A 72 ; Omega galvanometer is shunted by a res | vedclass

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(B) Let $G = 72 \; \Omega$ be the resistance of the galvanometer and $S = 8 \; \Omega$ be the shunt resistance.The current $I$ divides into $I_g$ (cur

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$10$

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(B) Let $G = 72 \; \Omega$ be the resistance of the galvanometer and $S = 8 \; \Omega$ be the shunt resistance.The current $I$ divides into $I_g$ (current through the galvanometer) and $I_s$ (current through the shunt).According to the principle of parallel circuits,the voltage across the galvanometer and the shunt is the same: $I_g G = I_s S$.Since $I = I_g + I_s$,we have $I_s = I - I_g$.Substituting this into the equation: $I_g G = (I - I_g) S$.Rearranging to find the ratio $\frac{I_g}{I}$: $I_g G = I S - I_g S \Rightarrow I_g (G + S) = I S$.Therefore,$\frac{I_g}{I} = \frac{S}{G + S}$.Substituting the values: $\frac{I_g}{I} = \frac{8}{72 + 8} = \frac{8}{80} = \frac{1}{10}$.The percentage of the total current passing through the galvanometer is $\frac{I_g}{I} 	imes 100 = \frac{1}{10} 	imes 100 = 10 \%$.

$A$ $72 \; \Omega$ galvanometer is shunted by a resistance of $8 \; \Omega$. The percentage of the total current which passes through the galvanometer is $.....$

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