$20\%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,then the resistance of the shunt will be

  • A
    $\frac{G}{50}$
  • B
    $\frac{G}{4}$
  • C
    $50\,G$
  • D
    $9\,G$

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$2^{nd}$ Way: By using a smaller coil.
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