If only $\frac{I^{\text {th }}}{51}$ of the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1^{\text {th }}}{11}$ of the main voltage is to be developed across the galvanometer, then the resistance required $R_2$. Then $\frac{R_2}{R_1}$

Question

If only $\frac{I^{\text {th }}}{51}$ of the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1^{\text {th }}}{11}$ of the main voltage is to be developed across the galvanometer, then the resistance required $R_2$. Then $\frac{R_2}{R_1}$, $\frac{1}{500}$, $\frac{50}{9}$, $\frac{500}{3}$, 500

MARKS App · Accepted Answer

According to the question, Case I If only $\frac{1}{51}$ th of the main current $i$ is to be passed through galvanometer $G$ then the shunt required is main current, $i=51$ $\begin{aligned} & i_g=1 \ & \therefore \quad R_1=\frac{G}{i-i_g} \Rightarrow R_1=\frac{G}{51-1}=\frac{G}{50} \quad \ldots .(i) \ \end{aligned}$ Case II If only $\frac{1}{11}$ th of the main voltage is developed across the $G$ then the resistance required, $R_2$. $\begin{aligned} & R_2=G\left(V_G-1\right) \ & R_2=G(11-1)=10 G \quad \ldots (ii) \end{aligned}$ Now, from Eqn. (i) and (ii), we get $\begin{aligned} & \frac{R_2}{R_1}=\frac{10 G}{\frac{G}{50}}=500 \ & \therefore \frac{R_2}{R_1}=500 \end{aligned}$

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