When the switch K is closed at t = 0,what can | vedclass

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(C) At $t = 0$,the capacitor acts as a short circuit (uncharged). The total resistance in the circuit is $1000\,\Omega$. Thus,the current $I = \frac{V

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At $t = 0$,$I = 2\,mA$ and as time passes,it becomes $1\,mA$

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(C) At $t = 0$,the capacitor acts as a short circuit (uncharged). The total resistance in the circuit is $1000\,\Omega$. Thus,the current $I = \frac{V}{R} = \frac{2\,V}{1000\,\Omega} = 2\,mA$.As $t 	o \infty$,the capacitor becomes fully charged and acts as an open circuit. The current now flows through the series combination of the two $1000\,\Omega$ resistors. The total resistance is $1000\,\Omega + 1000\,\Omega = 2000\,\Omega$. Thus,the current $I = \frac{V}{R_{total}} = \frac{2\,V}{2000\,\Omega} = 1\,mA$.

When the switch $K$ is closed at $t = 0$,what can be said about the current passing through the resistor $AB$ in the given circuit?

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