ENERGY AMPLIFICATION

[WhatIsIt]

Note: two 100uF caps in series equals 50uF, so we have 50uF at 20V): 10mJ

2x100uF = 200uf = in series 200/2 = 100uF, 20V

How you came from 200uF to 50uF?
Serial law states that in case 2 caps, their capacitance will be halved (200uF/2) and voltage doubled.

I don't wanna go down into this rabbit hole just as those members did and had nailed to cross.

So, for me this story is over, you are right and let it stays that way.

[Magluvin]

2x100uF = 200uf = in series 200/2 = 100uF, 20V

How you came from 200uF to 50uF?
Serial law states that in case 2 caps, their capacitance will be halved (200uF/2) and voltage doubled.

I don't wanna go down into this rabbit hole just as those members did and had nailed to cross.

So, for me this story is over, you are right and let it stays that way.

(2) 100uf caps in parallel = 200uf
(2) 100uf caps in series =50uf

You need more study time before you nail the story to the cross.

(2) 100uf caps 50v each in parallel = 200uf at 50v
(2) 100uf caps 50v each in series = 50uf at 100v

Now, as you may need to, find an online calculator for energy in a capacitor and run these figures...

(1) 100uf cap at 100v = ?
(1) 200uf cap at 50v = ?

Are they the same?

For a Sr. Member, you should know this....

Mags

[NickZ]

   Mags:    So now you are  making us work for  it...    Thanks for the reminder.

[Magluvin]

The story continues...

When I was told this by Sr. members years ago, I was in denial at first. And once the calculations were made, I was still in disbelief a bit. It was said that the 50% was lost due to resistance(no matter how small or large the resistance) and heat during the action of connecting the say 100v cap to the 0v cap. Seemed plausible, but something just didnt seem right still. 

Then the Sr. members gave an example that enabled me to argue that they were wrong and it was not the resistance that caused the loss at all. They said if we looked at it in an ideal world of no resistance, only then would we not experience the loss in the cap to cap experiment.  Well that had me going even more into disbelief. My mind was running like a full steam locomotive.  So doing some calculations, in the supposed ideal world, having 1 cap at 100v connected to a 0v cap, we would need to end up with 70.7v in each cap to still have 100% of the original energy of the first cap we started off with of 100v.

Now I was on the right track.  There was something terribly wrong here. Even more wrong than I had initially thought. I made analogies in my head. Then it hit me..  'Hmm. How could I take 10gal of water in 1 bucket and end up with 2 buckets that contain 7.07gal each?  We cannot.

There is a specific law in electricity that proves we cannot end up with 2 caps of 70.7v each from 1 cap of 100v in the experiment described, neither real world nor ideal world of no resistance.  ?

If we have a capacitor at 0v that was very precise as to its value of capacitance, and we were able to count the number of electrons pulled from the + plate and stored to the - plate, we should be able to calculate the exact voltage across that cap with extreme accuracy.

So.  with all this, I was able to argue that it was not the resistance and heat during the transfer of cap to cap that caused the losses. In the real world or ideal, we still lost 50% mostly because we didnt do anything with the transfer action. Whether we did anything with it like run a motor, a light bulb, heat from a resistor, or even ideally where no energy was expelled beyond what was stored in the caps, there will always be 50% of the initial energy gone in the end of the experiment.

So. How would we explain where the energy went in the ideal no resistance world??  Where did it convert to???

In the real world where resistance is applied, did we get that heat for free?   

Mags

[NickZ]

  No...  No free lunch in the real world. As free energy is killed by each and every component used. The more of them the less for us. Right?  So how do we get to OU. Remember OU...   As an old timer here, you should know how, by now. But,...