overunity problem or something else

[RIS]

I see this video been an extremely interesting and I need help in understanding
I'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 .

[Ian.M]

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

[RIS]

no really I do not wanna disrespect anyone,I'm not an electrician and this idea seemed to me a bit unusual
Now 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.

[miguelvp]

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.

[AlfBaz]

no really I do not wanna disrespect anyone,I'm not an electrician and this idea seemed to me a bit unusual
Now 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.

When 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.

[RIS]

one man said,capacitor is more powerful than dynamite
I 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.

[SeanB]

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.

[FrankBuss]

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.

No, 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.

[Ian.M]

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'.,

[RIS]

exactly two bacteria will always have 50%/50% Regardless of what type of load you take in between heat or whatever
and 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 difference
maybe I'm yust crazy  but I also see any use of energy between identical terminal is+ compared with only one battery

[Ian.M]

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.

[dom0]

and 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.

6 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.

[FrankBuss]

exactly two bacteria will always have 50%/50% Regardless of what type of load you take in between heat or whatever

No, 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

[Zero999]

Honestly?? Are you really attempting to troll this forum with such a lame effort?

Someone will bite. Good luck with it.

Don'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.

I see this video been an extremely interesting and I need help in understanding
I'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 .

Actually, 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 = 72J

However, 72J will also be lost in the resistance of the cable, the battery and capacitor, so 144J will be taken from the battery.

[RIS]

thank you boys It was the first league explanation  and  it is interesting how much energy disappears unbelievably

[RIS]

capacity sharing that brought me back to life -and of course the rest is a logical

[onlooker]

However, 72J will also be lost in the resistance of the cable, the battery and capacitor, so 144J will be taken from the battery.

This 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.

[RIS]

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

[FrankBuss]

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.

Interesting, 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?

[RIS]

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 frequency
If 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.

[Ian.M]

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.

[onlooker]

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?

For 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.

[onlooker]

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).

[RIS]

yust Dont hit me with this negative waves equations
common sense said voltage should go Infinity and they did not take into account Tesla work and how he disprove hertz
although I have no argument for hertz,also my recent study of resonance says whenever appears causes excessive voltage.

[dom0]

yust Dont hit me with this negative waves equations
common sense said voltage should go Infinity

No 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