The resistance of a galvanometer is $90\, ohms$. If only $10$ percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be used
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(b) ${i_g} = 10\% $ of $i = \frac{i}{{10}}$ $ \Rightarrow $ $S = \frac{G}{{(n - 1)}} = \frac{{90}}{{(10 - 1)}} = 10\,\Omega $
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