How does voltage sensitivity of a galvanometer change as number of turns of its coil are doubled?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Remains same
The voltage sensitivity of a moving coil galvanometer is the deflection per unit voltage.
\( \displaystyle \frac{\phi}{V} = \frac{N B A}{k R} \)
Here \( N \) is the number of turns, \( B \) the magnetic field, \( A \) the area of the coil, \( k \) the torsional constant and \( R \) the resistance of the coil.
Doubling \( N \) doubles the numerator. The coil now has twice the length of wire, so its resistance \( R \) also doubles. The two changes cancel, so the voltage sensitivity remains the same.
Why the other options are incorrect
Option (1) Halved — Incorrect. Nothing in the ratio is halved; both \( N \) and \( R \) double.
Option (2) Doubled — Incorrect. This is the change in the current sensitivity \( NBA/k \); the voltage sensitivity is also divided by the doubled resistance.
Option (4) Insufficient information — Incorrect. The relation above is enough to decide the change, so the information is sufficient.
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. … If \( N \rightarrow 2N \), i.e., we double the number of turns, then
\( \displaystyle \frac{\phi}{I} \rightarrow 2\frac{\phi}{I} \)
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), \( N \rightarrow 2N \), and \( R \rightarrow 2R \), thus the voltage sensitivity,
\( \displaystyle \frac{\phi}{V} \rightarrow \frac{\phi}{V} \)
remains unchanged.”