A galvanometer has 36,Omega resistance. If a | vedclass

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(D) Let $G = 36\,\Omega$ be the resistance of the galvanometer and $S = 4\,\Omega$ be the resistance of the shunt.When a shunt is connected in paralle

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$\frac{9}{10}$

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(D) Let $G = 36\,\Omega$ be the resistance of the galvanometer and $S = 4\,\Omega$ be the resistance of the shunt.When a shunt is connected in parallel to a galvanometer,the current $I$ divides into two parts: $I_g$ through the galvanometer and $I_s$ through the shunt.Since they are in parallel,the potential difference across both is the same: $I_g G = I_s S$.Therefore,the ratio of currents is $\frac{I_s}{I_g} = \frac{G}{S} = \frac{36}{4} = 9$.This means $I_s = 9 I_g$.The total current is $I = I_s + I_g = 9 I_g + I_g = 10 I_g$.The fraction of current passing through the shunt is $\frac{I_s}{I} = \frac{9 I_g}{10 I_g} = \frac{9}{10}$.

$A$ galvanometer has $36\,\Omega$ resistance. If a $4\,\Omega$ shunt is added to this,the fraction of current that passes through the shunt is

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