JEE Challenger
More from Units and Measurements
PhyMedMain 2026 • 5th Apr - Shift 2

Estimated Error in Resistance Measurement Using Ohms Law

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

Options

A

0.250.25

B

22

C

2.52.5

D

0.180.18

Correct

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsOhm's Law

Step-by-Step Solution

To find the estimated error in the measurement of resistance, we use Ohm's law: R=VIR = \frac{V}{I}

Taking the natural logarithm on both sides: ln⁡R=ln⁡V−ln⁡I\ln R = \ln V - \ln I

Differentiating to find the maximum fractional error (relative error): ΔRR=ΔVV+ΔII\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}

Given data:

  • Voltmeter reading, V=10 VV = 10\text{ V}
  • Ammeter reading, I=5 AI = 5\text{ A}
  • Least count of voltmeter (absolute error in voltage), ΔV=500 mV=0.5 V\Delta V = 500\text{ mV} = 0.5\text{ V}
  • Least count of ammeter (absolute error in current), ΔI=200 mA=0.2 A\Delta I = 200\text{ mA} = 0.2\text{ A}

First, calculate the measured resistance RR: R=10 V5 A=2 ΩR = \frac{10\text{ V}}{5\text{ A}} = 2\text{ }\Omega

Next, substitute the given values into the relative error formula: ΔRR=0.5 V10 V+0.2 A5 A\frac{\Delta R}{R} = \frac{0.5\text{ V}}{10\text{ V}} + \frac{0.2\text{ A}}{5\text{ A}} ΔRR=0.05+0.04=0.09\frac{\Delta R}{R} = 0.05 + 0.04 = 0.09

Now, calculate the estimated absolute error in resistance (ΔR\Delta R): ΔR=0.09×R\Delta R = 0.09 \times R ΔR=0.09×2 Ω=0.18 Ω\Delta R = 0.09 \times 2\text{ }\Omega = 0.18\text{ }\Omega

Thus, the estimated error in the resistance measurement is 0.18 Ω0.18\text{ }\Omega.

Estimated Error in Resistance Measurement Using Ohms Law | Physics PYQ Solution - JEE Challenger