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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
$\frac{500}{499}G$
Detailed Solution
Current through the galvanometer $= \frac{2I}{1000}$; through the shunt $= \frac{998I}{1000}$
$\left(\frac{2I}{1000}\right)G = \left(\frac{998I}{1000}\right)S \Rightarrow S = \frac{G}{499}$
Resistance of ammeter $R = \frac{SG}{S + G} = \frac{\frac{G}{499}\cdot G}{\frac{G}{499} + G} = \frac{G}{500}$