Is it possible to use the stored energy while the capacitor is being charged? In other words, use a capacitor almost as a conductor?

My gut says the answer is probably no (and my own experiments so far seems to be justifying my hunch) but if so, I'm very curious on what's actually happening inside the capacitor, in terms of what exactly is blocking the stored energy to not escape (discharge) until the capacitor is fully charged?

  • \$\begingroup\$ What for? The conductors it's connected to will already do a more than capable job of that. \$\endgroup\$ May 23, 2018 at 20:42
  • \$\begingroup\$ I suppose it would be dependant on the overall flow in the circuit. Current can't flow both ways at the same time in one conductor (wire). So if the load demands more than the supply can handle a capacitor will discharge but if the supply could handle the load then the capacitor charges ? \$\endgroup\$
    – SRR
    May 23, 2018 at 20:45
  • 3
    \$\begingroup\$ Discharging current happens in the opposite direction to charging current, and current can only flow in one direction! \$\endgroup\$
    – pjc50
    May 23, 2018 at 21:13
  • 1
    \$\begingroup\$ Can a glass of water be filled when you are drinking out of it and pouring more water into the glass simultaneously? \$\endgroup\$ May 23, 2018 at 23:11
  • \$\begingroup\$ @HarrySvensson not from the same "terminal", but if you are pouring water on one end and drinking from the other, it is quite possible! \$\endgroup\$
    – Peter Varo
    May 24, 2018 at 7:07

2 Answers 2


No, the charge on a capacitor is increasing (charging), decreasing (discharging) or remaining the same. There are no other possible states (assuming an ideal capacitor with no leakage).

When the capacitor is charging or discharging, there is a potential difference between the two terminals and apparent current flow.

This means a capacitor will appear to conduct an AC signal (above a critical frequency determined by the value of the capacitor and the impedance of the load/source) and will block a DC signal when used in series.

  • \$\begingroup\$ And can you elaborate on the second part of the question -- if this is the case, which I somewhat already suspected -- why is this exactly happening? What is the phenomenon behind it? \$\endgroup\$
    – Peter Varo
    May 23, 2018 at 20:43
  • \$\begingroup\$ Thought it can transistion from one state to another very fast. \$\endgroup\$
    – user105652
    May 23, 2018 at 20:44

No a passive part is either an insulator (dielectric) or a conductor with very weak character flaws of the other. We may ignore inductance and active parts.

When doing what you think it may be alternating pulse charging and then slow discharge in order to reduce voltage fluctuations or if near DC only getting the net difference. Like a weak laptop charger (say 30W instead of 65W)!that draws full power from the battery when running max power actually gets full power from the charger and only surplus from the battery. However chargers generally are sized to operate DC-DC converters and charge batteries with what’s left over at a slower rate.

Why ? Because the battery has capacitor characteristics so current (rate of charge flow) only flows in one direction at any time.

Conductors are somewhat different. You can have impedance controlled frequencies sent in opposite directions called full duplex modems or as in telephones control voltage in one direction and current in the other using a special hybrid transformer that isolates each direction.


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