When a shunt of 4,Omega is attached to a galv | vedclass

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(C) Let $G$ be the resistance of the galvanometer and $I$ be the total current. When a shunt $S_1 = 4\,\Omega$ is connected,the current through the ga

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$\frac{I}{9}$

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(C) Let $G$ be the resistance of the galvanometer and $I$ be the total current. When a shunt $S_1 = 4\,\Omega$ is connected,the current through the galvanometer becomes $I_g = \frac{I}{5}$.Using the principle of parallel circuits,the potential difference across the galvanometer and the shunt is equal:$I_g 	imes G = (I - I_g) 	imes S_1$$\frac{I}{5} 	imes G = (I - \frac{I}{5}) 	imes 4$$\frac{I}{5} 	imes G = \frac{4I}{5} 	imes 4$$G = 16\,\Omega$Now,an additional shunt of $4\,\Omega$ is connected in parallel to the first shunt. The equivalent resistance of the two shunts in parallel is:$S_{eq} = \frac{4 	imes 4}{4 + 4} = 2\,\Omega$Let the new current through the galvanometer be $I'$. The potential difference remains the same:$I' 	imes G = (I - I') 	imes S_{eq}$$I' 	imes 16 = (I - I') 	imes 2$$16I' = 2I - 2I'$$18I' = 2I$$I' = \frac{I}{9}$

When a shunt of $4\,\Omega$ is attached to a galvanometer,the deflection reduces to $\frac{1}{5}$th of its initial value. If an additional shunt of $4\,\Omega$ is attached in parallel to the first,what will be the new deflection?

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