JEE Advanced 2026 · Paper 2 · Physics Q13

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PhysicsNumerical+4Question 31 of 54

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 $R_2=4\ \Omega$ and thus the galvanometer resistance is found to be $6\ \Omega$ . In this half-deflection condition the current (in mA) through the resistor $R_1$ is:

A 10 V cell in series with resistor R₁ and galvanometer G; a variable resistor R₂ in series with key K is connected in parallel with G.

Your answer

Answer and worked solution

Official answer: 690

Find $R_1$ . With $K$ open, $I_g=\dfrac{10}{R_1+6}$ .

With $K$ closed, $G\parallel R_2=\dfrac{6\cdot4}{10}=2.4\ \Omega$ . The current through $R_1$ is $I=\dfrac{10}{R_1+2.4}$ , of which $G$ carries $\dfrac{4}{10}I$ .

Half deflection:
$0.4\cdot\frac{10}{R_1+2.4}=\frac12\cdot\frac{10}{R_1+6}\implies0.8(R_1+6)=R_1+2.4\implies R_1=12\ \Omega.$

Current.
$I=\frac{10}{12+2.4}=0.694\text{ A}\approx694\text{ mA}.$

Topics:Galvanometer experiments Series and parallel resistors Moving coil galvanometer

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JEE Advanced 2026 · Paper 2, Physics question 13