In an ammeter,5 % of the main current passes | vedclass

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(A) Let the main current be $I$. The current through the galvanometer is $I_g = 5 \% 	ext{ of } I = \frac{5}{100} I = \frac{I}{20}$.The current throu

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$\frac{G}{20}$

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(A) Let the main current be $I$. The current through the galvanometer is $I_g = 5 \% 	ext{ of } I = \frac{5}{100} I = \frac{I}{20}$.The current through the shunt resistance $S$ is $I_s = I - I_g = I - \frac{I}{20} = \frac{19I}{20}$.Since the galvanometer and shunt are in parallel,the potential difference across them is equal: $I_g G = I_s S$.Substituting the values: $(\frac{I}{20}) G = (\frac{19I}{20}) S$.Solving for $S$: $S = \frac{G}{19}$.The total resistance of the ammeter $R_A$ is the equivalent resistance of $G$ and $S$ in parallel: $R_A = \frac{G \cdot S}{G + S}$.Substituting $S = \frac{G}{19}$: $R_A = \frac{G \cdot (G/19)}{G + (G/19)} = \frac{G^2/19}{20G/19} = \frac{G}{20}$.

In an ammeter,$5 \%$ 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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