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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 $\mathrm{G}$, the shunt resistance $\mathrm{S}$ connected to the galvanometer is
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Verified Answer
The correct answer is:
$\frac{\mathrm{G}}{19}$
Current passing through galvanometer, $\mathrm{I}_{\mathrm{G}}=0.05 \mathrm{I}$
Current passing through shunt resistance.
$$
\begin{aligned}
& I_S=I-0.05 I \\
& =0.95 I \\
& I_S S=I_G G \Rightarrow \frac{S}{G}=\frac{I_G}{I_S} \\
& =\frac{0.05}{0.95}=\frac{1}{19} \Rightarrow S=\frac{G}{19}
\end{aligned}
$$
Current passing through shunt resistance.
$$
\begin{aligned}
& I_S=I-0.05 I \\
& =0.95 I \\
& I_S S=I_G G \Rightarrow \frac{S}{G}=\frac{I_G}{I_S} \\
& =\frac{0.05}{0.95}=\frac{1}{19} \Rightarrow S=\frac{G}{19}
\end{aligned}
$$
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