In a galvanometer $5 \%$ of the total current in the circuit passes through it. If the resistance of the…
In a galvanometer $5 \%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is $G$, the shunt resistance $S$ connected to the galvanometer is
$19 G$
$\frac{G}{19}$
$20 G$
$\frac{G}{20}$
Solution
Shunt of an ammeter,
$
S=\frac{I_g \times G}{I-I_g}
$
$\begin{aligned} & =\frac{5 \times G}{100-5} \\ & =\frac{G}{19}\end{aligned}$