The current sensitivity of a moving coil galv | vedclass

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Question

(A) The current sensitivity of a moving coil galvanometer is given by $I_s = \frac{NBA}{C}$,where $N$ is the number of turns,$B$ is the magnetic field

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$+25$

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(A) The current sensitivity of a moving coil galvanometer is given by $I_s = \frac{NBA}{C}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $C$ is the torsional constant.The voltage sensitivity is given by $V_s = \frac{I_s}{G}$,where $G$ is the resistance of the galvanometer coil.Given that the resistance $G$ is kept constant,the relationship between voltage sensitivity and current sensitivity is $V_s \propto I_s$.Since the current sensitivity $I_s$ increases by $25 \%$,the voltage sensitivity $V_s$ will also increase by $25 \%$ because the ratio $\frac{1}{G}$ is constant.Therefore,the percentage change in voltage sensitivity is $25 \%$.

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