[Prev][Next][Index][Thread]
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.