Question

The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is doubled. Calculate by what factor does the voltage sensitivity change?

• A. Becomes 3/5 times
• B. becomes 2/5 times
• C. No change
• D. decrease by a factor of 7/5

Answer 1

The correct option is A Becomes 3/5 times

Given $\Rightarrow I_s^{\prime }=I_s+\frac{20}{100}{I}_s=\frac{120}{100}{I}_s$

then, initial voltage sensitivity, $V_s=\frac{I_s}{R}$

new voltage sensitivity,

$V_s^{\prime }=\frac{I_s^{\prime }}{R^{\prime }}=\left(\frac{120}{100}{I}_s\right)\left(\frac{1}{2R}\right)$

$\Rightarrow V_s^{\prime }=\frac{60}{100}{V}_s=\frac{3}{5}{V}_s$

so $\left(A\right)$ is correct.