Current Electricity

JEE Main level, single correct, medium

4 of 380

Question 4 of 380

In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V10\text{ V} and 5 A5\text{ A}, respectively. The least counts of voltmeter and ammeter are 500 mV500\text{ mV} and 200 mA200\text{ mA}, respectively. The estimated error in the resistance measurement is _______ Ω\Omega.

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Answer: (D) 0.180.18

Given V=10 VV = 10\text{ V} and I=5 AI = 5\text{ A}

Least count of voltmeter, ΔV=500 mV=0.5 V\Delta V = 500\text{ mV} = 0.5\text{ V}

Least count of ammeter, ΔI=200 mA=0.2 A\Delta I = 200\text{ mA} = 0.2\text{ A}

Resistance is given by Ohm's law:

R=VI=105=2 ΩR = \dfrac{V}{I} = \dfrac{10}{5} = 2\text{ }\Omega

The maximum permissible relative error in resistance is:

ΔRR=ΔVV+ΔII\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I}

ΔR2=0.510+0.25\dfrac{\Delta R}{2} = \dfrac{0.5}{10} + \dfrac{0.2}{5}

ΔR2=0.05+0.04=0.09\dfrac{\Delta R}{2} = 0.05 + 0.04 = 0.09

ΔR=2×0.09=0.18 Ω\Delta R = 2 \times 0.09 = 0.18\text{ }\Omega

Answer: 0.180.18

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