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 its voltage sensitivity change?

Answer 1

Given I'_s = I_s + \(\frac{50}{100} \)Is = 1.5 I_s

And R' = 2R

Initial voltage sensitivity, V'_s = \(\frac{I_s}{R}\)

New, voltage sensitivity, V's = \(\frac{I'_s}{2R}\)

= 1.5 \(\frac{I_s}{2R}\) = 0.75 V_s

∴ % decrease in voltage sensitivity

\(\frac{V_s-V'_s}{V_s} \times 100 = \frac{V_s-0.75 V_s}{V_s} \times 100\) = 25%