JEE Main 2023 — Current Electricity Question with Solution

The formula to calculate the maximum current through the galvanometer, when a shunt resistance $r_1$ is connected is given by${\left(i_G\right)}_{\max }=\frac{r_1}{r_1+R_G}{i}_{\max } ...\left(1\right)$

When a series resistance $r_2$ is connected, the potential difference across the galvanometer is given by

$V={\left(i_G\right)}_{\max }\left[R_G+r_2\right] ...\left(2\right)$

From equations (1) and (2), it can be written that

$\begin{array}{rcl}V& =& \frac{r_1}{r_1+R_G}{i}_{\max }\left(R_G+r_2\right)\\ & \Rightarrow & V{r}_1+V{R}_G=i_{\max }{r}_1{R}_G+i_{\max }{r}_1{r}_2\\ & \Rightarrow & \left(V-i_{\max }{r}_1\right)R_G=i_{\max }{r}_1{r}_2-V{r}_1\\ & \Rightarrow & R_G=\frac{i_{\max }{r}_1{r}_2-V{r}_1}{V-i_{\max }{r}_1} ...\left(3\right)\end{array}$

Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance.

$\begin{array}{rcl}{R}_G& =& \frac{0.250 \mathrm{A}\times 5 \mathrm{\Omega }\times 1050 \mathrm{\Omega }-25 \mathrm{V}\times 5 \mathrm{\Omega }}{25 \mathrm{V}-0.250 \mathrm{A}\times 5 \mathrm{\Omega }}\\ & =& 50 \mathrm{\Omega }\end{array}$