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Author Topic: For Woopy: Explanation for why you lose 1/2 energy in a capacitor  (Read 23125 times)

Offline MileHigh

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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.

So you finally have your answer.

MileHigh

Free Energy | searching for free energy and discussing free energy


Offline Magluvin

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #1 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

Offline ibpointless2

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #2 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.


Here's a video i made two years ago about this topic [size=78%]http://www.youtube.com/watch?v=wbn4vede2us[/size]


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


Capacitors can be very mysterious devices.

Free Energy | searching for free energy and discussing free energy

Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #2 on: May 21, 2012, 02:09:50 PM »
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Offline IotaYodi

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #3 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?

Offline poynt99

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #4 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.  ;)

Free Energy | searching for free energy and discussing free energy

Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #4 on: May 21, 2012, 03:00:42 PM »
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Offline MileHigh

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #6 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

Free Energy | searching for free energy and discussing free energy

Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #6 on: May 21, 2012, 04:26:18 PM »
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Offline TinselKoala

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #7 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.

Offline NerzhDishual

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #8 on: May 22, 2012, 12:34:55 AM »
Please Gentlemen:

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

Free Energy | searching for free energy and discussing free energy

Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #8 on: May 22, 2012, 12:34:55 AM »
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Offline Magluvin

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #9 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

Offline DreamThinkBuild

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #10 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.

Link to datasheet
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.

Free Energy | searching for free energy and discussing free energy

Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #10 on: May 22, 2012, 04:17:47 AM »
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Offline TinselKoala

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #11 on: May 22, 2012, 05:08:14 AM »
Charge in series, discharge in parallel.

Offline Magluvin

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #12 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

Offline Magluvin

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #13 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

Offline MileHigh

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Re: For Woopy: Explanation for why you lose 1/2 energy in a capacitor
« Reply #14 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

 

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