Physics - Current Electricity Question with Solution | TestHub

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

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


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

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


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

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

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

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