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Re: Real data from real experiments.

Tesla List wrote:
> 
> >From hullr-at-whitlock-dot-comWed Nov  6 22:31:19 1996
> Date: Wed, 06 Nov 1996 11:38:55 -0800
> From: Richard Hull <hullr-at-whitlock-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Real data from real experiments.
> 
> All,
> 
> I took the time out to repeat two experiments which few seemed to do but
> a lot have commented on.  I performed the first experiment presented here
> over a year ago and the latter experiment 5 years ago.  So I repeated
> with fresh data and first rate instrumentation.
> 

<SNIP>

> Conclusion:
> 
> The voltage on any capacitor discharging into an identical capacitor with
> divide by 1/2 or equal 50% of its original voltage regardless of massive
> blasting of copper to molten beads in the exchange or a kinder, gentler
> flow of charge through limiting resistances.  1/2 of the original stored
> energy (1/2cv^2) is aboslutely and irrevocably lost in this situation to
> a plethora of forms of waste energy given off to the evironment and
> circuit components.  Charge is conversed as well as energy.  The energy
> transfer to the other capacitor will forever undergo a 50% loss factor.


> DO THE EXPERIMENT!  (Benjamin Franklin)
> 
> Richard Hull, TCBOR



Richard, et al:


 Thanks for the effort in duplicating the experiment. I do have one
comment, and that is on the subject of capacitive voltage division:

 I seem to recall that if you place two identical value capacitors
in series across a voltage source, then the charges on each capacitor
will be equal, and so will the voltage. Hence, each capacitor will
have 1/2 of the supply voltage across it.

 To me at least, there isn't one iota of difference between 2 caps
charged in series, or a single cap charged up to the full voltage,
then connected across a second capacitor. The charges are going to
level themselves out according to the capacitance value, period.
(BTW, if the capacitance values are *unequal*, the voltage across
each capacitor -- doing the two-capacitor charge experiment -- will
be equal, yet the amount of energy in each cap is now dependent upon
the capacitor's value. In this case, one cap might 'lose' 1/3 of it's 
'charge, and the other could lose 2/3... hmmmm, there's something to
muse over.)

 Not to beat this topic to death, but if anyone wants to, please
review my (rusty) algebra:

   given 100 volts on 1uF = 0.01 Joules of *potential* energy.

   connect identical capacitor to charged capacitor, resulting
    in 1/2 of original voltage as verfied by Richard. Hence, stored
`   charge in each capacitor is now 50 volts on 1 uF = 0.0025 Joules.
   
   since the two caps are in parallel, the net stored energy is now
    0.0025 + 0.0025 = 0.005 Joules, or exactly 1/2 of the beginning
    energy level.

   Here's the kicker, people:

   two *identical value* caps in series yields a total capacitance
    of 1/2 of the value. So, for a 1 uF cap, that gives us a total
    value of 0.5 uF.

   if we place 100 volts across the 0.5 uF 'cap', we get a total
    stored energy of only 0.005 Joules, which, amazingly enough, is
    1/2 of what was stored in a 1 uF cap with 100 volts.

Looks like everything adds up to me, which should make the physics
accountants mighty happy. I guess at this point, I can ask a related
question:

"When two identical capacitors are hooked in series, where does 1/2
of the capacitance go?"

No charge is actually 'lost' - rather, it is simply re-distributed.
The total stored potential energy is reduced though, according to the
square of the voltage. Seems to me that this is exactly the reason why
we use small value capacitors with a heaping amount of very high voltage
in the primary circuit!

- Brent