The value of shunt resistance, that allows only $10 \%$ of main current through the galvanometer of…

The value of shunt resistance, that allows only $10 \%$ of main current through the galvanometer of resistance $99 \Omega$ is
  1. $9 \Omega$
  2. $4 \Omega$
  3. $2 \Omega$
  4. $11 \Omega$

Solution


$\mathrm{G}=99 \Omega, \mathrm{I}_{\mathrm{g}}=\frac{1}{10}$ $\therefore \quad$ From figure, $\mathrm{I}_{\mathrm{g}} \cdot \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ $\Rightarrow \mathrm{S}=\left(\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\right) \mathrm{G}=\frac{\frac{\mathrm{I}}{10}}{\mathrm{I}-\frac{\mathrm{I}}{10}} \times 99=11 \Omega$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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