JEE Advanced 2026 — Paper 2 — PHY-III solution

JEE Advanced2026Paper 2PHY-III

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 $R_{2} = 4 Ω$ and thus the galvanometer resistance is found to be $6 Ω$ . In this half-deflection condition the current (in mA) through the resistor $R_{1}$ is:

Your answer

Solution

Key open. Only $R_{1}$ and $G$ are in the loop, so $I_{g} = \frac{10}{R _{1} + G} = \frac{10}{R _{1} + 6}$ .

Key closed. $R_{2}$ is in parallel with $G$ . Equivalent resistance of that parallel combination:

$R_{p} = \frac{G \cdot R _{2}}{G + R _{2}} = \frac{6 \times 4}{10} = 2.4 Ω$ .

Total current from the battery: $i = \frac{10}{R _{1} + 2.4}$ .

Current through the galvanometer (current-divider): $I_{g}^{'} = \frac{R _{2}}{G + R _{2}} i = \frac{4}{10} \cdot \frac{10}{R _{1} + 2.4} = \frac{4}{R _{1} + 2.4}$ .

Half-deflection condition. $I_{g}^{'} = \frac{I _{g}}{2}$ gives

$\frac{4}{R _{1} + 2.4} = \frac{1}{2} \cdot \frac{10}{R _{1} + 6} \Rightarrow 8 (R_{1} + 6) = 10 (R_{1} + 2.4) \Rightarrow R_{1} = 12 Ω$ .

Battery current with key closed. $i = \frac{10}{12 + 2.4} = \frac{10}{14.4} \approx 0.6944$ A $= 694.44$ mA.

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