$A$ galvanometer having a resistance $G$ and current $I_g$ flowing in it,produces full-scale deflection. If $S_1$ is the value of the shunt which converts it into an ammeter of range $0-I$ and $S_2$ is the value of the shunt for range $0-8I$,then the ratio $\frac{S_1}{S_2}$ will be:

  • A
    $\frac{8I - I_g}{I - I_g}$
  • B
    $1$
  • C
    $8$
  • D
    $\frac{1}{2} \left( \frac{I - I_g}{8I - I_g} \right)$

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