A galvanometer of resistance G is shunted wit | vedclass

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Question

(A) Let the total current be $I$ and the current through the galvanometer be $I_g$. The resistance of the galvanometer is $G$ and the shunt resistance

Vedclass · Accepted Answer

$\frac{1}{11} I$

Vedclass · Answer

(A) Let the total current be $I$ and the current through the galvanometer be $I_g$. The resistance of the galvanometer is $G$ and the shunt resistance is $S = 0.1 G$. According to the current divider rule,the current through the galvanometer is given by: $I_g = I \left( \frac{S}{S + G} \right)$ Substituting the value of $S$: $I_g = I \left( \frac{0.1 G}{0.1 G + G} \right)$ $I_g = I \left( \frac{0.1 G}{1.1 G} \right)$ $I_g = I \left( \frac{1}{11} \right)$ Therefore,the part of the total current that flows through the galvanometer is $\frac{1}{11} I$.

$A$ galvanometer of resistance $G$ is shunted with a resistance of $10 \%$ of $G$. The part of the total current that flows through the galvanometer is

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