Question 367 of 380

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:

A$S\left(R+G\right)=RG$B$2S\left(R+G\right)=RG$C$2G=S$D$2S=G$

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Answer: (A) $S\left(R+G\right)=RG$

Let resistance $S$ is connected with galvanometer of resistance $G$, than
${I}_{G}=\dfrac{V}{R+G}$


Now, as shown below figure a shunt $S$ is connected in parallel with the galvanometer for half deflection in it, that is ${I}_{G}'=\dfrac{{I}_{G}}{2}$, than
$I=\dfrac{V}{R+\dfrac{GS}{G+S}}$


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

$\Rightarrow {I}_{G}'\left(G+S\right)=IS$
$\Rightarrow \dfrac{{I}_{G}}{2}=\dfrac{IS}{G+S}\left(∵{I}_{G}'=\dfrac{{I}_{G}}{2}\right)$
On, substituting the values of ${I}_{G}$ and $I$ in above equation

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