Two moving coil galvanometers G1 and G2 have springs of the same torsional constant and coils of the same area. Compared with G1, the galvanometer G2 has twice the number of turns, a radial field twice as strong and twice the resistance. The ratio of the voltage sensitivity of G2 to that of G1 is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2:1\)
Voltage sensitivity is the deflection per unit voltage, \(\phi/V=NAB/kR\). It grows with the turns and the field, but falls as the resistance rises. With \(A\) and \(k\) the same for both meters,
\( \begin{aligned} \frac{(\phi/V)_2}{(\phi/V)_1} &= \frac{N_2B_2R_1}{N_1B_1R_2} \\ &= \frac{2\times2\times1}{1\times1\times2} \\ &= 2 \end{aligned} \)
The current sensitivity \(NAB/k\) becomes four times. The doubled resistance halves this, so the voltage sensitivity only doubles.
NCERT (2026-27):
The Moving Coil Galvanometer – “… We define the voltage sensitivity as the deflection per unit voltage. …”
\( \frac{\phi}{V}=\left(\frac{NAB}{k}\right)\frac{1}{R} \)\( \qquad (4.28) \)