A galvanometer of resistance G Omega is shun | vedclass

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Question

(A) Let the initial resistance of the galvanometer be $G$. When a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent r

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the initial resistance of the galvanometer be $G$. When a shunt resistance $S$ is connected in parallel with the galvanometer,the equivalent resistance of the combination is $R_{eq} = \frac{G 	imes S}{G+S}$.To keep the main current unchanged,the total resistance of the circuit must remain equal to the initial resistance $G$. Let a resistance $R$ be connected in series with this parallel combination.Therefore,the total resistance is $R_{total} = R + \frac{G 	imes S}{G+S}$.Setting $R_{total} = G$,we get:$G = R + \frac{GS}{G+S}$$R = G - \frac{GS}{G+S}$$R = \frac{G(G+S) - GS}{G+S}$$R = \frac{G^2 + GS - GS}{G+S}$$R = \frac{G^2}{G+S}$Thus,the required series resistance is $\frac{G^2}{S+G}$.Hence,the correct option is $A$.

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

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