Question

The current sensitivity of a moving coil galvanometer increases by \(35 \%\), when its resistance is increased by a factor of 3 . The voltage sensitivity of galvanometer changes by a factor

(a) \(35 \%\)

(b) \(45 \%\)

(c) \(55 \%\)

(d) None of thesea

Answer 1

The correct option of this question will be (c).

Solution —

Given $I_{s}^{\prime}=I_{s}+\frac{35}{100} I_{s}=\frac{135}{100} I_{s}$

Initial voltage sensitivity, $V_{s}=\frac{I_{s}}{R}$

New voltage sensitivity, $V_{s}^{\prime}=\frac{I_{s}^{\prime}}{R^{\prime}}$

$=\left(\frac{135}{100} I_{s}\right) \times \frac{1}{3 R}=\frac{9}{20} V_{s}$

% decrease in voltige sensitivity

$\left(\frac{V_{s}-V_{s}^{\prime}}{V_{s}}\right) \times 100 \%=\frac{V_{s}-\frac{9}{20} V_{s}}{V_{s}} \times 100 \%=55 \%$