Question

The ratio of voltage sensitivity $(V_S)$ and current sensitivity $(I_S)$ of a moving coil galvanometers:

• 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, $\frac{V_S}{I_S}=\frac{Q/V}{Q/I}=\frac{I}{V}=\frac{I}{IG}$ $\therefore$ $\frac{V_S}{I_S}=\frac{1}{G}$