Question

In an ammeter $0.2 \%$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of ammeter will be

• A. $\frac{1}{499}G$
• B. $\frac{499}{500}G$
• C. $\frac{1}{500}G$
• D. $\frac{500}{499}G$

Answer 1

Answer: (C)

Here, resistance of the galvanometer $=G$
Current through the galvanometer,
$I_{G}=0.2 \%$ 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
$\therefore I_{G} G=I_{s} S$
$\left(\frac{1}{500} I\right) G=\left(\frac{499}{500}\right) S \text { or } S=\frac{G}{499}$
Resistance of the ammeter $R_{A}$ is
$\frac{1}{R_{A}}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{\frac{1}{G}}{499}=\frac{500}{G}$
$R_{A}=\frac{1}{500} G$