In an ammeter,0.25 % of the main current pass | vedclass

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(B) Let $I$ be the main current and $I_g$ be the current through the galvanometer.Given that $I_g = 0.25 \% 	ext{ of } I = \frac{0.25}{100} I = \frac

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$\frac{1}{400} G$

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(B) Let $I$ be the main current and $I_g$ be the current through the galvanometer.Given that $I_g = 0.25 \% 	ext{ of } I = \frac{0.25}{100} I = \frac{1}{400} I$.The current through the shunt resistance $S$ is $I_s = I - I_g = I - \frac{1}{400} I = \frac{399}{400} I$.Since the galvanometer and shunt are in parallel,the potential difference across them is equal:$I_g G = I_s S$$\left( \frac{1}{400} I ight) G = \left( \frac{399}{400} I ight) S$$S = \frac{G}{399}$.The total resistance $R$ of the ammeter is the equivalent resistance of $G$ and $S$ in parallel:$R = \frac{G S}{G + S} = \frac{G \left( \frac{G}{399} ight)}{G + \frac{G}{399}} = \frac{\frac{G^2}{399}}{\frac{400 G}{399}} = \frac{G}{400}$.

In an ammeter,$0.25 \%$ 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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