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## Solid States Devices => Captret effect => Topic started by: MileHigh on May 21, 2012, 05:25:39 AM

Title: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 21, 2012, 05:25:39 AM
Woopy:

A long time ago you were curious about why you lose one-half the energy when you connect two capacitors together.

The experiment:  Start with a 1000 uF capacitor A charged to 10 volts and 1000 uF capacitor B fully discharged to zero volts.  Then when you connect them together capacitor A is charged to 5 volts and capacitor B is charged to 5 volts.  When you make the before and after energy calculation the results are that you lost 50% of the energy.

So the questions are, why did you lose 50% of the energy and where did it go?

We can answer the question by doing some simple experiments.

Put a 10K-ohm resistor between the two capacitors and repeat the test.  You will find that the two capacitors are charged to 5 volts at the end of the experiment.

Do the test again, with a 1K-ohm resistor, and you will get the same result.
Do the test again, with a 500-ohm resistor, and you will get the same result.
Do the test again, with a 100-ohm resistor, and you will get the same result.

So, what is this experiment telling you?

The answer is that for any resistor value you will get the same result.  So when you make a direct connection between two capacitors, it's the resistance in the wires that does the same thing, and gives you the same result.

Where did the lost energy go?  The answer is easy, you lost 50% of the initial energy when the current flowed through the resistor and this produced heat.

In the case where you make a direct connection and the energy is lost in the resistance of the wires themselves, perhaps that still sounds strange.  However, even though the resistance is very low, the current is very high, and the formula for the power dissipated in a resistor is = (i-squared x R).  R is very low but i is very high.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 21, 2012, 06:40:19 AM
Ok, well what if the resistance wasnt there? What if the caps and the leads and connections didnt have any resistance?

Cap A - 10v 1000uf discharged to Cap B - 0v 1000uf.  What will be the end result? ;]

And lets say that inductance is not in play also. ;] Just caps

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: ibpointless2 on May 21, 2012, 02:09:50 PM
@ MileHigh

Hi milehigh, I understand what your saying but this is not what really happens in real world. I understand where you coming from, if one capacitor has 10 volts and you connect them both the voltage in both will reach a even point where one doesn't have more than the other. In real world testing i have found that this doesn't always happen. I've found that if you have one capacitor that has 800mV in it and another capacitor that has 20mV they won't meet in the middle and instead will self charge off each other. I know it sounds crazy but I've done the test and created threads about it.

I also created a thread on the topic here too [size=78%]http://www.overunity.com/10216/parallel-charging-shows-overunity/ (http://www.overunity.com/10216/parallel-charging-shows-overunity/)[/size]

Capacitors can be very mysterious devices.
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: IotaYodi on May 21, 2012, 02:22:34 PM
Quote
I've found that if you have one capacitor that has 800mV in it and another capacitor that has 20mV they won't meet in the middle and instead will self charge off each other
Would the type and quality of the materiels along with the enviroment come into play on this?
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: poynt99 on May 21, 2012, 03:00:42 PM
Ok, well what if the resistance wasnt there? What if the caps and the leads and connections didnt have any resistance?

Cap A - 10v 1000uf discharged to Cap B - 0v 1000uf.  What will be the end result? ;]

And lets say that inductance is not in play also. ;] Just caps

Mags

With resistive wire and non-ideal capacitors, each cap will have 5V.

With ideal wire and ideal capacitors (made from ideal conductors and ideal dielectric), each capacitor would have 7.07V.

However, does or will ideal conductors (zero resistance) and capacitors ever exist? Not likely.

But let's imagine that the day has arrived, and we have produced the ideal wire and capacitor (and inductor), and there is 0 inductance in the circuit. The time required to transfer the charge from the charged capacitor to the discharged capacitor would be infinitely small (0 seconds), and the current would be infinitely large. Can infinity really be reached?

With an "infinitely-large" current flowing for that "infinitely-small" amount of time, anything ferromagnetic or electrostatic within "infinite" distance from this "event" would be adversely affected by the infinitely-large electric and magnetic fields produced, including the iron in your blood.

In other words, you'd likely destroy not only yourself, but the universe as well.  ;)
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: TinselKoala on May 21, 2012, 04:00:55 PM
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 21, 2012, 04:26:18 PM
ibpointless:

You are mixing up what happens when you short two capacitors together and the electrolytic capacitor self-charging effect.  They are two separate things.  Have you looked up the effect on Google?

I am pretty sure that the self-charging effect is related to a natural potential difference created by the thin membrane layer(s) of the electrolytic oil and the availability of positive and negative ions in the air.  The electrolytic capacitor leads pick up charge from the atmosphere.  So you can say that the sun is the ultimate power source for the self-charging effect in electrolytic capacitors.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: TinselKoala on May 22, 2012, 12:00:17 AM
ibpointless:

You are mixing up what happens when you short two capacitors together and the electrolytic capacitor self-charging effect.  They are two separate things.  Have you looked up the effect on Google?

I am pretty sure that the self-charging effect is related to a natural potential difference created by the thin membrane layer(s) of the electrolytic oil and the availability of positive and negative ions in the air.  The electrolytic capacitor leads pick up charge from the atmosphere.  So you can say that the sun is the ultimate power source for the self-charging effect in electrolytic capacitors.

MileHigh
Which can be significant. I left that 70000uF cap sitting disconnected overnight and today when I went to put it away I shorted the terminals with a keeper and there was enough self-charge that it gave a fair little spark when I made the contact.

A neat demo I sometimes do is to use a plastic pill bottle or film canister to make a capacitor by lining the inner side and the outer side with foil tape, then making a little spark gap at the top bridging the two "plates", and a frayed out fringe of braid or stranded wire to one plate to act as a pickup. This can be easily charged with a piece of PVC and some fur,  or by holding it up to a CRT monitor . The "self-charge" spark can get to 1/4 inch or more from just that simple little arrangement. Of course the charge doesn't stay on an improvised cap the way it will on a good electrolytic.
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: NerzhDishual on May 22, 2012, 12:34:55 AM

Do not use any resistor.
Just replace this resistor with a small motor geared to a small K7 recorder
mechanical counter (acting as a load).

Actually, in this case(?), you do not loose any Energy.
500  plus 250  plus  250 = 1000. (turns)

Very Best,
Jean
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 22, 2012, 03:21:12 AM
With resistive wire and non-ideal capacitors, each cap will have 5V.

With ideal wire and ideal capacitors (made from ideal conductors and ideal dielectric), each capacitor would have 7.07V.

However, does or will ideal conductors (zero resistance) and capacitors ever exist? Not likely.

But let's imagine that the day has arrived, and we have produced the ideal wire and capacitor (and inductor), and there is 0 inductance in the circuit. The time required to transfer the charge from the charged capacitor to the discharged capacitor would be infinitely small (0 seconds), and the current would be infinitely large. Can infinity really be reached?

With an "infinitely-large" current flowing for that "infinitely-small" amount of time, anything ferromagnetic or electrostatic within "infinite" distance from this "event" would be adversely affected by the infinitely-large electric and magnetic fields produced, including the iron in your blood.

In other words, you'd likely destroy not only yourself, but the universe as well.  ;)

Hey Poynt

Remember we discussed some of this. So how does the inductor and diode, used to try and transfer all from cap A to cap B, seem to overcome most of the resistance losses? Is it because of the odd voltage divisions in the circuit due to the inductors working characteristics?

Like if we have cap A and B.  A 1000uf 100v  and B 1000uf 0v. We connect them but with an inductor between 2 legs of the caps , like all in series loop. Assuming the lowest resistance, room temp, what would be the highest level of voltage, peak, seen in cap B once the oscillation begins?

Also, during the oscillation, same circuit, when the caps are at equal voltage in time, are they above 50v each? Lets say we are sampling the first cycles, from cap A to cap B, then back to cap A.

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: DreamThinkBuild on May 22, 2012, 04:17:47 AM
Hi All,

I've seen this self charging that IB is referring to in ultracaps. Take two 3000F ultracaps and short each one individually down to about 5mv each(10mv total). Place in series and leave them for an hour and re-measure, it usually ends up with a +50mv gain or around 60mv total(~30mv each cap). Did the same experiment with six 3000F caps in series short until 1.6mv each cap, place in series at around 10mv leave an hour and the voltage is up around 300mv or a gain of around +48 to +50mv per cap. If you look at the datasheet for the caps it has a ESR of ".29milliohms" which might factor into this self charging.

http://www.maxwell.com/products/ultracapacitors/docs/datasheet_k2_series_1015370.pdf

I'm wondering how far you can take this self charge effect before it breaks down. Is it better to stack x amount of capacitors in series to reduce the capacitance, increase the charge rate but also increase the ESR which will cause it to have diminishing returns. Or would it better to put them in parallel which will decrease the ESR but at the expense of increasing the capacitance, slowing the charge rate. Hmmm... something to test.
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: TinselKoala on May 22, 2012, 05:08:14 AM
Charge in series, discharge in parallel.
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 22, 2012, 05:38:56 AM
Charge in series, discharge in parallel.

Hey TK

Ive heard this before. A few times actually.

Here is something I found interesting.

The pic below shows a circuit that has a source, 5v and when you close the switch to the left, hold till the cap is at 5v then release.

Now, to the right, we have 2 options. Insert the wire or the cap into the empty space.

With the wire, when we close the switch to the right, the caps end up at 2.5v each. ;]

But, now here is the strange part.

With the 3rd cap in place, we charge the first cap from the source to 5v, then discharge that into the 2 caps in series. We end up with 3.3v in the first cap and 1.67 in each of the series caps to the right.  Got it? ;]  Now for the trick...

We remove the 3rd cap from the circuit(still holds 1.67v) and insert the wire once again.
The cap on the left has 3.3v and the one to the right has 1.67v.  When we hit the discharge switch once again, we end up with 2.5v in each.  :o :o :o :o :o :o

And we are left with the 3rd cap with 1.67v.

Now how did we end up with 2.5v in each, just like when we didnt have the 3rd cap in the first example?

Mind boggling.  ;)    So why did we not lose as much in the second example as we did in the first, considering we still have a charge in the 3rd cap when all is said and done? ;]

Does this have something to do with your post that I just quoted? ;]

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 22, 2012, 06:20:34 AM
Actually, I would like MH to answer why this occurs first, even though Im itching to hear from TK on this.

But I want to see if MH's reasoning adds up here, considering its his thread.  ;)

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 22, 2012, 07:13:46 AM
Magluvin:

Ever since I told you that your one-magnet no-bearing spinner was never going to work and your friend's clip was a fake you have gone ape-shit on me.  You have looked for opportunities to give me a hard time and you get argumentative to the point where most of what you say makes no sense and it's intentionally said just to add fuel to the fire.

The question that you posed to me that Poynt answered was a rigged question you intentionally made up to be as difficult as possible to start the same ridiculousness all over again.  You had no interest in getting the answer to that question, it was just an attempt to engage me in a fight.

So before I even discuss this, the question for you is this:

Is this over and are you going to stop this behaviour from now on or are you going to continue looking for opportunities to have "big fights" and throw insults at me and wig out?

So what's it going to be from this point onwards?

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 22, 2012, 07:58:10 AM
Magluvin:

Ever since I told you that your one-magnet no-bearing spinner was never going to work and your friend's clip was a fake you have gone ape-shit on me.  You have looked for opportunities to give me a hard time and you get argumentative to the point where most of what you say makes no sense and it's intentionally said just to add fuel to the fire.

The question that you posed to me that Poynt answered was a rigged question you intentionally made up to be as difficult as possible to start the same ridiculousness all over again.  You had no interest in getting the answer to that question, it was just an attempt to engage me in a fight.

So before I even discuss this, the question for you is this:

Is this over and are you going to stop this behaviour from now on or are you going to continue looking for opportunities to have "big fights" and throw insults at me and wig out?

So what's it going to be from this point onwards?

MileHigh

:o

Cmon M.  My first post was intended for you. Your thread. But Poynt nailed it. Not so complicated. He even went beyond.  The only things I would maybe disagree on is the 7.07 v results, because as he said, these would be ideal situations and they dont exist, sooo, are we sure. And its the ferromagnetic, end of the universe stuff, as we had no inductance and no resistance in the example. So currents should flow fast and smooth and no magnetic pulses or fields as inductance doesnt exist, and the closing of a switch should produce no sparks as there is no resistance.  ;)   But that wasnt a trick. I was just eliminating any criteria other than capacitance. Just to isolate the idea of capacitance.  ;)

As for my last experiment, I just came up with that after TK posted the series parallel cap post.  It doesnt even really match what he said, bit something clicked when he said it. So I tried it in sim, with a variation, and I was a little freaked out by it. I tried it again, different values and such, but look what I got. Is it not interesting?

Are we not creating a level of efficiency in what I have shown, on the whim, as compared to what this thread was intended to show in the first place? Think jim.

Now look. You blow up over what? Im ok. TK is ok. Poynt is ok.

I wanted you to answer because you are the one that is making a point here by starting the thread. My lil circuit is definitely on topic. No?  Im not here Shootin off at anyone. ;] Im showin real deal stuff also. No games.

U civil, me civil. Simple.  But if I smell a hint of insult, I will not remain in an imaginary corner. Does that make sense?

I think I know why the circuit gives the results, but I want to know if your explanation in the first post here can be applied to my lil version as well. Im just not sure how really.

Because, in my circuit, we didnt lose 50% total as in your first example, and we didnt use inductors, just caps.

So maybe it is a seed of something better than just a cap to cap exchange. Maybe it could be expanded upon to do better. Maybe TK's Post was a good hint. Can you explain why he posted that?

But maybe this doesnt sound good at all to you. We are always seemingly on the opposite sides of the table.

Anyways, take it for what you will.

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 22, 2012, 08:45:37 AM
Magluvin:

I told you the first "no bearing" setup was not going to work because I was speaking the truth and it was sound advice so you would not waste your time.  That was a crazy fight and subsequent to that the discussion about the setup with the two spinning magnets and the LEDs was borderline psycho.

So I am asking you for a straight answer to my question:  So what's it going to be from this point onwards?

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: woopy on May 22, 2012, 12:16:31 PM
Hi MileHigh

Thank's to come back to this topic, because as you know i made a lot of experiments some time ago with those cap transfer,
And of course each time we apply the famous stored energy formula  (1/2CV2), we always loose half the energy after the transfer of one charged cap in an other same but discharged one.

I am  not completely sastified with that situation, but i have to accept it because it seems to be a fact.

But today i am puzzled by the reply  number 8 by NerzhDishual on this thread, where he discharge the caps accross a motor and he counts the number of revolution per discharge , and it seems that he get no  loss in revolutions after the transfer. :o

And also Magluvin get some special result in the Falstad sim ??

Any explanations ?

Youp !! the thread will hopefully be of  great interest ;)

good luck at all

Laurent

Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: TinselKoala on May 22, 2012, 03:29:28 PM
@Mags: Your problem is an interesting one. But I don't agree with the numbers you get. Are your numbers measurements?

Here's how I am reasoning, and I haven't done this problem in a while so please check my work and tell me if I'm screwing up somewhere.

Energy is conserved. Charge is conserved. Not voltage. In a capacitor, the energy boils down to a product (technical term: means the result of multiplication) of the voltage on the cap and its capacitance. But the energy is directly proportional to the capacitance and to the _square_ of the voltage.

To find the capacitance of parallel capacitors you simply add the capacitances. And of course if caps are in parallel they are all at the same voltage. To find the capacitance of series capacitors, it's the same formula as _parallel_ resistors: 1/Ctotal = 1/C1 + 1/C2 + ... + 1/Cn.  And if two identical caps are in series and charged and then separated, they will each be at half the voltage of the series pair, just like matched batteries in series.

So.... now we can calculate. The CofE tells us that, barring losses, we have to have the same energy after as before. How much is that?
The Energy E on a cap is given by E = (CVV)/2, with the capacitance in Farads, the voltage in Volts and the answer is in Joules.
So after you charge the one 10uF cap to 5 volts I get 0.000125 J in the system. This energy will be conserved, I already know that much. Right?

Now I close the switch and charge the second lone 10uf cap. That original energy is now distributed in two identical caps, so half of it is in one, and half in the other. Since these caps are in parallel now, they are at the same voltage. (Yes, they are in series too, but we are going to "measure" their total voltage in parallel, aren't we?) But caps in parallel have total capacity equal to the sum of their capacitances. So we now have 0.000125 Joules in 20 uf. This means the voltage on the parallel pair, therefore on each cap, will be 3.53 V, from the cap energy equation.

Now we do it again, but instead of a single 10 uf cap in the second position we have two in series. Putting these in a black bag, we see a 5 uf cap (two 10s in series inside the black bag). So now we close the charging switch as before.

Now, the original 0.000125 J is distributed among the 10uF of the original cap and the 5 uF of the black bag. But since the 10 uf and the black bag are in parallel for the measurement, they are at the same voltage. So... calculating, we find that 0.000125 J in 15 uF will be at a voltage of 4.08 V , and since the 10 uf and the black bag are in parallel they both are at 4.08 V. And when we open the black bag we find two identical caps in series carrying the same charge, so they must be at 2.04 volts each.

So we have one 10uF cap at 4.08 V, and two 10uF caps at 2.04 V each. Computing backwards to the energy, it adds up to what we started with.

So now we take one of the 10 uF caps from the black bag  at 2.04 V and the original 10 uF cap at 4.08 V and close the switch. The charge equalizes across the caps again. Since the capacitances are equal the charge will be equal in each cap. The total energy will be E = 0.000125 J - 0.000021 J = 0.000104 J held in 20 uF of capacitance. This will give us a voltage of  V=sqrt((2xE)/C) = 3.25 Volts on each cap.

So now we have two caps at 3.25 volts and one at 2.04 volts. The two at 3.25 volts have 2 x (CVV)/2 = 0.000106 J
and the other one at 2.04 V has CVV/2 or 0.0000208 J and again we have the same total of 0.000125 J energy we started with.

In the real world there will be resistive and radiative losses at every stage so the resultant Voltages will be lower and so the energy totals will be lower at every stage.

But....if we hook our now charged ideal caps in series, we can measure almost 8.6 volts across the resulting 3.3 uF stack. What is its energy?
E= CVV/2 = 0.0000033 x 8.6 x 8.6 / 2 =  the same 0.000125 Joules we started with by charging the initial 10 uF cap to 5 volts.
Minus losses of course. (And we have just invented the "Mags Bank" voltage  multiplier.)

;D

Sorry, what was the question again?

Here's a way to think about charge, voltage and capacitance. Charge is the fundamental conserved quantity. Who cares what it is.... we just need to know about one property, and that is that like charges repel, and the force of repulsion is as the inverse square of the separation. So the closer you get like charges together the stronger they repel each other.
This repulsion of like charges IS voltage. ( Actually voltage is the result of the gradient of the E field, but it's simpler just to think of it as the "tension" of opposite charges attracting or the "pressure" of like charges repelling.)
And capacitance is like.... the space in a room. There is only so much space in any given room and the walls are only so strong. So now you start stuffing charge carriers (electrostatically charged ping pong balls, electrons, holes, cats,  whatever) into your room. The first one goes in easy. The second one not so easy, it is repelled by the one already in there, so to get the second one in you have to push it in harder (raise the voltage). The third one ditto. Eventually you reach the point where the _capacity_ of the room just can't hold any more of these mutually repelling charges and one of them punches through the wall and they all leak out fast through the hole in the dielectric wall.
So the bigger the room, the more charge it will hold at a given "pressure". And if you make the walls stronger the same sized room will hold more charge at higher pressure. There is more energy in the "pressure" than there is in the plain number of charges (the size of the room) so the higher the voltage, the "more more" higher the energy stored in the pressure.

I really like Jean's motor-counter idea. I'll have to build one myself, I have a counter like that in one of my boxes somewhere I think.
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 22, 2012, 04:03:45 PM
Magluvin:

Okay, in the spirit of cooperation I am going to assume that the crazy times are over and done with and it's water under the bridge.  Plus we need a simple explanation for this phenomenon.

I am going to assume that you have worked through the problem and you have confirmed that the simulation's numbers are correct.  If you haven't then I strongly advise that you do work through the problem.  If you crunch the numbers then you should see that it all checks out.  If you crunch the numbers the answer is basically staring right back at you.  Not trying to work through the problem is a mistake.  In other words saying, "If I short two capacitors together I get result A, so if I short some capacitors together together in a different sequence I am expecting result A but my simulation is giving me result B, what happened?"

The way to do the number crunching is that you always know that you have conservation of charge.  You know ahead of time that you don't have conservation of energy because of the presence of the resistor, but for sure you know that you have the conservation of charge.

I will leave it to anybody that is interested to crunch the numbers and I will give the answer in the next posting.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 22, 2012, 04:23:27 PM
Okay, so the root answer to explain this problem is deceptively simple.  You know that if you short  a 1000 uF capacitor at 10 volts together with a 1000 uF capacitor at 0 volts you lose half of the energy.  But who is to say that you should lose half of the energy when you short a 1000 uF capacitor to a much smaller capacitor?  If you do the number crunching you will quickly realize that if you short a 1000 uF capacitor at 10 volts to a much smaller capacitor at 0 volts then you lose only a very small amount of energy.  So as the size of the second capacitor goes towards zero the energy loss goes towards zero.

So in the example the first event is a 1000 uF capacitor being shorted with a 500 uF capacitor.  So right away without calculating anything you already know beforehand that you are not going to lose 50% of the energy in the first step.

So now going forward you should be much more confident that the result of the simulation is correct.  There is nothing here that can be worked on or used, it's another case of working with "less losses."   It's very easy to mix up "less losses" with "energy gain" and it happens all the time.

The mechanical analogy is actually very good in this case.  Imagine a 1000 Kg block of putty moving at 10 meters per second on a frictionless surface smashes into a 1000 Kg block of putty that is stationary.  When they hit they deform and stick together and become a 2000 Kg block moving at 5 meters per second.  You know that when two objects hit each other and stick together that you have conservation of momentum.

So if you rework the problem and substitute the capacitors for moving blocks of putty on a frictionless surface the results will be identical.   Obviously it's very intuitive to imagine that when a 1000 Kg block hits another 1000 Kg block that more energy is burned off in the collision as compared to when a 1000 Kg block hits a 50 Kg block.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 22, 2012, 06:32:36 PM
Laurent:

For reply number 8 by NerzhDishual the real issue is what happens when you have a coil (which is what the motor looks like) between the two capacitors?   You can remove the resistor also.

With the coil between the two capacitors there is no longer a "crash" when you connect the first capacitor to the second capacitor.  A coil is like a spring in the mechanical analogy, so you can think of a spring between the two masses of putty.  They don't hit each other any more, the spring absorbs the energy and stores it, and then transfers it into the second mass.

So if you ran some simulations with a capacitor connected to a coil connected to the second capacitor, you will see an oscillation when the connection is made.  An oscillation means no energy is being lost (in the ideal case).

So that's what is suggested with the motor.  The motor stores some of the first capacitor's energy in the motor coil and also in the rotation of the mass of the rotor.  As the rotor spins down the motor becomes a generator and turns the mechanical rotational energy back into electrical energy.  So you lose less energy with the motor setup because it resembles an inductor.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: woopy on May 22, 2012, 11:21:35 PM
Hi MileHigh

thank's for answering this way, and now i feel better regarding my problem with the " energy storing formula "

So it is a different way to see the transfer of energy , between the rigid  approach in the application of the "formula" due to the different mean of discharging a cap into  an other cap !

I mean if i short cut a charged cap in its twin uncharged cap , the formula is valuable, but if i use a different mean (as for instance the motor as per Nerzdishual proposition ) to do it, the formula is not fully applicable.

that's OK for me , now i am feeling much better concerning this problem

thank's very much for your explanation.

Good luck at all

Laurent
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 23, 2012, 01:16:01 AM
You are very welcome Laurent and good luck with your experiments.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 23, 2012, 01:28:28 AM
I was thinking the same thing with the motor as an inductor. So the inductance would go down with increased motor load?

As for my lil circuit, and the standard example of cap to cap transfer, are they ever used in any case?  Or is the cap to cap example just used for explanation?

I never considered the 1.67v cap extra as a gain, but just higher eff compared to the standard cap to cap, because we are still at a loss. I think more eff is all I stated. ;]  But I did say that what if we could find more or better ways to decrease losses, it would be good.

M, I can see that we only used a bit of the discharge cap into the 2 series, and then only some more when discharged from 3.33v into the lone 1.67v cap then levels out to 2.5v each. In some way some of the losses are severed in my version.  Your right, calculations will have to be made to see where and why the losses are different. I came up with a few ideas today to try and expand on this. Ya never know. ;] Till ya try ;]

It is an odd situation with my lil circuit, aint it? ;]

Im going to try the motor idea also.

Its funny how an idea can come full circle, as it was only months ago we were on this very subject and it has come alive again.

Mags

lol did anyone see Americas Got Talent last night?  Ooooo, Ouch And they made it to the next round!! :o :o :o :o :o :o What is this world coming to?  ;D

The use of safety goggles is hilarious.  ;)
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: MileHigh on May 23, 2012, 01:51:25 AM
Magluvin:

For the motor I think the idea is to not put a load on it.  I am being pretty loose calling a motor an "inductance" but certainly we know that there are coils in a motor and there is a flywheel in a motor.  So presumably both aspects will help.

In a way you can say all computer power supplies are based on this technology.  They are called switching power supplies.  You have a simple and "dumb" rectified sine wave + capacitor generating a relatively ugly DC waveform with a lot of AC ripple superimposed.  Then MOSFETs switch on and off at about 20 KHz and generate tiny little pulses that charge an output capacitor to keep it at a constant DC voltage under varying loads.  A servo control system decides how long the MOSFET pulses on to maintain the very clean DC output.  I may be off on some of the specifics but the basic idea is correct.

If someone runs a simulation for the cap connected to coil connected to cap scenario, at least you will see the energy sloshing around.  Monitoring the amount of energy in each component would be interesting.

MileHigh
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: NerzhDishual on May 25, 2012, 12:36:36 AM
@Woopy

You quoted my previous post. OK.
Actually, I did not want to blow my own horn.

Here is my web page about these caps (= Condos in French slang) experiments:
http://freenrg.info/Condos/ (http://freenrg.info/Condos/)
No cheating at all here.
These are old stuffs.  You can consult the OU dot Com records.

Now, anybody can criticize, mock, ridicule, scoff (at) me.
I really do not care. I'm 64 YO. I'm not a Scientist.
I do my best mostly using my Right Brain.

Very Best from Brest/Brittany/France,
Jean
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 25, 2012, 01:12:39 AM
Hey Nerzh

Do you have the setup?  The counter and rubber belt look pretty snug, offering some respectable resistance for the motor to work against. Mh, I believe was thinking that the motor had some inertial rundown as the charge transfer is on the second half of completion. But I see that the motor might stop fairly instantly once power is no longer driving it.

What Im wondering is, if you could do the tests over( just one that you were happy with) and instead of counting turns of the counter, just keep track of time to finish and voltages of the caps, but do the same test without the counter as a load on the motor. It would be interesting to see if the end voltages in the caps are different letting the motor freewheel as compared to loaded with the counter.  And even a third test would be a small flywheel on the motor, which would lend to what MH was saying about the motor acting as a generator during final wind down.

Went through your paper, and going to read it again. Nice work. ;]

Mags
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: NerzhDishual on May 25, 2012, 02:46:46 AM
Hi Magluvin,

I do not dismantle my setups.
One drawback: all these bl' "apparatus" are scattered over my house. :P
Fortunately, my wife is an angel... :)
-----------
The French man J.L Naudin site is : "The Quest For Overunity"
http://jnaudin.free.fr/ (http://jnaudin.free.fr/)
My site could be the: "The Mess for Over-Unity"
http://freenrg.info/The_Mess_4OU/ (http://freenrg.info/The_Mess_4OU/)
------------
Kidding appart:
Yes: I still have the setup.
Yes, the counter and rubber belt are pretty 'snug',
You mean 'tight'? Yes, It offers real resistance.
---------------
OK! I see what you mean. It is perfectly clear.
The idea of merely keeping 'track of time to finish' and caps voltage is, IMO, very creative.
No load needed. Just monitor the time.
Now: a small flywheel: just another creative idea. Thanks.

Does the motor act as a generator is a real question.
But? All these motors are (officially) 'UN' (Under Unity). No?

Anyway, in spite of my natural laziness (worsening with age) but according
to your suggestions I will make more experiments.

But, Finally? anybody can do it. No?

Very Best,
Jean
Title: Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
Post by: Magluvin on May 25, 2012, 03:43:49 AM
Hey Nerzh

Im not pushing you to do it. ;]  Its just that since you have done so much already, these tests would be a great comparison to what you have already done, using the same equipment. ;]

As for the motor generating, what I mean is, during the cycle from start to finish, maybe without the resistive drag of the belt and the counter, a freewheeling flywheel could give quite a different outcome.
Im really interested in the voltage outcomes in the caps mostly. The time it takes will help indicate higher or lower inductance in each case.  ;]   Im thinking any inductance in the motor will be different using the motor unloaded or flywheel added, giving us something other than just the inductance to store energy in a magnetic field but also store it in physical motion of the armature and flywheel, where with the counter, belt load, power must be lost in the mechanical resistance. ;]

Not that I consider the counter/belt resistance as useless. I just think these tests would give more detail as to what is happening and help others and myself use that knowledge to possibly expand on it for further discoveries, if anything interesting comes of it. ;]

Thanks Nerzh. I like your idea.  What you have shown is that you can transfer more energy cap to cap and get physical work done, as compared to just a cap to cap balance out in voltage and losing 50% while doing it.  ;)

Mags