Conservation of Charge violated?

[Sephiroth]

bugger...

the maths is really confusing! as I understand a coulomb is a "quantity" of "electrons"... just for simplicity let's say it is 10000 electrons (i know it is alot higher)

so if I started with a capacitor with 0.1245c there would be 1245 electrons seperated in the capacitor.

then I ended up with a capacitor with 724 electrons (0.0724c) and another with 874 electrons (0.0874c) seperated.

Logically, you should be able to add those quantities together ie. 1598 electrons. But I guess not.

Though pulsing the cap through the inductor and ending up being more efficient is counter intuitive as well. In the first experiment there was no load and resistance was negligable so you wouldn't expect there to be a high loss even though there is.

So surely you would expect that pulsing the cap through several hundred feet of magnet wire going through transistors and resistors and diodes would be FAR LESS efficient then simply connecting the caps together. So the second experiment seems to be a bit anomylous.

Is it possibly indicating that energy IS entering the system and that energy is compensating for the losses?

[Sephiroth]

oh no... I have just gone and made things complicated again.

What if I hooked the caps in parellel o_O

Q = VC

If we take our two capacitors and put them in series (assuming that in series the cap is now a 5000uf with 15v) then we have 0.075c of charge. Which is half what we started with.

BUT if we hook them in parellel I'll have a 20,000uf cap with 7.5v in it, right?

Well, that would give us a charge of 0.15 coulombs!

[Peter Lindemann]

Charge vs Energy

oh no... I have just gone and made things complicated again.

What if I hooked the caps in parellel o_O

Q = VC

If we take our two capacitors and put them in series (assuming that in series the cap is now a 5000uf with 15v) then we have 0.075c of charge. Which is half what we started with.

BUT if we hook them in parellel I'll have a 20,000uf cap with 7.5v in it, right?

Well, that would give us a charge of 0.15 coulombs!

Sephiroth,

I think you have "hit the nail on the head", so to speak. As you said, if we start with two capacitors of 10,000uf and each one is charged with 7.5 volts at the start, then no matter how we measure them or connect them (in series or parallel), the total QUANTITY of ENERGY in them is the same. That being 0.5625 Joules.

But, as you have also seen, the Charge, measured in Coulombs, is different in these two situations. WHY? Well, because "CHARGE" is only half of what electricity is! The other half is "FLOW". One Coulomb per second is the definition of an Ampere of current. So, the potential current flow from a 20,000uf capacitor charged to 7.5 volts is higher than the potential current flow from a 5,000uf capacitor charged to 15 volts. That is all the "Coulomb" numbers are telling you. And, in fact, that is true.

Does that help?

Peter

[Sephiroth]

Thanks Peter! Yes, I am beginning to understand it.

Though the conservation of charge isn't related to Joules is it? At least not directly...

What I need to know for sure is if I am correct in thinking this violates conservation of charge, which states:

The Law of conservation of charge states that the net charge of an isolated system remains constant.

Am I misinterpreting this? I think I must be...

[patmac]

I'm experimenting with capacitors....

By using software simulation apparently is possible creat more energy using capacitor as target of electrons, like ping pong, this circuit I've built based on Tom Bearden text.

I need Big capacitors to build it, I have no in this moment,

In scope yours can see that is with one signal two signals identical.

Twice energy ? Look scope input and output is taken using 1 ohm resitor.

Resistor output take energy when big capacitor is charged, after take energy again when capacitor is dicharged,
Using bridge full wave, I converted charging and discharging in DC. monopolar.
Blue line is input
Red line output.

C on output is nulled by 1 ohm resistor I putted to effectivelly measure output. Sorry I forgot delete it

[Peter Lindemann]

Change of Reference

Thanks Peter! Yes, I am beginning to understand it.

Though the conservation of charge isn't related to Joules is it? At least not directly...

What I need to know for sure is if I am correct in thinking this violates conservation of charge, which states:



Am I misinterpreting this? I think I must be...

Sephiroth,

I think this definition has to be taken LITERALLY!!!!! That means the the charges you are discussing MUST remain ISOLATED to remain at the same value. When you "combine" them, they are no longer isolated. By doing this, you have changed the reference, AND you may have changed the STATE of the charge. The state of the charge does not change if you do not "combine" the capacitors in various ways. I have looked at this situation for many years, and I do not believe there is a violation of any known rules or "Laws".

That is my opinion.

On the other hand, we have been lied to about a lot of things in science, so I am totally for re-looking at each and every situation like this. But right now, I don't see any violations.

Peter

[patmac]

I'll try to built the circuit, to see if possible use electrons catched by capacitors, I've tried two months ago, to use capacitors in series, and including the model simulated, no is possible recover differential potential from capacitors in series apparentely because potential is more high, but impendance on capacitors too, so, energy really util is the same by ohm law.

My cicuit posted above, use a capacitor to catch electrons, electrons activate the load (resistor 1 ohm), and capacitor is charged, when is full electrons no flow, after relay is switched, energy catched by capacitor, flow again on the load and electrons go to ground.

On the scope you can see blue scope is energy taken from the battery, and red line is energy used by resistor like load, there is two signals red per one blue signal....

The problem is demonstrate this theory on practical, on practical I think that is better use DC voltage some high, to guarantee fast charge on capacitor as minimal 36 volts DC.

[theremart]

Interesting thought...

If you had a 3rd capacitor of same rating could you take the 15V ( two in series) and charge the 3rd one higher than the 2nd one charged...

Just curious...

[patmac]

If you had a 3rd capacitor of same rating could you take the 15V ( two in series) and charge the 3rd one higher than the 2nd one charged...

Just curious...

No man, I've tried to send energy to other capacitor and no works, my theory is this: capacitor conduct energy (low impendance) when is dicharged, after is charged (high impendance) capacitor no conduct energy, then you can connect them in series but like no conduct, volt meter show voltage sum, but there is no sufficient amperage to pass to 3rd capacitor ( high impendance ).... The only way that is possible at least recover some potential is sending to 3rd capacitor by using a diode to avoid circulation, but energy is too much weak, ideal diode is theorically valid but real diode -0.6 volts in direct polarization, and you must conecting only by one second if you let it too much time, result is like without him. Remember the real model of diode, circuit is closed by diode.....

With this theory I've desing example above, I think that is like Tom Bearden explains about a colector. Energy moves the LOAD but electrons used by the load is catched by capacitor to reutilize it again.

[patmac]

Conservation Law can be used in real life...

This is basically how I have it set up though it isn't the exact schematic.

Sephiroth

I'm refining this schematics, I think that is possible recover energy "splitting positive" with capacitors; capacitors are highly efficient to catch energy.

Look this schematics, OUTPUT is Twice INPUT, using simple software simulation, is like the experiment with three batteries to split the positive (by Jonh Bedini) , but simplified.

I made a little test in real life and the circuit works, I need to find a way to measure the efficiency, I need a purpose to do it. Because I could be wrong.

I've simulated and tested capacitor paralel after in series and don't works in simulation and real life, I think that impendance change in capacitors depending of level charge, so using only one capacitor is like ping pong theory, if this works we can imaginate if use this experiment on inductors.
(Capacitor dicharged = closed circuit; Charged = Open) (Middle charged = middle closed or open????)
Efficiency; I want to measure without include relay consumption because the system must be proved on ideal case first. If works we can test with transistors mosfets etc. To see if completed real model can be near to ideal model.

I invete you to test this circuit with me... If you want we can keep in contact by IM, PM etc.

I have to work so much to survive, so if somebody test with me this we can avoid re invent the wheel..

[Sephiroth]

Hi Patmac,

I'm afraid I have a few too many projects on the go at once to take any more on at the moment sorry.

though I have been doing a fair bit of research the last few days about discharging caps into caps, and I agree with Peter that this isn't a violation of charge conservation... I was just misinterpretting the law

though it is an interesting phenomenon that I don't think has finished teaching me all it can yet.

The first question I needed an answer to is where do half the joules go when two capacitors are hooked together...

As has been said the answer appears to be heat through resistance and is completely regardless of resistor value. Even in a theroetical model of zero resistance.

I think I understand why. I infiltrated a popular physics forum but no one there can think of a way to to end up with two caps with greater than half the voltage of the original cap. All their formulas say that you will still end up with half the voltage in each cap even if you used an eleborate transformer. They agree it is probably possible since it won't violate energy conservation but they can't think how to do it

But it got me onto thinking about the cap pulser on bedini's set ups... wouldn't the same principle apply? I mean if you discharged a cap into a battery would you lose half the joules by conventional theory?

[Sephiroth]

I've just been doing the maths and it looks like this would explain why so many of us are reporting poor results discharing caps at high voltage as opposed to a few volts above the charging voltage.

If we look at this example:

Start with

10,000uf - 15v (1.25J)
20,000uf - 12v (1.44J)
Total Joules = 2.69

then discharge the 10,000 into the 20,000

10,000uf = 13v (0.845J)
20,000uf = 13v (1.69J)
Total Joules = 2.535

so there was a small loss

But then if we try this:

1,000uf - 100v ( 5J )
10,0000uf - 12v (0.72J)
Total Joules = 5.72

Then discharge the 1,000 into the 10,000

1,000uf = 20v ( 0.21632J )
10,000uf = 20v ( 2J )
Total Joules = 2.21632

So we lost over half the joules in this scenario!

I wonder if an inductor and diode could be used to transfer the energy more efficiently in a cap pulser

[Sephiroth]

It gets worse!

in the second example (with the 1000uf cap with 100v) 4.8 joules were sent out of the charged cap but the charging cap went up only 1.28 joules!

so that is 26% efficient transfer of energy!

then I tried

1,000uf = 100v (5 joules)
100,000uf = 12v (7.2 joules)
Total = 12.2 joules

and got

1,000uf = 12.87v (0.0828 joules)
100,000uf = 12.87v (8.281845 joules)
Total = 8.364645 joules

so 4.9172 joules left the charged cap and the charging cap recovered 1.081845 joules! Only 22% efficient transfer!

and the average battery has to be many many times more farads then the capacitors above so if this model applies to batteries the same way it does for capacitors then the charging efficiency should be appauling with the high voltage cap pulser!

[patmac]

It gets worse!

in the second example (with the 1000uf cap with 100v) 4.8 joules were sent out of the charged cap but the charging cap went up only 1.28 joules!

so that is 26% efficient transfer of energy!

then I tried

1,000uf = 100v (5 joules)
100,000uf = 12v (7.2 joules)
Total = 12.2 joules

and got

1,000uf = 12.87v (0.0828 joules)
100,000uf = 12.87v (8.281845 joules)
Total = 8.364645 joules

so 4.9172 joules left the charged cap and the charging cap recovered 1.081845 joules! Only 22% efficient transfer!

and the average battery has to be many many times more farads then the capacitors above so if this model applies to batteries the same way it does for capacitors then the charging efficiency should be appauling with the high voltage cap pulser!

Sephiroth

I would like to try my theory about impendance on capacitors. Why don't try test woth 3 capacitors, one, two and three.

One is charged fully.

Two discharged

Three discharged.

Then put One and Two capacitors in series and then connecte it to Three capacitor.

Can you ask me why?? ok. that is in my mind long time ago, because if you have capacitor "Two" fully charged, no conducts current, because is charged?? or can conducts current inverting polarity???.

Is that the reason that explains when you connect in series, you loose half capacitance.

I think that is more easy extract energy from "One" capacitor fully charged and " Two" discharged than "One & Two" half charged.

[patmac]

Sephiroth

I would like to try my theory about impendance on capacitors. Why don't try test woth 3 capacitors, one, two and three.

One is charged fully.

Two discharged

Three discharged.

Then put One and Two capacitors in series and then connecte it to Three capacitor.

Can you ask me why?? ok. that is in my mind long time ago, because if you have capacitor "Two" fully charged, no conducts current, because is charged?? or can conducts current inverting polarity???.

Is that the reason that explains when you connect in series, you loose half capacitance.

I think that is more easy extract energy from "One" capacitor fully charged and " Two" discharged than "One & Two" half charged.

I have anothre idea to try.... look this....

Many days ago, I was asking on forum about original Bedini's drawing where appear Bedini connected to ground earth. Appearentelly when motor is running on batteries is not necessesary.....

But what if... you use ground to get down differential voltage to discharge capacitor more efficientelly?

look this... Bedini system operate to 12 volts... then you get Back EMF from power coil, where negative is BACK EMF POSITIVE, AND negative is positive...

globally the circuit is normal positive. power coil is feeded by positive directally from the power source... then you charge capacitor and but terminal negative is really conected to positive power source, to create differential. THIS IS MY PURPOSE... WHEN IS FULLY CHARGED, USE DUAL RELAY TO CONNECT TERMINAL NEGATIVE (OF CAPACITOR) TO EARTH GROUND ROD. THIS CAN BE CREATE MORE DIFFERENTIAL IN RESPECT TERMINAL POSITIVE. AND TRY TO DISCHARGE THIS CAPACITOR ON ANOTHER.