Physics - Current Electricity Question with Solution | TestHub

The deflection in a galvanometer is directly proportional to the current passing through it, $\theta \propto I_g$.

Initially, the entire current $I$ passes through the galvanometer:
$I\propto 16^{\circ }$

When shunted by a resistance $S=8\Omega$, the current through the galvanometer becomes $I_g^{\prime }$:
$I_g^{\prime }\propto 4^{\circ }$

Taking the ratio of the two currents:
$\frac{I_g^{\prime }}{I}=\frac{4^{\circ }}{16^{\circ }}=\frac{1}{4}$

In a shunted galvanometer, the current through the galvanometer is given by:
$I_g^{\prime }=I\left(\frac{S}{S+G}\right)$

Substituting the ratio:
$\frac{S}{S+G}=\frac{1}{4}$

$4S=S+G$

$G=3S$

Given $S=8\Omega$:
$G=3\times 8=24\Omega$

Answer: $24\Omega$