In a galvanometer,5 % of the total current in | vedclass

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(B) Let the total current in the circuit be $I$. Given that the current through the galvanometer is $I_g = 5 \% 	ext{ of } I = 0.05 I$. The current t

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$\frac{G}{19}$

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(B) Let the total current in the circuit be $I$. Given that the current through the galvanometer is $I_g = 5 \% 	ext{ of } I = 0.05 I$. The current through the shunt resistance $S$ is $I_s = I - I_g = I - 0.05 I = 0.95 I$. Since the galvanometer and shunt are in parallel,the potential difference across them is equal: $I_g G = I_s S$ $0.05 I 	imes G = 0.95 I 	imes S$ $S = \frac{0.05 I 	imes G}{0.95 I} = \frac{5}{95} G = \frac{G}{19}$.

In a galvanometer,$5 \%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is $G$,the shunt resistance $S$ connected to the galvanometer is

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