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

Let resistance $S$ is connected with galvanometer of resistance $G$, than

${\mathrm{I}}_{\mathrm{G}}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{G}}$


Now, as shown below figure a shunt $S$ is connected in parallel with the galvanometer for half deflection in it, that is $I_G\text{'} = \frac{I_G}{2}$, than

 $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}+\frac{\mathrm{G}\mathrm{S}}{\mathrm{G}+\mathrm{S}}}$


As, shunt and galvanometer are in parallel  then $V_G=V_S$
${\mathrm{I}}_{\mathrm{G}}\text{'}\mathrm{G}=\left(\mathrm{I}-{\mathrm{I}}_{\mathrm{G}}\text{'}\right)\mathrm{S}$

$\Rightarrow {\mathrm{I}}_{\mathrm{G}}\text{'}\left(\mathrm{G}+\mathrm{S}\right)=\mathrm{IS}$
$\Rightarrow \frac{{\mathrm{I}}_{\mathrm{G}}}{2}=\frac{\mathrm{I}\mathrm{S}}{\mathrm{G}+\mathrm{S}} \left(\because {\mathrm{I}}_{\mathrm{G}}\text{'}=\frac{I_G}{2}\right)$

On, substituting the values of $I_G$ and $I$ in above equation

$\frac{\mathrm{V}}{2\left(\mathrm{R}+\mathrm{G}\right)}=\frac{\mathrm{V}}{\mathrm{R}+\frac{\mathrm{G}\mathrm{S}}{\mathrm{G}+\mathrm{S}}}\times \frac{\mathrm{S}}{\mathrm{G}+\mathrm{S}}$
$\Rightarrow \frac{1}{2\left(\mathrm{R}+\mathrm{G}\right)}=\frac{\mathrm{S}}{\mathrm{R}\left(\mathrm{G}+\mathrm{S}\right)+\mathrm{G}\mathrm{S}}$
$\Rightarrow \mathrm{R}\left(\mathrm{G}+\mathrm{S}\right)+\mathrm{G}\mathrm{S}=2\mathrm{S}\left(\mathrm{R}+\mathrm{G}\right)$
$\Rightarrow \mathrm{R}\mathrm{G}+\mathrm{R}\mathrm{S}+\mathrm{G}\mathrm{S}=2\mathrm{S}\left(\mathrm{R}+\mathrm{G}\right)$
$\Rightarrow \mathrm{R}\mathrm{G}=2\mathrm{S}\left(\mathrm{R}+\mathrm{G}\right)-\mathrm{S}\left(\mathrm{R}+\mathrm{G}\right)$
$\Rightarrow \mathrm{R}\mathrm{G}=\mathrm{S}\left(\mathrm{R}+\mathrm{G}\right)$