The value of shunt resistance, that allows only $10 \%$ of main current through the galvanometer of resistance $99 \Omega$ is

• A

  $9 \Omega$
• B

  $4 \Omega$
• C

  $2 \Omega$
• Correct answer
  D

  $11 \Omega$

Solution

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

Hence, the correct answer is option: 'D'.