- why does the voltage of the capacitor eventually go to 0 when discharging the capacitor? I heard that's because "current starts flowing when discharging", but how exactly does that lead to V going down? I know that I = C * dV/dt, but that doesn't seem to help me understand why V goes down (which is equivalent to dV/dt < 0).
- Alternatively, I heard this is due to Q = VC; discharging capacitor means reducing the charge on capacitor, so V must decrease as well. Okay, in that case, where does the charge go?
- should the capacitor be connected to the ground node in both directions if it's to be discharged?I saw a diagram/circuit with a short circuit and pre-charged capacitor only (), which would apparently lead to V decreasing, but the circuit didn't have any ground nodes labelled. Is it possible to discharge a capacitor without having a ground node?