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Imagine we have the circuit below:

Charging the capacitor

When the switch is connected to (a), we will be charging the capacitor. After the capacitor is fully charged, the left side of capacitor will be at 5V and right side at 0V. Then, we turn the switch to connect to (b):

Grounded capacitors

As we can see since there is a voltage across $R_1$, there will be a current from (b) to ground (i.e. to the negative pole of the voltage source). However, by KCL, there would also be a current flowing through $R_2$, which has no voltage across it. This seems to be a paradox to me.

What I think could be the reason for this is the turning of the switch, which cannot happen in an instant. I think something similar to what is described in this question is taking place but I am not sure.

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    $\begingroup$ The capacitor has charge on it. You have now shorted the two capacitor leads together (through ground). Why do you think no charge flows? $\endgroup$
    – Jon Custer
    Commented Sep 2, 2022 at 0:47

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After the switch, you've taken the power source and bottom left resister out of the equation. Assuming both leads are connected to a common ground, the capacitor will discharge through the two resisters. Just because something is labeled ground does not mean it is in all configurations.

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