Question

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 :

• A. $2S(R+G)=RG$
• B. $S(R+G)=RG$
• C. $2S=G$
• D. $2G=S$

Answer 1

Answer: (B)

$I_g=\frac{V}{R+G}$
$R_c=R+\frac{GS}{G+s}$
$I=\frac{V}{R+\frac{GS}{G+S}}$
$I_g^{\prime }G=\left(I-I_g^{\prime }\right)S$
$I_g^{\prime }\left(G+S\right)=IS$
$\frac{I_g}{2}=\frac{IS}{G+S}$
$\frac{V}{2\left(R+G\right)}=\frac{V}{R+\frac{GS}{G+S}}\times \frac{S}{G+S}$
$\frac{1}{2\left(R+G\right)}=\frac{S}{R\left(G+S\right)+GS}$
$R\left(G+S\right)+GS=2S\left(R+G\right)$
$RG+RS+GS=2S\left(R+G\right)$
$RG=2S\left(R+G\right)-S\left(R+G\right)$
$RG=S\left(R+G\right)$