A galvanometer has $36\,\Omega $ resistance. If a $4\,\Omega $ shunt is added to this, the fraction of current that passes through the shunt is
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$\mathrm{R}_{\mathrm{S}}: \mathrm{R}_{\mathrm{G}}=\frac{1}{9}$
$\therefore {{\rm{I}}_a}:{{\rm{I}}_g} = \frac{9}{1}$
$\frac{9}{10}$ th current goes through shunt
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