| General > General Technical Chat |
| Veritasium "How Electricity Actually Works" |
| << < (53/185) > >> |
| aetherist:
--- Quote from: electrodacus on May 08, 2022, 01:06:36 am --- --- Quote from: aetherist on May 08, 2022, 12:56:27 am --- --- Quote from: electrodacus on May 08, 2022, 12:42:26 am --- --- Quote from: aetherist on May 08, 2022, 12:22:31 am ---If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns. And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau. Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state. --- End quote --- A capacitor with a 1000mm gap between plates will require 3.3ns and one with 1mm gap will require 3.3ps LTspice as far as I know has no option to specify the gap between plates and all comercial capacitors even the very high voltage ones will typically have much less than 1mm gap as the smaller the gap the larger the capacity. So there is nothing strange or not understood about that 3.3ns. When you close the switch exactly at that contact interface one electron will move from one side of the switch to the other and if a capacitor is in series in the loop as is the case in Derek's experiment then it takes 3.3ns for an electron on the other side of the 1m thick dielectric (air) to feel the effect and vacate a space. In the battery the space is much smaller so when switch is closed and one electron moves on the other side of the switch the electron wave will travel in to battery say maybe 100mm from the switch then inside the battery the gap say is 1mm and so that electron that left from battery to switch will cause an electron from the wire connected on the other side of the battery to be accepted in the battery and so if total distance is 101mm in a straight line it may take as little as 0.33ns but likely it can not capacitively couple in straight line so it will be an ark maybe say 150mm so around 0.50ns then both sides are coupled at the same time with the wire above that is at about 1m thus 3.3nm + 0.5ns will be the total time from closing the switch until the first packet of energy is transferred from battery to the capacitor (transmission line is a capacitor) and since the lamp is in series between this two 1m capacitors all current used to charge the capacitor also will go through the lamp and be lost as heat and maybe photos. --- End quote --- Surely LTSpice is told the distance tween wires, ie 1000 mm. And the small gap tween plates can be ignored. In which case LTSpice should be able to model the initial transient delay & rise & plateau. --- End quote --- Spice is not the sort of simulation you are imagining. The transmission line is done by adding a bunch of inductors and capacitors together to simulate those two wires. And as mentioned that 3.3 or so ns delay at the beginning is irrelevant. I can add that delay but it will make no difference. What you need to understand is that two parallel wires are a capacitor thus an energy storage device. If you leave the ends open (the 1m pipes that connects the two parallel 10m pipes) the you will still have this first 65ns current through the lamp/resistor as the capacitor's one on each side will be charged but once they are charged you no longer have any current unless you reverse the battery polarity so that you discharge those capacitors and charge them in the other direction then it will again stop. So energy flows in and out of capacitors and not through capacitors. Energy flows in to capacitor is the key as that energy is not doing work it is stored and just because wires have resistance and a lamp is in series it means that charging is not efficient so not all energy from the source will end up in capacitor as some will be lost to power the lamp/heat the resistor. --- End quote --- I agree that the 1000 mm end connection is irrelevant to the initial transients. And i agree that parallel wires are a capacitor. Old electricity i think says that electrons passing the switch will repel electrons from the other wire. Some of these will go left (towards the bulb) & some will go right. Hence electrons will start to pass through the bulb at 3.3 ns. The rise ends at 21.1/24.8 ns. This means that the electron wavefront passing the switch has propagated say 7.3 m at 3.3 ns/m That propagation includes say 1 m crossing tween the wires, which leaves 6.3 m in the wires. That 6.3 m is say 3.2 m along the primary wire, plus 3.1 m to the bulb along the secondary wire. Hence the plateauing started when the wavefront was 3.2 m along past the switch. This is 1/3rd of the way to the end of that 10 m stretch of wire/tube. At 24 ns the induction process of squeezing of electrons from the secondary wire reaches a peak value of 3.7V, after which the voltage across the bulb drops from 3.7v to 3.1V. I think that the same kind of rise & plateau can be seen in the AlphaPhoenix-X. I had a look. Brian's rise takes about 75 ns, for a gap of say 250 mm, ie ΒΌ of Veritasium's gap of 1000 mm. AlphaPhoenix's L was say 250 m, & Veritasium's L was say 10 m. AlphaPhoenix's plateaux went up not down, ie it went from say 1.8V up to 2.2V at say 1600 ns. AlphaPhoenix's 75 ns indicates that the end of his rise was when the wavefront was 11.2 m past the switch. Add 1 ns for the 250 mm gap, plus 11.2 m going back to the bulb, & we have our 75 ns. I suspect that the height of the wires & tubes above the dirt would affect the rises etc. Anyhow, any lumped element transmission line model worth its salt will surely give these kinds of numbers. |
| electrodacus:
--- Quote from: IanB on May 08, 2022, 01:29:00 am ---This picture is widely understood to be an analogy. It does not represent what really happens, it is only a pictorial simplification. The complication here is that general relativity indicates that you can have gravity without mass, and an internal observer experiencing such gravity cannot tell whether the experienced gravitational field is due to mass or otherwise. --- End quote --- As any analogy it has limitations. The fact remains that energy cannot be transferred by the electric field else capacitors will not exist. It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor. All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer. Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged. |
| IanB:
--- Quote from: electrodacus on May 08, 2022, 01:36:04 am ---It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor. --- End quote --- Oh? What will you say about that? --- Quote ---All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer. Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged. --- End quote --- Congratulations. You just demonstrated that energy flows through the capacitor. |
| hamster_nz:
--- Quote from: electrodacus on May 08, 2022, 01:36:04 am ---The fact remains that energy cannot be transferred by the electric field else capacitors will not exist. It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor. All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer. Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged. --- End quote --- Are you saying that if you have a lamp between two capacitors (so all three components are in series) that it is impossible to get the lamp to glow? Because that would be "energy passing through the capacitors" to me. If not, where did the energy that makes the lamp glow come from? |
| electrodacus:
--- Quote from: IanB on May 08, 2022, 02:02:59 am ---Congratulations. You just demonstrated that energy flows through the capacitor. --- End quote --- You will need to understand what energy and energy storage is to make that claim. |
| Navigation |
| Message Index |
| Next page |
| Previous page |