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If only $2 \%$ of the main current is to be passed through a galvanometer of resistance $G$, then the resistance of shunt will be
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$\frac{G}{49}$
$i_g=2 \%$ of $i=\frac{i}{50} \Rightarrow S=\frac{G}{(n-1)}=\frac{G}{(50-1)}=\frac{G}{49}$
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