dump one capacitor into two empty connected in series, then in parallel
what is the result ?

Quote from: forest on April 17, 2011, 12:26:59 PM
dump one capacitor into two empty connected in series, then in parallel
what is the result ?

Specify the 3 values....then you provide the answer.

.99

Thank's p.99

I am happy that we are OK on the first point.

Now i will go on structuring my mind.

When i look at "hyperphysics.com", they explain the stored energy in a cap with a comparison with the  stored energy in an air tank. And i can very very well visualize this. So the voltage in the cap would compared at the pressure in the ai tank.

So 2 caps of same capacity would be compared to 2 air tanks of same capacity.

So the first  air tank(cap) is  at 99 BAR pressure and the receiver cap is empty (or has only the atmosphere pressure that is  about 1  BAR).

There is a tubing with a valve between the 2 tanks,

So when i open the valve on the tubing. depending of the opening  the tanks  will more or less fastly equalise to 50 BAR each.

So as the  pressure (exactly as the voltage  in energy stored formula for the caps) is SQUARED in the formuly 1/2 * C * V^2, so after the transfer, (of course if we operate carrefully ),  the pressure in the air tank is 25 % in each tank and together it is 50 % lost of energy as perfectly expected.

So we did not loose one molecule of air in the transfer, the quantity of air is the same but the pressure of this air is simply divided by 2 (or almost 2), and almost nothing was lost in the tubing.

The transfer was almost perfect (almost 100%) but the pressure lost is comparable at the Voltage lost in the direct cap transfer. And the end result is almost 50 % lost in energy .

Does it make sense ?

Thank's

Laurent

Assuming that the air tank analogy is a good one, then yes the combined potential energy in the tanks is reduced by 50%. Keep in mind also, that Hyperphysics does not touch on the efficiency or amount of transfer issues.

I would stay away from thinking that the process was 100% as you say. This is what I was trying to say before as it only adds confusion to the discussion, imho.

.99

Quote from: poynt99 on April 17, 2011, 09:36:42 AM
If you had two ideal capacitors (ESR=0 Ohms) of the same value connected together with an ideal wire (Z=0 Ohms) and no diode, then the transfer of energy from C1 to C2 would be instantaneous (with no oscillation), and there would be no energy loss.

So if C1 started with 1.00V and C2 with 0.00V, the end result would be with both caps at 0.707V.

.99

hmmm, well this is interesting.

If in the "normal" world,   100v on one cap and zero on the other, if we make direct transfer, we end up with 50v on each cap. Each cap is 1/4 and if we put them in parallel we have a total of 50% of the total energy left. 50% gone.

Now we go to superconductive world.....

100v in a cap, do a direct transfer, we end up with 70.7v in each cap. We combine them in parallel, we end up with what, 100% energy left?

100v 10uf = 70.7v 20uf   So if we skimmed 20.7v from 70.7 20uf to end up with 50v 20uf, we lost 50% of the energy in the 20uf cap?

I suppose thats correct. Then the 50v 20uf cap, we skim 15v to have 350v 20uf, we lose another 50%.  Then skim 10v to 250v, another 50% loss.

Its the same as my chart through the 4 stages.

Its hard to fathom, but seems correct.

It seems like a far throw for 70v 20uf =100% where 50v 20uf is only 50%. But it is what it is.

So in the BC if we start at 1kv and we do a cutoff of 703v, which gives 703v in each cap, we are just close to ideal world conversion. We have a bit of loss from 707v to 703v.  So with the BC we are very closely approaching ideal world conversions, without superconducting properties.

ps  An ideal wire has no inductance in superconducting world?

Mags