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Question 13

The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

$$I_s = \frac{NBA}{k}, \quad V_s = \frac{I_s}{R_g}$$

$$\text{Given that the resistance of the galvanometer coil } R_g \text{ is kept constant: } V_s \propto I_s$$

$$\% \Delta V_s = \% \Delta I_s = +25\%$$

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