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

The current flowing in a resistor of 1Ω is measured to be 25 A. But

1.	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» 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%.

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