JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

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.

A
1.05
B
0
C
0.95
D
2

Solution

Let original $R_{1} = R_{2} = R_{3} = R_{4} = R$ . After heating, $R_{3} = 1.1 R$ . Supply voltage $V = 40$ V.

Top branch ( $R_{1}$ and $R_{3} = 1.1 R$ in series) has total $2.1 R$ . Potential at junction $A$ (between $R_{1}$ and the new $R_{3}$ ): $V_{A} = V \cdot \frac{1.1 R}{2.1 R}$ . Actually it is cleaner to compute the drop from the top node: $V_{A} = V - V \cdot \frac{R}{2.1 R} = V (1 - \frac{1}{2.1})$ .

Bottom branch ( $R_{2}$ and $R_{4}$ each $R$ ) total $2 R$ : $V_{B} = V \cdot \frac{R}{2 R} = \frac{V}{2}$ .

$V_{A} - V_{B} = V [(1 - \frac{1}{2.1}) - \frac{1}{2}] = V [\frac{1}{2} - \frac{1}{2.1}] = 40 \cdot \frac{0.1}{2 \times 2.1} = \frac{4}{4.2} \approx 0.95$ V.

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