Question

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?

Answer 1

$R=\frac{V}{I}$
If $i$ increased by $50\%$ than
$I^{\prime }=1.5i$
The new resistance $\Rightarrow 2R$
$2R=1.5\frac{I}{V^{\prime }}$
$V^{\prime }=1.5\frac{I}{2R}$
$V^{\prime }=0.75\left(\frac{I}{R}\right)$
$V^{\prime }=0.75\left(V\right)$
The voltage changes by $0.25$
The percentage change $=25\%$