Where is energy dissipated on charging a capacitor?

Question

The energy lost on charging a capacitor can be easily found from the change in energy of the components of the circuit and the energy supplied by the battery.

On charging a capacitor I know that the energy loss appears as heat in the internal resistance of the battery and the wires.

But what if I take (Purely theoretically) a battery with 0 internal resistance and wires with 0 resistance?

I can't see why the capacitor won't be charged, so from the calculation there must be energy dissipated in the circuit.

So in that case, where is the heat dissipated?

Or is this ideal case faulty?

Thanks. But I couldn't find an appealing answer that told exactly what I needed — Vamsi Krishna, Jul 28 '20 at 08:27
The apparent loss lies in the over simplified formulation that leads to the result (Q^2)/2C. Ideally, the energy stored in a capacitor at full charge is exactly (Q^2)/C, where is the charge stored. — Madhubala, Jul 28 '20 at 10:22
Part of the intuitive part that goes into setting up the integral is that getting the first element of charge dq onto the capacitor plates takes much less work because most of the battery voltage is dropping across the resistance R and only a tiny energy dU = dqV is stored on the capacitor. Proceeding with the integral, which takes a quadratic form in q, gives a summed energy on the capacitor Q2/2C = CVb2/2 = QVb/2 where the Vb here is the battery voltage. — Madhubala, Jul 28 '20 at 10:27
energy supplied by the battery is E = CVb2, but only half that is on the capacitor - the other half has been lost to heat, or in the extremely low charging resistance case, to heat and electromagnetic energy. Because the inductance of the circuit is not 0, there will be osicllations and the energy will be dissipated as electromagnetic waves. — Madhubala, Jul 28 '20 at 10:29
I'm voting to close because the question was answered previously (and more clearly) at "What happens to half of the energy in a circuit with a capacitor?" and "Energy conservation when capacitor charges in a zero resistance circuit". Also see here. — Chemomechanics, Jul 28 '20 at 20:25

score 2 · Accepted Answer · answered Jul 28 '20 at 07:46

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When charging a capacitor the energy from the battery is transferred to the capacitor. If the wires have resistance, some of this energy is lost, i.e. dissipated. If the resistance is zero, there are no losses - but there is still the energy transfer from the battery to the capacitor.

answered Jul 28 '20 at 07:46

Roger V.

58,522

But charge supplied by the battery is QV$^2$(charge on capacitor, voltage of battery). And energy change in capacitor is QV$^2$/2. Where does this remaining QV$^2$/2 energy go in that ideal circuit? – Vamsi Krishna Jul 28 '20 at 07:52
How do you estimate the energy supplied by the buttery? – Roger V. Jul 28 '20 at 08:00
It moves the charge Q=CV to a potential of V(Or creates a potential difference V). So energy is Q×V which is CV$^2$ – Vamsi Krishna Jul 28 '20 at 08:22
@VamsiKrishna Indeed, but it doesn't move it in a single shot - it will have $1/2$ in exactly the same was as the capacitor charging energy, which also comes from moving charge $Q$. – Roger V. Jul 28 '20 at 08:42
Could you rephrase the "1/2 in exactly the same was as the capacitor charging energy" please? I'm sorry I couldn't understand it. – Vamsi Krishna Jul 28 '20 at 08:58
Do you know how the $QV^2/2$ expression for the capacitor energy is derived? – Roger V. Jul 28 '20 at 09:00
Yes it's found by integrating V.dq where V is q/c. (Work done in creating a charge separation of dq) – Vamsi Krishna Jul 28 '20 at 09:14
It is the same for the battery, except that it does work in the opposite direction. – Roger V. Jul 28 '20 at 09:21
But both don't do equal works right? There is a $QV^2/2$ difference that occurs. In a resistance it's produced as heat. What if I take this ideal case of 0 resistance? There should still be energy loss, but where? – Vamsi Krishna Jul 28 '20 at 09:31
The $QV^2/2$ difference comes from incorrect calculation. If the resistance is zero, the work is the same: the battery is doing work, the capacitor has work done on it. If there is resistance, the battery has to do more work, since not all of its work is stored as the capacitor energy. How much energy is lost depends on the resistance - it is not $CV^2/2$. – Roger V. Jul 28 '20 at 09:35
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hmm okay i got it! thanks for your time! – Vamsi Krishna Jul 28 '20 at 10:26

Where is energy dissipated on charging a capacitor?

1 Answers1