Physics - Magnetism Question with Solution | TestHub

New, current sensitivity, $I_s^{\text{'}}=I_s+\frac{50}{100}{I}_s$

$=\frac{150}{100}{I}_s=\frac{3}{2}{I}_s$

New resistance, $R^{\text{'}}=2 \mathrm{R}$

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

Now, new voltage sensitivity.

$V_s=\frac{i_s^{\text{'}}}{R^{\text{'}}}=\frac{\frac{3}{2} I_c}{2 \mathrm{R}}=\frac{3}{4}{V}_s$

$\%$ decrease in voltage sensitivity

$=\frac{V_s-V_s^{\text{'}}}{V_s}\times 100$

$=\left(1-\frac{V_s}{V_s}\right)\times 100$

$=\left(1-\frac{3}{4}\right)\times 100=25 \%$