Quote from: MarkE on February 14, 2014, 09:57:18 PM
Magluvin, take a 100uF cap.  Charge it to 10V.  Connect that to a 1uF cap.  Calculate the net energy.  It is close to the original.  The worst case is always when the capacitors are the same value.

Ugh, just wrote a long post and my laptop went for a shutdown.

Your right. And I was, well, half right. lol

"From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less"

Im trying to figure out what I was thinking when I wrote that. Was running late to pick up the pizza. I had gone over these things with Poynt a while back. I get what you are saying.

Anyways, thanks for catching that.

Mags

Quote from: Marsing on February 14, 2014, 10:02:22 PM
ok,  but you need energy to put back  only one cap  full of charge
the lose here is about to back to first condition.

btw, tesla have a great way to take advantages of this

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

As I said the "energy in a capacitor" thing is really just "potential energy" it can be wasted, or used wisely.

Same thing with two 10000 liter water tanks side by side, one empty one full, if you let the full one equalize with the empty one, the same amount of water is there. None wasted. If you want all the water in the first tank just leave it there.

With no reason to do a thing it is always a loss. It's like making a big high stack of bricks then leveling it out into two shorter stacks it takes energy to make the big stack into the two small stacks. It cannot do itself, something must do it.

Cheers

P.S. Imagine making one tall stack of bricks by hand then changing that to two shorter stacks, does that take no energy to do ? No you must expend energy to get from state one to state two. Then if you want to make the two stacks back into one tall one you must expend more energy again. It's really quite simple. The amount of brick remains the same.

Also the higher the stack gets the harder it becomes to add another brick to the top of the stack.  It depends on how you take them down and if you store the energy or not when you do it as to how much loss is involved.

...

Quote from: Farmhand on February 14, 2014, 10:51:32 PM

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

hey...    PM , oscillation .. etc
this is not about why,    But What and how

Did you forget this  forum name ? 

Quote from: Farmhand on February 14, 2014, 09:40:59 PM
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers

Charge is indeed the conserved quantity here, and capacitors do "store" charge... and just like springs or containers of compressed air, the more charge you put into a capacitor the more _energy_ it takes to do it and the more energy will be returned when you let the charge back out. To say that this is not storing energy is playing with words, I think.
Also, this is real energy, not potential energy we are talking about. The formula for the energy on the capacitor is, as we have seen, 1/2(CV²) in Joules, where C is capacitance in Farads and V is voltage in Volts. Note that this is the same form as the expression for Kinetic Energy of a moving mass: 1/2(mv²) where m is mass and v is velocity. It's not the "potential energy" form like mgh.
The relationship between energy on a cap and the voltage is, as the formula shows, quadratic, not "linear".  Voltage is "charge pressure". You can think of individual negative charges all trying to repel each other, in a fixed volume. The more charges you stuff in there the harder it is to stuff any individual charge in there. By the formula, if you take a 100 uF cap and charge it from 0 to 6 volts, you have to put 0.0018 Joules of work (energy) into the charging. If you then charge it from 6 to 12 volts ... the same number of volts increase as before ... you have to put 0.0054 Joules in to do it ... because you are pushing against the charge pressure of the charges already in there.
If I did the math right, that is.

My undertanding is stored energy is potential energy and energy at work is kinetic energy.

..

I made a P.S. to my last post.