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

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$\frac{1}{500} G$

$\frac{1}{499} G$

$\frac{500}{499} G$

$\frac{499}{500} G$

answer is C.

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Detailed Solution

$S (99.8) = G (0.2) \Rightarrow S = \frac{G}{499}$
Resistance of ammeter $= \frac{G S}{G + S}$
$= \frac{(\frac{G}{499}) G}{G / 499 + G} = \frac{G}{500}$

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