A galvanometer of resistance G is connected i | vedclass

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(A) Let the original resistance of the circuit be $G$. The current in the circuit is $I = \frac{V}{G}$.When a resistance $R$ is connected in series wi

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$\frac{G^2}{R} + G$

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(A) Let the original resistance of the circuit be $G$. The current in the circuit is $I = \frac{V}{G}$.When a resistance $R$ is connected in series with the galvanometer,the total resistance of this branch becomes $(G + R)$.To keep the main current unchanged,we connect a shunt resistance $S$ in parallel with the $(G + R)$ combination such that the equivalent resistance of this parallel network remains equal to the original resistance $G$.Therefore,the equivalent resistance $R_{eq} = \frac{(G + R)S}{(G + R) + S} = G$.Cross-multiplying gives $(G + R)S = G(G + R + S)$.Expanding the terms: $GS + RS = G^2 + GR + GS$.Subtracting $GS$ from both sides: $RS = G^2 + GR$.Solving for $S$: $S = \frac{G^2 + GR}{R} = \frac{G^2}{R} + G$.

$A$ galvanometer of resistance $G$ is connected in a circuit. Now,a resistance $R$ is connected in series with the galvanometer. To keep the main current in the circuit unchanged,the resistance $S$ to be put in parallel with the series combination of $G$ and $R$ is:

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