ENERGY AMPLIFICATION

[Magluvin]

That is 50% losses for 2 caps.

If we have 1x100uF, 20V and 10x100uF empty caps,
in each empty cap will be loss of 50%??? , 10x50% loss, if we try to equalize them parallel.
At the end empty parallel caps will eat all the energy from first one in the way you describe.
Is that so?
How much energy will remain in that case?

If I start with 100uF, 20V and I end up with 100uF, 20V, I will lost 50% of energy no matter that
values on the start and end are same?

So start 100uF, 20V and end 100uf, 20V does not contain same energy?

You are dividing 20 / 4 = 5
but then you adding 5 + 5 = 10 to get on the same point you started (20)
Can you mix operators ( * and / ) with ( + and - ) , because they are not the same order.

OF course, answer is, I don't know basic math.
For you 100uF, 20V and second caps serial 2x100uF, 10V (100uF, 20V in series) does not contain same energy?

But if you use step down transformer from 1A, 100V to 2A, 50V the energy is same (power)?
So in transformers it is the same, just in caps are not?

If you say so, then it must be true.

You are not figuring in what it takes in energy to fill the cap to 40v vs 20v.  At 20v there is an amount of energy there which is 20mj for 100uf.  And 100uf at 40v is 80mj.  So are you saying it doesnt take more energy to charge the 20v cap up to 40v than it did to just charge the cap to 20v in the first place?
I understand what you are thinking, but it is wrong. Just like in a car audio amplifier, if the rails are +20v and -20v and the rms sine output before clipping is 14.14v, then that divided by 4ohms speaker is rms 3.535amps, then the power sent to the speaker is 49.9849w..  But if the rails are 40v + and - we calculate the same. 28.28v sine just below clipping gives us 7.07amps rms to the speaker which is 199.9396w.   

So you cant think of energy and voltage numbers as linear.  get it?
Mags

[Magluvin]

Your great knowledge comes from Wikipedia which says Q/2 and V/2,
and you accepted that without thinking.

In our case there is NO Q/2,
there is just V/2.
Our capacitors did not change capacity, they remained 100uF at the start,
just as they were 100uF at the end.

No capacity change, so Q/2 can not be applied.
There is only voltage change.
Capacity of 2 capacitors remains constant, 100uF.

Nobody ever said nothing about capacity change, none.

Can you understand there is no change in capacity in our capacitors?

There is no conundrum.
For this case there is only law of parallel and serial capacitors connection with constant capacitance of 100uF.

Otherwise, parallel and serial connection law would not work.

Again you are missing the point.
If we have 1 cap 100uf at 100v and we calculate the energy, and then do the cap to cap into another 100uf cap at 0v then end up with 2 100uf caps at 50v each, there is no coming back from that. 50% was lost. No matter how you reconfigure the caps, parallel, series, the total amount of energy left after the fact is only 50% of what you started with. Lets say we connected the 100v cap to the 0v cap and we leave them connected, we then have 2 100uf caps in parallel, which equals 200uf total and 50v total.  There was 100% more energy in the 100uf cap at 100v than there is in a 200uf cap at 50v.  50% of 100 is 50. 100% of 50 is 50.  So 100uf 100v has 100% more energy than the total of 2 100uf caps at 50v, or to say also 200uf(2 100uf caps in parallel) at 50v.
Trust me. I have this down and imprinted. Unless you do your own real calculations and overall research, you 'cannot' change my mind on this. Cannot.
You come in like an all knowing being when really you are totally in the dark. That is not my problem any further here.

Mags

[WhatIsIt]

I am not here to make you think differently.

I am here for you to change my mind.

I start with 100uF, 20V.
I make parallel 2 caps 100uF, 10V.
Connect them in series, I end up with 2 caps which act as one 100uF, 20V,
according to law of serial connection.

Start = 100uF, 20V
End = 100uF, 20V

How the start and end have same values and different energy levels?

Something is wrong, isn't it.
I studied cap conundrum and it violates parallel, serial law of connection.

But if you say that is the case, let it be.

[WhatIsIt]

When you got more on output, it violates law of conservation of energy.

Now you lose energy in thin air and always by 50%, strange, and it does not violate law of conservation?

Stick has 2 ends, but for you only one is right.
For you it does not apply for loss, but only for gain?

[Magluvin]

I am not here to make you think differently.

I am here for you to change my mind.

I start with 100uF, 20V.
I make parallel 2 caps 100uF, 10V.
Connect them in series, I end up with 2 caps which act as one 100uF, 20V,
according to law of serial connection.

Start = 100uF, 20V
End = 100uF, 20V

How the start and end have same values and different energy levels?

Something is wrong, isn't it.
I studied cap conundrum and it violates parallel, serial law of connection.

But if you say that is the case, let it be.

No.
2 100uf caps in series becomes 50uf total.  If you cannot find that info anywhere, then you did not look. The only other thing with the series caps is that you have increased the voltage 'rating' to double.. 

2  100uf caps rated at 50v each. if put in parallel, then we have 200uf rated at 50v. In series we end up with 50uf that can 'handle' 100v, 50v each in series connection.Pretty simple. Those facts have not changed since I started in electronics at 10yrs old. Radioshack kits.


Mags