JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Current Electricity

As shown in the figure, the resistance of a galvanometer

G

can be found by the half-deflection method. Here the resistance

R_{2}

is adjusted such that when the key

K

is closed the deflection in the galvanometer becomes half of the value as compared to when

K

is open. Half-deflection is obtained at

R_{2} = 4Ω

and thus the galvanometer resistance is found to be

6Ω

. In this half-deflection condition the current (in mA) through the resistor

R_{1}

is:

Answer: 694.44

Solution

When key is open:

I_{g} = \frac{10}{R _{1} + 6}

.

When key is closed: current through battery

i = \frac{10}{R _{1} + 2.4}

(where 2.4 is the parallel combination of 6 and 4).

Current through galvanometer:

I_{g}^{'} = \frac{4}{10} \times \frac{10}{R _{1} + 2.4} = \frac{4}{R _{1} + 2.4}

.

Half-deflection condition:

I_{g}^{'} = \frac{I _{g}}{2}

, so

\frac{4}{R _{1} + 2.4} = \frac{1}{2} \times \frac{10}{R _{1} + 6}

.

Solving:

8 (R_{1} + 6) = 10 (R_{1} + 2.4)

, giving

R_{1} = 12Ω

.

Current through

R_{1}

i = \frac{10}{12 + 2.4} = \frac{10}{14.4} \approx 0.694

A = 694.44 mA.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer