When the key K is pressed at time t = 0, whic | vedclass

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(D) At time $t = 0$, the capacitor acts as a short circuit (zero resistance). The total resistance in the circuit is $1000 \, \Omega$. Thus, the curre

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At $t = 0$, $I = 2 \,mA$ and with time it goes to $1 \,mA$

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(D) At time $t = 0$, the capacitor acts as a short circuit (zero resistance). The total resistance in the circuit is $1000 \, \Omega$. Thus, the current $I = \frac{2 \,V}{1000 \, \Omega} = 2 \,mA$.As time passes, the capacitor charges up. When it is fully charged, it acts as an open circuit. The current then flows through the series combination of the $1000 \, \Omega$ resistor and the $1000 \, \Omega$ resistor. The total resistance becomes $2000 \, \Omega$. Thus, the steady-state current $I = \frac{2 \,V}{2000 \, \Omega} = 1 \,mA$. Therefore, the current decreases from $2 \,mA$ to $1 \,mA$ over time.

When the key $K$ is pressed at time $t = 0$, which of the following statements about the current $I$ in the resistor $AB$ of the given circuit is true?

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