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RE: 50%
Please indulge all,
This one's directed to me. I've commented on this
already so this is the last time....
> > 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?
In my humble opinion, no. The circuit initially oscillates if there
is insufficient energy loss _to the system_. This is the classic
underdamped case. No inductor couples energy to itself 100%. If it
did, you have a circuit ringing forever. NB the equivalent circuit
for the lossless case is a _single_ capacitor and a single inductor
(2 caps in series = a single cap) voila - a tuned circuit!. You can
discharge one cap through a diode to the other and if the circuit is
lossless, _all_ energy ends up in the second cap. You can measure
this. In my humble opinion this is well established theory. In the
real world, you _cannot_ have a lossless circuit like this, but you
can come close. Our coupled circuits rely on underdamped oscillations
to effect an almost complete energy transfer from one capacitance to
another.
I am done,
Malcolm