JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Current Electricity

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.

Answer: (C)

Solution

The bridge is supplied with

40

V across its top and bottom nodes. Point

a

is the junction of

R_{1}

(top) and

R_{2}

(bottom), and point

b

is the junction of

R_{3}

(top) and

R_{4}

(bottom); each branch acts as a potential divider. Initially all arms equal

R

, so

V_{a} = V_{b} = 20

V and the bridge is balanced. Heating

R_{3}

raises it to

1.1 R

, so

V_{b} = 40 \times \frac{R _{4}}{R _{3} + R _{4}} = 40 \times \frac{R}{1.1 R + R} = \frac{40}{2.1} \approx 19.05

V, while

V_{a}

stays at

20

V. Hence

V_{a} - V_{b} = 20 - 19.05 = 0.95

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option