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RE: 50%
>Message-ID: <199610310526.WAA12887-at-poodle.pupman-dot-com>
>Date: Wed, 30 Oct 1996 22:26:02 -0700
>From: Tesla List <tesla-at-poodle.pupman-dot-com>
>From MALCOLM-at-directorate.wnp.ac.nzWed Oct 30 21:56:15 1996
>Date: Thu, 31 Oct 1996 13:10:55 +1200
>From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
>To: tesla-at-pupman-dot-com
>Subject: RE: 50%
[ snip ]
> If the circuit settles down to a steady state (no oscillation),
>50% of the energy is indeed lost, but charge is indeed conserved.
>Elaborating: E=0.5CV^2 You can see that a cap at half its original
>voltage has only 1/4 of its original energy. Two identical caps at
>half the original voltage of one has still only half the original
>energy.
> BUT, since Q=CV, you can see that if you double C and halve V,
>Q (charge) remains conserved.
>
>Malcolm
Well, I am a firm believer in the law of Conservation of Mass and
Energy. Pure capacitors do not use energy. One capacitor
discharging through a perfect conductor in a totally isolated
environment. Where did it go? EM radiation? The magnetic field
around the conductors rises and collapses. Energy in, energy out.
YOu CANNOT destroy it, so where the heck did it go? Isn't there
something terribly wrong with our thinking here somewhere?
Fred W. Bach , Operations Group | Internet: music-at-triumf.ca
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