We have a galvanometer of resistance 25,Omega | vedclass

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(A) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.Given: $G = 25\,\Omega$ and $S = 2.5\,\Omega$.Let $I$ be the total c

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$\frac{I_g}{I} = \frac{1}{11}$

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(A) Let $G$ be the resistance of the galvanometer and $S$ be the shunt resistance.Given: $G = 25\,\Omega$ and $S = 2.5\,\Omega$.Let $I$ be the total current and $I_g$ be the current flowing through the galvanometer.According to the current divider rule,the current through the galvanometer is given by:$I_g = I 	imes \frac{S}{G + S}$Therefore,the ratio of the current through the galvanometer to the total current is:$\frac{I_g}{I} = \frac{S}{G + S}$Substituting the given values:$\frac{I_g}{I} = \frac{2.5}{25 + 2.5} = \frac{2.5}{27.5}$$\frac{I_g}{I} = \frac{25}{275} = \frac{1}{11}$

We have a galvanometer of resistance $25\,\Omega$. It is shunted by a $2.5\,\Omega$ wire. The part of the total current that flows through the galvanometer is given as:

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