Q. In conversion of a galvanometer to ammeter if $2 \%$ of the main current is to be passed through the galvanometer of resistance $G$, the resistance of shunt required is :

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A

$\frac{G}{49}$

B

$\frac{G}{50}$

C

$49 \,G$

D

$50\, G$

Solution:

For the currents through galvanometer and shunt resistance we have
$\frac{I_{G}}{I_{S}}=\frac{S}{G}$
$\Rightarrow S=\left(\frac{I_{G}}{I_{S}}\right) G$
$\Rightarrow S=\left(\frac{2}{98}\right) G=\frac{G}{49}$

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