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A setup is used to measure resistance R. The ammeter and voltmeter resistance are 0.01 Ω and 2000 Ω respectively. Their readings are 2 A and 180 V respectively, giving a measured resistance of 90 Ω. The percentage error in the measurement is?(a) 2.25 %(b) 2.35 %(c) 4.5 %(d) 4.71 %This question was posed to me in quiz.Asked question is from Advanced Miscellaneous Problems on Measurement of Resistance in division Measurement of Resistance of Electrical Measurements

Answer»

Correct choice is (d) 4.71 %

Easy explanation: Current through the voltmeter Iv = \(\frac{180}{2000}\)

Current through R, IR = 2 – 9/100 = 1.91 A

Since, 1.91 R = 180

∴ R = 94.24

PERCENTAGE ERROR = \(\frac{94.24-90}{90}\) × 100 = 4.71 %.



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