Re: 50%

To: tesla-at-pupman-dot-com
Subject: Re: 50%
From: Richard Hull <hullr-at-whitlock-dot-com>
Date: Wed, 30 Oct 1996 11:52:16 -0800
Tesla List wrote:
> 
> >From music-at-triumf.caTue Oct 29 22:46:00 1996
> Date: Tue, 29 Oct 1996 09:40:00 PST
> From: "Fred W. Bach, TRIUMF Operations" <music-at-triumf.ca>
> To: tesla-at-pupman-dot-com
> Cc: music-at-triumf.ca
> Subject: RE: 50%
> 
> >Message-ID: <199610290525.WAA02026-at-poodle.pupman-dot-com>
> >Date: Mon, 28 Oct 1996 22:25:20 -0700
> >From: Tesla List <tesla-at-poodle.pupman-dot-com>
> >To: Tesla-list-subscribers-at-poodle.pupman-dot-com
> >Subject: 50%
> >
> >From hullr-at-whitlock-dot-comMon Oct 28 21:48:14 1996
> >Date: Mon, 28 Oct 1996 12:43:40 -0800
> >From: Richard Hull <hullr-at-whitlock-dot-com>
> >To: tesla-at-pupman-dot-com
> >Subject: 50%
> >
> >All,
> >
> >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 ,

Fred,

Think about this,  In the circuit you propose you have lost 1/2 of the 
energy present in capacitor #1 when all the smoke clears!  This was 
posted early this week.  Take a 2uf cap charged to 10Kv-(1/2 C V^2)-
100 Joules, and then remove the supply.

Place an identical uncharged capacitor across it.  They now both have 
half the original voltage appearing across the pair.  This is 5KV across 
a 4uf for a total of 50 joules!  Or, 2 X 2uf at 5KV.  this represents 50% 
circuit, internal, arcing, etc, total losses.  It matters not whether you 
use a 50 megohm resistor in the circuit or a 2" copper buss bar to make 
the connection, you will suffer a minimum of 50% losses.  Thus my 50% 
figure still stands.

Richard Hull, TCBOR