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Re: 50%

Hi Jack,
          Last one from me on this subject.

> This capacitor problem is a tricky question and and covers many
> possibilities and is sometimes given by electrical/electronic instructors
> because it has to do with both capacitor theory and unfamiliar algebra.
> 
>    Your Q = CV = coulombs is correct
>    My   J = 1/2 CV^2  = joules (energy) is also correct
> 
> Coulombs are a quantity of electricity (not energy). One coulomb passing a
> fixed point in one second is one ampere. You have to multiply this by
> voltage, etc., to get energy. When you multiply coulombs by voltage you get
> CV^2.
> 
> Joules (energy) is a unit of work (something a customer is willing to pay
> for as I have said before). One joule is one watt second. Energy is always
> conserved in a test when correctly performed and calculated.
> 
> There are several ways to solve this problem. However, the voltage across
> the two capacitors after reconnecting them must be found by the conservation
> of energy equation.
> 
> I will start with both capacitors equal to 1 farad and the voltage at 10 volts.
> 
> At start - energy in one capacitor = 1/2 CV^2 = 1/2(1)(10)^2 = 50 joules
> Connecting the two capacitors  J = 1/2 CV^2    50 = (1/2)(2)(V)^2 
> Voltage across both capacitors V = sqrt(2*(50)/2) = sqrt(100/2) = sqrt(50) =
> 7.07 volts not 5 volts.
> 
> Energy in both caps = (1/2)(1)(7.07)^2 + (1/2)(1)(7.07)^2 = 25 + 25 = 50
> joules. Energy is conserved.
>     
> Note that Q = CV in the energy equation gives  J = (1/2)(Q)(V). The
> difficulty  is that the V is in two of the variables of the equation. This
> means you would have to find the volts by some other method such as the
> conservation of energy equation.
> 
> This brings up the question. How did you measure the voltage in your test
> and come up with V/2? Was it a rough estimate? Was the voltage on the two
> caps a steady voltage or a reducing voltage? Were the caps equal, see below.
> I used 200000 uf and a digital meter and got a little under V/sqrt2 and
> greater than V/2. I found that smaller capacitors did not work very well to
> get the accuracy required using equal capacitors..
> 
> Testing requires a person well versed in the test possibilities. For
> example, in this test note that when the second capacitor is made smaller
> the two caps voltage approches the original voltage!! Also, when the second
> cap is made larger the two caps voltage approches zero!! You can come up in
> the test with the two cap voltage of almost zero to 100% original voltage! 
> 
> Can you find the combination of unequal caps to come up with V/2 and still
> have conservation of energy?? You could get 5 volts with certain unequal
> capacitors in the test. The two capacitor voltage would not be V/sqrt2, this
> only applies to equal capacitance capacitors. 
> 
> Whenever anyone comes up with a test that does not conserve energy you must
> realize, as you did, that something is wrong with the testing even though
> you do not have an immediate answer. This occurs often with tests that
> supposedly proves free energy.
> 
> Try calculating the two "capacitor voltage" using unequal capacitances. If
> you can do the algebra correctly (tricky) you have a good understanding of
> this type of problem. 
> 
> Note that this has very little to do with Tesla coils but it is an
> interesting problem that gives engineers bigger problems.

I measured using two identical capacitors and a hi-Z DVM. I tried 3 
different cap values ranging from 1uF (polyester) to 2200uF 
(electrolytic). I can do no better at this stage than to refer you to 
Ed Harris's post. I stand by my previous posts on this subject.

Malcolm