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Author Topic: Silly question about capacitors  (Read 16938 times)

Offline dieter

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Silly question about capacitors
« on: February 14, 2014, 04:48:52 PM »
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.


Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?

Free Energy | searching for free energy and discussing free energy

Silly question about capacitors
« on: February 14, 2014, 04:48:52 PM »

Offline MarkE

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Re: Silly question about capacitors
« Reply #1 on: February 14, 2014, 05:04:43 PM »
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.


Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?
Almost of it goes into heat, the rest which is quite small goes into radiation.  Model each capacitor with a small series resistance, ESR.   All capacitors exhibit some ESR, and then of course the wiring has some resistance.  Most of the lost energy heats the wiring and ESR resistance.

Offline vasik041

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Re: Silly question about capacitors
« Reply #2 on: February 14, 2014, 05:20:38 PM »
Quote
Almost of it goes into heat, the rest which is quite small goes into radiation.

I agree.

Here an illustration of the process.

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Re: Silly question about capacitors
« Reply #2 on: February 14, 2014, 05:20:38 PM »
Sponsored links:




Offline Cadman

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Re: Silly question about capacitors
« Reply #3 on: February 14, 2014, 06:17:36 PM »


@dieter
Since you asked, the formula for caps energy is joules = 0.5 * voltage^2 * C



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Offline Farmhand

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Re: Silly question about capacitors
« Reply #4 on: February 14, 2014, 07:34:59 PM »
I think there is some element of impedance matching involved, for instance if we use a very low resistance inductor in the positive line and no switch in the current loop between the capacitors as shown in my drawing, as well as we are doing it for a reason so there is a load involved as well. Then the efficiency of the transfer can be better I think.

So the idea is the system is charged to 10 volts to begin with, the inductior is considered ideal and we switch the RL load in for 1.5 mS only once then calculate the power dissipated by the resistor as legitimate load power and calculate or measure the voltage on the two capacitors and calculate the energy missing.
Anyone want to put that in a simulator ?

Poynt99 did a good paper on this kind of thing. But I cannot find it.  :-[

..


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Re: Silly question about capacitors
« Reply #4 on: February 14, 2014, 07:34:59 PM »
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Offline dieter

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Re: Silly question about capacitors
« Reply #5 on: February 14, 2014, 08:01:56 PM »
Cadman, so I was right with F*v*v, the *0.5 just uses the half, but the relation is the same.


Now, you guys say that, like a leathal strike of Watts within a cap is lost in heat creation? I never notized a cap getting hot by unloading it. Do they?


Even more absurd: when you add an LED, so cap A has to flow trough it to go to cap B, you still get 8v in both, but the LED was lit.


When the cap loses 50% when unloaded, what happens during loading? Or, does both lose 25% and when 2 caps are used as described, the loss adds to 50%?


Anyway it seems, using caps can be a very wasteful factor. At least, if you don't take care of what's in the previous posting. Well I  thought this is the proof for the conservation of energy being wrong  ;D   . Which would have been pretty weird.

Offline Farmhand

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Re: Silly question about capacitors
« Reply #6 on: February 14, 2014, 08:13:30 PM »
It's good to know the formula for this and other things but I find it is simpler, faster and more accurate to just use a calculator that tells us the energy in a capacitor. When installing be sure to uncheck any toolbars ect. that the installation package may offer, unless you want them of course.

All you need to know is the voltage across the capacitor and it's capacitance, then the calculator will tell you the Charge in Coulombs and the Energy in and Joules.

Download Electronics Assistant.
http://www.electronics-lab.com/downloads/calculators/002/

..

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #6 on: February 14, 2014, 08:13:30 PM »
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Offline mscoffman

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Re: Silly question about capacitors
« Reply #7 on: February 14, 2014, 08:25:19 PM »
Yes everyone is correct. When you connect two capacitors together in the real world one needs to reduce the complex
impedance to a resistive impedance. This real impedance will be some number because pure capacitive impedance
is only a theoretical construct. So even if this resistive component is a decimal with a number of zeroes after it is likely
that both identical capacitors have the same amount. The means that half the energy has been left behind dissipated in
the source impedance and half is dissipated in the destination impedance. The residual voltage is what happens when
capacitive impedance "dissipates" energy. Inductive impedance does the same but exchange word "current" for "voltage"
in the previous discussion.

:S:MarkSCoffman

Offline TinselKoala

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Re: Silly question about capacitors
« Reply #8 on: February 14, 2014, 11:31:38 PM »
Everyone? About everything?
 ::)

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #8 on: February 14, 2014, 11:31:38 PM »
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Offline Magluvin

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Re: Silly question about capacitors
« Reply #9 on: February 15, 2014, 12:57:55 AM »
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.


Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?

Lets say we have 2 equal size air tanks, 1 filled to 300psi and the other with 0 psi.

We connect the 2 together with a hose and we open the valves on the tanks. We should end up with 150psi in each. But we lost 50% of the total energy stored the first 300psi tank by the simple transfer, expansion and equalization of the 2 tanks.  Now, each tank contains half of the total air pressure that was in the 300psi tank. How did we lose 50%? The total amount of air(electrical charge) is still there in the 2 tanks, but we lost pressure. ???

We lost it stupidly. ;)   It wasnt lost due to resistance. It wasnt lost due to air leaking out. We lost it by reducing the pressure that was in the first tank and expanding it into 2 cambers, each the size of the first full tank.

Now, if we had a pump motor that uses air pressure to provide mechanical output(turbine may be a bit lossy due to blow by of the turbine fan) and we connect the air motor inline with the hose between the 2 tanks(1 full, 1 empty), then we are capturing that lost 50%(or so) of the total energy of the first tank during the transfer. So now we are not losing that energy stupidly. ;)

And in the end, we still have 150 psi in each tank, just like when we transfer the pressure from 1 tank to the other directly without the air powered motor, but without the huge stupid loss. ;)

So I dont believe that resistance is the issue with losing 50% of the initial energy of the full tank equalizing to the second tank.  Its just the fact that we reduced the energy of 300psi air pressure of 1 tank and expanded it into a tank that is twice as large, without doing anything with the actions of the transfer. And once the 2 tanks are equalized, there is no way that the 2 tanks of 150psi can be used to compress the total amount of air back into 1 tank at 300psi. But, if we mounted a large flywheel to the air motor, the energy that was lost in in the direct connect example is stored in the flywheel, and once the 2 tanks become equal(150psi), the flywheel is fully charged and would continue to pump the originally empty tank, pretty close to what was in the originally full tank, minus losses in the mechanical motor and resistances of the air being pushed through the hoses and fittings, etc.

Now with the caps, we can use an inductor as the air motor and get most all the energy of the full cap into the empty cap. Using a diode(check valve) to keep the previously empty cap, now nearly full, from going back to the first tank. ;) A bit of loss in the diode due to voltage drop inherent in the diode.

Mags

Offline poynt99

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Re: Silly question about capacitors
« Reply #10 on: February 15, 2014, 01:45:28 AM »
How about one more time.  :)

See the attachment for more info on capacitor energy transfer and how to maximize it.

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #10 on: February 15, 2014, 01:45:28 AM »
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Offline Farmhand

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Re: Silly question about capacitors
« Reply #11 on: February 15, 2014, 02:10:37 AM »
Thanks Poynt99, I have that filed appropriately now so i can link it in future.

....

To anyone, does anyone have the time to simulate my circuit example drawing in Falstad simulator or something ? I am trying to learn LT spice, but I don't have a lot of time lately to learn new stuff with my medical issues and chores ect. If I had time to learn it I would simulate it myself. All we need is a DC one shot pulse to the mosfet or switch, the pulse width can vary, but should not allow full current through the series inductor, the 10 Ohm load would limit the current to 1 Ampere max anyway with the circuit pre-charged to 10 volts.

Maybe a longer pulse than 1.5 mS might work better or more or less load resistance to vary peak current

..

Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.

Cheers

Offline Magluvin

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Re: Silly question about capacitors
« Reply #12 on: February 15, 2014, 02:11:59 AM »
To expand on it a bit further, how can we get an actual 50% of the 300psi(total energy) into each tank? Meaning, the 2 tanks together equal round abouts the amount of energy of the original 300psi tank.

We would need the 2 tanks to round about 212psi each, and used together can do as much work as the single 300psi tank. Even though the pressures are different, 300psi(1 tank) and 212psi(2 tanks), the same amount of work can be done with each. The 2 tanks of 212psi may seem like less than 300psi, but we have more total air in the 2 tanks than the 1 tank at 300psi. Like the difference between a 12v batt at 12ah and a 24v batt at 6ah.

But how can this be done?   ??? ;)

We have our 2 tanks, 1 full(300psi) and the other empty, and we have our air motor with flywheel setup. Now we open the valves. The pressure from the full tanks drives the motor/flywheel and begins to pressurize the empty tank. But, we have an added redirect valve between the source tank and the air motor. When we see the source tank get down to 212psi, we turn the added valve. This valve closes off the source tank and opens the input to the air motor to the open air instead. Now, at this point, we stopped the source tank from depressurizing and it now contains 1/2 the energy of the original 300psi. But our flywheel is still going, and pumping the originally empty tank with outside air. And if there wasnt losses in the motor and such, it would be able to pump the second tank to right about 212psi. Now we have 2 tanks that have the same energy potential of the original single tank that was 300psi.   ;)   But one thing has changed. There is more total air in the 2 tanks than was in the single 300psi tank, because we pulled in air from the outside once we turned our newly added redirect valve to do so.

With caps, in this situation, we are basically doing the same thing, except once we turn our redirecting valve, the flywheel(inductor) is just transferring from one plate of the cap to the other.
So my analogy would be better to have 2 air tanks for the source and 2 tanks for the empty cap. The problem would be that the source tanks would need to be 1 full at 150psi and the other source tank at a negative 150 psi to have a clearer comparison between the 2 ideas. But the single tank to single tank gives a basic and simple understanding. ;)

Mags

Offline Magluvin

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Re: Silly question about capacitors
« Reply #13 on: February 15, 2014, 02:32:40 AM »

Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.



Hey Farmhand

The inductor/flywheel is a way of capturing energy that is lost because of a stupid way of losing it. ;)
Think. Do you believe that the energy lost in the direct transfer between 2 caps can be captured by storing the heat created during that transfer? Or by capturing the 'radiation' ? Maybe a little will be recovered, but definitely not all or near 50%, no way no how. Lets say the air tank analogy produces heat in the hose.  Does that in any way explain how we took 1 tank full, directly to 1 tank empty till they are equal, has anything to do with the 50% loss in initial source energy?? ;) ;D

There is something a miss there with the ideals of the so called losses of direct cap to cap transfer. I have some new ideas on this lately that may blow your mind, but I will reveal them when I have it all down pat myself. ;)   It took me a while the get this far to be able to explain what I have here.

Mags

Offline Marsing

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Re: Silly question about capacitors
« Reply #14 on: February 15, 2014, 03:00:25 AM »
hi all..  nice explanation

what is formula the time needed to transfer from c1( full cap ) to c2 ( empty cap)?
according to VAsic041 pict, i took about 200ms

 

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