A galvanometer of resistance G is shunted by | vedclass

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(A) Let the resistance of the galvanometer be $G$ and the shunt resistance be $S = 5 \Omega$.To keep the main current unchanged,the equivalent resista

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$\frac{G^2}{5+G}$

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(A) Let the resistance of the galvanometer be $G$ and the shunt resistance be $S = 5 \Omega$.To keep the main current unchanged,the equivalent resistance of the circuit must remain equal to the original resistance of the galvanometer,$G$.Let $R$ be the resistance connected in series with the galvanometer.The parallel combination of the galvanometer and the shunt resistance is in series with $R$.Thus,the equivalent resistance is $R_{eq} = R + \frac{G \cdot S}{G + S}$.Setting $R_{eq} = G$,we get:$G = R + \frac{G \cdot S}{G + S}$$R = G - \frac{G \cdot S}{G + S} = \frac{G(G + S) - GS}{G + S} = \frac{G^2 + GS - GS}{G + S} = \frac{G^2}{G + S}$.Substituting $S = 5 \Omega$,the required series resistance is $R = \frac{G^2}{G + 5}$.

$A$ galvanometer of resistance $G$ is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged,the resistance to be put in series with the same galvanometer is

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