In the circuit diagrams (A, B, C and D) shown below, R is a high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as:

(a)

(b)

(c)

(d)

In Circuit (d), the high resistance $$R$$ is placed in series with the main circuit branch. The low resistance shunt $$S$$ (along with key $$K_2$$) is connected in parallel directly across the galvanometer $$G$$. This is the correct configuration.

$$I_g = \frac{V}{R + G}$$

$$\frac{I_g}{2} = I \cdot \left(\frac{S}{G+S}\right) = \left(\frac{V}{R + \frac{GS}{G+S}}\right) \cdot \frac{S}{G+S} = \frac{V \cdot S}{R(G+S) + GS}$$

$$\frac{1}{2}\left(\frac{V}{R+G}\right) = \frac{V \cdot S}{R \cdot G + R \cdot S + G \cdot S}$$

$$R \cdot G + R \cdot S + G \cdot S = 2 \cdot S(R + G)$$

$$R \cdot G - G \cdot S = R \cdot S \implies G(R - S) = R \cdot S$$

$$G = \frac{R \cdot S}{R - S}$$