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| Veritasium "How Electricity Actually Works" |
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| hamster_nz:
--- Quote from: electrodacus on May 15, 2022, 02:14:49 am --- --- Quote from: vad on May 15, 2022, 02:00:37 am ---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… --- End quote --- 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. --- End quote --- 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. |
| IanB:
There is an interesting thought experiment which was hinted at earlier in the thread I think. You are allowed to observe two wires floating in free space, which stretch out of sight in both directions. There is nothing near either wire except you and any instruments you care to have about your person. Each wire is identical in all physical respects, and each wire is carrying exactly 1 amp DC (which you can verify by measuring the magnetic field, or by measuring the voltage drop along a short length of the wire, or by measuring the heat being radiated). One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done? |
| electrodacus:
--- Quote from: vad on May 15, 2022, 02:29:59 am --- 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… --- End quote --- Are you trying to say magnetic energy going in to creating the magnetic field around a wire or inductor (same thing) will be energy lost ? Because that is not according to any evidence. That magnetic field is conservative so when you disconnect the circuit all that energy will be put back in to the circuit. It seems you are not the only one having this incorrect view about an inductor/wire. If you looked at any of my simulations maybe the most relevant is that simulating a transmission line you will see that all energy is accounted for. The one where switch is turned ON for just 30ns is particularly relevant for your concern. |
| electrodacus:
--- Quote from: hamster_nz on May 15, 2022, 02:31:31 am --- 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. --- End quote --- The lumped element model will not have been used if it did not provide accurate predictions of what happens. And it is accurate because it represents reality just a reduced form of that in terms of resolution. Where is the electric field inside a charged capacitor transferring energy ? The switch itself is also a capacitor and there is an electric field before closing the switch so how come you need to actually close the switch to transfer energy. |
| Sredni:
--- Quote from: electrodacus on May 15, 2022, 01:34:05 am ---Not sure I understand what you want to say. If resistance is zero you still have inductance... --- End quote --- No, I was talking of a circuit where the only dissipation mechanism is that through Joule loss. The reasoning is that the loss is the same irregardless of the value of R, so all dissipation is accounted for when R= 1 ohm, or 0.1 ohm, or 0.00001 ohm or... and then we take the limit for R->0 and say that the loss is still the same - so it is all accounted for by ohmic losses even when R->0. The first paper in my list above (the paper that was posted earlier by someone else here) avoid this problem by computing the radiation loss contribute considering NOT and oscillating LC circuit, but simply the radiation associated with the accelerated charges. It shows that for a circuit with IIRC a diameter of 10cm but still negligible self-inductance, the radiation loss becomes relevant only when the resistance of the loop falls under a handful of microohm (I don't have it at hand now, but the numbers are in that ballpark). So, ohmic loss till a certain value of resistance, but under that it's radiation that takes over. At first they share the losses, then radiation becomes dominant and account for nearly all losses. The current does not oscillate: the solution is a decaying exponential - and with a more advanced model - there are also solutions where the current dies off in a finite time (paper 2 or 3 in the list above). No more paradox of the missing energy (it is extracted from the circuit) No more paradox of ohmic loss with R=0 (it is taken care of by radiation) No more need for an inductance that makes the circuit oscillate at a frequency 1/Sqrt[L Ceq]. |
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