Question

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. $G^2$

Answer 1

Answer: (A)

Voltage sensitivity $=\frac{Q}{V}$
Current sensitivity $=\frac{Q}{I}$
Also, potential difference
$V=IG$
Hence,
$\therefore \frac{V_s}{I_s}=\frac{1}{G}$