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. Increases by 15%
B. Decreases by 15%
C. Increases by 25%
D. Decreases by 25%

Answer

D

Explanation

New, current sensitivity, `l_(s)= l_(s)+ (50)/(100)l_(s)= (150)/(100)l_(s)= (3)/(2) l_(s)`
New resistance, R.= 2R
Initial voltage sensitivity, `V_(s)= (I_(s))/(R )`
Now, new voltage sensitivity `V_(S)= (I_(s))/(R.)= ((3)/(2)l_(s))/(2R)= (3)/(4)V_(s)`
% decrease in voltage sensitivity `=(V_(S)-V_(S))/(V_(S))xx 100`
`=(1-(V_(S))/(V_(S)))xx 100`
`=(1-(3)/(4))xx100=25%`