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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
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Verified Answer
The correct answer is:
$\frac{G}{19}$
Shunt of an ammeter,
$$
S=\frac{I_g \times G}{I-I_g}
$$
$\begin{aligned} & =\frac{5 \times G}{100-5} \\ & =\frac{G}{19}\end{aligned}$
$$
S=\frac{I_g \times G}{I-I_g}
$$
$\begin{aligned} & =\frac{5 \times G}{100-5} \\ & =\frac{G}{19}\end{aligned}$
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