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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(b) ${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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