Q. 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 $S$ are related by the equation :
Solution:
$I_{g}=\frac{V}{R+G}$
$R_{c}=R+\frac{GS}{G+s}$
$I=\frac{V}{R+\frac{GS}{G+S}}$
$I '_{g}G=\left(I-I '_{g}\right)S$
$I '_{g}\left(G+S\right)=IS$
$\frac{I_{g}}{2}=\frac{IS}{G+S}$
$\frac{V}{2\left(R+G\right)}=\frac{V}{R+\frac{GS}{G+S}}\times\frac{S}{G+S}$
$\frac{1}{2\left(R+G\right)}=\frac{S}{R\left(G+S\right)+GS}$
$R\left(G+S\right)+GS=2S\left(R+G\right)$
$RG+RS+GS=2S\left(R+G\right)$
$RG=2S\left(R+G\right)-S\left(R+G\right)$
$RG=S\left(R+G\right)$
