$20\%$ of main current passes through the galvanometer. If the resistance of the galvanometer is $G$ , then the resistance of the shunt will be
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$\frac{20}{100} \mathrm{IG}=\frac{80}{100} \mathrm{IS} \quad \Rightarrow \mathrm{S}=\frac{\mathrm{G}}{4}$
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