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RE: 50%

Fred, all,
            I think there might be some possible confusion between 
conservation of charge and conservation of energy in your reply.....

> >In theory,  We should never be more than 50% efficient in energy transfer 
> >from one capacitor to another!  i.e. the Cprimary to the Csecondary 
> >Usually the resonator load capacitance plus Ion cloud loading.  This 
> >assumes 100% coupling and zero other losses!  If fact, we are much lower 
> >than that with the finest system in operation.
> >
> >50% of the capacitively stored energy always disappears in circuit loses 
> >(resistive and magnetic) even with direct wired connections.  There is a 
> >lot of additional wasteful garbage going on in between the primary and 
> >secondary capacitors.
> >
> >Richard Hull, TCBOR
> 
>    What are you talikng about?  Some specific circuit I presume.  Your
>    post lacks quoted text and the second paragraph is rather general,
>    so I take some exception.
> 
>    For instance,
> 
>    If I take a 0.1ufd cap charged at 10KV and another identical one
>    but uncharged, and I connect them together (as you said, "direct
>    wired connections"), even with a small resisitor in series, I will
>    get a final voltage of very close to 5kV, will I not?  This is
>    nearly 50% of the energy in the first capacitor transferred to the
>    second with very little losses to the circuit.  There are tiny
>    losses in the dielectric and the wires.  Typical dielectric
>    'efficiencies' exceed 99% and the resistive losses are trivial. 
>    This is close to 100% transmission of 50% of the charge.  The
>    remaining charge stayed on the first capacitor.  The losses to the
>    **circuit** resistive and magnetic are tiny.  Most of the energy is
>    still stored *in* the two capacitors.
> 
>  Fred W. Bach ,    Operations Group        | Internet: music-at-triumf.ca

     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