When a 12,Omega resistor is connected in para | vedclass

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(C) The deflection of a galvanometer is directly proportional to the current passing through it,i.e.,$	heta \propto I$.Initially,the current $I$ corr

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$48$

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(C) The deflection of a galvanometer is directly proportional to the current passing through it,i.e.,$	heta \propto I$.Initially,the current $I$ corresponds to $50$ divisions. When a shunt resistor $S = 12\,\Omega$ is connected in parallel,the current $I_g$ through the galvanometer corresponds to $10$ divisions.Using the current divider rule for a galvanometer with resistance $G$ and shunt $S$:$I_g = I \left( \frac{S}{S + G} ight)$Given $	heta_1 = 50$ and $	heta_2 = 10$,we have $\frac{I_g}{I} = \frac{10}{50} = \frac{1}{5}$.Substituting the values: $\frac{1}{5} = \frac{12}{12 + G}$.$12 + G = 60$.$G = 60 - 12 = 48\,\Omega$.Therefore,the resistance of the galvanometer is $48\,\Omega$.

When a $12\,\Omega$ resistor is connected in parallel with a moving coil galvanometer,its deflection reduces from $50$ divisions to $10$ divisions. The resistance of the galvanometer is ............. $\Omega$.

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