CHAPTER 10 ELECTRICAL MEASURING INSTRUMENTS AND
MEASUREMENTS
EXERCISE 51, Page 123
1. A moving-coil instrument gives f.s.d. for a current of 10 mA. Neglecting the resistance of the
instrument, calculate the approximate value of series resistance needed to enable the instrument
to measure up to (a) 20 V (b) 100 V (c) 250 V
(a) If ra = 0 , then when V = 20 V, series resistance, R M 
V
20

= 2 k
I 10 103
(b) If ra = 0 , then when V = 100 V, series resistance, R M 
V
100

= 10 k
I 10 103
(c) If ra = 0 , then when V = 250 V, series resistance, R M 
V
250

= 25 k
I 10 103
2. A meter of resistance 50  has a f.s.d. of 4 mA. Determine the value of shunt resistance required
in order that f.s.d. should be (a) 15 mA (b) 20 A (c) 100 A
(a) When I = 15 mA, IS  15  4  11mA
Then V = Ia ra  IS R S
3
Ia ra  4 10   50 

from which, shunt resistance, R S 
= 18.18 
IS
11103 
(b) When I = 20 A, IS  20  4 103  19.996 A
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3
Ia ra  4 10   50 

from which, shunt resistance, R S 
= 10.00 m
IS
19.996 
Then V = Ia ra  IS R S
(c) When I = 100 A, IS  100  4 103  99.996 A
Then V = Ia ra  IS R S
3
Ia ra  4 10   50 

from which, shunt resistance, R S 
= 2.00 m
IS
 99.996 
3. A moving-coil instrument having a resistance of 20 Ω gives a f.s.d. when the current is 5 mA.
Calculate the value of the multiplier to be connected in series with the instrument so that it can be
used as a voltmeter for measuring p.d.’s up to 200 V.
In diagram,
V  Va  VM  Ia ra  IR M
i.e.
200 =  5 103   20    5 103   R M 
i.e.
200 = 0.1 +  5 103  R M
from which,
RM 
200  0.1
= 39.98 k in series
5  10 3
4. A moving-coil instrument has a f.s.d. of 20 mA and a resistance of 25 . Calculate the values of
resistance required to enable the instrument to be used (a) as a 0 – 10 A ammeter, and (b) as a
0 – 100 V voltmeter. State the mode of resistance connection in each case.
(a) In diagram (i), IS  10  20 103 = 9.98 A
Then Ia ra  IS R S
from which,
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I r  20 10   25 
shunt resistance, R S  a a 
= 50.10 m in parallel
IS
 9.98 
3
(b) In diagram (ii), V  Va  VM  Ia ra  IR M
i.e.
100 =  20 103   25    20 103   R M 
i.e.
100 = 0.5 +  20  103  R M
from which,
RM 
100  0.5
= 4.975 k in series
20 103
5. A meter has a resistance of 40  and registers a maximum deflection when a current 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) In diagram (i), IS  50  15 103  49.985 A
Then Ia ra  IS R S
from which,
shunt resistance, R S 
3
Ia ra 15 10   40 

= 0.01200 = 12.00  in parallel
IS
 49.985
(i)
(b) In diagram (ii), V  Va  VM  Ia ra  IR M
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(ii)
i.e.
250 = 15 103   40   15 103   R M 
i.e.
250 = 0.6 + 15 10 3  R M
from which,
RM 
250  0.6
 16626.7  = 16.63 k in series
15 103
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EXERCISE 52, Page 126
1. A 0 – 1 A ammeter having a resistance of 50  is used to measure the current flowing in a 1 k
resistor when the supply voltage is 250 V. Calculate: (a) the approximate value of current
(neglecting the ammeter resistance), (b) the actual current in the circuit, (c) the power dissipated
in the ammeter, (d) the power dissipated in the 1 k resistor.
(a) Approximate value of current =
(b) Actual current =
V 250

= 0.250 A
R 1000
V
250

= 0.238 A
R  ra 1000  50
(c) Power dissipated in ammeter, P = I 2 ra   0.238   50  = 2.832 W
2
(d) Power dissipated in the 1 k resistor, P = I 2 ra   0.238  1000  = 56.64 W
2
2. (a) A current of 15 A flows through a load having a resistance of 4 . Determine the power
dissipated in the load. (b) A wattmeter, whose current coil has a resistance of 0.02 , is
connected to measure the power in the load. Determine the wattmeter reading assuming the
current in the load is still 15 A.
(a) Power in load, P = I 2 R  15   4  = 900 W
2
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(b) Total resistance in circuit, R T  4  0.02  4.02 
Wattmeter reading, P = I 2 R T  15   4.02  = 904.5 W
2
3. A voltage of 240 V is applied to a circuit consisting of an 800  resistor in series with a 1.6 k
resistor. What is the voltage across the 1.6 k resistor? The p.d. across the 1.6 k resistor is
measured by a voltmeter of f.s.d. 250 V and sensitivity 100 /V. Determine the voltage
indicated.
 1.6 103 
Voltage, V1  
240  = 160 V
3 
800

1.6

10


Resistance of voltmeter = 250V  100 /V = 25 k
The circuit is now as shown in (a) below.
25 k in parallel with 1.6 k = 1.5038 k and circuit (a) simplifies to circuit (b).
 1.5038 103 
Now voltage indicated, V1  
240  = 156.7 V
3 
 800  1.5038 10 
4. A 240 V supply is connected across a load resistance R. Also connected across R is a voltmeter
having a f.s.d. of 300 V and a figure of merit (i.e. sensitivity) of 8 k/V. Calculate the power
dissipated by the voltmeter and by the load resistance if (a) R = 100  (b) R = 1 M. Comment
on the results obtained.
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(a) Resistance of voltmeter, R V = 8 k/V  300 = 2.4 M
From the circuit shown, current in voltmeter, I V 
V
240

 100 A
rV 2.4 106
Power dissipated by the voltmeter, P = V I V   240  100 106  = 24 mW
When R = 100 , current in load resistor, I R 
240
= 2.4 A
100
Power dissipated by the load resistor, P = V IR   240 2.4 = 576 W
The power dissipated by the voltmeter is very small in comparison to the power dissipated
by the load resistor.
(b) When R = 1 M, power dissipated by voltmeter is the same as above, i.e. 24 mW
Current in load resistor, I R 
240
= 240 A
1 106
Power dissipated by the load resistor, P = V I R   240   240 106  = 57.6 mW
In this case, the larger load resistor reduces the power dissipated such that the voltmeter uses a
comparable amount of power as the load resistor R.
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EXERCISE 53, Page 131
1. For the square voltage waveform displayed on an oscilloscope shown below, find (a) its frequency,
(b) its peak-to-peak voltage
(a) The width of one complete cycle is 4.8 cm
Hence the periodic time, T = 4.8 cm  5  10-3 s/cm = 24 ms
Frequency, f =
1
1
=
= 41.7 kHz
24 10 3
T
(c) The peak-to-peak height of the display is 4.4 cm, hence the
peak-to-peak voltage = 4.4 cm  40 V/cm = 176 V
2. For the pulse waveform shown below, find (a) its frequency, (b) the magnitude of the pulse
voltage.
(a) Time for one cycle, T = 3.6 cm  500 ms/cm = 1.8 s
Hence, frequency, f =
1
1

= 0.56 Hz
T 1.8
(b) Magnitude of the pulse voltage = 4.2 cm  2V/cm = 8.4 V
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3. For the sinusoidal waveform shown below, determine (a) its frequency, (b) the peak-to-peak
voltage, (c) the r.m.s. voltage.
(a) Periodic time, T = 2.8 cm  50 ms/cm = 0.14 s
Hence, frequency, f =
1
1

= 7.14 Hz
T 0.14
(b) Peak-to-peak voltage = 4.4 cm  50 V/cm = 220 V
(c) Peak voltage =
220
1
= 110 V and r.m.s. voltage =
 110 = 77.78 V
2
2
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EXERCISE 54, Page 139
1. The ratio of two powers is (a) 3 (b) 10 (c) 20 (d) 10000
Determine the decibel power ratio for each.
(a) Decibel power ratio = 10 lg 3 = 4.77 dB
(b) Decibel power ratio = 10 lg 10 = 10 dB
(c) Decibel power ratio = 10 lg 20 = 13 dB
(d) Decibel power ratio = 10 lg 10000 = 40 dB
2. The ratio of two powers is (a)
1
1
1
(b)
(c)
10
3
40
(d)
1
100
Determine the decibel power ratio for each.
1
(a) Decibel power ratio = 10 lg   = - 10 dB
 10 
1
(b) Decibel power ratio = 10 lg   = - 4.77 dB
3
 1 
(c) Decibel power ratio = 10 lg   = - 16.02 dB
 40 
 1 
(d) Decibel power ratio = 10 lg 
 = - 20 dB
 100 
3. The input and output currents of a system are 2 mA and 10 mA respectively. Determine the
decibel current ratio of output to input current assuming input and output resistances of the
system are equal.
Decibel current ratio = 20 lg
10
= 13.98 dB
2
4. 5% of the power supplied to a cable appears at the output terminals. Determine the power loss in
decibels.
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If P1 = input power, and P2 = output power then
Decibel power ratio = 10 lg
5
P2
=
= 0.05
100
P1
P2
= 10 lg 0.05 = -13 dB
P1
Hence, the power loss, or attenuation, is 13 dB
5. An amplifier has a gain of 24 dB. Its input power is 10 mW. Find the output power.
P = 10 lg
i.e.
P2
P1
hence, 24 = 10 lg
lg
P2
24
 2.4
=
10 10
102.4 
Then
P2
where P2 is in mW
10
P2
10
from which, output power, P2 = 10 10 
2.4
= 2512 mW or 2.51 W
6. Determine, in decibels, the ratio of the output power to input power of a four stage system, the
stages having gains of 10 dB, 8 dB, -5 dB and 7 dB. Find also the overall power gain.
The decibel ratio may be used to find the overall power ratio of a chain simply by adding the decibel
power ratios together.
Hence the overall decibel power ratio = 10 + 8 – 5 + 7 = 20 dB gain.
Thus
P 
P 
20 = 10 lg  2  from which, 2 = lg  2 
 P1 
 P1 
102 =
and
Thus the overall power gain,
P2
= 100
P1
P2
= 100
P1
7. The output voltage from an amplifier is 7 mV. If the voltage gain is 25 dB calculate the value of
the input voltage assuming that the amplifier input resistance and load resistance are equal.
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Voltage gain = 20 lg
Thus,
V2
V1
lg
hence, 25 = 20 lg
25
7
 1.25
=
20
V1
7
= 101.25
V1
i.e.
7
where V1 is in mV
V1
and the input voltage, V1 
7
= 0.39 mV
101.25
8. The voltage gain of a number of cascaded amplifiers are 23 dB, -5.8 dB, -12.5 dB and 3.8 dB.
Calculate the overall gain in decibels assuming that input and load resistances for each stage are
equal. If a voltage of 15 mV is applied to the input of the system, determine the value of the
output voltage
Overall gain in decibels = 23 – 5.8 – 12.5 + 3.8 = 8.5 dB
Voltage gain = 20 lg
Hence,
and
V2
V1
V
8.5
 lg 2
20
15
100.425 
V2
15
hence,
8.5 = 20 lg
i.e.
0.425 = lg
V2
15
where V2 is in mV
V2
15
from which, output voltage, V2  15 100.425 = 39.91 mV
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 is dissipated in a 600  resistor. Determine the voltage at
(a) 0 dB (b) 1.5 dB and (c) -15 dB. (d) What decibel reading corresponds to 0.5 V?
Power, P 
V2
V2
hence 1103 
from which, V = 0.775 V
R
600
 V 
(a) Number of dBm = 20 lg 

 0.775 
 V 
Hence, at 0 dB, then 0 = 20 lg 

 0.775 
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 V 
from which, 0 = lg 

 0.775 
(b) At 1.5 dB,
V
0.775
and V = 0.775 V
 V 
1.5 = 20 lg 

 0.775 
1.5
 V 
from which,
= lg 

20
 0.775 
(c) At -15 dB,
100 
and
and
10
1.5
20

V
0.775
1.5
and V = 0.775 10  20 = 0.921 V
 V 
-15 = 20 lg 

 0.775 
from which, 
15
 V 
= lg 

20
 0.775 
and
100.75 
V
0.775
and V = 0.775 10 
0.75
= 0.138 V
 0.5 
(d) When V = 0.5 V, then the decibel reading = 20 lg 
 = - 3.807 dB
 0.775 
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EXERCISE 55, Page 141
1. In a Wheatstone bridge PQRS, a galvanometer is connected between Q and S and a voltage
source between P and R. An unknown resistor R X 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 R X
From the diagram,
and
10  R X = 150  200
unknown resistor, R X =
150  200
= 3 k
10
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.
E1 l1

E 2 l2
hence,
1.0186 31.2

E2
46.7
 46.7 
from which, e.m.f. of dry cell, E 2  1.0186 
 = 1.525 V
 31.2 
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EXERCISE 56, Page 142
1. A Maxwell bridge circuit ABCD has the following arm impedances: AB, 250  resistance; BC,
15 F capacitor in parallel with a 10 k resistor; CD, 400  resistor; DA, unknown inductor
having inductance L and resistance R. Determine the values of L and R assuming the bridge is
balanced.
The bridge circuit is similar to the diagram below, R 1 = 250 , R 2 = 400 , R 3 = 10 k and
C = 15 F
From equation (2), page 142, inductance, L = R1R 2C  250  400 15 106 = 1.5 H
From equation (3), page 142, resistance, R =
R1R 2 250  400

= 10 
R3
10 103
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EXERCISE 57, Page 143
1. A Q-meter measures the Q-factor of a series L-C-R circuit to be 200 at a resonant frequency of
250 k. If the capacitance of the Q-meter capacitor is set to 300 pF determine (a) the inductance
L, and (b) the resistance R of the inductor.
(a) From Problem 21, page 143, inductance, L =
1
 2f r 
2

C
1
 2 250 10   300 10 
3 2
12
= 1.351 mH
(b) Also from Problem 21, page 143, resistance, R =
2f r L 2 250 103 1.351103

Q
200
= 10.61 
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EXERCISE 58, Page 145
1. The p.d. across a resistor is measured as 37.5 V with an accuracy of  0.5%. The value of the
resistor is 6 k  0.8% . Determine the current flowing in the resistor and its accuracy of
measurement.
Current flowing, I =
V 37.5

= 6.25 mA
R 6000
Maximum possible error is 0.5 + 0.8 = 1.3%
1.3% of 6.25 = 0.08 mA
Hence,
I = 6.25 mA  1.3%
or 6.25 mA  0.08 mA
2. The voltage across a resistor is measured by a 75 V f.s.d. voltmeter which gives an indication of
52 V. The current flowing in the resistor 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 both instruments
have an accuracy of  2% of f.s.d.
Resistance, R =
V 52

= 4.16 
I 12.5
Voltage error =  2% of 75 V =  1.5 V
As a percentage of the voltage reading, this is 
1.5
100% =  2.88%
52
Current error =  2% of 20 A =  0.4 A
As a percentage of the current reading, this is 
0.4
100% =  3.20%
12.5
Maximum relative errors = 2.88 + 3.20 = 6.08%
6.08% of 4.16  = 0.25 
Hence, resistance, R = 4.16   6.08% or 4.16   0.25 
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3. A Wheatstone bridge PQRS has the following arm resistances:
PQ, 1 k  2% ; QR, 100   0.5% ; RS, unknown resistance; SP, 273.6   0.1% .
Determine the value of the unknown resistance and its accuracy of measurement.
From the diagram below,
and
1000  R X = 100  273.6
unknown resistor, R X =
100  273.6
= 27.36 
1000
Maximum relative error of R X = 2% + 0.5% + 0.1% = 2.6%
2.6% of 27.36 = 0.71 
Thus,
R X = 27.36   2.6%
or 27.36   0.71 
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