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Re: 50%
Tesla List wrote:
>
> >From couturejh-at-worldnet.att-dot-netSun Nov 10 21:35:49 1996
> Date: Sun, 10 Nov 1996 06:27:45 +0000
> From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
> To: tesla-at-pupman-dot-com
> Subject: Re: 50%
>
> At 05:25 AM 11/9/96 +0000, you wrote:
> >From sroys-at-umabnet.ab.umd.eduFri Nov 8 22:16:35 1996
> >Date: Fri, 8 Nov 1996 09:04:03 -0500 (EST)
> >From: Steve Roys <sroys-at-umabnet.ab.umd.edu>
> >To: tesla-at-pupman-dot-com
> >Subject: Re: 50%
> >
> >On Thu, 7 Nov 1996, Tesla List wrote:
> >
> >> None of the tests gave end voltages of V/sqrt2. However, this is probably
> >> due to the sparking losses which would cause an additional voltage drop when
> >> the second cap is connected. It appears that this test will give a V/2 end
> >> voltage but because of the sparking voltage loss the true voltage is between
> snip --
> >>
> >> Jack C.
> >
> >Interesting...where do you think the extra charge came from?
> >
> >Steve Roys.
> >
> --------------------------------------------
>
> Steve -
>
> I tried to change the laws of the conservation of energy and the
> conservation of charge without success. Fortunately. The next time I will do
> the experiment first before I comment. I now realize that when doing this
> equal capacitor test that after the transfer 50% of the energy is in the
> capacitors and 50% is lost in noise and spark. This gives a final V/2
> voltage across each cap.
>
> One thing the experimenters did not mention is that the energy loss by noise
> and spark can be eliminated by connecting a large resistor between the two
> capacitors when the transfer is made. The transfer will be made without bang
> and spark. However, the resistor energy loss is the same as the noise and
> spark energy loss so the energy conditions are the same. This also gives a
> final V/2 voltage across each cap.
>
> For the example with two equal capacitors of 1 farad and a 10 volt source.
>
> First cap J = 1/2 CV^2 = 1/2(1)(10^2) = 50 joules
> 25 of the joules (energy) are lost in noise and spark
>
> 25 of the joules are in the caps
> The voltage on each cap V = sqrt(2J/C) = sqrt(2)(25)/2) = 5 volts = V/2
>
> Total energy in caps
> Joules = 1/2C(V/2)^2 + 1/2C(V/2)^2 = 1/2(1)(V/2)^2 + 1/2(1)(V/2)^2
> Joules = 1/2(1)(V^2/4) + 1/2(1)(V^2/4) = V^2/8 + V^2/8 = 200/8 = 25 total
> joules in the caps.
> Energy is conserved
>
> Coulombs in first cap = CV = (1)(10) 10 coulombs
> Total coulombs in caps = 2CV = 2(1)(5) = 10 coulombs
> Coulombs are conserved
>
> Does anyone know how to find what the voltage would be if the second
> capacitor was 1/2 farad instead of 1 farad? The voltage would be greater
> than V/2. Note that the losses would now be less than the above problem. Do
> you want to give it a try, R. Hull, to compare with answers from other coilers?
>
> This does have a tie in to Tesla coils. The toroid is charged by the Tesla
> coil. If it is discharged to a second similar toroid the voltage would be
> V/2 maybe??
> It may be possible to measure the secondary voltage this way. Any comments?
>
> Jack C.
Jack,
In metrology, specifically in the use of an electrometer as a
coulombmeter, (one of its many roles), The rule of thumb is that a
differential of 100 times the charging capacitor is needed to reduce the
transfer losses to insignificance. Much like the old saw about RC
timeconstants needing 5-7 periods to effect completion for all practical
purposes. In theory, a capacitor is never fully charged! In fact, it
virtually is, in about 7 time constants.
Richard Hull, TCBOR