Question

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:

Hint:

Remember that in the half-deflection method, $R_1$ is typically very large compared to $G$, but if values are small like in this problem, you must use the exact formula $R_1 = \frac{G R_2}{G - R_2}$ if applicable.
Always distinguish between the total current and the galvanometer current in parallel circuits.
Check the final units carefully as the question asks for mA.

Answer 1