To increase the current sensitivity of a moving coil galvanometer by 75%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decreases by 12.5%
The current sensitivity of a moving coil galvanometer is the deflection per unit current, \( I_s = \theta/I \), and the voltage sensitivity is the deflection per unit voltage, \( V_s = \theta/V \). Since \( V = IR \), the two are related by:
\( \displaystyle V_s = \frac{I_s}{R} \)
The current sensitivity increases by 75%, so \( I_s' = 1.75\,I_s \), and the resistance becomes \( R' = 2R \).
\( \displaystyle \frac{V_s'}{V_s} = \frac{I_s'/R'}{I_s/R} = \frac{1.75}{2} = 0.875 \)
The voltage sensitivity falls to 87.5% of its value, that is it decreases by 12.5%.