In order to pass $10\,\%$ of main current through a moving coil galvanometer of $99\, ohm$, the resistance of the required shunt is ............ $\Omega $
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(c) Shunt resistances $S = \frac{{{i_g}G}}{{(i - {i_g})}} = \frac{{10 \times 99}}{{(100 - 10)}} = 11\,\Omega $
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