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| Veritasium "How Electricity Actually Works" |
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| electrodacus:
--- Quote from: hamster_nz on April 30, 2022, 09:36:20 pm ---I fully agree half the energy is lot from the circuit into the environment it one way or another. The caps are at half the voltage. The disagreement I have is with those who say that the caps are at 0.71 of the initial voltage, and all the energy is still in the circuit. They are missing an insight. --- End quote --- For an ideal capacitor setup that 0.71 of initial voltage will be correct as there will be no losses. And even in real world you can build capacitors with superconducting plates and they will also not have any loss. But whatever you take that loss in to consideration real world or not the fact still is that a transmission line has capacitance. The claim Derek makes as energy not traveling trough wires is wrong. That small current and energy transfer to the lamp is just because the transmission line capacitance is charging. If you ignore that energy storage you get to the wrong conclusion that energy is not traveling trough wires and that is wrong. Same problem with energy storage applies to the direct down wind faster than wind vehicle and there the concussion Derek makes as as outrageous basilica claiming (not directly) that system is an overunity device (more than a perpetuum mobile). |
| Naej:
--- Quote from: HuronKing on April 29, 2022, 05:34:34 pm ---It's still too bad though that Heaviside's work with coaxial cables didn't have time to get mentioned - that's the most immediately practical application of Poynting Theory beyond it being a mere theoretical curiosity. --- End quote --- How did he use the mere theoretical curiosity, and to prove what? |
| hamster_nz:
--- Quote from: electrodacus on April 30, 2022, 11:22:44 pm --- --- Quote from: hamster_nz on April 30, 2022, 09:36:20 pm ---I fully agree half the energy is lot from the circuit into the environment it one way or another. The caps are at half the voltage. The disagreement I have is with those who say that the caps are at 0.71 of the initial voltage, and all the energy is still in the circuit. They are missing an insight. --- End quote --- For an ideal capacitor setup that 0.71 of initial voltage will be correct as there will be no losses. --- End quote --- Nope, if there are no losses it will oscillate forever. How could it not? there is nothing to dampen it down and the system starts in an unstable state. There is no steady state where the paired capacitors have the same total energy and total charge as the original charged plus uncharged capacitors. --- Quote ---And even in real world you can build capacitors with superconducting plates and they will also not have any loss. --- End quote --- Are you implying that a superconducting system would not radiate EM waves? Even with 'ideal' circuit the energy is lost: https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation By treating radiative effects in the simplest approximation, we show that the paradox is really nothing more than an inappropriately applied lumped-parameter model. In particular, we show that in the zero-resistance circuit, radiation fully accounts for all of the energy lost. |
| eti:
He should leave this alone. Lol. |
| electrodacus:
--- Quote from: hamster_nz on May 01, 2022, 12:57:30 am ---Are you implying that a superconducting system would not radiate EM waves? Even with 'ideal' circuit the energy is lost: By treating radiative effects in the simplest approximation, we show that the paradox is really nothing more than an inappropriately applied lumped-parameter model. In particular, we show that in the zero-resistance circuit, radiation fully accounts for all of the energy lost. --- End quote --- Quite some years ago experiments with super conductors where performed where a current flow was induced in a super conductor ring and that current was there with no losses. Of course each time you make a measurement you will influence the current conditions but you account for those. Instructors are also energy storage devices (maybe harder to understand but still energy storage devices). After you charge a capacitor there is no loss other than the small leakage current inside the capacitor. Same after you charge an inductor and there is a constant current flowing trough it all the losses are restive and same as with capacitor there are some small leakage losses. When you disconnect the supply from a capacitor the stored energy will be provided to the load. Same as when you reduce the current the field will collapse providing that stored energy to the circuit. Any capacitor will have some small amounts of distributed inductance and the same is true for an inductor that will have small amounts of distributed capacitance. Both the inductor and capacitor are energy storage devices. If Derek did not closed the circuit at the ends when closing the switch he will still have seen that small amount of transient current that is due to energy storage being charged but nothing after that. Energy is transferred trough the wires both when circuit is open at the ends and when it is closed. When it is open only the amount of energy needed to charge the transmission line will flow (trough wires) and when circuit is closed energy will continue to flow trough wires. His main claim is that energy is not delivered to lamp (load) trough wires and that is just not the case. Is also easy to demonstrate that the energy transfer in real world is not 100% efficient thus you will need to see heat generate in the medium that transfers the energy. A thermal camera can easily show that the wire is where that loss happens as it will heat up and not the medium around the wire. |
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