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(C) When the capacitor is fully charged,it acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit si

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(C) When the capacitor is fully charged,it acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit simplifies to a series combination of the $1\, k\Omega$ resistor and the $2\, k\Omega$ resistor connected to the $9\, V$ battery.The equivalent resistance of the circuit is $R_{eq} = 1\, k\Omega + 2\, k\Omega = 3\, k\Omega$.Using Ohm's law,the current $I$ drawn from the cell is:$I = \frac{V}{R_{eq}} = \frac{9\, V}{3\, k\Omega} = 3\, mA$.

When the capacitor is fully charged,find the current drawn from the $9\, V$ cell (in $mA$).

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