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Can someone explain how the capacitor behaves (how it charges/discharges) when the switch is opened/closed?

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Assuming you start with 0V across the capacitor. When the switch closes, the capacitor will charge through the 1k resistor, and will have an RC time constant using the C and the 1k. Eventually the capacitor will charge to 5V.

When the switch opens, the capacitor will discharge through the 1k and 10k resistors with an RC time constant using the C and the 11k, until it eventually fully discharges.

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    \$\begingroup\$ a capacitor is only charged, by definition, so long as there is a voltage difference across its terminals. In order for it to have no voltage difference across it, which is the obvious steady state condition of this circuit, because a capacitor looks like an open circuit to DC, it must discharge. The mechanics of this, or a way to think about it, is that the charge on one plate must migrate to the other plate through a conductive path, in order to balance the stored charge, and that action is "discharging". \$\endgroup\$
    – vicatcu
    Nov 2, 2019 at 22:54

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