To increase the current sensitivity of a moving coil galvanometer by `50%` its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Options

A. Increased by 50%
B. increased by 25%
C. decreased by 25%
D. decreased by 50%

Answer

C

Explanation

The initial voltage senstivity,`S_V=("current senstivity "(S_i))/R=(S_i)/(R)` When the resistance is increased to 2R, the current senstivity becomes `S_i=3/2S_i` and the new voltage senstivity `S'_v=(S'_i)/(2R)=(3/2S_i)/(2R)=3/4(S_V)` Thus the voltage senstivity becomes `3/2(S_V)` or it decreases by 25%.