Question

The current flowing in a resistor of 1Ω is measured to be 25 A. But it was discovered that ammeter reading was low by 1% and resistance was marked high by 0.5%. Find true power as a percentage of the original power.

(a) 95%

(b) 101.5%

(c) 100.1%

(d) 102.4%

The question was posed to me by my college director while I was bunking the class.

I want to ask this question from Advanced Problems on Error Analysis in Electrical Instruments topic in section Measurement of Resistance of Electrical Measurements

Answer 1

The correct option is (b) 101.5%

To explain: True current = 25(1 + 0.01) = 25.25 A

True resistance R = 1(1 – 0.005) = 0.995Ω

∴ True power = I^2R = 634.37 W

Measured power = (25)^2 × 1 = 625 W

∴ \( \frac{True \,power}{Measured \,power}\) × 100 = \(\frac{634.37}{625}\) × 100 = 101.5%.