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

The correct option is A $\frac{{\sigma }_i}{G}={\sigma }_v$

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 $V=IR=IG$ as given $G$is the resistance of the galvanometer.

Therefore, ${\sigma }_v=\frac{\theta }{IG}=\frac{{\sigma }_i}{G}$

i.e.,$\text{Voltage sensitivity}=\frac{\text{current sensitivity}}{\text{Resistance}}$

The above relation implies that if current sensitivity increases as well as the resistance increases in same order, the voltage sensitivity will remain unchanged.