overunity problem or something else - Page 1
Forum section: electronics / beginners

author: RIS
author_rank: Regular Contributor
timestamp: December 11, 2015, 10:03:34 pm
content: I see this video been an extremely interesting and I need help in understandingI'll try to explain my small problem.we have a battery with a rated capacity that means can deliver a precise quantity of energy.so when I connect a capacitor in a circuit as in this video through switch through a resistor and of course its filling capacitor,Now we take that charged capacitor and connect resistor across the capacitor.Now would not, like that constantly charging and discharging through same resistor increased energy efficiency of batterie for purely mathematicaly 100%.in other words resistor would receive twice as much energy rather than from independent battery source .

author: Ian.M
author_rank: Super Contributor
timestamp: December 11, 2015, 10:24:40 pm
content: No. TANSTAAFL applies.In the limit, if you switch quickly enough, the resistor's power dissipation will be 1/4 of the directly connected figure. As you slow down the switching the mean dissipation tends towards an asymptote of zero.If you doubt this, and you have a high school maths diploma and are still competent at integration, you should be able to prove it from first principles, Ohm's law and Coulomb's law.If you believe otherwise, you are either mentally incompetent or severely undereducated

author: RIS
author_rank: Regular Contributor
timestamp: December 11, 2015, 10:40:56 pm
content: no really I do not wanna disrespect anyone,I'm not an electrician and this idea seemed to me a bit unusualNow that you know that I'm not an electrician or engineer whether you can explain it simpler.Of course instead of resistors may be an inductor or bulb.

author: miguelvp
author_rank: Super Contributor
timestamp: December 11, 2015, 11:53:14 pm
content: Water analogy to help you understand.The battery is a water tower reservoir.A wire is a large pipe with a big diameter attached to that reservoir. It has some resistance just like wires do, but can handle a decent amount of water flow.A resistor is a pipe with a small diameter, it offers a lot of resistance and restricts the water flow.A capacitor is a balloon with walls thick enough that can hold certain water pressure.If you turn off the water main, there is no pressure on the resistor (small pipe) only on the capacitor (balloon)No pressure is ever stored in the small pipe.

author: AlfBaz
author_rank: Super Contributor
timestamp: December 12, 2015, 01:29:53 am
content: Quote from: RIS on December 11, 2015, 10:40:56 pmWhen charging the capacitor through a bulb the battery has to do 2 things, use energy to light the bulb and use energy to charge the capacitor (charging a battery or capacitor isn't free). When the capacitor lights the bulb it only has to do one thing so essentially the battery has to do twice the work of the capacitor.

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 11:18:28 am
content: one man said,capacitor is more powerful than dynamiteI agree with your idea that the batteries must do two things it seems logical -but as far as I am familiar with capacitors They have suction power or something like a vacuum,because of this the effect when we filling large bank of capacitors or individual capacitor we have to have some kind of resistance which corresponds to our power supply otherwise we would destroy power supply or source.so capacitor has interesting properties and I believe that the battery does not have to do two things.Another example perhaps simpler than capacitor but in general is same.we have a charged battery 12V , 100Ah when we take another same battery but completely empty,If we connect the two batteries in parallel it will happen 50% of capacity equalization ,before that happens we can connect some kind of load between terminals + + or - - to make things even more interesting it can be some switching transformer that can charge a third battery,and it has to work to the point of equalization of capacities and because of the potential differences.at the end of the process we are left with the initial capacity(two batteries with 50% capacity, which can be for later use connected in series) +some amount of energy in the third battery.now it is more than crazy and I really do not want to make any problems here(on forum) I just drain all standard ideas how to solve this problem and really looking for someone with a higher education for help,and this forum it seemed like a good place.

author: SeanB
author_rank: Super Contributor
timestamp: December 12, 2015, 11:50:25 am
content: Taking your 2 batteries, you end up with 2 batteries charged to 25%, as half the energy is dissipated in heat in the ESR of the 2 batteries. If you use the current flow as a power source to charge a third battery, and have some converter that has zero voltage drop at any current flow ( pretty hard, even an AC current transformer on a solid wire will cause a measurable voltage drop in the wire in addition to the existing one due to the wire resistance just from it's effect on the phase of the current WRT the voltage) it would only have a short pulse of current to charge the third battery, so you would merely be moving the heat generation out of the ESR of the 2 cells slightly, and the third battery would only get a very small charge.

author: FrankBuss
author_rank: Unknown
timestamp: December 12, 2015, 12:05:44 pm
content: Quote from: RIS on December 12, 2015, 11:18:28 amNo, the two batteries will be not 50% charged. Forget about RMS and other details, just think of the water and ballon analogy: The first battery is fully charged, full ballon. Then you connect the second battery, same ballon but empty. After some time both ballons are half full. If you connect a 3rd ballon in between while the water is flowing and later disconnect it, the 3rd ballon might be 10% full, the other two ballons 45%.You can try it with capacitors yourself. Buy some 10000 uF capacitors, use a 1k series resistor (will give you 10 seconds time for charging and discharging) and test it with a multimeter.

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2015, 12:41:10 pm
content: You need to make the distinction between battery 'charge' as a percentage of the maximum energy capacity, and electrical charge in iCoulombs. The electrical charge will equalise - it has nowhere else to go and if the batteries or capacitors are identical will split 50/50 like water finding its own level in two identical tanks connected by a pipe when you open the valve. However in the process, the charged corrispondattery voltage drops, so half the electrical charge at a lower voltage is NOT as much as half the energy that was originally in the charged battery. The current flowing through the resistance of the wire and internal resistance of the batteries produced heat, and the loss of stored energy taking the two batteries as a pair exactly corresponds to the heat produced, resulting in two batteries with <50% 'charge'.,

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 12:59:58 pm
content: exactly two bacteria will always have 50%/50% Regardless of what type of load you take in between heat or whateverand let that the voltage also drop 6 volts and 50Ah but connected in series is 12 volts 100Ah We again hawe original state 12 V 100Ah +what we pulled between the terminal when sharing capacity.I also do not believe in overunity-but also do not understand why the usuall empty containers make so much differencemaybe I'm yust crazy but I also see any use of energy between identical terminal is+ compared with only one battery

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2015, 01:33:55 pm
content: A battery that has dropped from 12V to 6V is effectively dead and will have a massively increased internal resistance. Connect two in series and you are back to 12V NO LOAD. Connect a normal load, and you'll be lucky if the terminal voltage of the series pair is a couple of volts.

author: dom0
author_rank: Super Contributor
timestamp: December 12, 2015, 01:44:43 pm
content: Quote6 V / 50 Ah in series produces 12 V / 50 Ah (not 100 Ah). This is exactly half the original energy (12 V / 100 Ah). Exactly what the "capacitor paradox" predicted (it is not so much about capacitors, but is true in general for linear charge redistribution processes).Of course in practice it will be even less, since the battery charge/discharge process has losses beyond ESR.

author: FrankBuss
author_rank: Unknown
timestamp: December 12, 2015, 01:58:55 pm
content: Quote from: RIS on December 12, 2015, 12:59:58 pmNo, and it is very easy to understand. Just connect one resistor to one battery. The resistor gets hot (ony a little bit if it is a high resistor) and the battery gets discharged. With your two batteries, the resistor gets hot, too, and the two batteries gets to maybe 45% both, and 10% was emitted by the hot resistor (of course, only for ideal components, in practice there will be more losses). Same if you use another battery or capacitor instead of the resistor. As mentioned before, don't trust anyone, measure it yourself

author: Zero999
author_rank: Super Contributor
timestamp: December 12, 2015, 02:05:33 pm
content: Quote from: wilfred on December 11, 2015, 10:11:14 pmDon't be like that. This is the beginners section for goodness sake!Don't put people off asking questions. Just because someone talks about over unity, it doesn't mean they're trolling. The could be asking a legitimate question to fill a gap in their understanding.Quote from: RIS on December 11, 2015, 10:03:34 pmActually, if a capacitor is charged via a resistor (be it just the resistance of the cable, battery and the effective series resistance of the battery) half of the power will always be dissipated in the resistor. If you connect a 1F capacitor to a 12V battery, the energy in the capacitor will be:E =1/2CV2= 0.5×144 = 72JHowever, 72J will also be lost in the resistance of the cable, the battery and capacitor, so 144J will be taken from the battery.

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 02:17:03 pm
content: thank you boys It was the first league explanationand it is interesting how much energy disappears unbelievably

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 02:28:41 pm
content: capacity sharing that brought me back to life -and of course the rest is a logical

author: onlooker
author_rank: Frequent Contributor
timestamp: December 12, 2015, 02:36:39 pm
content: QuoteThis is a good explanation. I like to add the part that Hero999 did not explicitly say: in an ideal serial battery, resistor and capacitor charging model, the energy dissipated through the resistor is the same as the energy stored in the capacitor and is independent of the value of the resistor.

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 02:51:40 pm
content: and when we are filling capacitors ,in the initial state we have a potential difference as a standalone battery but during the filling this difference declining so it is practically the same with or without capacitor.my problem is I took the battery purely quantitatively one full battery is the same as two half full and I did not take into account the voltage drop

author: FrankBuss
author_rank: Unknown
timestamp: December 12, 2015, 03:30:35 pm
content: Quote from: onlooker on December 12, 2015, 02:36:39 pmInteresting, I didn't know this either and was confused by the voltages. I did an experiment: a 220 uF capacitor charged to 3 V. Then connected another empty 220 uF capacitors, waited a few seconds, and disconnected. Measured 1.5 V for each capacitor. This might be surprising for programmers like me or overunity experts, but it is right, because the stored energy of a capacitor is 0.5*C*V^2. A 220 uF capacitor, charged to 3 V, stores 990 mJ. A 220 uF capacitor, charged to 1.5 V, stores 247.5 mJ. So half of it, 495 mJ, was lost in the resistance (just a wire in this case, and of course, I measured a bit less than 1.5 V).But this is only true because of the voltage difference. Can you discharge a capacitor without loss (in theory), if you connect it to a coil?

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 04:56:31 pm
content: I think capacitor does not lose power because of resistance-I think it has to do with, charge density of the surface area of a distance.creating more voltage through the coil will not increase power it will create additional dissipation in one discharge.but there is one catch 22 resonant frequencyIf capacitor is discharged through the coil at resonant frequency with some Q factor it would not increase power but would increase the time appearance of that power.

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2015, 05:06:22 pm
content: An ideal capacitor would have superconducting plates separated by vacuum and the only energy losses would be EM radiation from the edges.A practical capacitor has non-zero plate resistance and various other losses due to leakage, dielectric dissipation and absorption.

author: onlooker
author_rank: Frequent Contributor
timestamp: December 12, 2015, 05:20:39 pm
content: QuoteFor an ideal inductor (also ideal capacitor and battery), the circuit will oscillates forever. Any serial resistance will dump the oscillation. Since I do not remember a quick way to do this calculation, LTSpice comes to the help.Interesting enough, again, at the end and for all I can tell, the energy wasted on the resistor is the same as that stored in the capacitor and is independent of the values of the resistor and the inductor.

author: onlooker
author_rank: Frequent Contributor
timestamp: December 12, 2015, 08:09:58 pm
content: It turns out it is not that hard to mathly prove the invariant statement in my last post (an exercise for 2nd or 3rd year college student).In any case, "https://en.wikipedia.org/wiki/RLC_circuit" has the solution listed. The only thing one needs is to show that integral(R*I^2) (t=0 to infinity) is independent of both L and R, and depends only on C and V (or =0.5*C*V^2 to be precise).

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 08:56:32 pm
content: yust Dont hit me with this negative waves equationscommon sense said voltage should go Infinity and they did not take into account Tesla work and how he disprove hertzalthough I have no argument for hertz,also my recent study of resonance says whenever appears causes excessive voltage.

author: dom0
author_rank: Super Contributor
timestamp: December 12, 2015, 09:25:29 pm
content: Quote from: RIS on December 12, 2015, 08:56:32 pmNo it doesn't. It never does. Because there are- no inductors with zero resistance and zero stray capacitance- no connections without mutual capacitance- no switches with zero off-state capacitance- no switches with infinitesimal fast switching speed

author: RIS
author_rank: Regular Contributor
timestamp: December 12, 2015, 09:39:16 pm
content: hmmm maybe you're right but statements from Tesla 20 000 000V even 20 000V is enoughso something somewhere is wrong and I have mechanicall feeling that it is possible.

author: RIS
author_rank: Regular Contributor
timestamp: December 13, 2015, 04:29:24 pm
content: I again and There is something very wrongfor example-when We discharge battery let say 50% voltage remains unchanged Only capacitance is variedso I believe that we are again at the beginning of my problem

author: miguelvp
author_rank: Super Contributor
timestamp: December 13, 2015, 07:34:06 pm
content: Quote from: RIS on December 13, 2015, 04:29:24 pmTo answer that you have to be able to understand this graph:You can use the voltage output to find out the remaining charge as long as the battery is "under load" Just measuring the battery cell to a DMM wont do.Look for Dave's videos regarding batteries, he does explain it in detail.

author: mikerj
author_rank: Super Contributor
timestamp: December 13, 2015, 08:29:14 pm
content: Quote from: RIS on December 13, 2015, 04:29:24 pmExcept non of that is actually true. When you discharge a battery by 50%, you reduce the loaded voltage and increase the equivalent series resistance.

author: RIS
author_rank: Regular Contributor
timestamp: December 13, 2015, 08:32:50 pm
content: yes I fully agree with itbut if you take two batteries one full and the other completely empty if you connect them in series you can take the "free" energy between two of the same positive terminals.If you take enough thick cable equalization of capacity will happen instantly,If you take a thin cable equalization will happen after some time,if you take bulb Equalization will happen over time, depending on the resistancealso let we consider light bulb which has a dissipation factor that is heat,but I think that this waste is only applicable for the bulb not battery,battery is not interested in the dissipation factor of bulb and logically electrons or whatever they simply have nowhere to go they must end up in second battery and second battery energy can be used later.when we use battery in classic way connecting plus and minus between we use electricity for whatever that is the end of storie once when we lose a potential difference there is no turning back, only with chargeing.My mode if only if possible will allow another passage of electricityOf course this is only debate and really do not see what could be hiding behind I think on problems.

author: IanB
author_rank: Super Contributor
timestamp: December 13, 2015, 09:13:36 pm
content: Quote from: RIS on December 13, 2015, 08:32:50 pmWhat you are missing is the law of conservation of energy.Otherwise stated as "You can't get something for nothing."Also known as "There is no such thing as a free lunch."Electricity itself does not light bulbs, or run motors, or charge batteries. What happens is that when electricity flows through a bulb it does work, and doing work uses up energy. If that energy was in the battery to start with it is depleted and less of it remains after the bulb has been running.Just like if you push a car down the road you do work moving the car. After you have moved the car, 100% of the car remains and none of it is used up. However, you have used up energy and your muscles ache. You have to see electricity the same way. You can move it around, and moving it around takes energy. It requires energy to push electricity into a battery to charge it, and when you take the electricity out of the battery it gives up energy to do useful things.

author: RIS
author_rank: Regular Contributor
timestamp: December 13, 2015, 09:34:02 pm
content: yes I agree with that-although I have a slightly different view(The only difference is that I think we need lot less quantity off energi than standard model says)but it's different subject.feel free to look at my example and find a reasonable solution for problem.

author: IanB
author_rank: Super Contributor
timestamp: December 13, 2015, 09:56:34 pm
content: Quote from: RIS on December 13, 2015, 09:34:02 pmIf your example produces a result that does not match experiment, then you have made a mistake in your calculations, or you have a mistake in your understanding.It is important to understand that thought experiments are just that; experiments you do in your head. Whatever you do in your head can come up with any kind of unreasonable result, because it is not real.If you want to consider what happens when you connect two batteries, with thick wires, or with bulbs in between, you must do this experiment for real on a workbench and measure the results carefully. If you do that you will find that nothing mysterious happens.

author: RIS
author_rank: Regular Contributor
timestamp: December 13, 2015, 10:19:16 pm
content: I thought before any actions make a reasonable discussion with experts so to eliminate possible errorsI believe such a way of thinking would solve most of the problems in today's world.

author: Ian.M
author_rank: Super Contributor
timestamp: December 13, 2015, 11:07:53 pm
content: Scientists and engineers do research and experimental work to validate their theories. Philosophers in the ancient Greek tradition sit on the forum steps and waffle, preferring elegant discourse to reality.Which do you want to be?

author: onlooker
author_rank: Frequent Contributor
timestamp: December 14, 2015, 02:53:18 am
content: The problem is that OP's mental experiments were full of holes according to physics theory as others have explained. But, the OP did not understand or accept them.I can see the hopelessness here. Apparently, the OP needs some basic grasp about simple physics theories.If the OP did not learn such simple theory and calculations before, he should just take any well established physics laws, such as the "energy conservation law" in such a traditional discipline (or any field) as a given fact. Any thing inconsistent should be caused by errors of OP's arguments.On the other hand, if the OP assumed the "energy conservation law" may not be true in this case, then, essentially, he is rejecting any established theory that is partly based on the "energy conservation law" and no theory or calculation will help. For OP's such extraordinary claim, he is better to show experimental evidences. And we all know any such experiment will surely prove he is wrong.

author: RIS
author_rank: Regular Contributor
timestamp: December 14, 2015, 09:41:41 am
content: there is no need for sarcasm boys,have not I already proved This is a small insignificant problem and should be dismissed instantaneously according to standard model-in fact, we do not even need to do measurements simple paper and pencil right.Of course I'm not an expert that is why in the first place I came here.if you do not want to help it's OK , but you could tell at the beginning You are dumb ass you have not learned anything and therefore you do not deserve help and learning right.but look even I have an excuse improper education,but you boys you have no excuse if you stand behind what you claim who you are.and in the end we are left with only one answer inappropriate behavior regardless of education.Merry Christmas

author: onlooker
author_rank: Frequent Contributor
timestamp: December 14, 2015, 11:46:00 am
content: I am sorry if you felt that way, maybe, its because I have had bad flu and fever and stayed at home for several days now.My problem is not if the question is basic or not, but, many times, your reply to other's explanation did not make sense and showed the lack of understanding. In such case, try to understand and ask pointed questions is more helpful to you and others.And you changed the setup from the original battery+capacitor to battery+battery. But why? Did you feel the battery+capacitor problem is solved and accept that no overunity is possible there?Then, when you talked about bettery charge/discharge, your premises and arguments are wrong:To simplify the explanation, we can assume the batteries are "ideal" in the sense that the amount of energy to charge it is the same as the energy the battery will output when discharging (lossless). In this case, yes, at equal voltage, each bettery will be charged to 50% rated Ah.But, you forget that50% rated Ah means<50%rated Whsince the same amount of Ah delta represents more energy at higher voltage (near full) than at lower voltage (near empty). That is, adding up two batteries charged to 50% rated Ah will have less energy than one bettery charged to 100% rated Ah. The energy difference is dissipated on the "resistor" that connectes the 2 batteries. The same invariant is true here too: the total wasted energy on the resistor is independent of the value of the resistor, be it a bulb or a thick wire.Note here, "Ideal" is not to mean a constant voltage source. If the battery is modelled to have its voltage independent of the charge level, the full battery will never charge the empty battery.If the battery is not "ideal", you would have more energy loses. The amount of energy you can output from a battery is always less than the energy you put into the bettery. That is, the battery you charged with 50% rated Ah (or Wh) as input will actually hold less than 50% rated Ah (or Wh).

author: RIS
author_rank: Regular Contributor
timestamp: December 14, 2015, 02:09:52 pm
content: flu and sickness that I did not take into account so you have my discount on that.I took out a capacitor due to the complexity of charging and discharging I did not want to create additional misunderstandings Therefore the battery is simpler example.yess I agree that the battery can never give 100% It will therefore be perhaps 95% out The same will be for two batteries at 50% capacity maybe even more.when We connect two batteries in parallel one is 100% full second is empty and during the equalization of potential both battery will have losses and it is an internal problem of the battery.but if We try take power when the same ++ poles are connected through some load dissipation will happen again but I think I will not affect the equalization of capacitiesThere is a simple answer current must flow between two terminal and go to the lower potential of empty battery if you extract from the first battery some amount of energy let say 15% and blame resistance as that amount of energy did not come to empty battery then we have the energy disappearance as like terminals are not connected.

author: Omicron
author_rank: Regular Contributor
timestamp: December 14, 2015, 02:57:04 pm
content: You are confusingcapacitywithenergy. They are not the same things! Capacity is the ability of the battery to deliver a current for a given time. To convert that into energy you still need to multiply with the batteryvoltage. In your example the battery voltage willnot stay constant. As onlooker has tried to explain to you, if you do the maths properly you will find that two batteries charged to 50% capacity containlessenergy than one battery charged to 100% capacity (assuming identical batteries).

author: Nerull
author_rank: Frequent Contributor
timestamp: December 14, 2015, 03:51:08 pm
content: Frankly I'm not sure what anyone can gain by arguing over made up rules that even the OP doesn't understand. If you're going to just ignore any physics you don't like, make up rules in your head, and refuse to do any experimental validation you can "prove" the earth floats through space on the back of a giant turtle if you like. If you refuse to use a telescope in case you might find out you were wrong, it doesn't show a lot of confidence in your conclusion, though.

author: Ian.M
author_rank: Super Contributor
timestamp: December 14, 2015, 05:15:19 pm
content: +1. Life's too short to argue with inadequately educated, or 'hard of thinking' 'philosophers'!@O.P. Take a pair of well matched NiMH D cells, whatever circuits, components etc. you like. Set up the circuit to run unattended and put the whole lot in a sealed tin totally immersed in a known volume of water in a Dewar flask with a mag stirrer and a sensitive RTD thermometer probe. You have then got what is basically a 'bomb' calorimeter, and if you even were to get 1% excess energy it would be *BLOODY* *OBVIOUS*. If you get any interesting and reproducible results that appear to be not fully compliant with the laws of thermodynamics, *PUBLISH*, as a letter to a respectable journal that hilights the inconsistency and asks for suggestions to resolve it rather than making wild claims.If you diligently pursue this course of action, you are most likely to end up in 20 years time as a well-respected 'debunker' of over-unity claims!

author: RIS
author_rank: Regular Contributor
timestamp: December 14, 2015, 05:54:04 pm
content: yes I might do just that