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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 % moreI 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»

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.



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