Silly question about capacitors

[dieter]

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     . Which would have been pretty weird.

[Farmhand]

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/

..

[mscoffman]

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

[TinselKoala]

Everyone? About everything?

[Magluvin]

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