10

Electrical measuring instruments and

measurements

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

ž recognize the importance of testing and measurements in electric circuits

ž appreciate the essential devices comprising an analogue instrument
ž explain the operation of an attraction and a repulsion type of moving-iron instrument
ž explain the operation of a moving-coil rectifier instrument
ž compare moving-coil, moving-iron and moving coil rectifier instruments
ž calculate values of shunts for ammeters and multipliers for voltmeters
ž understand the advantages of electronic instruments
ž understand the operation of an ohmmeter/megger
ž appreciate the operation of multimeters/Avometers
ž understand the operation of a wattmeter
ž appreciate instrument ‘loading’ effect
ž understand the operation of a C.R.O. for d.c. and a.c. measurements
ž calculate periodic time, frequency, peak to peak values from waveforms on a C.R.O.
ž recognize harmonics present in complex waveforms
ž determine ratios of powers, currents and voltages in decibels
ž understand null methods of measurement for a Wheatstone bridge and d.c. poten-
tiometer
ž understand the operation of a.c. bridges
ž understand the operation of a Q-meter
ž appreciate the most likely source of errors in measurements
ž appreciate calibration accuracy of instruments

quantities such as current, voltage, resistance or

10.1 Introduction power, it is necessary to transform an electrical
Tests and measurements are important in designing, quantity or condition into a visible indication. This
evaluating, maintaining and servicing electrical is done with the aid of instruments (or meters) that
circuits and equipment. In order to detect electrical indicate the magnitude of quantities either by the

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 105

position of a pointer moving over a graduated scale

(called an analogue instrument) or in the form of a
decimal number (called a digital instrument).

10.2 Analogue instruments

All analogue electrical indicating instruments
require three essential devices:

(a) A deflecting or operating device. A mechanical

force is produced by the current or voltage
which causes the pointer to deflect from its zero
position.
(b) A controlling device. The controlling force acts
in opposition to the deflecting force and ensures
that the deflection shown on the meter is always
the same for a given measured quantity. It also
prevents the pointer always going to the max-
imum deflection. There are two main types of
controlling device – spring control and gravity
control. Figure 10.2

(c) A damping device. The damping force ensures

current flows in the solenoid, a pivoted soft-
that the pointer comes to rest in its final position
iron disc is attracted towards the solenoid and
quickly and without undue oscillation. There
the movement causes a pointer to move across
are three main types of damping used – eddy-
a scale.
current damping, air-friction damping and fluid-
friction damping. (b) In the repulsion type moving-iron instrument
shown diagrammatically in Fig. 10.2(b), two
There are basically two types of scale – linear and pieces of iron are placed inside the solenoid, one
non-linear. A linear scale is shown in Fig. 10.1(a), being fixed, and the other attached to the spin-
where the divisions or graduations are evenly dle carrying the pointer. When current passes
spaced. The voltmeter shown has a range 0–100 V, through the solenoid, the two pieces of iron are
i.e. a full-scale deflection (f.s.d.) of 100 V. A non- magnetized in the same direction and therefore
linear scale is shown in Fig. 10.1(b) where the scale repel each other. The pointer thus moves across
is cramped at the beginning and the graduations are the scale. The force moving the pointer is, in
uneven throughout the range. The ammeter shown each type, proportional to I2 and because of
has a f.s.d. of 10 A. this the direction of current does not matter. The
moving-iron instrument can be used on d.c. or
a.c.; the scale, however, is non-linear.

10.4 The moving-coil rectifier

instrument
Figure 10.1
A moving-coil instrument, which measures only
d.c., may be used in conjunction with a bridge
rectifier circuit as shown in Fig. 10.3 to provide an
10.3 Moving-iron instrument indication of alternating currents and voltages (see
Chapter 14). The average value of the full wave
(a) An attraction type of moving-iron instrument is rectified current is 0.637 Im . However, a meter being
shown diagrammatically in Fig. 10.2(a). When used to measure a.c. is usually calibrated in r.m.s.

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106 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Type of instrument Moving-coil Moving-iron Moving-coil rectifier

Suitable for Direct current and Direct and alternating Alternating current
measuring voltage currents and voltage and voltage (reads
(reading in rms value) average value but
scale is adjusted to
give rms value for
sinusoidal waveforms)
Scale Linear Non-linear Linear
Method of control Hairsprings Hairsprings Hairsprings
Method of damping Eddy current Air Eddy current
Frequency limits 20–200 Hz 20–100 kHz
–
Advantages 1 Linear scale 1 Robust construction 1 Linear scale
2 High sensitivity 2 Relatively cheap 2 High sensitivity
3 Well shielded 3 Measures dc and ac 3 Well shielded from
from stray 4 In frequency range stray magnetic fields
magnetic fields 20–100 Hz reads 4 Lower power
4 Low power rms correctly consumption
consumption regardless of supply 5 Good frequency
wave-form range
Disadvantages 1 Only suitable for 1 Non-linear scale 1 More expensive
dc 2 Affected by stray than moving iron
2 More expensive magnetic fields type
than moving iron 3 Hysteresis errors in 2 Errors caused when
type dc circuits supply is
3 Easily damaged 4 Liable to non-sinusoidal
temperature errors
5 Due to the
inductance of the
solenoid, readings
can be affected by
variation of
frequency

10.5 Comparison of moving-coil,

moving-iron and moving-coil
rectifier instruments

See Table above. (For the principle of operation of

a moving-coil instrument, see Chapter 8, page 89).

Figure 10.3

values. For sinusoidal quantities the indication is 10.6 Shunts and multipliers
⊲0.707Im ⊳/⊲0.637Im ⊳ i.e. 1.11 times the mean value.
Rectifier instruments have scales calibrated in r.m.s. An ammeter, which measures current, has a low
quantities and it is assumed by the manufacturer that resistance (ideally zero) and must be connected in
the a.c. is sinusoidal. series with the circuit.

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 107

A voltmeter, which measures p.d., has a high

resistance (ideally infinite) and must be connected
in parallel with the part of the circuit whose p.d. is
required.
There is no difference between the basic instru-
ment used to measure current and voltage since both
use a milliammeter as their basic part. This is a Figure 10.5
sensitive instrument which gives f.s.d. for currents
of only a few milliamperes. When an ammeter is current flowing in instrument D 40 mA D 0.04 A,
required to measure currents of larger magnitude, a Is D current flowing in shunt and I D total circuit
proportion of the current is diverted through a low- current required to give f.s.d. D 50 A.
value resistance connected in parallel with the meter.
Such a diverting resistor is called a shunt. Since I D Ia C Is then Is D I
Ia
From Fig. 10.4(a), VPQ D VRS . D 50
0.04 D 49.96 A.
Hence Ia ra D IS RS . Thus the value of the shunt,

Ia ra V D Ia ra D Is Rs , hence
RS = ohms Ia ra ⊲0.04⊳⊲25⊳
IS Rs D D D 0.02002 
IS 49.96
The milliammeter is converted into a voltmeter by = 20.02 mZ
connecting a high value resistance (called a mul-
tiplier) in series with it as shown in Fig. 10.4(b). Thus for the moving-coil instrument to be used as
From Fig. 10.4(b), an ammeter with a range 0–50 A, a resistance of
value 20.02 m needs to be connected in parallel
V D Va C VM D Ira C IRM with the instrument.

Thus the value of the multiplier, Problem 2. A moving-coil instrument

having a resistance of 10 , gives a f.s.d.
V − Ira when the current is 8 mA. Calculate the value
RM = ohms of the multiplier to be connected in series
I with the instrument so that it can be used as
a voltmeter for measuring p.d.s. up to 100 V

The circuit diagram is shown in Fig. 10.6, where

ra D resistance of instrument D 10 , RM D
resistance of multiplier I D total permissible instru-
ment current D 8 mA D 0.008 A, V D total p.d.
required to give f.s.d. D 100 V
V D Va C VM D Ira C IRM
Figure 10.4
i.e. 100 D ⊲0.008⊳⊲10⊳ C ⊲0.008⊳RM
or 100
0.08 D 0.008 RM , thus
Problem 1. A moving-coil instrument gives
a f.s.d. when the current is 40 mA and its 99.92
RM D D 12490  D 12.49 kZ
resistance is 25 . Calculate the value of the 0.008
shunt to be connected in parallel with the
meter to enable it to be used as an ammeter
for measuring currents up to 50 A

The circuit diagram is shown in Fig. 10.5, where

ra D resistance of instrument D 25 , Rs D
resistance of shunt, Ia D maximum permissible Figure 10.6

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108 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Hence for the moving-coil instrument to be used as input resistance (some as high as 1000 M) and can
a voltmeter with a range 0–100 V, a resistance of handle a much wider range of frequency (from d.c.
value 12.49 k needs to be connected in series with up to MHz).
the instrument. The digital voltmeter (DVM) is one which
provides a digital display of the voltage being mea-
sured. Advantages of a DVM over analogue instru-
Now try the following exercise ments include higher accuracy and resolution, no
observational or parallex errors (see section 10.20)
and a very high input resistance, constant on all
Exercise 48 Further problems on shunts ranges.
and multipliers A digital multimeter is a DVM with additional
circuitry which makes it capable of measuring a.c.
1 A moving-coil instrument gives f.s.d. for a voltage, d.c. and a.c. current and resistance.
current of 10 mA. Neglecting the resistance Instruments for a.c. measurements are generally
of the instrument, calculate the approximate calibrated with a sinusoidal alternating waveform to
value of series resistance needed to enable the indicate r.m.s. values when a sinusoidal signal is
instrument to measure up to (a) 20 V (b) 100 V applied to the instrument. Some instruments, such
(c) 250 V [(a) 2 k (b) 10 k (c) 25 k] as the moving-iron and electro-dynamic instruments,
2 A meter of resistance 50  has a f.s.d. of give a true r.m.s. indication. With other instruments
4 mA. Determine the value of shunt resis- the indication is either scaled up from the mean
tance required in order that f.s.d. should be value (such as with the rectified moving-coil instru-
(a) 15 mA (b) 20 A (c) 100 A ment) or scaled down from the peak value.
[(a) 18.18  (b) 10.00 m (c) 2.00 m] Sometimes quantities to be measured have com-
3 A moving-coil instrument having a resistance plex waveforms (see section 10.13), and whenever a
of 20 , gives a f.s.d. when the current is quantity is non-sinusoidal, errors in instrument read-
ings can occur if the instrument has been calibrated
5 mA. Calculate the value of the multiplier to
for sine waves only. Such waveform errors can be
be connected in series with the instrument so
largely eliminated by using electronic instruments.
that it can be used as a voltmeter for measuring
p.d.’s up to 200 V [39.98 k]
4 A moving-coil instrument has a f.s.d. of 20 mA
and a resistance of 25 . Calculate the val- 10.8 The ohmmeter
ues of resistance required to enable the instru-
ment to be used (a) as a 0–10 A ammeter, An ohmmeter is an instrument for measuring
and (b) as a 0–100 V voltmeter. State the electrical resistance. A simple ohmmeter circuit
mode of resistance connection in each case. is shown in Fig. 10.7(a). Unlike the ammeter or
[(a) 50.10 m in parallel voltmeter, the ohmmeter circuit does not receive the
(b) 4.975 k in series] energy necessary for its operation from the circuit
under test. In the ohmmeter this energy is supplied
5 A meter has a resistance of 40  and reg- by a self-contained source of voltage, such as a
isters a maximum deflection when a cur- battery. Initially, terminals XX are short-circuited
rent of 15 mA flows. Calculate the value of
resistance that converts the movement into
(a) an ammeter with a maximum deflection of
50 A (b) a voltmeter with a range 0–250 V
[(a) 12.00 m in parallel
(b) 16.63 k in series]

10.7 Electronic instruments

Electronic measuring instruments have advantages
over instruments such as the moving-iron or
moving-coil meters, in that they have a much higher Figure 10.7

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 109

and R adjusted to give f.s.d. on the milliammeter. If supplied to a load. The instrument has two coils:
current I is at a maximum value and voltage E is
constant, then resistance R D E/I is at a minimum (i) a current coil, which is connected in series with
value. Thus f.s.d. on the milliammeter is made zero the load, like an ammeter, and
on the resistance scale. When terminals XX are
(ii) a voltage coil, which is connected in parallel
open circuited no current flows and R ⊲D E/O⊳ is
with the load, like a voltmeter.
infinity, 1.
The milliammeter can thus be calibrated directly
in ohms. A cramped (non-linear) scale results and is
‘back to front’, as shown in Fig. 10.7(b). When cal- 10.11 Instrument ‘loading’ effect
ibrated, an unknown resistance is placed between
terminals XX and its value determined from the Some measuring instruments depend for their oper-
position of the pointer on the scale. An ohmme- ation on power taken from the circuit in which
ter designed for measuring low values of resis- measurements are being made. Depending on the
tance is called a continuity tester. An ohmmeter ‘loading’ effect of the instrument (i.e. the current
designed for measuring high values of resistance taken to enable it to operate), the prevailing circuit
(i.e. megohms) is called an insulation resistance conditions may change.
tester (e.g. ‘Megger’). The resistance of voltmeters may be calculated
since each have a stated sensitivity (or ‘figure of
merit’), often stated in ‘k per volt’ of f.s.d. A volt-
meter should have as high a resistance as possible
10.9 Multimeters (– ideally infinite). In a.c. circuits the impedance of
the instrument varies with frequency and thus the
Instruments are manufactured that combine a loading effect of the instrument can change.
moving-coil meter with a number of shunts and
series multipliers, to provide a range of readings
on a single scale graduated to read current and Problem 3. Calculate the power dissipated
voltage. If a battery is incorporated then resistance by the voltmeter and by resistor R in
can also be measured. Such instruments are Fig. 10.9 when (a) R D 250 
called multimeters or universal instruments or (b) R D 2 M. Assume that the voltmeter
multirange instruments. An ‘Avometer’ is a typical sensitivity (sometimes called figure of merit)
example. A particular range may be selected either is 10 k/V
by the use of separate terminals or by a selector
switch. Only one measurement can be performed at
a time. Often such instruments can be used in a.c. as
well as d.c. circuits when a rectifier is incorporated
in the instrument.

10.10 Wattmeters Figure 10.9

A wattmeter is an instrument for measuring electri- (a) Resistance of voltmeter, Rv D sensitivity ð

cal power in a circuit. Fig. 10.8 shows typical con- f.s.d. Hence, Rv D ⊲10 k/V⊳ ð ⊲200 V⊳ D
nections of a wattmeter used for measuring power 2000 k D 2 M. Current flowing in voltmeter,
V 100
Iv D D D 50 ð 10
6 A
Rv 2 ð 106
Power dissipated by voltmeter

D VIv D ⊲100⊳⊲50 ð 10
6 ⊳ D 5 mW.

When R D 250 , current in resistor,
V 100
Figure 10.8 IR D D D 0.4 A
R 250

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Power dissipated in load resistor R D VIR D

⊲100⊳⊲0.4⊳ D 40 W. Thus the power dissipated Problem 5. A voltmeter having a f.s.d. of
in the voltmeter is insignificant in comparison 100 V and a sensitivity of 1.6 k/V is used
with the power dissipated in the load. to measure voltage V1 in the circuit of
Fig. 10.11 Determine (a) the value of voltage
(b) When R D 2 M, current in resistor, V1 with the voltmeter not connected, and (b)
the voltage indicated by the voltmeter when
V 100 connected between A and B
IR D D D 50 ð 10
6 A
R 2 ð 106
Power dissipated in load resistor R D VIR D
100ð50ð10
6 D 5 mW. In this case the higher
load resistance reduced the power dissipated
such that the voltmeter is using as much power
as the load. Figure 10.11

Problem 4. An ammeter has a f.s.d. of (a) By voltage division,

100 mA and a resistance of 50 . The

40

ammeter is used to measure the current in a V1 D 100 D 40 V
40 C 60
load of resistance 500  when the supply
voltage is 10 V. Calculate (a) the ammeter (b) The resistance of a voltmeter having a 100 V
reading expected (neglecting its resistance), f.s.d. and sensitivity 1.6 k/V is 100 V ð
(b) the actual current in the circuit, (c) the 1.6 k/V D 160 k. When the voltmeter is
power dissipated in the ammeter, and (d) the connected across the 40 k resistor the circuit
power dissipated in the load. is as shown in Fig. 10.12(a) and the equivalent
resistance of the parallel network is given by


From Fig. 10.10, 40 ð 160
k i.e.
40 C 160


40 ð 160
k D 32 k
200

The circuit is now effectively as shown in

Fig. 10.12(b). Thus the voltage indicated on the
voltmeter is


32
Figure 10.10 100 V D 34.78 V
32 C 60

(a) expected ammeter reading D V/R D 10/500 D A considerable error is thus caused by the load-
20 mA. ing effect of the voltmeter on the circuit. The error
is reduced by using a voltmeter with a higher
(b) Actual ammeter reading D V/⊲R C ra ⊳ D sensitivity.
10/⊲500 C 50⊳ D 18.18 mA. Thus the ammeter
itself has caused the circuit conditions to change
from 20 mA to 18.18 mA.

(c) Power dissipated in the ammeter D I2 ra D

⊲18.18 ð 10
3 ⊳2 ⊲50⊳ D 16.53 mW.

(d) Power dissipated in the load resistor D I2 R D

⊲18.18 ð 10
3 ⊳2 ⊲500⊳ D 165.3 mW. Figure 10.12

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 111

3 A voltage of 240 V is applied to a circuit

Problem 6. (a) A current of 20 A flows
consisting of an 800  resistor in series with
through a load having a resistance of 2 .
a 1.6 k resistor. What is the voltage across
Determine the power dissipated in the load.
the 1.6 k resistor? The p.d. across the 1.6 k
(b) A wattmeter, whose current coil has a
resistor is measured by a voltmeter of f.s.d.
resistance of 0.01  is connected as shown in
250 V and sensitivity 100 /V. Determine the
Fig. 10.13 Determine the wattmeter reading.
voltage indicated. [160 V; 156.7 V]

10.12 The cathode ray oscilloscope

The cathode ray oscilloscope (c.r.o.) may be used
in the observation of waveforms and for the mea-
surement of voltage, current, frequency, phase and
periodic time. For examining periodic waveforms
Figure 10.13 the electron beam is deflected horizontally (i.e. in
the X direction) by a sawtooth generator acting as
a timebase. The signal to be examined is applied to
(a) Power dissipated in the load, P D I2 R D the vertical deflection system (Y direction) usually
⊲20⊳2 ⊲2⊳ D 800 W after amplification.
Oscilloscopes normally have a transparent grid
(b) With the wattmeter connected in the circuit the of 10 mm by 10 mm squares in front of the screen,
total resistance RT is 2 C 0.01 D 2.01 . The called a graticule. Among the timebase controls is
wattmeter reading is thus I2 RT D ⊲20⊳2 ⊲2.01⊳ D a ‘variable’ switch which gives the sweep speed as
804 W time per centimetre. This may be in s/cm, ms/cm
or µs/cm, a large number of switch positions being
available. Also on the front panel of a c.r.o. is a
Now try the following exercise Y amplifier switch marked in volts per centimetre,
with a large number of available switch positions.

Exercise 49 Further problems on (i) With direct voltage measurements, only the
instrument ‘loading’ effects Y amplifier ‘volts/cm’ switch on the c.r.o. is
1 A 0–1 A ammeter having a resistance of 50  used. With no voltage applied to the Y plates
is used to measure the current flowing in a the position of the spot trace on the screen is
1 k resistor when the supply voltage is 250 V. noted. When a direct voltage is applied to the
Calculate: (a) the approximate value of current Y plates the new position of the spot trace is
(neglecting the ammeter resistance), (b) the an indication of the magnitude of the voltage.
actual current in the circuit, (c) the power For example, in Fig. 10.14(a), with no voltage
dissipated in the ammeter, (d) the power dis- applied to the Y plates, the spot trace is in the
sipated in the 1 k resistor. centre of the screen (initial position) and then
[(a) 0.250 A (b) 0.238 A the spot trace moves 2.5 cm to the final position
(c) 2.83 W (d) 56.64 W] shown, on application of a d.c. voltage. With the
‘volts/cm’ switch on 10 volts/cm the magnitude
2 (a) A current of 15 A flows through a load of the direct voltage is 2.5 cm ð 10 volts/cm, i.e.
having a resistance of 4 . Determine the 25 volts.
power dissipated in the load. (b) A wattmeter,
whose current coil has a resistance of 0.02  is (ii) With alternating voltage measurements, let a
connected (as shown in Fig. 10.13) to measure sinusoidal waveform be displayed on a c.r.o.
the power in the load. Determine the wattmeter screen as shown in Fig. 10.14(b). If the time/cm
reading assuming the current in the load is still switch is on, say, 5 ms/cm then the periodic
15 A. time T of the sinewave is 5 ms/cm ð 4 cm, i.e.
[(a) 900 W (b) 904.5 W] 20 ms or 0.02 s. Since frequency

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Turning it to zero ensures no signal is

applied to the X-plates. The Y-plate input
is left open-circuited.
(iii) Set the intensity, X-shift and Y-shift con-
trols to about the mid-range positions.
(iv) A spot trace should now be observed on
the screen. If not, adjust either or both
of the X and Y-shift controls. The X-shift
control varies the position of the spot trace
in a horizontal direction whilst the Y-shift
control varies its vertical position.
(v) Use the X and Y-shift controls to bring the
spot to the centre of the screen and use the
focus control to focus the electron beam
into a small circular spot.
(b) To obtain a continuous horizontal trace on the
screen the same procedure as in (a) is initially
adopted. Then the timebase control is switched
to a suitable position, initially the millisecond
timebase range, to ensure that the repetition rate
Figure 10.14 of the sawtooth is sufficient for the persistence
of the vision time of the screen phosphor to hold
1 1 a given trace.
fD , frequency = = 50 Hz
T 0.02
If the ‘volts/cm’ switch is on, say, 20 volts/cm Problem 8. For the c.r.o. square voltage
then the amplitude or peak value of the waveform shown in Fig. 10.15 determine (a)
sinewave shown is 20 volts/cmð2 cm, i.e. 40 V. the periodic time, (b) the frequency and (c)
Since the peak-to-peak voltage. The ‘time/cm’ (or
peak voltage timebase control) switch is on 100 µs/cm and
r.m.s. voltage D p , (see Chapter 14), the ‘volts/cm’ (or signal amplitude control)
2 switch 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 7. Describe how a simple c.r.o. is Figure 10.15

adjusted to give (a) a spot trace, (b) a
continuous horizontal trace on the screen,
explaining the functions of the various (In Figures 10.15 to 10.18 assume that the squares
controls. shown are 1 cm by 1 cm)

(a) The width of one complete cycle is 5.2 cm.

(a) To obtain a spot trace on a typical c.r.o. screen: Hence the periodic time,
(i) Switch on the c.r.o. T D 5.2 cm ð 100 ð 10
6 s/cm D 0.52 ms.
(ii) Switch the timebase control to off. This
control is calibrated in time per centime- 1 1
tres – for example, 5 ms/cm or 100 µs/cm. (b) Frequency, f D D D 1.92 kHz.
T 0.52 ð 10
3

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 113

(c) The peak-to-peak height of the display is 3.6 cm, (a) The width of one complete cycle is 4 cm. Hence
hence the peak-to-peak voltage the periodic time, T is 4 cm ð 500 µs/cm, i.e.
2 ms.
D 3.6 cm ð 20 V/cm D 72 V 1 1
Frequency, f D D D 500 Hz
T 2 ð 10
3
Problem 9. For the c.r.o. display of a pulse (b) The peak-to-peak height of the waveform is
waveform shown in Fig. 10.16 the ‘time/cm’ 5 cm. Hence the peak-to-peak voltage
switch is on 50 ms/cm and the ‘volts/cm’ D 5 cm ð 5 V/cm D 25 V.
switch is on 0.2 V/cm. Determine (a) the
1
periodic time, (b) the frequency, (c) the (c) Amplitude D 2 ð 25 V D 12.5 V
magnitude of the pulse voltage.
(d) The peak value of voltage is the amplitude, i.e.
12.5 V, and r.m.s.
peak voltage 12.5
voltage D p D p D 8.84 V
2 2

Problem 11. For the double-beam

oscilloscope displays shown in Fig. 10.18
determine (a) their frequency, (b) their r.m.s.
values, (c) their phase difference. The
Figure 10.16
‘time/cm’ switch is on 100 µs/cm and the
‘volts/cm’ switch on 2 V/cm.
(a) The width of one complete cycle is 3.5 cm.
Hence the periodic time, T D 3.5 cm ð
50 ms/cm D 175 ms.
1 1
(b) Frequency, f D D D 5.71 Hz.
T 0.52 ð 10
3
(c) The height of a pulse is 3.4 cm hence the magni-
tude of the pulse voltage D 3.4 cmð0.2 V/cm D
0.68 V.
Figure 10.18
Problem 10. A sinusoidal voltage trace
displayed by a c.r.o. is shown in Fig. 10.17
If the ‘time/cm’ switch is on 500 µs/cm and (a) The width of each complete cycle is 5 cm for
the ‘volts/cm’ switch is on 5 V/cm, find, for both waveforms. Hence the periodic time, T, of
the waveform, (a) the frequency, (b) the each waveform is 5 cm ð 100 µs/cm, i.e. 0.5 ms.
peak-to-peak voltage, (c) the amplitude, Frequency of each waveform,
(d) the r.m.s. value. 1 1
fD D D 2 kHz
T 0.5 ð 10
3
(b) The peak value of waveform A is
2 cm ð 2 V/cm D 4 V, hence the r.m.s. value of
waveform A
p
D 4/⊲ 2⊳ D 2.83 V
The peak value of waveform B is
2.5 cm ð 2 V/cm D 5 V, hence the r.m.s. value
of waveform B
Figure 10.17 p
D 5/⊲ 2⊳ D 3.54 V

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114 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

(c) Since 5 cm represents 1 cycle, then 5 cm rep- 3 For the sinusoidal waveform shown in
resents 360° , i.e. 1 cm represents 360/5 D 72° . Fig. 10.21, determine (a) its frequency, (b) the
The phase angle  D 0.5 cm peak-to-peak voltage, (c) the r.m.s. voltage
D 0.5 cm ð 72° /cm D 36° . [(a) 7.14 Hz (b) 220 V (c) 77.78 V]
Hence waveform A leads waveform B by 36°

Now try the following exercise

Exercise 50 Further problems on the

cathode ray oscilloscope
1 For the square voltage waveform displayed
on a c.r.o. shown in Fig. 10.19, find (a) its
frequency, (b) its peak-to-peak voltage
[(a) 41.7 Hz (b) 176 V]
Figure 10.21

10.13 Waveform harmonics

(i) Let an instantaneous voltage v be represented

by v D Vm sin 2ft volts. This is a waveform
which varies sinusoidally with time t, has a
frequency f, and a maximum value Vm . Alter-
nating voltages are usually assumed to have
wave-shapes which are sinusoidal where only
one frequency is present. If the waveform is
Figure 10.19 not sinusoidal it is called a complex wave,
and, whatever its shape, it may be split up
2 For the pulse waveform shown in Fig. 10.20, mathematically into components called the fun-
find (a) its frequency, (b) the magnitude of the damental and a number of harmonics. This
pulse voltage process is called harmonic analysis. The funda-
[(a) 0.56 Hz (b) 8.4 V] mental (or first harmonic) is sinusoidal and has
the supply frequency, f; the other harmonics
are also sine waves having frequencies which
are integer multiples of f. Thus, if the supply
frequency is 50 Hz, then the third harmonic fre-
quency is 150 Hz, the fifth 250 Hz, and so on.
(ii) A complex waveform comprising the sum of
the fundamental and a third harmonic of about
half the amplitude of the fundamental is shown
in Fig. 10.22(a), both waveforms being initially
in phase with each other. If further odd har-
monic waveforms of the appropriate amplitudes
are added, a good approximation to a square
wave results. In Fig. 10.22(b), the third har-
monic is shown having an initial phase dis-
Figure 10.20 placement from the fundamental. The positive

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 115

and negative half cycles of each of the com- a mirror image of the positive cycle about
plex waveforms shown in Figures 10.22(a) and point B. In Fig. 10.22(f), a complex wave-
(b) are identical in shape, and this is a feature form comprising the sum of the fundamen-
of waveforms containing the fundamental and tal, a second harmonic and a third harmonic
only odd harmonics. are shown with initial phase displacement. The
positive and negative half cycles are seen to be
dissimilar.

The features mentioned relative to Figures 10.22

(a) to (f) make it possible to recognize the harmon-
ics present in a complex waveform displayed on
a CRO.

10.14 Logarithmic ratios

In electronic systems, the ratio of two similar quan-
tities measured at different points in the system, are
often expressed in logarithmic units. By definition, if
the ratio of two powers P1 and P2 is to be expressed
in decibel (dB) units then the number of decibels,
X, is given by:


P2
X = 10 lg dB ⊲1⊳
P1

Thus, when the power ratio, P2 /P1 D 1 then the

decibel power ratio D 10 lg 1 D 0, when the
power ratio, P2 /P1 D 100 then the decibel power
ratio D 10 lg 100 D C20 (i.e. a power gain), and
Figure 10.22
when the power ratio, P2 /P1 D 1/100 then the
decibel power ratio D 10 lg 1/100 D
20 (i.e. a
(iii) A complex waveform comprising the sum of power loss or attenuation).
the fundamental and a second harmonic of Logarithmic units may also be used for voltage
about half the amplitude of the fundamen- and current ratios. Power, P, is given by P D I2 R
tal is shown in Fig. 10.22(c), each waveform or P D V2 /R. Substituting in equation (1) gives:
being initially in phase with each other. If
further even harmonics of appropriate ampli-  
I22 R2
tudes are added a good approximation to a X D 10 lg dB
triangular wave results. In Fig. 10.22(c), the I21 R1
negative cycle, if reversed, appears as a mir-  
ror image of the positive cycle about point A. V22 /R2
In Fig. 10.22(d) the second harmonic is shown or X D 10 lg dB
V21 /R1
with an initial phase displacement from the fun-
damental and the positive and negative half If R1 D R2 ,
cycles are dissimilar.  
I22
(iv) A complex waveform comprising the sum then X D 10 lg dB or
of the fundamental, a second harmonic and I21
a third harmonic is shown in Fig. 10.22(e),  
each waveform being initially ‘in-phase’. The V22
X D 10 lg dB
negative half cycle, if reversed, appears as V21

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116 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY


I2
i.e. X = 20 lg dB
I1


V2
or X = 20 lg dB
V1

Figure 10.23
(from the laws of logarithms).
From equation (1), X decibels is a logarithmic
ratio of two similar quantities and is not an absolute From above, the power ratio in decibels, X, is given
unit of measurement. It is therefore necessary to by: X D 10 lg ⊲P2 /P1 ⊳
state a reference level to measure a number of
decibels above or below that reference. The most P2
widely used reference level for power is 1 mW, and (a) When D 3,
P1
when power levels are expressed in decibels, above
or below the 1 mW reference level, the unit given X D 10 lg ⊲3⊳ D 10⊲0.477⊳
to the new power level is dBm. D 4.77 dB
A voltmeter can be re-scaled to indicate the power
level directly in decibels. The scale is generally cal- P2
ibrated by taking a reference level of 0 dB when a (b) When D 20,
P1
power of 1 mW is dissipated in a 600  resistor (this
being the natural impedance of a simple transmis- X D 10 lg ⊲20⊳ D 10⊲1.30⊳
sion line). The reference voltage V is then obtained D 13.0 dB
from
P2
V2 (c) When D 400,
PD , P1
R
X D 10 lg ⊲400⊳ D 10⊲2.60⊳

3 V2
i.e. 1 ð 10 D D 26.0 dB
600
from which, V D 0.775 volts. In general, the number P2 1
(d) When D D 0.05,
of dBm, P1 20
X D 10 lg ⊲0.05⊳ D 10⊲
1.30⊳


V
X D 20 lg
0.775 D −13.0 dB


0.2
 (a), (b) and (c) represent power gains and (d) repre-
Thus V D 0.20 V corresponds to 20 lg sents a power loss or attenuation.
0.775
D
11.77 dBm and

 Problem 13. The current input to a system
0.90 is 5 mA and the current output is 20 mA.
V D 0.90 V corresponds to 20 lg
0.775 Find the decibel current ratio assuming the
input and load resistances of the system are
D C1.3 dBm, and so on.
equal.
A typical decibelmeter, or dB meter, scale is shown
in Fig. 10.23. Errors are introduced with dB meters
From above, the decibel current ratio is
when the circuit impedance is not 600 .



I2 20
20 lg D 20 lg
Problem 12. The ratio of two powers is I1 5
(a) 3 (b) 20 (c) 4 (d) 1/20. Determine the D 20 lg 4 D 20⊲0.60⊳
decibel power ratio in each case.
D 12 dB gain

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 117

power ratio D 12 C 15
8 D 19 dB gain.

Problem 14. 6% of the power supplied to a
cable appears at the output terminals.


P2
Determine the power loss in decibels. Thus 19 D 10 lg
P1


P2
If P1 D input power and P2 D output power then from which 1.9 D lg
P1
P2 6 P2
D D 0.06 and 101.9 D D 79.4
P1 100 P1


Decibel P2 P2
D 10 lg D 10 lg ⊲0.06⊳ Thus the overall power gain, = 79.4
power ratio P1 P1
D 10⊲
1.222⊳ D
12.22 dB [For the first stage,


Hence the decibel power loss, or attenuation, is P2
12.22 dB. 12 D 10 lg
P1

Problem 15. An amplifier has a gain of from which

14 dB and its input power is 8 mW. Find its P2
output power. D 101.2 D 15.85
P1

Decibel power ratio D 10 lg ⊲P2 /P1 ⊳ where P1 D Similarly for the second stage,
input power D 8 mW, and P2 D output power.
P2
Hence D 31.62

 P1
P2
14 D 10 lg and for the third stage,
P1

from which P2
D 0.1585

 P1
P2
1.4 D lg The overall power ratio is thus
P1
15.85 ð 31.62 ð 0.1585 D 79.4]
P2 from the definition
and 101.4 D
P1 of a logarithm
Problem 17. The output voltage from an
P2 amplifier is 4 V. If the voltage gain is 27 dB,
i.e. 25.12 D
P1 calculate the value of the input voltage
assuming that the amplifier input resistance
Output power, P2 D 25.12 P1 D ⊲25.12⊳⊲8⊳ D and load resistance are equal.
201 mW or 0.201 W

Voltage gain in decibels D 27 D 20 lg ⊲V2 /V1 ⊳ D

Problem 16. Determine, in decibels, the 20 lg ⊲4/V1 ⊳. Hence
ratio of output power to input power of a 3
stage communications system, the stages


27 4
having gains of 12 dB, 15 dB and
8 dB. D lg
20 V1
Find also the overall power gain.

4
i.e. 1.35 D lg
V1
The decibel ratio may be used to find the overall
power ratio of a chain simply by adding the decibel 4
power ratios together. Hence the overall decibel Thus 101.35 D
V1

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118 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

4 3.8 dB. Calculate the overall gain in decibels

from which V1 D
101.35 assuming that input and load resistances for
4 each stage are equal. If a voltage of 15 mV is
D applied to the input of the system, determine
22.39 the value of the output voltage.
D 0.179 V [8.5 dB, 39.91 mV]

Hence the input voltage V1 is 0.179 V. 9 The scale of a voltmeter has a decibel scale
added to it, which is calibrated by taking a
reference level of 0 dB when a power of 1 mW
Now try the following exercise is dissipated in a 600  resistor. Determine
the voltage at (a) 0 dB (b) 1.5 dB (c)
15 dB
(d) What decibel reading corresponds to
0.5 V?
[(a) 0.775 V (b) 0.921 V
Exercise 51 Further problems on (c) 0.138 V (d)
3.807 dB]
logarithmic ratios
1 The ratio of two powers is (a) 3 (b) 10 (c) 20
(d) 10 000. Determine the decibel power ratio
for each.
[(a) 4.77 dB (b) 10 dB (c) 13 dB (d) 40 dB] 10.15 Null method of measurement
1 1 1
2 The ratio of two powers is (a) 10 (b) (c)
3 40
1 A null method of measurement is a simple, accu-
(d) 100 . Determine the decibel power ratio for rate and widely used method which depends on an
each. instrument reading being adjusted to read zero cur-
[(a)
10 dB (b)
4.77 dB
(c)
16.02 dB (d)
20 dB] rent only. The method assumes:

3 The input and output currents of a system are (i) if there is any deflection at all, then some current
2 mA and 10 mA respectively. Determine the is flowing;
decibel current ratio of output to input current
assuming input and output resistances of the (ii) if there is no deflection, then no current flows
system are equal. [13.98 dB] (i.e. a null condition).
4 5% of the power supplied to a cable appears
at the output terminals. Determine the power Hence it is unnecessary for a meter sensing current
loss in decibels. [13 dB] flow to be calibrated when used in this way. A sensi-
tive milliammeter or microammeter with centre zero
5 An amplifier has a gain of 24 dB and its input position setting is called a galvanometer. Examples
power is 10 mW. Find its output power. where the method is used are in the Wheatstone
[2.51 W] bridge (see section 10.16), in the d.c. potentiometer
(see section 10.17) and with a.c. bridges (see sec-
6 Determine, in decibels, the ratio of the output tion 10.18)
power to input power of a four stage system,
the stages having gains of 10 dB, 8 dB,
5 dB
and 7 dB. Find also the overall power gain.
[20 dB, 100]
10.16 Wheatstone bridge
7 The output voltage from an amplifier is 7 mV.
If the voltage gain is 25 dB calculate the value Figure 10.24 shows a Wheatstone bridge circuit
of the input voltage assuming that the amplifier which compares an unknown resistance Rx with
input resistance and load resistance are equal. others of known values, i.e. R1 and R2 , which have
[0.39 mV] fixed values, and R3 , which is variable. R3 is varied
until zero deflection is obtained on the galvanometer
8 The voltage gain of a number of cascaded G. No current then flows through the meter, VA D
amplifiers are 23 dB,
5.8 dB,
12.5 dB and VB , and the bridge is said to be ‘balanced’. At

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 119

balance,
10.17 D.C. potentiometer
R2 R3 The d.c. potentiometer is a null-balance instru-
R1 Rx D R2 R3 i.e. Rx = ohms
R1 ment used for determining values of e.m.f.’s and
p.d.s. by comparison with a known e.m.f. or p.d. In
Fig. 10.26(a), using a standard cell 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 .

Figure 10.24

Problem 18. In a Wheatstone bridge

ABCD, a galvanometer is connected between
A and C, and a battery between B and D. A
Figure 10.26
resistor of unknown value is connected
between A and B. When the bridge is
balanced, the resistance between B and C is
100 , that between C and D is 10  and The standard cell is now replaced by a cell of
that between D and A is 400 . Calculate the unknown e.m.f. E2 (see Fig. 10.26(b)) and again
value of the unknown resistance. balance is obtained (shown as l2 ). Since E1 / l1
and E2 / l2 then

The Wheatstone bridge is shown in Fig. 10.25 where E1 l1

D
Rx is the unknown resistance. At balance, equating E2 l2
the products of opposite ratio arms, gives:


l2
⊲Rx ⊳⊲10⊳ D ⊲100⊳⊲400⊳ and E2 = E1 volts
l1
⊲100⊳⊲400⊳
and Rx D D 4000 
10
A potentiometer may be arranged as a resistive two-
element potential divider in which the division ratio
is adjustable to give a simple variable d.c. supply.
Such devices may be constructed in the form of a
resistive element carrying a sliding contact which
is adjusted by a rotary or linear movement of the
control knob.

Problem 19. In a d.c. potentiometer,

balance is 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 if balance
is obtained with a length of 650 mm
Figure 10.25

Hence, the unknown resistance, Rx D 4 kZ. E1 D 1.0186 V, l1 D 400 mm and l2 D 650 mm

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120 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

With reference to Fig. 10.26, When the potential differences across Z3 and
Zx (or across Z1 and Z2 ) are equal in magnitude
E1 l1 and phase, then the current flowing through the
D
E2 l2 galvanometer, G, is zero. At balance, Z1 Zx D Z2 Z3
from which
from which,
Z2 Z3
Zx = Z



l2 650
E 2 D E1 D ⊲1.0186⊳ Z1
l1 400
D 1.655 volts There are many forms of a.c. bridge, and these
include: the Maxwell, Hay, Owen and Heaviside
bridges for measuring inductance, and the De Sauty,
Now try the following exercise Schering and Wien bridges for measuring capaci-
tance. A commercial or universal bridge is one
which can be used to measure resistance, inductance
Exercise 52 Further problems on the or capacitance. A.c. bridges require a knowledge
p of
Wheatstone bridge and d.c. potentiometer complex numbers (i.e. j notation, where j D
1).
A Maxwell-Wien bridge for measuring the induc-
1 In a Wheatstone bridge PQRS, a galvanometer
tance L and resistance r of an inductor is shown in
is connected between Q and S and a voltage
Fig. 10.28
source between P and R. An unknown resistor
Rx is connected between P and Q. When the
bridge is balanced, the resistance between Q
and R is 200 , that between R and S is 10 
and that between S and P is 150 . Calculate
the value of Rx [3 k]
2 Balance is obtained in a d.c. potentiometer at a
length of 31.2 cm when using a standard cell of
1.0186 volts. Calculate the e.m.f. of a dry cell
if balance is obtained with a length of 46.7 cm
[1.525 V]

10.18 A.C. bridges

Figure 10.28
A Wheatstone bridge type circuit, shown in
Fig. 10.27, may be used in a.c. circuits to determine
unknown values of inductance and capacitance, as At balance the products of diagonally opposite
well as resistance. impedances are equal. Thus
Z1 Z2 D Z3 Z4
Using complex quantities, Z1 D R1 , Z2 D R2 ,


R3 ⊲
jXC ⊳ product
Z3 D i.e.
R3
jXC sum
and Z4 D r C jXL . Hence
R3 ⊲
jXC ⊳
R 1 R2 D ⊲r C jXL ⊳
R3
jXC
i.e. R1 R2 ⊲R3
jXC ⊳ D ⊲
jR3 XC ⊳⊲r C jXL ⊳
Figure 10.27 R1 R2 R3
jR1 R2 XC D
jrR3 XC
j2 R3 XC XL

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 121

i.e. R1 R2 R3
jR1 R2 XC D
jrR3 XC C R3 XC XL If the frequency is constant then R3 / L/r / ωL/r /

Q-factor (see Chapters 15 and 16). Thus the bridge
(since j2 D
1⊳. can be adjusted to give a direct indication of Q-factor.
Equating the real parts gives: A Q-meter is described in section 10.19 following.
R1 R2 R3 D R3 XC XL
R 1 R2 Now try the following exercise
from which, XL D
XC
R 1 R2 Exercise 53 Further problem on a.c.
i.e. 2fL D D R1 R2 ⊲2fC⊳ bridges
1
2fC 1 A Maxwell bridge circuit ABCD has the fol-
lowing arm impedances: AB, 250  resistance;
Hence inductance, BC, 15 µF capacitor in parallel with a 10 k
resistor; CD, 400  resistor; DA, unknown
L D R1 R2 C henry ⊲2⊳ inductor having inductance L and resistance
R. Determine the values of L and R assuming
Equating the imaginary parts gives: the bridge is balanced. [1.5 H, 10 ]

R1 R2 XC D
rR3 XC

from which, resistance,

10.19 Q-meter
R 1 R2
rD ohms ⊲3⊳ The Q-factor for a series L–C–R circuit is the
R3 voltage magnification at resonance, i.e.
voltage across capacitor
Problem 20. For the a.c. bridge shown in Q-factor D
Fig. 10.28 determine the values of the supply voltage
inductance and resistance of the coil when Vc
R1 D R2 D 400 , R3 D 5 k and C D 7.5 µF D (see Chapter 15).
V
The simplified circuit of a Q-meter, used for mea-
From equation (2) above, inductance suring Q-factor, is shown in Fig. 10.29. Current
from a variable frequency oscillator flowing through
L D R1 R2 C D ⊲400⊳⊲400⊳⊲7.5 ð 10
6 ⊳ a very low resistance r develops a variable fre-
quency voltage, Vr , which is applied to a series
D 1.2 H L–R–C circuit. The frequency is then varied until
resonance causes voltage Vc to reach a maximum
From equation (3) above, resistance,
value. At resonance Vr and Vc are noted. Then
R 1 R2 ⊲400⊳⊲400⊳ Vc Vc
rD D = 32 Z Q-factor D D
R3 5000 Vr Ir
In a practical Q-meter, Vr is maintained constant and
the electronic voltmeter can be calibrated to indicate
From equation (2), the Q-factor directly. If a variable capacitor C is
L used and the oscillator is set to a given frequency,
R2 D then C can be adjusted to give resonance. In this
R1 C way inductance L may be calculated using
and from equation (3), 1
fr D p
R1 2 LC
R3 D R2 2fL
r Since QD ,
R1 L L R
Hence R3 D D then R may be calculated.
r R1 C Cr

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122 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

(b) Q-factor at resonance D 2fr L/R from which

resistance
2fr L
RD
Q
2⊲400 ð 103 ⊳⊲0.396 ð 10
3 ⊳
D
100
D 9.95 Z

Now try the following exercise

Figure 10.29

Q-meters operate at various frequencies and Exercise 54 Further problem on the

instruments exist with frequency ranges from 1 kHz Q-meter
to 50 MHz. Errors in measurement can exist with 1 A Q-meter measures the Q-factor of a series L-
Q-meters since the coil has an effective parallel self C-R circuit to be 200 at a resonant frequency
capacitance due to capacitance between turns. The of 250 kHz. If the capacitance of the Q-meter
accuracy of a Q-meter is approximately š5%. capacitor is set to 300 pF determine (a) the
inductance L, and (b) the resistance R of the
Problem 21. When connected to a Q-meter inductor. [(a) 1.351 mH (b) 10.61 ]
an inductor is made to resonate at 400 kHz.
The Q-factor of the circuit is found to be 100
and the capacitance of the Q-meter capacitor
is set to 400 pF. Determine (a) the
inductance, and (b) the resistance of the 10.20 Measurement errors
inductor.
Errors are always introduced when using instru-
ments to measure electrical quantities. The errors
Resonant frequency, fr D 400 kHz D 400 ð 103 Hz,
most likely to occur in measurements are those
Q-factor = 100 and capacitance, C D 400 pF D
due to:
400 ð 10
12 F. The circuit diagram of a Q-meter is
shown in Fig. 10.29
(i) the limitations of the instrument;
(a) At resonance, (ii) the operator;
(iii) the instrument disturbing the circuit.
1
fr D p
2 LC (i) Errors in the limitations of the instrument
for a series L–C–R circuit. Hence The calibration accuracy of an instrument
1 depends on the precision with which it is
2fr D p constructed. Every instrument has a margin of
LC
error which is expressed as a percentage of the
from which instruments full scale deflection. For example,
1 industrial grade instruments have an accuracy of
⊲2fr ⊳2 D
LC š2% of f.s.d. Thus if a voltmeter has a f.s.d. of
and inductance, 100 V and it indicates 40 V say, then the actual
voltage may be anywhere between 40š(2% of 100),
1
LD or 40 š 2, i.e. between 38 V and 42 V.
⊲2fr ⊳2 C When an instrument is calibrated, it is compared
1 against a standard instrument and a graph is drawn
D H of ‘error’ against ‘meter deflection’. A typical graph
⊲2 ð 400 ð 103 ⊳2 ⊲400 ð 10
12 ⊳ is shown in Fig. 10.30 where it is seen that the
D 396 mH or 0.396 mH accuracy varies over the scale length. Thus a meter

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 123

with a š2% f.s.d. accuracy would tend to have an

accuracy which is much better than š2% f.s.d. over
much of the range.

Figure 10.31

Voltage, V D IR D ⊲2.5 ð 10
3 ⊳⊲5 ð 103 ⊳ D 12.5 V.

The maximum possible error is
Figure 10.30 0.4% C 0.5% D 0.9%.
Hence the voltage, V D 12.5 V š 0.9% of 12.5 V
(ii) Errors by the operator 0.9% of 12.5 D 0.9/100 ð 12.5 D 0.1125 V D
It is easy for an operator to misread an instrument. 0.11 V correct to 2 significant figures.
With linear scales the values of the sub-divisions Hence the voltage V may also be expressed
are reasonably easy to determine; non-linear scale as 12.5 ± 0.11 volts (i.e. a voltage lying between
graduations are more difficult to estimate. Also, 12.39 V and 12.61 V).
scales differ from instrument to instrument and some
meters have more than one scale (as with multime- Problem 23. The current I flowing in a
ters) and mistakes in reading indications are easily resistor R is measured by a 0–10 A ammeter
made. When reading a meter scale it should be which gives an indication of 6.25 A. The
viewed from an angle perpendicular to the surface voltage V across the resistor is measured by
of the scale at the location of the pointer; a meter a 0–50 V voltmeter, which gives an
scale should not be viewed ‘at an angle’. indication of 36.5 V. Determine the
(iii) Errors due to the instrument disturbing resistance of the resistor, and its accuracy of
the circuit measurement if both instruments have a limit
Any instrument connected into a circuit will of error of 2% of f.s.d. Neglect any loading
affect that circuit to some extent. Meters require effects of the instruments.
some power to operate, but provided this power
is small compared with the power in the measured Resistance,
circuit, then little error will result. Incorrect posi- V 36.5
tioning of instruments in a circuit can be a source RD D D 5.84 
of errors. For example, let a resistance be mea- I 6.25
sured by the voltmeter-ammeter method as shown Voltage error is š2% of 50 V D š1.0 V and
in Fig. 10.31 Assuming ‘perfect’ instruments, the expressed as a percentage of the voltmeter reading
resistance should be given by the voltmeter read- gives
ing divided by the ammeter reading (i.e. R D š1
V/I). However, in Fig. 10.31(a), V/I D R C ra ð 100% D š2.74%
and in Fig. 10.31(b) the current through the amme- 36.5
ter is that through the resistor plus that through Current error is š2% of 10 A D š0.2 A and express-
the voltmeter. Hence the voltmeter reading divided ed as a percentage of the ammeter reading gives
by the ammeter reading will not give the true
value of the resistance R for either method of š0.2
ð 100% D š3.2%
connection. 6.25
Maximum relative error D sum of errors D
Problem 22. The current flowing through a 2.74% C 3.2% D š5.94%. 5.94% of 5.84  D
resistor of 5 k š 0.4% is measured as 0.347 . Hence the resistance of the resistor may
2.5 mA with an accuracy of measurement of be expressed as:
š0.5%. Determine the nominal value of the 5.84 Z ± 5.94% or 5.84 ± 0.35 Z
voltage across the resistor and its accuracy.
(rounding off)

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flowing in the resistor and its accuracy of

Problem 24. The arms of a Wheatstone
measurement.
bridge ABCD have the following resistances:
[6.25 mA š 1.3% or 6.25 š 0.08 mA]
AB: R1 D 1000  š 1.0%; BC:
R2 D 100  š 0.5%; CD: unknown resistance 2 The voltage across a resistor is measured by a
Rx ; DA: R3 D 432.5  š 0.2%. Determine 75 V f.s.d. voltmeter which gives an indication
the value of the unknown resistance and its of 52 V. The current flowing in the resistor
accuracy of measurement. is measured by a 20 A f.s.d. ammeter which
gives an indication of 12.5 A. Determine the
resistance of the resistor and its accuracy if
The Wheatstone bridge network is shown in both instruments have an accuracy of š2% of
Fig. 10.32 and at balance: f.s.d. [4.16  š 6.08% or 4.16 š 0.25 ]
R1 Rx D R2 R3 , 3 A 240 V supply is connected across a load
R 2 R3 ⊲100⊳⊲432.5⊳ resistance R. Also connected across R is a
i.e. Rx D D D 43.25  voltmeter having a f.s.d. of 300 V and a figure
R1 1000 of merit (i.e. sensitivity) of 8 k/V. Calculate
the power dissipated by the voltmeter and by
the load resistance if (a) R D 100  (b) R D
1 M. Comment on the results obtained.
[(a) 24 mW, 576 W (b) 24 mW, 57.6 mW]
4 A Wheatstone bridge PQRS has the following
arm resistances: PQ, 1 k š 2%; QR, 100  š
0.5%; RS, unknown resistance; SP, 273.6  š
0.1%. Determine the value of the unknown
resistance, and its accuracy of measurement.
[27.36  š 2.6% or 27.36  š 0.71 ]

Figure 10.32
Exercise 56 Short answer questions on
The maximum relative error of Rx is given by the electrical measuring instruments and
sum of the three individual errors, i.e. 1.0%C0.5%C measurements
0.2% D 1.7%. Hence 1 What is the main difference between an ana-
logue and a digital type of measuring instru-
Rx D 43.25 Z ± 1.7% ment?

1.7% of 43.25  D 0.74  (rounding off). Thus Rx 2 Name the three essential devices for all ana-
may also be expressed as logue electrical indicating instruments
3 Complete the following statements:
Rx D 43.25 ± 0.74 Z (a) An ammeter has a . . . . . . resistance and
is connected . . . . . . with the circuit
(b) A voltmeter has a . . . . . . resistance and
Now try the following exercises is connected . . . . . . with the circuit
4 State two advantages and two disadvantages
of a moving coil instrument
Exercise 55 Further problems on 5 What effect does the connection of (a) a
measurement errors shunt (b) a multiplier have on a milliamme-
ter?
1 The p.d. across a resistor is measured as 37.5 V
with an accuracy of š0.5%. The value of the 6 State two advantages and two disadvantages
resistor is 6 k š 0.8%. Determine the current of a moving coil instrument

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 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS 125

7 Name two advantages of electronic measur-

ing instruments compared with moving coil Exercise 57 Multi-choice questions on
or moving iron instruments electrical measuring instruments and
8 Briefly explain the principle of operation of measurements (Answers on page 375)
an ohmmeter 1 Which of the following would apply to a
9 Name a type of ohmmeter used for measur- moving coil instrument?
ing (a) low resistance values (b) high resis- (a) An uneven scale, measuring d.c.
tance values (b) An even scale, measuring a.c.
(c) An uneven scale, measuring a.c.
10 What is a multimeter? (d) An even scale, measuring d.c.
11 When may a rectifier instrument be used in 2 In question 1, which would refer to a moving
preference to either a moving coil or moving iron instrument?
iron instrument? 3 In question 1, which would refer to a moving
12 Name five quantities that a c.r.o. is capable coil rectifier instrument?
of measuring 4 Which of the following is needed to extend
13 What is harmonic analysis? the range of a milliammeter to read voltages
of the order of 100 V?
14 What is a feature of waveforms containing (a) a parallel high-value resistance
the fundamental and odd harmonics? (b) a series high-value resistance
(c) a parallel low-value resistance
15 Express the ratio of two powers P1 and P2 (d) a series low-value resistance
in decibel units
5 Fig. 10.33 shows a scale of a multi-range
16 What does a power level unit of dBm indi- ammeter. What is the current indicated when
cate? switched to a 25 A scale?
17 What is meant by a null method of measure- (a) 84 A (b) 5.6 A (c) 14 A (d) 8.4 A
ment?
18 Sketch a Wheatstone bridge circuit used for
measuring an unknown resistance in a d.c.
circuit and state the balance condition
19 How may a d.c. potentiometer be used to
measure p.d.’s
20 Name five types of a.c. bridge used for
measuring unknown inductance, capacitance
or resistance Figure 10.33
21 What is a universal bridge?
A sinusoidal waveform is displayed on a
22 State the name of an a.c. bridge used for c.r.o. screen. The peak-to-peak distance is
measuring inductance 5 cm and the distance between cycles is 4 cm.
23 Briefly describe how the measurement of Q- The ‘variable’ switch is on 100 µs/cm and
factor may be achieved the ‘volts/cm’ switch is on 10 V/cm. In ques-
tions 6 to 10, select the correct answer from
24 Why do instrument errors occur when mea- the following:
suring complex waveforms? (a) 25 V (b) 5 V (c) 0.4 ms
25 Define ‘calibration accuracy’ as applied to a (d) 35.4 V (e) 4 ms (f) 50 V
measuring instrument (g) 250 Hz (h) 2.5 V (i) 2.5 kHz
(j) 17.7 V
26 State three main areas where errors are most
likely to occur in measurements 6 Determine the peak-to-peak voltage
7 Determine the periodic time of the waveform

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8 Determine the maximum value of the voltage 15 R.m.s. value of waveform P

9 Determine the frequency of the waveform 16 R.m.s. value of waveform Q
10 Determine the r.m.s. value of the waveform 17 Phase displacement of waveform Q relative
to waveform P
Fig. 10.34 shows double-beam c.r.o. wave-
form traces. For the quantities stated in ques- 18 The input and output powers of a system are
tions 11 to 17, select the correct answer from 2 mW and 18 mW respectively. The decibel
the following: power ratio of output power to input power
(a) 30 V (b) 0.2 s (c) 50 V is:
15 250 (a) 9 (b) 9.54 (c) 1.9 (d) 19.08
(d) p (e) 54° leading (f) p V 19 The input and output voltages of a system are
2 2
50 500 µV and 500 mV respectively. The deci-
(g) 15 V (h) 100 µs (i) p V bel voltage ratio of output to input voltage
2 (assuming input resistance equals load resis-
(j) 250 V (k) 10 kHz (l) 75 V tance) is:
3 (a) 1000 (b) 30 (c) 0 (d) 60
(m) 40 µs (n) rads lagging
10 20 The input and output currents of a system are
25 30 3 mA and 18 mA respectively. The decibel
(o) p V (p) 5 Hz (q) p V
2 2 ratio of output to input current (assuming the
75 input and load resistances are equal) is:
(r) 25 kHz (s) p V (a) 15.56 (b) 6 (c) 1.6 (d) 7.78
2
3 21 Which of the following statements is false?
(t) rads leading (a) The Schering bridge is normally used for
10
measuring unknown capacitances
(b) A.C. electronic measuring instruments
can handle a much wider range of fre-
quency than the moving coil instrument
(c) A complex waveform is one which is
non-sinusoidal
(d) A square wave normally contains the
fundamental and even harmonics
22 A voltmeter has a f.s.d. of 100 V, a sensitivity
of 1 k/V and an accuracy of š2% of f.s.d.
When the voltmeter is connected into a cir-
cuit it indicates 50 V. Which of the following
statements is false?
(a) Voltage reading is 50 š 2 V
(b) Voltmeter resistance is 100 k
(c) Voltage reading is 50 V š 2%
Figure 10.34 (d) Voltage reading is 50 V š 4%
23 A potentiometer is used to:
11 Amplitude of waveform P
(a) compare voltages
12 Peak-to-peak value of waveform Q (b) measure power factor
(c) compare currents
13 Periodic time of both waveforms (d) measure phase sequence
14 Frequency of both waveforms

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