Question:
In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is :
In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is :
Updated On: Apr 12, 2026
- 0.25 $\Omega$
- 2 $\Omega$
- 2.5 $\Omega$
- 0.18 $\Omega$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The resistance is $R = V/I$. The absolute error in $R$ is calculated using the relative errors of $V$ and $I$ based on the formula $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Step 2: Detailed Explanation:
Given $V = 10 V, I = 5 A$.
Least counts (errors): $\Delta V = 500 mV = 0.5 V$, $\Delta I = 200 mA = 0.2 A$.
Value of Resistance $R = \frac{V}{I} = \frac{10}{5} = 2 \Omega$.
Relative error: $\frac{\Delta R}{R} = \frac{0.5}{10} + \frac{0.2}{5}$
$\frac{\Delta R}{2} = 0.05 + 0.04 = 0.09$
Absolute error $\Delta R = 2 \times 0.09 = 0.18 \Omega$.
Step 3: Final Answer:
The estimated error is 0.18 $\Omega$.
The resistance is $R = V/I$. The absolute error in $R$ is calculated using the relative errors of $V$ and $I$ based on the formula $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Step 2: Detailed Explanation:
Given $V = 10 V, I = 5 A$.
Least counts (errors): $\Delta V = 500 mV = 0.5 V$, $\Delta I = 200 mA = 0.2 A$.
Value of Resistance $R = \frac{V}{I} = \frac{10}{5} = 2 \Omega$.
Relative error: $\frac{\Delta R}{R} = \frac{0.5}{10} + \frac{0.2}{5}$
$\frac{\Delta R}{2} = 0.05 + 0.04 = 0.09$
Absolute error $\Delta R = 2 \times 0.09 = 0.18 \Omega$.
Step 3: Final Answer:
The estimated error is 0.18 $\Omega$.
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