To increase the current sensitivity of a moving coil galvanometer by `50%` its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Options

A. decreasesd by `75%`
B. Increased by `75%`
C. decreased by `25%`
D. Increase by `25%`

Answer

3

Explanation

Let `CS` and `VS` be the original current sensitivity and voltage sensitivity of `MCG`
Changed current sensitivity
`CS^(1)=CS+(50)/(100)CS=(3)/(2)CS,`Since `VS=(CS)/(R)` changed voltage sensitivity, `i.e.,`
`VS^(1)=(CS^(1))/(2R)=((3//2)CS)/(2R)`
`=(3)/(4)((CS)/(R))=0.75VS=75%=75%VS`
Thus, voltage sensitivity decreases by` 25%`