Energy Loss - diyAudio
 Energy Loss
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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 docu-system was not really high end
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diyAudio Member

Join Date: Dec 2003
Location: Munich
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diyAudio Member

Join Date: Dec 2003
Location: Munich
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 27th August 2006, 05:21 AM #4 diyAudio Member   Join Date: Dec 2003 Location: Munich ...so for me this charging-changing-event 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     Join Date: Oct 2002 Location: Chicagoland Blog Entries: 2 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... __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 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     Join Date: Oct 2002 Location: Chicagoland Blog Entries: 2 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. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

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