Only 4 % of the total current in the circuit | 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 $I_g = 4\% 	ext{ of } I = 0.04 I$.Theref

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$\frac{G}{24}$

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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 $I_g = 4\% 	ext{ of } I = 0.04 I$.Therefore,the ratio $\frac{I}{I_g} = \frac{1}{0.04} = 25$.The shunt resistance $S$ connected in parallel to a galvanometer of resistance $G$ is given by the formula:$S = \frac{I_g 	imes G}{I - I_g}$.Dividing the numerator and denominator by $I_g$,we get:$S = \frac{G}{\frac{I}{I_g} - 1}$.Substituting the value $\frac{I}{I_g} = 25$:$S = \frac{G}{25 - 1} = \frac{G}{24}$.

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is $G$,then the shunt resistance connected to the galvanometer is

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