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I understand that when the switch is open, the capacitor acts like an open circuit & therefore the current in the capacitor branch would be equal to 0. When I solved the circuit using nodal analysis, I realized that the 2k ohm and the 4k ohm are in series.

How would I look at the circuit and automatically realize they're in series? Because realizing they're in series would result in a much easier answer using KVL. 5(-10) + Vc - 4(6/6) = 0

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If the capacitor is an open circuit then you can remove the capacitor from the circuit and the behavior of the circuit does not change. Once you remove the capacitor you can see that the two resistors are connected together at a single point and nothing else is connected at that point. Therefore, the current through the two resistors must be the same simply because of how they are connected. Therefore, the two resistors are in series.

By the way, the current through the capacitor is zero not because the switch is open; it is zero because the circuit is effectively a dc circuit because the switch has not changed position for an infinite amount of time before \$t=0\$. This is a mathematically convenient assumption but it is also a reasonable approximation to many real-life situations.

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