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(C) Let the total current be $I$ and the resistance of the galvanometer be $G$.The current through the galvanometer is $I_G = 0.2\% 	ext{ of } I = \f

Vedclass · Accepted Answer

$\frac{1}{500} G \, \Omega$

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(C) Let the total current be $I$ and the resistance of the galvanometer be $G$.The current through the galvanometer is $I_G = 0.2\% 	ext{ of } I = \frac{0.2}{100} I = \frac{1}{500} I$.The current through the shunt resistance $S$ is $I_S = I - I_G = I - \frac{1}{500} I = \frac{499}{500} I$.Since the galvanometer and shunt are in parallel, the potential difference across them is the same:$I_G G = I_S S$$\left( \frac{1}{500} I ight) G = \left( \frac{499}{500} I ight) S$$S = \frac{G}{499}$.The resistance of the ammeter $R_A$ is the equivalent resistance of the parallel combination of $G$ and $S$:$\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S} = \frac{1}{G} + \frac{499}{G} = \frac{500}{G}$.Therefore, $R_A = \frac{G}{500}$.

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

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