Chapter 10

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 +


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)


(b) In diagram (ii),

(ii)

i.e. 250 =

i.e. 250 = 0.6 +




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


(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


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.



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.


(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



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


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


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


(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



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.


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


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


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


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



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.


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


Documents

Chapter 10