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 percentage does its voltage sensitivity change?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → It decreases by 25%.
The voltage sensitivity of a galvanometer is the current sensitivity divided by the resistance of the coil:
\( \displaystyle \frac{\theta}{V} = \frac{\theta/I}{R} \)
Let the initial current sensitivity be \( S_I \) and the initial resistance \( R \). A 50% increase makes \( S_I' = 1.5\,S_I \), and the new resistance is \( R' = 2R \).
\( \displaystyle \frac{S_V'}{S_V} = \frac{S_I'/R'}{S_I/R} = \frac{1.5}{2} = 0.75 \)
The voltage sensitivity becomes 75% of its original value, so it decreases by 25%.