You're searching for a mechanism, where the charging energy is lost?
As far as I remember my Electrodynamics lectures, decades ago, the answer is not obvious, but simple in the end.
The capacitor builds an electrostatic field between his plates.
Every time, you CHANGE this field, an electro-magnetic wave is emitted, as the capacitor acts as an antenna; therefore energy from the charging / discharging process, which always changes the electrostatic field, is 'lost' due to the electro magnetic wave. Therefore, to solve this problem, you have to apply Maxwells equations, also.
So what is the amount of energy lost via this phenomenon you are discussing, and is it a constant amount, or does it vary (with charging time, for example)? Because this is something you cannot see in practical electronics (at least I cannot) - you can get something like 99.9% efficiency of the capacitor with proper current-mode DC/DC charging and discharging, and that 0.1% is clearly coming from capacitor's ESR, which has a simple physical explanation (no antennas). Capacitors can be used, for example, in storing and reusing braking energy in urban transit systems, and these give great efficiency; the AC motor VFDs give the current control ("DC/DC") charging and discharging automatically.
The 50% loss described by the OP is very simple and has nothing to do with your explanation, but maybe your point works out to be important with extremely high (theoretical) currents resulting in from approaching zero ohms? So, please elaborate.
But the magnetic field created by a current doesn't necessarily generate losses unless there's another circuit magnetically coupled in which to induce a current. For example a transformer's primary with the secondary opened draws ~nil (just the losses due to alternating magnetization of the core) but with the secondary shorted the primary looks like a short too.
The 50% loss described by the OP is very simple and has nothing to do with your explanation, but maybe your point works out to be important with extremely high (theoretical) currents resulting in from approaching zero ohms? So, please elaborate.
But the magnetic field created by a current doesn't necessarily generate losses unless there's another circuit magnetically coupled in which to induce a current. For example a transformer's primary with the secondary opened draws ~nil (just the losses due to alternating magnetization of the core) but with the secondary shorted the primary looks like a short too.
It is an electromagnetic field.
If you think that cannot carry energy away from the source, then I suggest you go outside and look up, anytime day or night, preferably when it is clear.
But a coil generates a magnetic field too. Won't that energy return to the circuit when the field collapses? Unless a secondary robs it?
Hello again,
The energy in the magnetic field may be stored energy, but the energy that leaves the system is radiant energy that never returns unless it can be reflected back somehow, which usually isnt the case. With a capacitor we probably have something like a dipole antenna where each plate acts as one side of the dipole.
Using a formula for the E and B fields we would come up with a value for E and B vectors, and then the power radiated per unit area would be the cross product of E and B times a universal constant. This would never return to the source but would someday get absorbed by a surface that absorbs it, or partially absorbed and partially reflected. The levels of the E and B field are probably due to the physical dimensions of the capacitor in the same way that a dipole radiator is dependent on dimensions relative to the frequency. If we knew the impulse response of the dipole we might be able to calculate the fields and thus the power emitted.
What i think we are surmising in this thread is that if there is no resistance then it has to radiate, but how much we havent determined yet. We cant use general circuit analysis because it's not really applicable to a physical system unless we include parameters in that analysis that includes the dimensions. In general circuit analysis we usually assume a lumped condition where we only have to worry about current and voltage and time and not anything like length or width of the physical circuit elements.
What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.
What's indeed striking, is the independence of this phenomenon from the resistance .. that's a very strong indication, that there is a deeper physical effect buried in the grains, something like a "universal" physical effect.
There is. The deeper physical principle is thermodynamics: the difference between heat and work, and the consideration of entropy. When you equalize two capacitors, or when you charge a capacitor from a voltage source, entropy increases.
But it should be possible to derive that strictly from the electrodynamics theory, so that you can exactly assign these thermodynamic entities to their electrodynamic counterparts or -effects.
That means, the entropy is increased in this system, but where does this Entropy go to?
Entropic considerations on the Two-Capacitor Problem
V.O.M. Lara, A. P. Lima, and A. Costa
Instituto de F´?sica - Universidade Federal Fluminense
Av. Litorˆanea s/n
24210-340 Niter´oi - RJ Brazil
(Dated: June 12, 2012)
In the present work we study the well-known Two Capacitor Problem from a new perspective.
Although this problem has been thoroughly investigated, as far as we know there are
no studies of the thermodynamic aspects of the discharge process. We use the Free Electron
Gas Model to describe the electrons’ energy levels in both capacitors in the low temperature
regime. We assume that the capacitors and the resistor can exchange energy freely with a
heat reservoir. We assume that the resistance is large enough to consider an isothermal heat
exchange between the resistor and the heat reservoir. Thereby we obtain a positive entropy
variation due to the discharge process, corroborating its irreversibility.
Yes, that's the 2nd theorem of Thermodynamics.
Helmholtz Free Energy equals Inner Energy minus Temperature times Entropy: F = U - T*S
Fits here, because this problem is equivalent to every other irreversible process, comparable to dilution, for example.
But it should be possible to derive that strictly from the electrodynamics theory, so that you can exactly assign these thermodynamic entities to their electrodynamic counterparts or -effects.
That means, the entropy is increased in this system, but where does this Entropy go to?
In the resistive case, it's directly the Ohm power dissipation, but with zero resistance, it is not evident.
This is just like the "two buckets of water" problem, where a pipe connects the bottoms of the buckets. In the total absence of friction (which ain't gonna happen in real life), water would slosh between the buckets forever.
But I've left out one other important point so far: The inductors and capacitors are storing energy in electrical and magnetic fields, and these fields are constantly changing. Maxwell tells us that changing electrical and magnetic fields will radiate energy away, as electromagnetic radiation. This loss of energy will eventually damp the oscillations, causing the apparatus to reach the state of each capacitor having equal charge, just as though we had placed a resistor in the wire. The energy left the system via EM radiation.
Even if you build the capacitor out of superconducting plates in vacuum, and the inductor out of superconductor and vacuum, and house everything within a superconducting shield chamber -- or better still, construct all three together as a resonator -- the energy decays, because superconductors aren't ideal at AC.
An ideal ideal superconductor would conserve energy for all time, but there is no physical realization of it.
Practical superconducting resonators (used for linear particle accelerators) have Q factors in the 3 x 10^8 range, which is better than a quartz crystal, but still results in a time constant of seconds at best.
The conclusion is, even if radiation (to infinity) is blocked by as ideal a shield as possible, there's still loss in the system somewhere, dissipating that energy as heat, so that in the end, thermodynamics wins.
Tim
Even if you build the capacitor out of superconducting plates in vacuum, and the inductor out of superconductor and vacuum, and house everything within a superconducting shield chamber -- or better still, construct all three together as a resonator -- the energy decays, because superconductors aren't ideal at AC.
An ideal ideal superconductor would conserve energy for all time, but there is no physical realization of it.
Practical superconducting resonators (used for linear particle accelerators) have Q factors in the 3 x 10^8 range, which is better than a quartz crystal, but still results in a time constant of seconds at best.
The conclusion is, even if radiation (to infinity) is blocked by as ideal a shield as possible, there's still loss in the system somewhere, dissipating that energy as heat, so that in the end, thermodynamics wins.
Tim
The three laws of thermodynamics:
1. You can't win.
2. You can't even break even.
3. You can't get out of the game.
-Some have suggested in this thread that slow charging trough an inductor can reduce the losses below that 50%, is it true?-
Those are the same people who have absolutely no clue what they are advocating.
The central point here is that the resistive loss in any r/c charging circuit is only independent of the resistance in the circuit. Actually, it is only dependent of the supply voltage and capacitance.
That means the circuit is as efficient the the resistance goes to zero (then where is the energy lost?) or infinity (the capacitor is charged by a constant current source).