Question

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 the voltage sensitivity change?

• A. decreases by $75$ %
• B. increases by $75$ %
• C. decreases by $25$ %
• D. increases by $25$ %

Answer 1

The correct option is C decreases by $25$ %

Let $CS$ and $VS$ be the original current sensitivity and voltage sensitivity of the moving coil galvanometer respectively.

New current sensitivity is given by,

$C{S}^{\prime }=CS+\frac{50}{100}CS=\frac{3}{2}CS$

Since, $VS=\frac{CS}{R}$

The new voltage sensitivity is given by,

$V{S}^{\prime }=\frac{C{S}^{\prime }}{2R}=\frac{\left(3/2\right)CS}{2R}$

$\Rightarrow V{S}^{\prime }=\frac{3}{4}\frac{CS}{R}=0.75VS$

Thus, the voltage sensitivity decreases by $25$ %

Hence, option $\left(C\right)$ is the correct answer.