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Say we have a battery at a fixed voltage of $V$ charging a capacitor with capacitance $C$.

When the capacitor is fully charged, the energy stored in it will be $\frac{1}{2}QV$, where $V$ is the final voltage of the capacitor (and the voltage of the battery).

However, I'm a little confused as to the work the battery does...

To me, it seems that if the battery pushed $Q$ onto the capacitor, and the voltage of the battery is $V$, it must've done $QV$ of work, which means that the energy stored in the capacitor only accounts for half of it.

Where did the rest of the work the battery did go?

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    $\begingroup$ If you use the search engine for this website you will find a more detailed explanation of the fact that the rest of the energy is dissipated as heat in the resistance which is in the circuit (and electromagnetic radiation which is emitted because the the accelerating unbound mobile charges in the circuit). $\endgroup$
    – Farcher
    Commented Mar 5, 2020 at 23:02
  • $\begingroup$ @Farcher thank you. Could you pherhaps link that answer? I can't find it....Thanks again. $\endgroup$
    – joshuaronis
    Commented Mar 5, 2020 at 23:12
  • $\begingroup$ @JoshuaRonis Check this out: en.wikipedia.org/wiki/Two_capacitor_paradox Its the same idea that Farcher pointed out, but using one capacitor to charge another. $\endgroup$
    – Bob D
    Commented Mar 5, 2020 at 23:23
  • $\begingroup$ A very detailed paper A capacitor paradox by Kirk T McDonald and from this website Energy lost in capacitors. $\endgroup$
    – Farcher
    Commented Mar 5, 2020 at 23:33
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    $\begingroup$ Does this answer your question? What happens to half of the energy in a circuit with a capacitor? $\endgroup$
    – BowlOfRed
    Commented Mar 6, 2020 at 6:48

1 Answer 1

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When the battery 'charges' the capacitor, it overshoots by a factor of $\frac 12 CV$. This produced a ringing signal which is absorbed by the wire. Eventually, this extra energy is dissipated in the impedance of the circuit. Hence the extra energy from the battery is lost as heat.

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