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Resistance is measured by the voltmeter-ammeter method employing DC excitation and a voltmeter is connected directly across the unknown resistance. If the voltmeter and ammeter readings are subject to maximum possible errors of ±2.4 % and ±1% respectively, then the magnitude of the maximum possible percentage error in the value of resistance deduced from the measurement is?(a) 1.4 %(b) 1.7 %(c) 2.4 %(d) 3.4 %The question was asked during an internship interview.I'm obligated to ask this question of Advanced Miscellaneous Problems on Measurement of Resistance topic in section Measurement of Resistance of Electrical Measurements

Answer»

Correct OPTION is (d) 3.4 %

For explanation I would SAY: Ammeter error ∆I = ± 1%

VOLTMETER error ∆V = ± 2.4%

We know that \(\frac{∆R}{R} = \frac{∆V}{V} + \frac{∆I}{I}\)

∴ MAXIMUM percentage error = 2.4% + 1% = 3.4%.


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