The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
Of course. And you absolutely need to stop saying that other people on this thread are claiming conservation of energy is violated. That is just an outright lie and it isn't ok. Stop it.
The point of the paradox is to show that the initial setup is non physical. Postulating zero loss and zero inductance and requiring that you reach a steady state is a non physical setup and the equivalent to dividing by zero. Any conclusions drawn from it are meaningless. Your entire premise is based on something that cannot exist and is meaningless.
With no inductance or resistance the voltage is discontinuous and the current flow is infinite. That is impossible. If you add any finite resistance or inductance -- even an attohenry everything becomes finite and does not agree with your claim. This is trivially verifiable analytically, by simulation, or by experent.
I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).
A question (for Tim, ejeffrey, or whoever):
How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not?
I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).
A question (for Tim, ejeffrey, or whoever):
But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.
But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?
But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?
Yes but it may take hundreds of years if the distance of that loop is relatively far. And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.
like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.
You seem to be saying:
"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."
Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?
You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?
Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0. Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero. Here, "limit" is the normal mathematical meaning of the term.
My biggest gripe with the R->0 limit is that it necessarily lead to an unphysical model. If charge needs to move from cap1 to cap 2 at a finite distance, the zero transfer time implies faster than light charge motion. Radiation takes you out of this untenable position: the charges move from here to there, but they need to accelerate and decelerate - hence there will be radiation.
The two-cap problem is a model breaker. One way or the other, something's gotta give.
So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away"
Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.
A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).
Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.
PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…
There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.
Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.
A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).
Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.
PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…
There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.
There is a reason a transmission line in spice is simulated as a series of LC elements as that is the best approximation of what happens and it is confirmed by the results that are not in contradiction to experimental results including the one Derek made.
I probably needed to express myself more clear. I asked you to create an ideal transformer by adding the second inductor.
Anyway, you are wrong. Changing magnetic field of the inductor induces changing electric field that in its turn induces changing magnetic field, and so on. And all that radiates into the Universe at a speed of light…
If you watch long enough you will see get the general pattern.
- Nobody is allowed an electric field, except in the dielectric of capacitors
- Nobody is allowed a magnetic field, except in for in inductors or transformers
- Nobody is allowed to have magnetic or electric interaction between wires, but transmission lines are allowed
... leading the result that if you deny any existence of the electric and magnetic fields and their ability to actually transfer the energy, then it is a fait accompli that energy flows only in the wires.
The argument from electrodacus all along is that Lumped Element model the "level 0" description of electrical reality, and the underlying physics (e.g. electro magnetic field simulations) are misguided, do not reflect reality, and can be ignored, because against intuition they show that the energy does not flow in the conductors.
Not sure I understand what you want to say.
If resistance is zero you still have inductance...