To know the resistance $$G$$ of a galvanometer by half deflection method, a battery of emf $$V_E$$ and resistance $$R$$ is used to deflect the galvanometer by angle $$\theta$$. If a shunt of resistance $$S$$ is needed to get half deflection the $$G$$, $$R$$ and $$S$$ are related by the equation:

First let us note the condition for full deflection. When the galvanometer of resistance $$G$$ is placed in series with the known resistance $$R$$ and the battery of emf $$V_E$$, the current through the circuit is $$I_g$$ (the subscript “g” stands for “galvanometer”). Ohm’s law gives

$$V_E = I_g\,(R + G).$$

Because the deflection is proportional to the current, this current $$I_g$$ produces the full-scale deflection $$\theta$$.

Now we connect a shunt of resistance $$S$$ in parallel with the galvanometer and adjust its value so that the observed deflection becomes exactly one-half, i.e. $$\dfrac{\theta}{2}$$. Therefore the current that now passes through the galvanometer is only half of the original value:

$$I_g' = \dfrac{I_g}{2}.$$

The galvanometer and the shunt are in parallel, so they have the same potential difference across them. The current in the shunt will be the remainder of the total current $$I$$ supplied by the battery. Hence, using current division we have

$$I - I_g' = I - \dfrac{I_g}{2}$$

through the shunt, and the equality of the two parallel branch voltages gives

$$\text{(current in galvanometer)} \times G \;=\; \text{(current in shunt)} \times S.$$

Substituting the actual currents, we obtain

$$\left(\dfrac{I_g}{2}\right)G = \left(I - \dfrac{I_g}{2}\right)S.$$

Solving this relation for the total current $$I$$:

$$\dfrac{I_g}{2}\,G = S\,I - \dfrac{S\,I_g}{2}$$

$$\Rightarrow\; S\,I = \dfrac{I_g}{2}\,G + \dfrac{S\,I_g}{2}$$

$$\Rightarrow\; I = \dfrac{I_g}{2}\left( \dfrac{G}{S} + 1 \right)$$

$$\Rightarrow\; I = \dfrac{I_g}{2}\left( \dfrac{G + S}{S} \right).$$

The battery emf has not changed, so $$V_E$$ must still equal the drop across the external resistance $$R$$ plus the drop across the parallel combination of $$G$$ and $$S$$. The equivalent resistance of the parallel pair is

$$R_{\text{eq}} = \dfrac{G\,S}{G + S}.$$

Hence, for the half-deflection condition

$$V_E = I\left(R + \dfrac{G\,S}{G + S}\right).$$

But from the initial (full-deflection) arrangement we already have $$V_E = I_g\,(R + G)$$. Equating the two expressions for $$V_E$$ gives

$$I_g\,(R + G) = I\left(R + \dfrac{G\,S}{G + S}\right).$$

Substituting the earlier expression for $$I$$, namely $$I = \dfrac{I_g}{2}\left(\dfrac{G + S}{S}\right)$$, we get

$$I_g\,(R + G) = \dfrac{I_g}{2}\left(\dfrac{G + S}{S}\right)\left(R + \dfrac{G\,S}{G + S}\right).$$

Cancel the common factor $$I_g$$ on both sides:

$$(R + G) = \dfrac{1}{2}\left(\dfrac{G + S}{S}\right)\left(R + \dfrac{G\,S}{G + S}\right).$$

Multiply both sides by $$2S$$ to clear the fraction:

$$2S(R + G) = (G + S)\left(R + \dfrac{G\,S}{G + S}\right).$$

Now distribute on the right. First term:

$$(G + S)\,R = R\,G + R\,S.$$

Second term:

$$(G + S)\left(\dfrac{G\,S}{G + S}\right) = G\,S$$

because the factor $$(G + S)$$ cancels. Adding the two results, the right-hand side becomes

$$R\,G + R\,S + G\,S.$$

So we have

$$2S(R + G) = R\,G + R\,S + G\,S.$$

Expand the left-hand side:

$$2S(R + G) = 2S\,R + 2S\,G.$$

Bring every term to one side to simplify:

$$2S\,R + 2S\,G - R\,S - R\,G - G\,S = 0.$$

Group like terms:

$$\bigl(2S\,R - R\,S\bigr) + \bigl(2S\,G - G\,S\bigr) - R\,G = 0$$

$$\Rightarrow S\,R + S\,G - R\,G = 0.$$

Finally, factor out the common terms:

$$S(R + G) = R\,G.$$

This is exactly the relation demanded in option A.

Hence, the correct answer is Option A.