Q. The ratio of voltage sensitivity ($V_{S}$) and current sensitivity ($I_{s}$) of a moving coil galvanometer is
A) $\frac{1}{G}$
B) $\frac{1}{G^{2}}$
C) G
D) $G^{2}$
Explanation
(a) Voltage sensitivity $= \frac{Q}{V}$
Current sensitivity $= \frac{Q}{I}$
Also, potential difference
$V = IG$
Hence, $\frac{V_{S}}{I_{s}} = \frac{Q/V}{Q/I} = \frac{I}{V} = \frac{I}{IG}$
$\therefore\frac{V_{s}}{I_{s}} = \frac{1}{G}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.