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 = 0 , then when V = 20 V, series resistance, = 2 k

(b) If = 0 , then when V = 100 V, series resistance, = 10 k

(c) If = 0 , then when V = 250 V, series resistance, = 25 k

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,

Then V = from which, shunt resistance, = 18.18

(b) When I = 20 A, © John Bird Published by Taylor and Francis 86

Then V = from which, shunt resistance, = 10.00 m

(c) When I = 100 A,

Then V = from which, shunt resistance, = 2.00 m

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,

i.e. 200 =

i.e. 200 = 0.1 +

from which, = 39.98 k in series

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), = 9.98 A

Then from which,

© John Bird Published by Taylor and Francis 87

shunt resistance, = 50.10 m in parallel

(b) In diagram (ii),

i.e. 100 =

i.e. 100 = 0.5 +

from which, = 4.975 k in series

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),

Then from which,

shunt resistance, = 0.01200 = 12.00 in parallel

(i)

© John Bird Published by Taylor and Francis 88

(b) In diagram (ii),

(ii)

i.e. 250 =

i.e. 250 = 0.6 +

from which, = 16.63 k in series

© John Bird Published by Taylor and Francis 89

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 = = 0.250 A

(b) Actual current = = 0.238 A

(c) Power dissipated in ammeter, P = = 2.832 W

(d) Power dissipated in the 1 k resistor, P = = 56.64 W

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 = = 900 W

© John Bird Published by Taylor and Francis 90

(b) Total resistance in circuit,

Wattmeter reading, P = = 904.5 W

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.

Voltage, = 160 V

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).

Now voltage indicated, = 156.7 V

4. A 240 V supply is connected across a load resistance R. Also connected across R is a voltmeter

© John Bird Published by Taylor and Francis 91

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.

(a) Resistance of voltmeter, = 8 k/V 300 = 2.4 M

From the circuit shown, current in voltmeter,

Power dissipated by the voltmeter, P = V = 24 mW

When R = 100 , current in load resistor, = 2.4 A

Power dissipated by the load resistor, P = V = 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, = 240 A

Power dissipated by the load resistor, P = V = 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.

© John Bird Published by Taylor and Francis 92

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 = = = 41.7 kHz

(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.

© John Bird Published by Taylor and Francis 93

(a) Time for one cycle, T = 3.6 cm 500 ms/cm = 1.8 s

Hence, frequency, f = = 0.56 Hz

(b) Magnitude of the pulse voltage = 4.2 cm 2V/cm = 8.4 V

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 = = 7.14 Hz

(b) Peak-to-peak voltage = 4.4 cm 50 V/cm = 220 V

(c) Peak voltage = = 110 V and r.m.s. voltage = 110 = 77.78 V

© John Bird Published by Taylor and Francis 94

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) (b) (c) (d)

Determine the decibel power ratio for each.

(a) Decibel power ratio = 10 lg = - 10 dB

(b) Decibel power ratio = 10 lg = - 4.77 dB

(c) Decibel power ratio = 10 lg = - 16.02 dB

(d) Decibel power ratio = 10 lg = - 20 dB

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

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system are equal.

Decibel current ratio = 20 lg = 13.98 dB

4. 5% of the power supplied to a cable appears at the output terminals. Determine the power loss in

decibels.

If = input power, and = output power then = = 0.05

Decibel power ratio = 10 lg = 10 lg 0.05 = -13 dB

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 hence, 24 = 10 lg where is in mW

i.e. lg =

Then

from which, output power, = = 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 20 = 10 lg from which, 2 = lg

© John Bird Published by Taylor and Francis 96

and 102 = = 100

Thus the overall power gain, = 100

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.

Voltage gain = 20 lg hence, 25 = 20 lg where is in mV

Thus, lg =

i.e. = and the input voltage, = 0.39 mV

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, 8.5 = 20 lg where is in mV

Hence, i.e. 0.425 =

and from which, output voltage, = 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

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(a) 0 dB (b) 1.5 dB and (c) -15 dB. (d) What decibel reading corresponds to 0.5 V?

hence from which, V = 0.775 V

(a) Number of dBm = 20 lg

Hence, at 0 dB, then 0 = 20 lg

from which, 0 = lg and and V = 0.775 V

(b) At 1.5 dB, 1.5 = 20 lg

from which, = lg and and V = 0.775 = 0.921 V

(c) At -15 dB, -15 = 20 lg

from which, = lg and and V = 0.775 = 0.138 V

(d) When V = 0.5 V, then the decibel reading = 20 lg = - 3.807 dB

© John Bird Published by Taylor and Francis 98

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 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

From the diagram, 10 = 150 200

and unknown resistor, = = 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.

© John Bird Published by Taylor and Francis 99

hence, from which, e.m.f. of dry cell, = 1.525 V

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, = 250 , = 400 , = 10 k and

C = 15 F

From equation (2), page 142, inductance, L = = 1.5 H

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From equation (3), page 142, resistance, R = = 10

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 =

= 1.351 mH

(b) Also from Problem 21, page 143, resistance, R =

= 10.61

© John Bird Published by Taylor and Francis 101

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 = = 6.25 mA

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.

© John Bird Published by Taylor and Francis 102

Resistance, R = = 4.16

Voltage error = 2% of 75 V = 1.5 V

As a percentage of the voltage reading, this is = 2.88%

Current error = 2% of 20 A = 0.4 A

As a percentage of the current reading, this is = 3.20%

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

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, 1000 = 100 273.6

and unknown resistor, = = 27.36

Maximum relative error of = 2% + 0.5% + 0.1% = 2.6%

2.6% of 27.36 = 0.71 © John Bird Published by Taylor and Francis 103

Thus, = 27.36 2.6% or 27.36 0.71

© John Bird Published by Taylor and Francis 104