Chapter 42 Electrical Measuring Instruments and Measurements

CHAPTER 42 ELECTRICAL MEASURING INSTRUMENTS AND MEASUREMENTS

EXERCISE 200, Page 448

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.

V250 (a) Approximate value of current =  = 0.250 A R1000 V 250 (b) Actual current =  = 0.238 A R ra 1000 50

2 2 (c) Power dissipated in ammeter, P = I ra   0.238 50 = 2.832 W

2 2 (d) Power dissipated in the 1 k resistor, P = I ra   0.238 1000 = 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 = IR2   152 4 = 900 W

440 © John Bird Published by Taylor and Francis

(b) Total resistance in circuit, RT  4 0.02  4.02 

2 2 Wattmeter reading, P = I RT   15 4.02  = 904.5 W

441 © John Bird Published by Taylor and Francis

EXERCISE 201, Page 452

1. For the square voltage waveform displayed on a c.r.o. shown below, find (a) its frequency, (b) its

peak-to-peak voltage

(a) Periodic time, T = 4.8 cm  5 ms/cm = 24 ms

11 Hence, frequency, f =  = 41.7 Hz T2410 3

(b) 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

11 Hence, frequency, f =  = 0.56 Hz T1.8

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

442 © John Bird Published by Taylor and Francis

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

11 Hence, frequency, f =  = 7.14 Hz T0.14

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

220 1 (c) Peak voltage = = 110 V and r.m.s. voltage =  110 = 77.78 V 2 2

EXERCISE 202, Page 459

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, 10  R X = 150  200

150 200 and unknown resistor, R = = 3 k X 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.

El11 1.0186 31.2 46.7  hence,  from which, e.m.f. of dry cell, E2  1.0186 = 1.525 V El22 E2 46.7 31.2

EXERCISE 203, Page 459

Answers found from within the text of the chapter, pages 445 to 459.

EXERCISE 204, Page 459

1. (f) 2. (c) 3. (a) 4. (i) 5. (j) 6. (g) 7. (c) 8. (b) 9. (p) 10. (d) 11. (o) 12. (n)

13. (a)