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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?

Qmechanic
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joshuaronis
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1 Answers1

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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.

Sam
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