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
$\frac{1}{R_A}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{\frac{1}{G}}{499}=\frac{500}{G}$
$R_A=\frac{1}{500}G$