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:

Using the half-deflection condition relation: $$G = \frac{R_1 R_2}{R_1 - R_2}$$

$$6 = \frac{R_1 \times 4}{R_1 - 4} \implies 6R_1 - 24 = 4R_1 \implies 2R_1 = 24 \implies R_1 = 12\ \Omega$$

Evaluating the equivalent circuit resistance when key $$K$$ is closed:

$$R_{\text{eq}} = R_1 + \frac{G R_2}{G + R_2} = 12 + \frac{6 \times 4}{6 + 4} = 12 + 2.4 = 14.4\ \Omega$$

Evaluating total current from the source passing through $$R_1$$:

$$I = \frac{V}{R_{\text{eq}}} = \frac{10}{14.4} = \frac{100}{144} = \frac{25}{36}\text{ A}$$

$$I = \frac{25}{36} \times 1000\text{ mA} = \frac{25000}{36}\text{ mA} \approx 694.44\text{ mA}$$