| General > General Technical Chat |
| Veritasium "How Electricity Actually Works" |
| << < (7/185) > >> |
| electrodacus:
--- Quote from: hamster_nz on April 30, 2022, 08:27:05 pm ---I suggest exactly the same, but you focus on charge (the actual conserved quantity), not energy. Draw +s and -s on the plates it that helps. Half the charge in each cap gives half the voltage, giving each capacitor a quarter of the energy of the original capacitor's charge. --- End quote --- This is where you are wrong. The important thing is energy as energy is what needs to be conserved. If you claim that in an ideal capacitor case the energy is not conserved you need to explain where it was lost. In real case half of the energy (half in case of two identical capacitors one fully charged and one fully discharged so just special symmetrical case) ends up as heat due to ESR. Like I mentioned already do a test with different capacitors then try to explain the results. You will not be able to do so if you do not understand what happens. |
| rfeecs:
--- Quote from: rfeecs on April 29, 2022, 06:17:53 pm ---One point that sounded odd to me is at 8:00 he talks about a python simulation of a DC circuit. He talks about there is an electric field in the center of the wire. Maybe I misunderstood what he is trying to say. --- End quote --- The python simulation is here: http://tinyurl.com/SurfaceCharge I had a look at it and I now realize that it is simulating a resistive wire, hence the electric field in the center of the wire is not zero. I thought he was assuming a perfectly conducting wire, as he stated in the first video. Now I see in this video he is assuming a real wire with some resistance. |
| hamster_nz:
I fully understand your position, but it is the wrong one. Energy is not conserved in an electric circuit. Do the math as you approach ESR = 0. For all values of ESR except 0 half the energy is lost. At zero everything is vaporized by an infinitely short current pulse of infinite amps. This debate is offering you an 'Aha!' moment. It is up to you if you take it. |
| electrodacus:
--- Quote from: hamster_nz on April 30, 2022, 09:03:33 pm ---I fully understand your position, but it is the wrong one. Energy is not conserved in an electric circuit. Do the math as you approach ESR = 0. For all values of ESR except 0 half the energy is lost. This debate is offering you an 'Aha!' moment. It is up to you if you take it. --- End quote --- Energy is always conserved. In the real case (as you seen from your own test) half of the energy you started with remains stored in the two capacitors and the other half ends up as heat. So all energy is accounted for. It is irrelevant what the ESR is as long as it is different from zero as energy will be lost and the smaller the ESR value the faster the charging takes place so high current for a short period or smaller current for a longer period if ESR value is higher. If you disagree that energy (half in this symmetrical example) is lost in the ESR as heat then please let me know where that energy ended up ? |
| hamster_nz:
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. |
| Navigation |
| Message Index |
| Next page |
| Previous page |