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