A galvanometer has a current sensitivity of $1 \mathrm{~mA}$ per division. A variable shunt is connected…
- $47.1 \mathrm{~V}$
- $57.1 \mathrm{~V}$
- $67.1 \mathrm{~V}$
- $77.1 \mathrm{~V}$
Solution

Here, $I=\frac{E}{R+r+\frac{G S}{G+S}}$ and $I_g=\frac{I S}{G+S}$ $\begin{aligned} & I_g=\frac{E}{(R+r)+\frac{G S}{(G+S)}} \times \frac{S}{(G+S)} \\ & \therefore \quad I_g=\frac{E S}{(R+r)(G+S)+G S} \end{aligned}$ For $S=5 \mathrm{ohm}, I_g=5 \times 10^{-3} \mathrm{~A}$ and for $S=25 \mathrm{ohm}$ $I_g=20 \times 10^{-3} \mathrm{~A}$ Hence, $5 \times 10^{-3}=\frac{E \times 5}{501(G+5)+5 G}$ and $20 \times 10^{-3}=\frac{E \times 25}{501(G+25)+25 G}$ Dividing and solving, $G=88.2 \Omega$ From (i), we get $\begin{aligned} E & =10^{-3}[501(88.2+5)+5 \times 88.2] \\ & =47.1 \text { volt } \end{aligned}$ ^
Asked in: NEET 2021