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In an ammeter $10 \%$ of main current is passing through the galvanometer. If the resistance of the galvanometer is $G$, then the shunt resistance (in $\Omega$ ) is
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The correct answer is:
$\frac{G}{9}$
If $I$ be the main current, then current passing through galvanometer,
$I_{g}=10 \% \text { of } I=\frac{I}{10}$
Resistance of galvanometer, $R_{g}=G$
Shunt resistance, $R_{S}=\frac{I_{g} R_{g}}{I-I_{g}}=\frac{\frac{I}{10} \times G}{I-\frac{I}{10}}=\frac{G}{9}$
$I_{g}=10 \% \text { of } I=\frac{I}{10}$
Resistance of galvanometer, $R_{g}=G$
Shunt resistance, $R_{S}=\frac{I_{g} R_{g}}{I-I_{g}}=\frac{\frac{I}{10} \times G}{I-\frac{I}{10}}=\frac{G}{9}$
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