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| Veritasium "How Electricity Actually Works" |
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| rfeecs:
--- Quote from: Sredni on May 04, 2022, 09:27:31 pm --- --- Quote from: rfeecs on May 02, 2022, 08:44:59 pm --- --- Quote from: Sredni on May 02, 2022, 07:59:18 pm ---If so: HOW MUCH ENERGY does travel along the path? --- End quote --- If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path. Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B. The amount of energy is the energy converted to heat. So it depends on which hole it goes down. --- End quote --- This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference. --- End quote --- I don't think this is a problem if you consider the energy is relative to a reference. So zero relative energy may still be nonzero "absolute" energy. Instead of assuming the energy is located in the rock, you can eliminate the problem perhaps by using the gravitational potential: https://en.wikipedia.org/wiki/Gravitational_potential If the gravitational potential is a property of location, then when you move the mass you are not "transporting energy". As Naej said, you are only converting energy at the two end points. We have heard the same point of view applied to electromagnetic systems. |
| rfeecs:
--- Quote from: hamster_nz on May 04, 2022, 12:18:26 am ---The unasked question so far is: Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe. --- End quote --- What if you only connected pipe at one end. The wire is connected at both ends. Then the current only flows through the wire. But since the pipe is at the same voltage as the wire, you have no electric field around the wire. You moved the electric field to the outside of the pipe. Now according to the Poynting vector, the energy flow is on the outside of the pipe, not even at the surface of the wire. The current flows through the wire, but the energy flows on the outside of the pipe. You could make the pipe 1 meter in diameter and the energy flow would be half a meter away from the current flow. Yet another odd result of the Poynting vector applied to DC. |
| aetherist:
--- Quote from: electrodacus on May 05, 2022, 03:10:15 pm --- --- Quote from: EEVblog on May 05, 2022, 08:06:18 am --- --- Quote from: electrodacus on May 05, 2022, 04:31:52 am ---2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m. This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires. --- End quote --- If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire. --- End quote --- Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire. If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line). With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance. --- End quote --- Dave is correct(ish). If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch. This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly). The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns. Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 35 ns, nearly the same as for the early (air) signal. If the wire is insulated then this 35 ns would be 50% greater, ie 52.5 ns. But the 34 ns for the early signal would remain at 34 ns. Much of what electrodacus says is ok i think (if i could understand it). But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube. I will have to have another look to check. Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay. |
| electrodacus:
--- Quote from: aetherist on May 05, 2022, 09:56:28 pm ---Dave is correct(ish). If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch. This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly). The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns. Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal. Much of what electrodacus says is ok i think (if i could understand it). But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube. I will have to have another look to check. Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay. --- End quote --- There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned. The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at Vi The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire. Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field. Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points. Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions. The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved. The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge. The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage. It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets. |
| aetherist:
--- Quote from: electrodacus on May 05, 2022, 10:26:12 pm --- --- Quote from: aetherist on May 05, 2022, 09:56:28 pm ---Dave is correct(ish). If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch. This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly). The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns. Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal. Much of what electrodacus says is ok i think (if i could understand it). But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube. I will have to have another look to check. Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay. --- End quote --- There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned. The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at Vi The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire. Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field. Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points. Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions. The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved. The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge. The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage. It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets. --- End quote --- Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal. I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal. It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns. I have been looking at the arguments re capacitors, & everyone has been wrong in every way. |
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