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)

$\frac{1}{500}G$

Answer 2

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
$\therefore I_G G=I_s S$
$\big(\frac{1}{500}I\big)G=\big(\frac{499}{500}\big)S$ or $S=\frac{G}{499}$
Resistance of the ammeter $R_A$ is
$\frac{1}{R_A}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{\frac{1}{G}}{499}=\frac{500}{G}$
$R_A=\frac{1}{500}G$