A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R 1 = R 2 = R 3 = R 4 ). When R 3 resistance is heated to some temperature, its resistance value has gone up by 10 \% . The potential difference ( V a - V b ) (after R 3 is heated) is \ \ \ \ V.

Physics · Current Electricity

JEE Main 2026 — 28 January, Evening Shift — Question 31

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}=\mathrm{R}_{4}$ ). When $\mathrm{R}_{3}$ resistance is heated to some temperature, its resistance value has gone up by $10 \%$ . The potential difference $\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)$ (after $\mathrm{R}_{3}$ is heated) is $\_\_\_\_$ V.

Option A:
1.05
Option B:
0
Option C:
0.95
Correct
Option D:
2

Answer: C

Step-by-step solution

$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{V}}{2}$ $\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{V}}{2.1 \mathrm{R}} \times \mathrm{R}=\frac{\mathrm{V}}{2.1}$ $\therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\mathrm{V}\left[\frac{1}{2}-\frac{1}{2.1}\right]$ $V_{A}-V_{B}=\frac{0.1}{2 \times 2.1} \times 40$ $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\frac{4}{4.2}=0.95$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge