JEE Main 2023 — Current Electricity Question with Solution

The current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$$I_s=\frac{nBA}{C}$$

where $$n$$ is the number of turns in the coil, $$B$$ is the magnetic field, $$A$$ is the area of the coil, and $$C$$ is the torsional constant of the suspension wire.

We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $$n^{\prime }$$, and the new area be $$A^{\prime }$$.

Since the current sensitivity is increased by 25%, we have:

$$I_s^{\prime }=1.25I_s=\frac{n^{\prime }B{A}^{\prime }}{C}$$

The resistance ($R$) of the galvanometer coil is given by:

$$R=\rho \frac{l}{A}$$

where $$\rho$$ is the resistivity of the wire and $$l$$ is the length of the wire. Since the resistance is kept constant, we can write:

$$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=R$$

Now, the voltage sensitivity ($V_s$) of the galvanometer is given by:

$$V_s=I_s\times R$$

We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $$V_s^{\prime }$$:

$$V_s^{\prime }=I_s^{\prime }\times R^{\prime }$$

Since $$R=R^{\prime }$$, we can write:

$$V_s^{\prime }=1.25I_s\times R=1.25V_s$$

The percentage change in the voltage sensitivity is:

$$\frac{V_s^{\prime }-V_s}{V_s}\times 100\%=\frac{1.25V_s-V_s}{V_s}\times 100\%=25\%$$