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Re: 50%

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> 
> 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%
> 
snip snip

> 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?
> 
Once, you have confidence in the physics, try a calculation -
we find it saves time! (see below) EH


> 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

Dear Jack,

For two capacitors C1, C2:
----------------------------------------------------------------------------
-----------------
Initial conditions ( i stands for initial values of voltage or charge):

cap 1: Q1i=C1V1i 

cap 2: Q2i=C2V2i let's not assume neither is zero for gernerality

Total Initial Energy:

	Ei= 0.5( C1V1i^2 + C2V2i^2 )
----------------------------------------------------------------------------
------------------
After connecting them in parallel (f stands for final quanitity):

V1f=V2f=Vf
Q1f+Q2f=Q1i+Q2i 	:charge conservation

or upon substituting q=cv ---> (C1+C2)Vf=C1V1i+C2V2i


					C1V1i+C2V2i
so the final voltage on both caps is: Vf = -------------------------
					  C1+C2


Final charges:
		C1V1i+C2V2i
cap1 Q1f = C1 -------------------------
		C1+C2

		C1V1i+C2V2i
cap2 Q2f = C2 -------------------------
		C1 + C2


Final Energy:
( from 2 parallel caps voltage Vf)

 	     (C1V1i +C2V2i)^2
Ef = 0.5  --------------------------
		C1+C2	

----------------------------------------------------------------------------
---
Limiting Case 1 ---
If you set V2i=0 and C1=C2, you get the expected:

Ef/Ei=1/2 and Vf/V1i=1/2
----------------------------------------------------------------------------
--
Limiting Case 2 ----
A More interesting limit:
set V2i=0 C1 <> C2:


Ef     C1                               Vf        C1
--- = ----------   		and ----- = ---------------
Ei    C1 + C2                        Vi      C1+ C2


Conclusion, the tranfer efficiency approaches 100% ( or loss ---> 0 )
as C2 becomes much less than C1

AND the transfer efficiency approches 0%  (loss --> 100%)
as C2 becomes much larger than C1

Note again, for C1=C2=C we get 50% transfer as before!

----------------------------------------------------------------------------
---

Now if all that hasn't made everyone sick, I'll add one more
point: by using large INDUCTORS and switches which could be timed
very precisely, you COULD split the charge from the cap 1 equally into
cap2 and cap 2 (both C) with Very Little Energy Loss (as Jack was
originally hoping for). The inductor must be large and not lossy so that
the Q is high and you must be able to discharge cap 1 into the inductor
until the voltage goes exactly to zero. Then precisely at this moment, you
switch in cap  2 in parallel with cap1 to let the inductor charge them both
back up to half the energy of the original cap 1. When the inductor current
goes to zero, disconnect them both -- viola!


Sorry Chip,

-Ed Harris