We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

  • A
    $\frac{I_g}{I} = \frac{1}{11}$
  • B
    $\frac{I_g}{I} = \frac{1}{10}$
  • C
    $\frac{I_g}{I} = \frac{3}{11}$
  • D
    $\frac{I_g}{I} = \frac{4}{11}$

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