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In an ammeter $0.2 \%$ of main current passes through the galvanometer. If resistance of galvanometer is $\mathbf{G}$, the resistance ammeter will be-Option A $\quad \frac{1}{499}$ GOption B $\quad \frac{499}{500} \mathrm{G}$Option C $\quad \frac{1}{500}$ GOption D $\quad \frac{500}{499} \mathrm{G}$
In an ammeter $0.2 \%$ of main current passes through the galvanometer. If resistance of galvanometer is $\mathbf{G}$, the resistance ammeter will be-
Option A $\quad \frac{1}{499}$ G
Option B $\quad \frac{499}{500} \mathrm{G}$
Option C $\quad \frac{1}{500}$ G
Option D $\quad \frac{500}{499} \mathrm{G}$
Physics
In an ammeter $0.2 \%$ of main current passes through the galvanometer. If resistance of galvanometer is $\mathbf{G}$, the resistance ammeter will be-
Option A $\quad \frac{1}{499}$ G
Option B $\quad \frac{499}{500} \mathrm{G}$
Option C $\quad \frac{1}{500}$ G
Option D $\quad \frac{500}{499} \mathrm{G}$

The correct option is C

Potential drop is same for both, so on equating,

$0.002 \mathrm{I}_{0} \times \mathrm{G}=0.998 \mathrm{I}_{0} \times \mathrm{S}$

$\left(\frac{21}{1000}\right) \mathrm{G}=\left(\frac{9981}{1000}\right) \mathrm{S}$

$\Rightarrow \mathrm{S}=\frac{\mathrm{G}}{499}$

Total resistan ce of ammeter R can be calculated as,

$\mathrm{R}=\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}=\frac{\left(\frac{\mathrm{G}}{499}\right) \mathrm{F}}{\left(\frac{\mathrm{G}}{499}\right)+\mathrm{G}}=\frac{\mathrm{G}}{500}$

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