The current sensitivity of a moving coil galvanometer is doubled by making the number of turns double. Then its voltage sensitivity will be :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Remain unchanged
For a moving coil galvanometer the current sensitivity and the voltage sensitivity are
\( \displaystyle \frac{\theta}{I} = \frac{NBA}{k}, \qquad \frac{\theta}{V} = \frac{NBA}{kR} \)
where \( N \) is the number of turns, \( B \) the field, \( A \) the area of the coil, \( k \) the torsion constant and \( R \) the resistance of the coil.
Doubling \( N \) doubles the current sensitivity. But the coil now has twice as much wire, so its resistance also doubles.
\( \displaystyle \frac{\theta}{V}\Big|_{\mathrm{new}} = \frac{(2N)BA}{k(2R)} = \frac{NBA}{kR} \)
The voltage sensitivity remains unchanged.
Note: A similar question was also asked in CUET (UG) 2025 on 02-06-2025 (Slot 2), and in CUET (UG) 2022 on 17-08-2022 (Slot 2).
NCERT (2026-27):
The Moving Coil Galvanometer – “… An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.27) which provides a measure of current sensitivity. If \( N \to 2N \), i.e., we double the number of turns, then \( \frac{\phi}{I} \to 2\frac{\phi}{I} \). Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), \( N \to 2N \), and \( R \to 2R \), thus the voltage sensitivity, \( \frac{\phi}{V} \to \frac{\phi}{V} \) remains unchanged.”