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A galvanometer of resistance $\mathrm{G}$ is shunted with a resistance of $10 \%$ of $\mathrm{G}$. The part of the total current that flows through the galvanometer is
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$\frac{1}{11} \mathrm{I}$
$\begin{aligned} \frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}} & =\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}=\frac{0.1 \mathrm{G}}{0.1 \mathrm{G}+\mathrm{G}}=\frac{1}{11} \\ \therefore \quad \mathrm{I}_{\mathrm{g}} & =\frac{1}{11} \mathrm{I}\end{aligned}$
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