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 then \(G, R\)  and  \(S\) related by the equation

\(\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{G}}\)

where, \(I \)\(_{g}\) - Galvanometer current, \(G-\)Galvonometer resistance

When shunt of resistance \(S\) is connected parallel to the Galvanometer then

\(\mathrm{G}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}\)

\(\therefore I=\frac{V}{R+\frac{G S}{G+S}}\)

Equal potential difference is given by

\(\mathrm{I}_{\mathrm{g}}^{\prime} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{B}}^{\prime}\right) \mathrm{S}\)

\(\mathrm{I}_{\mathrm{g}}^{\prime}(\mathrm{G}+\mathrm{S})=\mathrm{IS}\)

\(\Rightarrow \frac{\mathrm{I}_{\mathrm{g}}}{2}=\frac{\mathrm{IS}}{\mathrm{G}+\mathrm{S}}\)

\(\Rightarrow \frac{\mathrm{v}}{2(\mathrm{R}+\mathrm{G})}=\frac{\mathrm{v}}{\mathrm{R}+\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}} \times \frac{\mathrm{S}}{\mathrm{G}+\mathrm{S}}\)

\( \Rightarrow \frac{1}{{2(R + G)}} = \frac{S}{{R(G + S) + GS}}\)

\(\Rightarrow R(G+S)+G S=2 S(R+G)\)

\(\Rightarrow \mathrm{RG}+\mathrm{RS}+\mathrm{GS}=2 \mathrm{S}(\mathrm{R}+\mathrm{G})\)

\(\Rightarrow \mathrm{RG}=2 \mathrm{S}(\mathrm{R}+\mathrm{G})-\mathrm{S}(\mathrm{R}+\mathrm{G})\)

\(\therefore \mathrm{RG}=\mathrm{S}(\mathrm{R}+\mathrm{G})\)