Two capacitors, no paradox

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OverUnity Research > General - Free Energy Related > Miscellaneous Discussions (Moderator: Allcanadian) > Two capacitors, no paradox

2024-11-26, 22:43:45
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F6FLT	Two capacitors, no paradox « on: 2021-12-04, 10:08:48 »
Group: Experimentalist Hero Member Posts: 2072	When we charge a capacitor from a voltage source, half of the energy is lost in the circuit resistance. Efficiency: 50%. Decreasing or increasing the resistance does not change anything. And if the device were superconducting, the charging current would be infinite, so it doesn't work either. To charge a capacitor without loss, you need a current generator, because the ideal is to add the electrons one by one, so the charge voltage must increase step by step just a little above the instantaneous capacitor voltage. This can be done from a voltage generator (e.g. another capacitor) but by putting a choke in series with the capacitor, with a long time constant so that the impedance of the circuit is large compared to the resistance of the circuit. This allows an adiabatic charge. An approximation is given here, with an efficiency of 94%: https://www.sciencedirect.com/science/article/pii/S2211379717306757 (You can use sci-hub to get the paper). A few years ago, I made a very simple LTSpice model of the charge of a capacitor C2 from a capacitor C1 initially charged at 100v. Thanks to a diode in series, the system remains stable when C2 is charged (otherwise, we would have an oscillation). We see that when C1 is completely discharged, C2 is charged with the same voltage as the starting voltage of C1, no loss, no paradox, see attachment: ------------------------ Two capacitors, no paradox capToCapLoad.png (45.38 kB, 1772x1296 - viewed 452 times.) --------------------------- "Open your mind, but not like a trash bin"

muDped	Re: Two capacitors, no paradox « Reply #1 on: 2021-12-04, 17:13:42 »
Group: Tinkerer Hero Member Posts: 3055	Congratulations! Similar to the resonance condition in a parallel LC circuit where the energy shuttles back and forth with little loss. Are there any other solutions? --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

Allcanadian	Re: Two capacitors, no paradox « Reply #2 on: 2021-12-04, 17:45:11 »
Group: Moderator Hero Member Posts: 2735	F6FLT I think you may have missed the point. The paradox was not whether all the energy from one capacitor can be transferred to another. It has been known for decades that adding an inductor can help transfer more energy and increase efficiency. I did a real experiment and proved as much on my bench. https://www.overunityresearch.com/index.php?topic=3994.0 The real paradox relates to "where and how" the energy dissipates when one charged capacitor balances with an empty capacitor. In my opinion this real experiment and science paper resolve the paradox... The Paradox of Two Charged Capacitors – A New Perspective https://arxiv.org/pdf/1309.5034.pdf Simply put, it was shown that once switching, conduction and radiation losses were removed a change in charge density was the only correct solution to explain the energy loss. So you are correct that the paradox has been resolved. Regards AC --------------------------- Comprehend and Copy Nature... Viktor Schauberger “The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman

Magluvin	Re: Two capacitors, no paradox « Reply #3 on: 2021-12-04, 22:35:28 »
Group: Mad Scientist Hero Member Posts: 549	As in the pdf, starting the last sentence of page 5 suggests the end result of 2 water tanks that the energy is lost in heat, if we didnt do anything with the action of the transfer from tank to tank. Say 10gal in tank 1 and 0gal in tank 2. we should end up with 5gal in each tank when done. Same a caps, 10v willl divide into 5v in each. Same as we can count the gal in the water tanks, we can count(calculate) the number of indifference of electrons between + and - plates according to the cap value and the resultant voltage for each. No matter the resistance involved with the transfer, heat developed, energy generated by making use of the transfer till complete, we still end up with 5gall in each container or the same voltage in each cap. Even if we could eliminate all losses, heat and otherwise, in the end we will have 5gal and 5v.... The loss is only because we let it disappear in a case of zero measurable or calculable losses by letting the pressure be reduced, nothing more. If you end up with heat, then yay, we have avoided that stupid loss. If we gen electricity by use of the transfer, then yay, we have avoided the loss again. But in no way did we lose 50% by way of resistance, or heat, etc, because in the end the number of gal or electron count imbalance of the cap plates will not change no matter how we look at it. if there were no loses evident, the 5gal and 5v would remain the same result. The same as the pdf gives an alternate view of the same experiment but of zero resistance wires to avoid loss, he explains that the loss can be had by just increasing the cap plates to 2 times the area.. The capacitance increases or the container of water increases in width, which lowers the water level thus decreasing the bottom pressure, or the voltage in the cap which is reduced pressure. I like that example in place of the ideal super conducting example as it is more realistic but accomplishes the same idea. So.. If conservation of energy says it cannot be destroyed or created.. But in this case, can it be destroyed? If so, then can it be created also? Think. If after all this time, we are just now becoming aware of the actual results of this widely known paradox, then it may be just a bit longer to figure out how to create it. Mags

Magluvin	Re: Two capacitors, no paradox « Reply #4 on: 2021-12-04, 23:18:32 »
Group: Mad Scientist Hero Member Posts: 549	What brought me to these conclusions after being in disbelief of the issues of the paradox was an argument presented to me using an ideal scheme where there is no resistance thus no heat losses and it was explained to me that the end result would be 7.07v in each cap from the initial 10v cap. I already had the analogy of water tanks or air pressure in my mind and thoughts, how could we end up with 7.07gal of water in 2 buckets from 1 with 10gal to begin with? Same with air pressure. Well that cant be.. So we look at Coulombs Law to work out the same for the charges of a capacitor and we know that this cannot happen with the cap to cap. We cannot end up wit 7.07v in each cap from 1 cap of 10v,even without any known losses, unless we were to introduce or inject more electrons in the system, as we would have to do with water or air to get the illusion of positive results. Now, if in a simulator using ideal components, we use an ideal inductor and ideal switching and we close the switch and wait for the voltage of the source cap go down to 7.07v, then open that switch from the source cap and simultaneously close another switch that would connect the charged series inductor across the receiving cap to finish the charge of cap 2 until cap 2 is also 7.07v then disconnect that switch. All avoiding diode junction voltage drops and such. Now the energy in both caps total is equal to the energy of the initial source cap of 10v. Did it on SIM circuit simulator. So the ideal version of the experiment will result in the same 50% loss because we did not do anything with the transfer, The energy was just lost very stupidly by doing such. Mags

muDped	Re: Two capacitors, no paradox « Reply #5 on: 2021-12-05, 02:27:34 »
Group: Tinkerer Hero Member Posts: 3055	At the instant the two capacitors are connected closely examine the resultant current impulse. What is the pulse shape, the pulse duration and the current distribution curve? Is any form of resonance observed as a consequence of the very brief pulse? Is energy radiated as a consequence of the phenomenon? We must remember that every capacitor has some inductance present in its makeup. The interconnecting wiring has stray inductance as well. Is a Radiant Impulse developed? Some form of Energy Transformation is obviously taking place. Energy never goes away quietly. It always makes some sort of noise. --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

Magluvin	Re: Two capacitors, no paradox « Reply #6 on: 2021-12-07, 05:06:32 »
Group: Mad Scientist Hero Member Posts: 549	Here is the part that you are missing... In this exercise, if we were to eliminate all losses, there would still be a 50% loss. No matter what, the loss is the same with known losses or with none at all. With any other circuit that the losses were eliminated, say with a source and a motor, we would end up with 100% output and zero loss. But here, the loss remains no matter what. By doubling the container size, for either water, air pressure or even electrons, we have released the initial pressure to half. By doing so, we lose 50%. Now, lets say this. Instead of increasing the container size, we shrink it. We somehow reduce the size of the cap plates by 1/2. Do we get an increase in energy? Mags

muDped	Re: Two capacitors, no paradox « Reply #7 on: 2021-12-07, 07:09:45 »
Group: Tinkerer Hero Member Posts: 3055	Quote from: Magluvin ...We somehow reduce the size of the cap plates by 1/2. Do we get an increase in energy?... This, in essence, is what takes place within the Parametric Amplifier when the VariCaps are pumped. As the plates of the Capacitor are electrically moved there is an instantaneous voltage change across the Capacitor proportional to its instantaneous charge level. In this process is energy lost or is it gained? --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

Magluvin	Re: Two capacitors, no paradox « Reply #8 on: 2021-12-07, 07:22:36 »
Group: Mad Scientist Hero Member Posts: 549	Quote from: muDped on 2021-12-07, 07:09:45 This, in essence, is what takes place within the Parametric Amplifier when the VariCaps are pumped. As the plates of the Capacitor are electrically moved there is an instantaneous voltage change across the Capacitor proportional to its instantaneous charge level. In this process is energy lost or is it gained? Well what amount of change happened to the cap? Did the capacitance reduce? And if the voltage increased, then calculate the energy before the cap change and the energy after the cap change. Mags

muDped	Re: Two capacitors, no paradox « Reply #9 on: 2021-12-07, 14:52:15 »
Group: Tinkerer Hero Member Posts: 3055	The WikiPedia article which explains the operation of the Varactor Capacitor Parametric Amplifier is quite good. The principle of decreasing capacitance, by moving the plates further apart electrically, is exploited to produce a signal voltage gain. The over-all operation of the amplifier is much more complex in that it operates as an oscillator-mixer to produce its amplified output as a consequence of that voltage gain. Are we to conclude then that decreasing the capacitance transforms the context of the charge without loss? --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

Magluvin	Re: Two capacitors, no paradox « Reply #10 on: 2021-12-08, 00:01:39 »
Group: Mad Scientist Hero Member Posts: 549	Anyway, considering, I believe it is still a bit of a paradox. Mags

F6FLT	Re: Two capacitors, no paradox « Reply #11 on: 2021-12-11, 15:24:58 »
Group: Experimentalist Hero Member Posts: 2072	Quote from: Allcanadian on 2021-12-04, 17:45:11 F6FLT I think you may have missed the point. The paradox was not whether all the energy from one capacitor can be transferred to another. It has been known for decades that adding an inductor can help transfer more energy and increase efficiency. I did a real experiment and proved as much on my bench. https://www.overunityresearch.com/index.php?topic=3994.0 The real paradox relates to "where and how" the energy dissipates when one charged capacitor balances with an empty capacitor. In my opinion this real experiment and science paper resolve the paradox... The Paradox of Two Charged Capacitors – A New Perspective https://arxiv.org/pdf/1309.5034.pdf Simply put, it was shown that once switching, conduction and radiation losses were removed a change in charge density was the only correct solution to explain the energy loss. So you are correct that the paradox has been resolved. Regards AC I'm not sure if you interpreted this paper correctly, because it says, in other words, the same thing I said. Heat is kinetic energy. This is what we find when the resistance of the circuit heats up because of the current: the electrons have transferred their kinetic energy to the crystal lattice of the resistance conductor. If the resistance of the circuit is zero, then the kinetic energy of the electrons would be conserved with their movement. Of course the dimensions of the circuit make it have an inductance, so we find ourselves like in my diagram with the self-inductance, but without the diode. In these conditions the system would oscillate indefinitely at the LC resonant frequency, preserving the kinetic energy of the electrons. There could be loss by electromagnetic radiation, but it's another question, an adequate setup could remedy this. The energy is only lost if the kinetic energy of the electrons is dissipated somewhere. All this is absolutely trivial, I don't even see the shadow of the slightest paradox, but Singal expresses it in a rather twisted way. I am afraid that it is not the best method for a paper which wants to be educational. --------------------------- "Open your mind, but not like a trash bin"

F6FLT	Re: Two capacitors, no paradox « Reply #12 on: 2021-12-11, 16:01:55 »
Group: Experimentalist Hero Member Posts: 2072	Quote from: muDped on 2021-12-07, 14:52:15 The WikiPedia article which explains the operation of the Varactor Capacitor Parametric Amplifier is quite good. The principle of decreasing capacitance, by moving the plates further apart electrically, is exploited to produce a signal voltage gain. The over-all operation of the amplifier is much more complex in that it operates as an oscillator-mixer to produce its amplified output as a consequence of that voltage gain. Are we to conclude then that decreasing the capacitance transforms the context of the charge without loss? Yes, you can say it like that. The energy in the capacitor being E=0.5*Q²/C, we can see that if the capacitor is charged with constant Q, then dividing its capacity by two, for example, by moving the plates away, doubles the energy. A clever way to use this method is to do it in synchronism with a signal to be amplified. The mechanical energy supplied is found in the increase of the stored electrical energy. Of course the mechanical energy is obtained from electrical energy, which allows from a control signal (the "pump"), to modify the capacitance of the varactor. This is what we call a parametric amplifier. At a time when conventional microwave receiver amplifiers did not exist, as in the early 1960s, this is what was used instead, often cooled to very low temperatures to reduce noise. This type of amplifier, associated with a 34 m horn antenna, protected by a huge radome, allowed the first France-USA TV link by satellite in 1962 from the Pleumeur-Bodou station, in Brittany. Like all parametric systems, the useful energy to be amplified increases, but only at the cost of the energy used to vary the "parameter", like the capacitance. The free energy is not there. --------------------------- "Open your mind, but not like a trash bin"

muDped	Re: Two capacitors, no paradox « Reply #13 on: 2021-12-11, 17:14:55 »
Group: Tinkerer Hero Member Posts: 3055	Quote from: F6 Like all parametric systems, the useful energy to be amplified increases, but only at the cost of the energy used to vary the "parameter", like the capacitance. The free energy is not there. Salient portion with bold emphasis Aye, and could we not expand the scope of this true statement to encompass virtually all electrical and electronic circuitry? The "Free Energy" is not there. --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

F6FLT	Re: Two capacitors, no paradox « Reply #14 on: 2021-12-11, 18:54:29 »
Group: Experimentalist Hero Member Posts: 2072	Quote from: muDped on 2021-12-11, 17:14:55 Salient portion with bold emphasis Aye, and could we not expand the scope of this true statement to encompass virtually all electrical and electronic circuitry? The "Free Energy" is not there. I think we are more likely to find it in extremely nonlinear systems, or by a Maxwell demon. Concerning parametric systems, it is perhaps possible to get free energy if the parameter can be modified by a natural phenomenon at no cost, for example if the heat makes the plates of a capacitor move apart, we can perhaps create a thermoelectric generator (I already thought about it, the problem is that it is too slow to have a useful energy). --------------------------- "Open your mind, but not like a trash bin"

Magluvin	Re: Two capacitors, no paradox « Reply #15 on: 2021-12-12, 03:40:47 »
Group: Mad Scientist Hero Member Posts: 549	Quote from: F6FLT on 2021-12-11, 15:24:58 The energy is only lost if the kinetic energy of the electrons is dissipated somewhere. All this is absolutely trivial, I don't even see the shadow of the slightest paradox, but Singal expresses it in a rather twisted way. I am afraid that it is not the best method for a paper which wants to be educational. Im sure there are many that are highly intelligent, that have really dug deep into this subject in the past, even before it was labeled a paradox. I can see the problem that is faced here. The problem I had with it all was the explanation that the energy was lost in some fashion or another, and that it was the function of the energy that escaped the system(cap to cap) is how the loss occurred. Again, I began to comprehend that the 50% loss was not due to the energy or energies that escaped the system. The cap to cap, water tank to water tank, air tanks to air tank, are probably the only examples that if you apply IDEAL wisdom to the system, that if you can eliminate the heat, resistance, even get rid of inductance, that in the end, when each bucket, each air tank and each capacitor, level out to the same pressure for each of the pairs, we still lose 50%. Now I propose that we consider the identity of the initial (stored)energy to be called Pressure Energy. whether its Lb per sq in, or emf of an electrical source. Once we relieve that initial pressure from the first cap or tank, we only have from 10v to 7.07v or 100lb to 70.7lb to stop the transfer at that point to end up with only 1/2 of the initial energy that we started with, and even less than 5v or 50Lb in the second cap or tank. But, we didnt lose 50% total. Mags

Allcanadian	Re: Two capacitors, no paradox « Reply #16 on: 2021-12-12, 23:44:08 »
Group: Moderator Hero Member Posts: 2735	magluvin I would agree and a paradox is simply a failure to fully understand the nature of the problem in question. I did some more research/testing on the two capacitor paradox and learned some more. It's like anything, a work in progress and learning and moving forward is what's important. It's when our thoughts become static with no clear path forward that we seem to have problems. I also read some interesting articles on the nature of science which really helped. That in many respects there is no absolute proof of anything only evidence to suggest a way forward to better answers. So it is evidence we seek not so much proof as a be all end all to anything. This hit home in my mind and as we learn more we acquire more evidence always building our case towards a better understanding. As such I'm not sure I believe anything and I'm simply following along wherever the supposed evidence leads me... I'm also gearing up for a couple new experiments on the two capacitor problem and may have discovered something. If in fact the charges do move or disperse on the plate surface then we should be able to measure it. So I reconfigured my charge detection array to see if I can detect this motion. I will probably have to write some new code in LabView which may take some time. Oh, the life of a nerd and I find this kind of stuff really exciting, lol. Regards AC « Last Edit: 2021-12-13, 06:12:35 by Allcanadian » --------------------------- Comprehend and Copy Nature... Viktor Schauberger “The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman

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