A moving coil galvanometer has a current sensitivity of 20 divisions per mA. The resistance of its coil is 50 Ω. The potential difference across the coil that gives a deflection of 30 divisions is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 75 mV
Voltage sensitivity is the deflection per unit voltage. Since \( V=IR \), it equals the current sensitivity divided by the resistance \( R \) of the coil. Once it is known, the voltage for any deflection follows.
Given:
\( \phi/I=20\ \mathrm{div/mA} \)
\( R=50\ \Omega \)
\( \phi=30\ \mathrm{div} \)
\( \begin{aligned} \frac{\phi}{V} &= \frac{1}{R}\left(\frac{\phi}{I}\right) \\ &= \frac{20\ \mathrm{div/mA}}{50\ \Omega} \\ &= 0.4\ \mathrm{div/mV} \end{aligned} \)
\( \begin{aligned} V &= \frac{\phi}{\phi/V} \\ &= \frac{30\ \mathrm{div}}{0.4\ \mathrm{div/mV}} \\ &= 75\ \mathrm{mV} \end{aligned} \)
Here \( 1\ \mathrm{mA}\times1\ \Omega=1\ \mathrm{mV} \). As a check, 30 divisions need a current of 1.5 mA, and \( 1.5\ \mathrm{mA}\times50\ \Omega=75\ \mathrm{mV} \).
NCERT (2026-27):
The Moving Coil Galvanometer – “We define the voltage sensitivity as the deflection per unit voltage. …”
\( \displaystyle \frac{\phi}{V}=\left(\frac{NAB}{k}\right)\frac{I}{V}=\left(\frac{NAB}{k}\right)\frac{1}{R} \)\( \qquad (4.28) \)