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

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

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$\frac{G}{19}$

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(B) Let the total current in the circuit be $I$ and the current through the galvanometer be $I_g$.Given that $I_g = 5 \% 	ext{ of } I = 0.05 I$.The shunt resistance $S$ is connected in parallel to the galvanometer $G$.Using the current division rule,the current through the galvanometer is given by $I_g = I 	imes \frac{S}{S+G}$.Substituting the values: $0.05 I = I 	imes \frac{S}{S+G}$.$0.05 = \frac{S}{S+G} \implies 0.05(S+G) = S$.$0.05 S + 0.05 G = S$.$0.05 G = 0.95 S$.$S = \frac{0.05 G}{0.95} = \frac{G}{19}$.

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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