To increase the current sensitivity of a moving coil galvanometer by 75%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decreases by 12.5%
$R=\frac{V}{I}$ [By Ohm's law]
if $i$ is increased by 75%
$I'=1.75I$
Also, $R_{new}=2R$
$∴V'=\frac{1.75I}{2R}$
$⇒V'=0.875V$
∴ it decreases by 12.5%