| General > General Technical Chat |
| Veritasium "How Electricity Actually Works" |
| << < (35/185) > >> |
| T3sl4co1l:
--- Quote from: electrodacus on May 04, 2022, 03:06:53 am ---Are you saying those 20nF capacitors are made of superconductor material ? Even the wires ? Do you also have the means to cool them down to whatever temperature is needed to become superconductors ? You will not be able to use a mosfet switch in that case anyway. --- End quote --- FETs work fine at superconducting temperatures, just need the right ones. ;) --- Quote ---The effect can be observed for superconducting wires (this need to include the switch and the capacitor plates). Unless you have access to some university lab with this sort of equipment I doubt this is an option for you. You did not wired as shown in diagram as you added significant inductance connecting the super small 20nF capacitors with wires. There is no inductor drawn in the schematic for a good reason. --- End quote --- But you specifically requested superconductors; there is no resistance in the circuit. And it is, well, wired -- I'm not sure where you suggest I go and get noninductive wires from, I mean, that would be preposterous! ...Wouldn't it? So I've done the nearest thing to the diagram I can do, and assumed the wire links mean wires, that, well, sometimes have inductance to them. I can't exactly help it, okay! --- Quote ---And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires. --- End quote --- Negligibly small in relation to......what? :o It's not like there's an RC time constant in there. You said no resistance, I went to quite extreme lengths to eliminate it! But then what can be left?! --- Quote ---So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors. I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct. Main equation is the one for Energy stored in a capacitor = 0.5 * C * V2 --- End quote --- Yes yes yes I get the refrain, if I wanted perfect energy balance I could do that. But you don't understand, this isn't an application -- this is a curiosity. Surely you have a curiosity about this conundrum as well? -- Else, why stick in this thread so long? Your claim that a sqrt(2) should pop up, is quite a curious one, and as a scientist, you should be as interested to disprove it, as I am to prove it! Yes, to disprove ones' own ideas, such a strange notion to some people, but it is the scientific method; there is none easier to fool than the self, as a famous scientist once said. If you're curious, I reconstructed the experiment with much shorter, non-superconducting wires, and obtained this waveform: It again does not show a sqrt(2) ratio, but shows the perfect 0.5 as predicted by charge balance. Energy is of course conserved because the excess is dissipated in the resistance, and no worries about whether the energy flowed around or through the wires, it got where it needed to go all the same. Tim |
| ejeffrey:
--- Quote from: electrodacus on May 04, 2022, 03:06:53 am ---And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires. --- End quote --- I think this is the core of the problem. You cannot make a circuit with current flowing where both inductance and resistance are negligible. It is not physical to ignore both at the same time, and when you try to you get incorrect answers. There are two things that can happen: The circuit will have significant resistance and the circuit will reach a steady state at half the initial voltage with a time constant RC. Or the circuit will not have significant resistance and the circuit will oscillate at it's resonance frequency given by w= 1/sqrt(LC). Reducing L will increase the resonant frequency but the overall behavior will stay the same. The first case does not conserve electrical energy because it has resistance. The second does conserve electrical energy but never reaches a steady state, instead it (ideally) oscillates forever. Both conserve charge of course, globally and locally (i.e., the net charge on each metal island remains the same). --- Quote ---I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct. Main equation is the one for Energy stored in a capacitor = 0.5 * C * V2 --- End quote --- Of course the correct answer has to conserve both charge and (total) energy, not just one or the other. |
| electrodacus:
Is late and I do not want to answer all the silly questions you (all) have. You have no understanding on the basics so my answers are done considering you at least understand some basic parts. Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit. If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem. The mosfet can work at low temperatures but it will not be a superconductor so even if you have superconductor wires and capacitors adding the mosfet that will have a resistance will cancel everything. I mentioned no resistance in the circuit and superconductors just to inform you that is not just theoretical but also possible in real life to setup such a test setup but for probably very few physics labs in the world not for any of you to attempt. Not to mention none of you even have the knowledge necessary to attempt this. The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me. The correct answer is just energy conservation not charge conservation as I already showed. If any of you have an engineering degree then you should be ashamed for not knowing what energy is. Do not think I will have any time in the next few day's. I may check once a day and replay if there is a serious question for someone that understand the subject. |
| ejeffrey:
--- Quote from: electrodacus on May 04, 2022, 06:30:29 am ---Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit. --- End quote --- Any indicance and capacitance will (at least in an ideal situation) form a resonant circuit. --- Quote ---If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem. --- End quote --- Only to the extent that the parasitic L and R of a real world 1 mF capacitor are likely to be larger. But an ideal 1 mF capacitor plus a 10 nH inductor will resonate at about 50 kHz. A more realistic 100 nH inductance will lower the frequency to 15 kHz. --- Quote ---The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me. --- End quote --- Everybody here agrees that the DC DC converter will do exactly as you have described. Nobody is trying because there isn't any disagreement on this topic so no point arguing. Our contention is that simply shorting a capacitor out will do something different and Tim did experiments to demonstrate that the circuit either oscillates or decays to V/2. |
| cj:
Removed as it didn't really answer Daves question. CJ |
| Navigation |
| Message Index |
| Next page |
| Previous page |