In Electroboom's "How to turn an LED On" video he mentions that (ideal) capacitors do not consume real power. I was trying to reason through this and would like to gauge the accuracy of my mental model.

Preface: I am not an engineer

For an AC power source connected to a resistor, the resistor will get hot and dissipate heat; power=voltage * current, or electrons flowing through the resistor, and the resistor gets hot.

For a ideal capacitor, could we say that no electrons actually flow through the capacitor so no heat is generated, rather a rising voltage charges the capacitor (work done on the capacitor), but when the negative voltage cycle comes around the capacitor discharges (capacitor does work on the surroundings) so there is no net work done and thus no real power consumption? In that case, would the general process of only charging a capacitor consume real power?

  • \$\begingroup\$ For an AC circuit, the capacitor is always discharged immediately after it is charged, so there is no net storage of energy. In a DC circuit, charging the capacitor does consume real power (which you eventually get back when you discharge it). \$\endgroup\$ Oct 17, 2022 at 2:35
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    \$\begingroup\$ The capacitor stores energy. The resistor converts it to heat. \$\endgroup\$
    – user57037
    Oct 17, 2022 at 2:43
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    \$\begingroup\$ Electrons DO flow through the capacitor. Or, I mean, for every electron that enters on one side, another electron exits on the other side. \$\endgroup\$
    – user57037
    Oct 17, 2022 at 2:52
  • \$\begingroup\$ The discussion I see about charging and recovering energy is misleading. The process by which you electronically charge up a capacitor simply isn't a reversible process. (It is reversible if and only if it can be reversed by making only an infinitesimal change in conditions -- which almost never applies in electronics.) It's also not slow in any sense. So there will be a change in entropy in the larger system and therefore non-recoverable energy must be expended. You may want to read some of this, as well. \$\endgroup\$
    – jonk
    Oct 17, 2022 at 3:11
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    \$\begingroup\$ @jonk the question is about an ideal cap and an ideal AC source. You are trying to make it into some esoteric thing that I don't even actually understand. But putting two capacitors together with or without a resistor between them is not the same as driving a cap with an AC source. I = Vo + C dv/dt. There is a whole class of questions to do with ideal components. But now here you are demanding I satisfy some strange burden of proof. Can you not just see the steady state situation with an AC voltage and an ideal capacitor? \$\endgroup\$
    – user57037
    Oct 17, 2022 at 5:56

3 Answers 3


Yes - charging a capacitor (or inductor) consumes energy; conversely discharging it delivers energy from the component to the rest of the circuit.

In an AC circuit or analysis, this cyclic exchange of energy (power) is called 'apparent power'.

  • \$\begingroup\$ Fantastic answer, looking up apparent power on led me to wikipedia: "The portion of instantaneous power that results in no net transfer of energy but instead oscillates between the source and load in each cycle due to stored energy, is known as instantaneous reactive power...Apparent power is the product of the RMS values of voltage and current. Apparent power is taken into account when designing and operating power systems, because although the current associated with reactive power does no work at the load, it still must be supplied by the power source." I am now enlightened! \$\endgroup\$ Oct 17, 2022 at 3:41
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    \$\begingroup\$ @YousifAlniemi, That Wikipedia article may be misleading you. It talks about "Apparent Power" which is a topic of importance in designing power distribution systems so that they can handle the necessary currents and voltages. But if a purely reactive load (e.g., a capacitor) consumes no net power, then the source delivers no net power... to the load. There is a whole other issue though: The current flowing back and forth in the transmission line between the source and load causes power to be lost in the line because the line has electrical resistance. \$\endgroup\$ Oct 17, 2022 at 14:08
  • \$\begingroup\$ Interesting! Is that where things like the skin effect comes in? \$\endgroup\$ Oct 17, 2022 at 18:03
  • \$\begingroup\$ No, skin effect is completely different -- it is because the magnetic fields generated by (any) currents flowing in a wire cause the current to crowd to the outer surface of the wire, thus the RESISTIVE losses are higher than would normally be expected. \$\endgroup\$
    – jp314
    Oct 17, 2022 at 22:38

capacitors are a bit like springs, you can do work on them, but the work is stored and will be released as the force is reduced.

By pushing charge through capacitors you do work on them, but they will push the charge back the other way when the electromotive force is reduced, so (ideally) all the work done to charge them is returned when they discharge.

In that case, would the general process of only charging a capacitor consume real power?

Yes it does, free energy is a myth.

"the general process of charging a capacitor" is not really suited to complex power analysis as it is not a cyclical activity. Use the time domain instead, integrate work over time.


Considering the situation where an ideal capacitor is connected to an ideal AC source, I think the most intuitive way to look at it is that a charged capacitor temporarily stores energy.

With an AC source, during part of the cycle energy is transferred into the capacitor where it is stored. During the other part of the cycle, energy is delivered from the capacitor to the AC source. The energy transferred to the capacitor is equal to the energy transferred back to the AC source. So the net amount of energy, tracked over a full cycle, is zero. This holds true for an ideal capacitor connected to an ideal AC voltage source.

Note that from a circuit analysis perspective, AC current flows through the capacitor. The current flowing into the capacitor is exactly equal and opposite to the current flowing out. Sometimes people talk about capacitors "storing charge." But they don't really do that, not in the way beginners often think. They DO store energy though.

This idea, that capacitors always store energy and return it, it does not necessarily apply to other cases (like the two capacitors connected by a switch). But it definitely applies to ideal capacitors stimulated by an AC voltage. And it applies to DC-DC converters which typically use an output filter formed by an inductor and a capacitor.


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