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| "Veritasium" (YT) - "The Big Misconception About Electricity" ? |
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| aetherist:
--- Quote from: SandyCox on February 18, 2022, 11:22:16 am ---I looked at the "Electronics World" paper by Ivor Catt and applied old-school transmission-line theory. The setup An 18 m long 75 Ohm transmission line. The reed switch is located at z=0m and the open end of the transmission line is at z=-18m. The transmission line is pre-charged to 8V. The reed switch is open-ended at t=0s. I chose a propagation speed of c=2.55e8 m/s. What transmission line theory tells us * The voltage wave is the sum of a wave travelling in the positive z direction (indicated by v+) and a wave travelling in the negative z direction (indicated by v-). * The reflection coefficient at the open end is equal to 1. The reflection coefficient at the reed switch is 0. * Prior to t=0s, the voltage across the entire length (z=-18m to z=0m) of the transmission line is equal to 8V. There is no current in the transmission line. We can use this information to calculate v+ and v- for t<0s. It turns out that both are constant and equal to 4V. * Closing the reed switch at t=0s initiates the transient. At z=0, v+ enters the 75 Ohm resistor, where it is terminated. v- is reflected from the open end of the transmission line and becomes part of v+ I attach figures for comparison with the results in Catt's paper. You do the comparison. A video showing the waves is available at The error in Catt's paper is that he views the transmission line as a capacitor, which it most certainly is not. A more detailed explain and the mathematical derivations will follow. There is nothing strange about this setup. I remember doing experiments like this when I was an undergraduate student and using transmission line theory to explain the measurements. Note that Catt's paper was published in the April 2013 issue. --- End quote --- (1) I don’t see how u can split the electricity into 2 half currents, ie 2 half voltages. I can see that on one side of the switch we have +4V & on the other side of the switch we have -4V. But when the switch is closed we immediately have 8V. Looking simply at drifting electrons producing a wave, there will be an 8V wave going right (to the terminating resistor), this will be a depletion wave, ie that wire (which includes the outer sheath of the coax) is electron rich, & the conduction electrons will begin to spread out. And, there will be an 8V wave going left (along the core wire of the 18m long coax), this will be an enrichment wave, ie the core wire is electron poor, & the electrons will begin flow into it & begin to bunch up. If the core wire was neutral, ie with 00V, then in that case there would be a 4V wave going right & a 4V wave going left. But it aint neutral, it has +4V. (2) Nextly i could explain why it is an impossibility for an electron wave to reflect at a dead end of a wire. Or, putting it another way, why such a reflexion would not add to the gradual addition of electrons into that wire, ie it would not add to the gradual bunching up of electrons in that wire. But not today, i might explain later. (3) Here is my main objection. U chose the propagation speed of light, but in a part of your explanation u invoke an infinite speed of light for one of your 4V waves. U have this wave starting to wave at t=0 s, at the end of the coax, which is 18m from the switch. (4) I noticed a few things, as an interesting aside, not necessarily a criticism of your explanation. And u might address some of these things in your detailed explanation later. I noticed that u did not mention drifting electrons (& what parts they played), surface charge (ie surface electrons)(& surface charge distribution), Poynting, & whether electrical energy is in or on or near the wires. (5) Re a capacitor not being a transmission line. Perhaps so, in a way. But a transmission line is a capacitor. Perhaps it depends on whether ends are open or shorted. Perhaps it depends on the kind of source producing the electricity (ie the charge)(eg a lead acid battery). Anyhow i don’t see how arguing about that stuff would help us today. |
| aetherist:
--- Quote from: SandyCox on February 18, 2022, 01:02:28 pm ---Can new electricity answer the following question? What is the equation for the voltage across the capacitor, as a function of time, as it discharges through the resistor? --- End quote --- I don’t know what the old (electron) electricity equation(s) is for discharge of a capacitor. But the new (electon) electricity equation(s) would be almost identical, except that it would have to show the correct steady half voltage for the correct double the distance, ie for double the time (at least it would for the case of zero reflexions). And it would need an additional equation for the additional voltage from electons leaving the length of the wire from the capacitor to the resistor. This extra voltage would be for a doubled time, ie electons have to go the wrong way along that wire & later return along that wire, hence a doubled distance & a doubled time. For a giant capacitor & a short or thin wire this voltage might be insignificant. And it would need an additional equation for the very small extra voltage happening for a very long time due to surface electrons gradually slowly entering the positive plate. But this voltage might be insignificant. |
| aetherist:
I just noticed today. Erik Margan did a 2m long coax version of Wakefield's 18m X. And Margan confirmed Wakefield's half voltage for double time. But, Margan erred in his own explanation (page 11). He said that photons had to go up & back in his 2m test piece. No. The photons (my electons actually) came from his 12.8m intermediate coax. Yes, they had to propagate up & back in the 2m test coax, & hence he got his desired trace for the pulse, but his pulse came from his negative wire, ie the coax sheath of the 12.8m coax, not from the positive core wire of his 2m test piece. Hence i think that the sign of his voltage for his pulse might be wrong too (perhaps he fudged his graphs)(perhaps non-intentionally)(i might have a think about that some other day)(my brain hurts). Anyhow, his mistake shows that u have to follow the electons. Electons live on the negative wire-plate-terminal. Margan (& everybody else but me) think that photons (or whatever) live on both plates, positive plate & negative plate. Or, they think that photons live in the space tween plates, & that the plates themselves dont matter much. No. The energy is on the surface of the negative plate-wire-terminal, in the electons (& in their radiation). U will notice that both Wakefield & Margan both connected the negative terminal of their lead acid battery to the outer sheath of their coax. They thought that this was a non-critical detail, outer sheath, inner core, who cares. But, i care, it makes (or can make) lots of difference (depending on the problem in question). U have to follow the electon. Its not just a matter of drawing silly Poynting lines on a bit of paper. http://www.ivorcatt.co.uk/x37p.htm Erik Margan repeated Wakefield's X. http://www.ivorcatt.co.uk/x726.pdf |
| adx:
From an abandoned post a few days ago: The voltage on a capacitor you might use (ie, a real one you can buy) rises linearly when a constant current (coulombs per second) is injected into it, and falls when that charge is removed. The rate at which the voltage rises is predicted by the capacitance value, it can vary for non-quality capacitors (a well known set of problems) but is generally stable and almost exact for good ones. Not 100% out. There isn't a "charging time" independent of what I just described. So there is no stair step change in voltage. There is no steady half voltage. There is no distance. There is no double the time. There are no "reflexions". This is all complete nonsense. There also no polarity effects except electrolytic capacitors which have been formed to a particular polarity - many can be reformed (carefully) and used in reverse. There are no differences between positive and negative beyond the sign. There is no "very small extra voltage happening for a very long time" in a vacuum capacitor and the small amount that occurs in usual capacitors is due to dielectric absorption, ie the insulator taking a 'set'. That goes away if the insulating material is removed. |
| SandyCox:
--- Quote from: aetherist on February 19, 2022, 03:33:14 am ---(1) I don’t see how u can split the electricity into 2 half currents, ie 2 half voltages. I can see that on one side of the switch we have +4V & on the other side of the switch we have -4V. But when the switch is closed we immediately have 8V. Looking simply at drifting electrons producing a wave, there will be an 8V wave going right (to the terminating resistor), this will be a depletion wave, ie that wire (which includes the outer sheath of the coax) is electron rich, & the conduction electrons will begin to spread out. And, there will be an 8V wave going left (along the core wire of the 18m long coax), this will be an enrichment wave, ie the core wire is electron poor, & the electrons will begin flow into it & begin to bunch up. If the core wire was neutral, ie with 00V, then in that case there would be a 4V wave going right & a 4V wave going left. But it aint neutral, it has +4V. (2) Nextly i could explain why it is an impossibility for an electron wave to reflect at a dead end of a wire. Or, putting it another way, why such a reflexion would not add to the gradual addition of electrons into that wire, ie it would not add to the gradual bunching up of electrons in that wire. But not today, i might explain later. (3) Here is my main objection. U chose the propagation speed of light, but in a part of your explanation u invoke an infinite speed of light for one of your 4V waves. U have this wave starting to wave at t=0 s, at the end of the coax, which is 18m from the switch. (4) I noticed a few things, as an interesting aside, not necessarily a criticism of your explanation. And u might address some of these things in your detailed explanation later. I noticed that u did not mention drifting electrons (& what parts they played), surface charge (ie surface electrons)(& surface charge distribution), Poynting, & whether electrical energy is in or on or near the wires. (5) Re a capacitor not being a transmission line. Perhaps so, in a way. But a transmission line is a capacitor. Perhaps it depends on whether ends are open or shorted. Perhaps it depends on the kind of source producing the electricity (ie the charge)(eg a lead acid battery). Anyhow i don’t see how arguing about that stuff would help us today. --- End quote --- I will attempt to address your misconceptions in the note I am writing. In the meantime, let me set a challenge: Let's change the termination resistor in Catt's paper from 75Ohm to 47Ohm. How, according to "new electricity", will the measured pulses look if we do that? |
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