A galvanometer of resistance $\mathrm{G}$ is shunted with a resistance of $10 \%$ of $\mathrm{G}$. The part…

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
  1. $\frac{1}{11} \mathrm{I}$
  2. $\frac{2}{11} \mathrm{I}$
  3. $\frac{1}{10} \mathrm{I}$
  4. $\frac{1}{5} \mathrm{I}$

Solution

$\begin{aligned} \frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}} & =\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}=\frac{0.1 \mathrm{G}}{0.1 \mathrm{G}+\mathrm{G}}=\frac{1}{11} \\ \therefore \quad \mathrm{I}_{\mathrm{g}} & =\frac{1}{11} \mathrm{I}\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

Practice more Current Electricity questions on Aicharya