Question

A circuit is used to measure the power consumed by the load. The current coil and the voltage coil of the watt-meter have 0.02 Ω and 1000 Ω resistances respectively. The measured power (as compared to the load power) will be?

(a) 0.4 % less

(b) 0.2 % less

(c) 0.2 % more

(d) 0.4 % more

I had been asked this question by my school teacher while I was bunking the class.

Origin of the question is Advanced Miscellaneous Problems on Measurement of Power in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer 1

The correct choice is (c) 0.2 % more

For explanation: Power indicated by watt-meter = 400 × 20 + 20^2 × 0.02 = 4008 W

Percentage increase = \(\frac{4008-4000}{4000}\) × 100 = 0.2 % more.