing systems device?
(a) a microswitch (b) a relay
1. Define a system. (c) a variable resistor (d) a rotary switch
2. State, and briefly explain, four common types of engi-
6. Which one of the following is NOT an advantage of
neering systems, drawing the system block diagram using negative feedback?
for each.
3. Define a transducer and state four examples. (a) the overall gain produced by the system is
4. State three methods of system control. increased
5. Briefly explain (a) open-loop control (b) closed-loop
control. (b) the system is less prone to variations in the
characteristics of individual components
(c) the performance of the system is more predictable
(d) the stability of the system is increased