A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R₁ = R₂ = R₃ = R₄). When R₃ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ( Va - Vb) (after R₃ is heated) is _________ V. (a) 1.05 (b) 0.95 (c) 0 (d) 2

Answer: (b). Take the bottom node at 0 V and top at 40 V; let each arm be R, with R₃ = 1.1R. Va = 40 × (R₂)/(R₁+R₂) = 20 V Vb = 40 × (R₄)/(R₃+R₄) = 40 × 1/2.1 = 19.05 V Va - Vb = 0.95 V So the answer is (b) 0.95.

A Wheatstone bridge is initially at room temperature and all arms of… | JEE Main 2026 (28 Jan, Shift 2)

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Wheatstone Bridge & Metre Bridge

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.

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Answer: (b)

Take the bottom node at $0$ V and top at $40$ V; let each arm be $R$ , with $R_3 = 1.1R$ .
$V_a = 40 \times \dfrac{R_2}{R_1+R_2} = 20$ V
$V_b = 40 \times \dfrac{R_4}{R_3+R_4} = 40 \times \dfrac{1}{2.1} = 19.05$ V
$V_a - V_b = 0.95$ V
So the answer is (b) $0.95$ .

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