A $36 \,\Omega$ galvanometer is shunted by resistance of $4\,\Omega$. The percentage of the total current, which passes through the galvanometer is
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(c) $\frac{{{i_g}}}{i} = \frac{S}{{G + S}} = \frac{4}{{36 + 4}} = \frac{1}{{10}}\,\,i.e.\,10\% .$
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