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| electrodacus:
--- Quote from: TimFox on May 14, 2022, 02:04:30 am ---Spell check error: “calorimetry”. No, I don’t think it’s necessary. Just an example of how energy conservation is not the most relevant calculation method in certain situations. --- End quote --- The relevant part is that energy conservation can not be violated. If you were to measure energy going out of the source in Derek's experiment and energy received by the lamp/resistor you will see the situation in this graph That shows that much more power exits the source than gets to the lamp and that is explained by the fact that energy is stored in the line capacitance. To make things even more clear the second graph shows what happens if the switch is only closed circuit for 30ns so less than half it is required for the electron wave to travel the transmission line. In both graphs the green is the power from source so the integral of that is the energy and the purple line is the lamp/resistor power. The difference that is missing ends as heat on the line but most of the energy gets to lamp after the time needed for the electron wave to get there. |
| Sredni:
A selection of references for the two-capacitor problem (one of which has already been given in this very thread) The two-capacitor problem with radiation Timothy B. Boykin, Dennis Hite, and Nagendra Singh Department of Electrical and Computer Engineering, The University of Alabama in Huntsville, Huntsville, Alabama 35899 ~Received 12 June 2001; accepted 8 November 2001; Published Online: 11 March 2002 American Association of Physics Teachers. @DOI: 10.1119/1.1435344 available upon request to the author from this ResearchGate page: https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation Capacitors can radiate: Some consequences of the two-capacitor problem with radiation 14 May 2003 T.C. CHOY arXiv:physics/0305062v1 [physics.class-ph] https://www.researchgate.net/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation freely available at ResearchGate (direct link) https://www.researchgate.net/profile/Tuck-Choy/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation/links/5935528daca272fc5556a317/Capacitors-can-radiate-some-consequences-of-the-two-capacitor-problem-with-radiation.pdf Radiative effects and the missing energy paradox in the two capacitor problem Gilberto A Urzua, Omar Jimenez, Fernando Maass and Alvaro Restuccia Departamento de Fisica, Universidad de Antofagasta, Casilla 170, Antofagasta, Chile E-mail: gilberto.urzua@uantof.cl, omar.jimenez@uantof.cl, fernando.maass@uantof.cl, alvaro.restuccia@uantof.cl Journal of Physics: Conference Series 720 (2016) 012054 doi:10.1088/1742-6596/720/1/012054 https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem freely available at ResearchGate (direct link) https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem/fulltext/57a4bae308ae455e8539f85d/Radiative-effects-and-the-missing-energy-paradox-in-the-ideal-two-capacitors-problem.pdf The two-capacitor problem revisited: a mechanical harmonic oscillator model approach Keeyung Lee Department of Physics, Inha University, Incheon, 402-751, Korea Received 17 July 2008, in final form 4 September 2008; Published 6 November 2008 EUROPEAN JOURNAL OF PHYSICS Eur. J. Phys. 30 (2009) 69–74 doi:10.1088/0143-0807/30/1/007 Online at stacks.iop.org/EJP/30/69 Preprint available from ArXiv: https://arxiv.org/abs/1210.4155 Entropic Considerations of the Two-Capacitor Problem January 19, 2012 V.O.M. Lara, A. P. Lima, and A. Costa Instituto de Fısica - Universidade Federal Fluminense https://www.researchgate.net/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem freely available at ResearchGate (direct link) https://www.researchgate.net/profile/Ap-Lima/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem/links/55e9aa6208ae21d099c302fb/Entropic-considerations-on-the-Two-Capacitor-Problem.pdf The Paradox of Two Charged Capacitors -- A New Perspective August 2013 Authors: Ashok K. Singal Indian Space Research Organization arXiv https://www.researchgate.net/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective (direct link) https://www.researchgate.net/profile/Ashok-Singal/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective/links/559e349008ae76bed0bb6d46/The-Paradox-of-Two-Charged-Capacitors--A-New-Perspective.pdf The idea I have come to so far: You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same. The transformation is irreversible, therefore some energy MUST be lost and entropy must increase. Turns out that energy it can be lost in different ways, depending on the actual circuit: if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate). |
| electrodacus:
--- Quote from: Sredni on May 14, 2022, 03:30:18 am ---A selection of references for the two-capacitor problem (one of which has already been given in this very thread) The idea I have come to so far: You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same. The transformation is irreversible, therefore some energy MUST be lost and entropy must increase. Turns out that energy it can be lost in different ways, depending on the actual circuit: if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate). --- End quote --- While I have not read any of those papers the fact that there are so many is ridiculous. If you read any of those paper can you tell me if any of them actually did the test ? Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance. I guess none of them had the budget to do the superconductor experiment as they will have found out all energy remains stored and it is not radiated as some of the titles of those papers you link may suggest. Even my simple simulation shows that losses in the case of two identical capacitors can be reduced from 50% to something like 12% or less by using an inductor and a diode to transfer the energy. I can even provide you with a simple way to test that all loss is thermal loss not radiated (unless you call infrared photons radiated energy). Experiment will be fairly simple: Take to large capacitors either large electrolytic or super capacitors (just because is easier to measure the energy lost as heat not that is needed as it can be calculated). I will give this example 3V 1F for the charged capacitor identical 1F discharged capacitor. Take this for example https://www.mouser.com/datasheet/2/40/AVX_SCC_3_0V-1128335.pdf SCCR12E105PRB it has a max DC ESR of 860mOhm I will rounded up to 1Ohm. Use a 8Ohm series resistor just that total resistance in the circuit is a round 10Ohm the 1Ohm from the two capacitors than the 8Ohm you add. Now just use a multimeter with data logging to measure the voltage drop across the 8Ohm resistor. You know for sure that voltage drop divided by resistance will give you the power dissipated as heat on the resistor so no need to measure any temperature and you can have very accurate measurement. Make 5 or 10 measurements per second depending on how capable your logging multimeter is as experiment will likely need to run for about one minute to properly charge the discharged capacitor from the charged one. Then integrate that power measurement over time to get energy and divide by 0.8 to get the entire energy dissipated as heat including the heat dissipated on the capacitor's plates that 1Ohm DC ESR. You will get exactly all the missing energy so half of 4.5Ws that you started as as heat and the other half is split between the two capacitors as stored energy. With accurate equipment and careful measurement you will find that not even a faction of a percent of energy is unaccounted for. But for anyone that understand physics this experiment will not even be needed. And yes it is sad to see all those universities and research institutions wasting resources and making fools of themselves. |
| Sredni:
--- Quote from: electrodacus on May 14, 2022, 04:29:05 am ---While I have not read any of those papers the fact that there are so many is ridiculous. --- End quote --- Not necessarily: the first one made the analysis for a magnetic dipole radiator, obtaining under certain simplifications a nonlinear differential equation of third order that gives an exponentially decaying solution. IIRC, the second one generalized the results by making the analysis more general, obtaining a fifth order differential equation and finding a new type of solution that goes to zero in a finite time ('sudden death'). The third one considers radiation by electric dipole, as well. The others consider the problem from a more general point of view (increase in entropy and loss of energy). But get the same conclusion about the necessity to lose half the initial energy. So, they are not necessarily in contrast with one another, but rather explore different ways to see the same problem. Incidentally, there is a video by youtuber SimplyPut that reaches the same conclusions about entropy in a rather intuitive way. The guy is a bus driver, but in my opinion he got the problem right. --- Quote ---If you read any of those paper can you tell me if any of them actually did the test ? --- End quote --- I guess they had a little problem in finding ideal capacitor and getting a grant to set up the superconducting wires. So, no, I don't think you can test the ideal problem in this world. But the first paper has a very interesting graph that gives you how much of the energy goes into heat and how much into radiation, based on the value of resistance in the wires. --- Quote ---Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance. --- End quote --- Did you perform the experiment with an ideal capacitor and superconducting wires? |
| electrodacus:
--- Quote from: Sredni on May 14, 2022, 05:08:36 am --- Did you perform the experiment with an ideal capacitor and superconducting wires? --- End quote --- No ideal capacitor or superconducting wires are needed. To prove that all energy that is not in capacitors at the end of the test is found as heat on the resistive elements (basically all conductors in the circuit). There will be no energy left that is not accounted for either as energy stored in capacitor or as heat so no energy is "radiated away" other than as heat. Hope you understand that supercapacitors have nothing to do with superconductors as I mentioned supercapacitors for the test just because they have higher capacity so discharge is slower and you can use a multimeter but you can use any other type of capacitors and an oscilloscope tho you lose the vertical resolution compared to a multimeter so accuracy may be a bit lower. If you did not heard of supercapacitors before you can check the spec that I listed for the example. If you understand the mechanisms you do not need to do the test to know what the results will be. This is my main hobby (energy generation and energy storage) so it will be boring to do a test for witch I know exactly the result. If you measure a voltage drop across a resistor you know that all that energy calculated form that ended up as heat.is not like an LED where some was radiated as visible photons there will be infrared radiated from the resistor but that is not the type of radiation discussed in those papers. I feel frustrated not being able to properly explain what looks like such a simple problem. |
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