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)

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 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.