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27th August 2006, 05:12 AM  #1 
diyAudio Member
Join Date: Dec 2003
Location: Munich

Energy Loss
Hi Ho !
... just strolled through some of my old records and found a nice phenomenon, which might make fun to discuss here... Imagine two caps. Both have the capacitance. One is charged (U1), the second is uncharged. The energy in the first is evidently E=1/2 x C x U1^2. If we now connect ideal the second cap, then the charge will split half and half on both caps.... balancing at U2=1/2 x U1. If we now calculate the energy, we will find that half of the energy is lost.. ... somehow.. each cap has just a quarter of the original energy.. Curious thing, only ideal and lossfree assumed components involved and still energy is lost. Some years back I started some calculation on this. I started with an Resistor in between both caps and then planned to move the R mathematically to zero.... hoping to get reasonable results with the sentence of de l'hopital or similar... Aehem well, of course before I could move R to zero.... I came to the well know result from capacitve switching devices , that the losses are independent from impedance of the switching device. As soon as you calculate the dissipated power the resistor drops out of the calculation.... Sorry for poor scan quality. In the last millenium my docusystem was not really high end 
27th August 2006, 05:14 AM  #2 
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Join Date: Dec 2003
Location: Munich

second page

27th August 2006, 05:15 AM  #3 
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third page

27th August 2006, 05:21 AM  #4 
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Join Date: Dec 2003
Location: Munich

...so for me this chargingchangingevent looks similar to a Dirac pulse.
Means: Indefinite high current pulse, indefinite short time, well defined energy. E = t x P = t x I^2 x R = 0 x (indefinite)^2 x 0 
27th August 2006, 05:24 AM  #5 
diyAudio Member
Join Date: Dec 2003
Location: Munich

In case of two identical caps:
E_loss = 0^2 x (indefinite)^2 = 1/4 x C x U^2 
27th August 2006, 05:45 AM  #6 
diyAudio Member
Join Date: Nov 2005

Dude... I was planning on sleeping tonight... good one!

27th August 2006, 06:41 AM  #7 
diyAudio Member
Join Date: Dec 2003
Location: Munich

...Arizona... must be something like midnight...
OK: Access to bed granted. And dream with your wife , not about caps 
27th August 2006, 07:51 AM  #8 
diyAudio Moderator

Here's an analogy that (at least for me) clarifies things. Imagine your first situation, a cap with a voltage V and a charge Q. Now, move one of the plates closer to the other to increase the capacitance...
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Remember: life is ten per cent what happens to you, ten per cent how you respond to it, and eighty per cent how good your reflexes are when the Tall Ones come at your throat with their pincers. 
27th August 2006, 08:07 AM  #9 
diyAudio Member
Join Date: Nov 2005

Choc,
aaaarrggghhhhhhh! Something about no resistance, infinite current, no inductance bugs me. Consider the problem chronologically reversed... where does the energy COME from. 
27th August 2006, 10:21 AM  #10 
diyAudio Moderator

poobah, you can still conserve energy and charge with idealized conductors. Try answering my question first: what happens when you move capacitor plates closer together to double the capacitance while conserving charge?
BTW, 2nd Law makes the reversal unphysical.
__________________
Remember: life is ten per cent what happens to you, ten per cent how you respond to it, and eighty per cent how good your reflexes are when the Tall Ones come at your throat with their pincers. 
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