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We have a galvanometer of resistance $25\,\Omega $. It is shunted by a $2.5\,\Omega $ wire. The part of total current that flows through the galvanometer is given as
  • A$\frac{I}{{{I_0}}} = \frac{1}{{11}}$
  • B$\frac{I}{{{I_0}}} = \frac{1}{{10}}$
  • C$\frac{I}{{{I_0}}} = \frac{3}{{11}}$
  • D$\frac{I}{{{I_0}}} = \frac{4}{{11}}$
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