In an ammeter $0.2\%$ of main current passes through the galvanometer. If resistance of galvanometer is $G,$ the resistance of ammeter will be
AIPMT 2014, Medium
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Here, resistance of the galvanometer $=G$
Current through the galvanometer,
$I_{G}=0.2 \% \text { of } I=\frac{0.2}{100} I=\frac{1}{500} I$
$\therefore$ Current through the shunt,
$I_{S}=I-I_{G}=I-\frac{1}{500} I=\frac{499}{500} I$
As shunt and galvanometer are in parallel
$\therefore \quad I_{G} G=I_{S} S$
$\left(\frac{1}{500} I\right) G=\left(\frac{499}{500}\right) S \text { or } S=\frac{G}{499}$
Resistance of the ammeter $R_{A}$ is
${\frac{1}{R_{A}}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{1}{G}=\frac{500}{G}}$
${R_{A}=\frac{1}{500} \,G}$
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