Question

The relation between voltage sensitivity $\left(\sigma_{ v }\right)$ and current sensitivity $\left(\sigma_{1}\right)$ of a moving coil galvanometer is (resistance of galvanometer is G).

• A. $\frac{\sigma_{i}}{G}=\sigma_{ V }$
• B. $\frac{\sigma_{ V }}{ G }=\sigma_{ i }$
• C. $\frac{ G }{\sigma_{i}}=\sigma_{ V }$
• D. $\frac{ G }{\sigma_{ V }}=\sigma_{ v }$

Answer 1

Answer: (A)

$\frac{\sigma_{i}}{G}=\sigma_{ V }$

Answer 2

We know in a galvanometer, Current sensitivity, $\sigma_{i}=\frac{\theta}{ I }$, where $\theta$ is the deflection produced. Voltage sensitivity, $\sigma_{ v }=\frac{\theta}{ V }$ But $V=I R=I G$ as given Gis the resistance of the galvanometer. Therefore, $\sigma_{ v }=\frac{\theta}{ IG }=\frac{\sigma_{ i }}{ G }$ i.e., Voltage sensitivity $=\frac{\text { current sensitivity }}{\text { Resistance }}$ The above relation implies that if current sensitivity increases as well as the resistance increases in same order, the voltage sensitivity will remain unchanged.