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Re: 50%
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Jack Couture Replies to Malcolm>
> From couturejh-at-worldnet.att-dot-netMon Nov 4 21:57:51 1996
> Date: Tue, 5 Nov 1996 01:36:25 +0000
> From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
> To: tesla-at-pupman-dot-com
> Subject: Re: 50%
>
> At 03:25 PM 11/4/96 +0000, you wrote:
> >From MALCOLM-at-directorate.wnp.ac.nzMon Nov 4 06:45:29 1996
> >Date: Mon, 4 Nov 1996 20:09:43 +1200
> >From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
> >To: tesla-at-pupman-dot-com
> >Subject: Re: 50%
> >
>
> Malcolm -
>
> V/2 New zealand
> V/sqrt2 San Diego
>
> The reason the reciprocals are different in New Zealand than in San Diego
is
> because we are on different sides of the planet. (:>) (:]) (:<)
>
> 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.
>
Jack,
Energy is conserved in a thermodynamic sense ONLY if you pick a system in
which energy is guaranteed NOT to escape. Sometimes the only system which
satisfies this requirement is the entire universe! In the capacitor
problem, energy can only be considered to be conserved if the energy loss
is accounted for: ie, resistive heating, light and sound generation.
> 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.
>
You are, unfortunately, only "proving" that energy is conserved when you
actually STARTED with the assumption that energy is conserved! That's a
circular argument.
> 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.
This is a bit patronizing, I think. You should check your expermient....
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 is assuming that one understands the correct application of the
conservation of energy. I will bet 2 cases of beer that Malcolm's
experiment was done correctly and gave the result which IS predicted by the
correct application of the physics.
Let me turn your result back around on you:
If the final voltage is Vf=0.707Vi for both of the capacitors of
capacitance C as you state, then what is the final total charge? Well since
q=Cv, then we get:
Qtot(final)=CVf+CVf = 2C*(0.707Vi)=1.414CVi
but since the initial Qi=CVi
the final total charge is Qtot(final) = 1.414Qi
You have managed, with your algebra, to generate 41% new charge out of thin
air!!!
-Ed Harris
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.
>
> Jack C.
>
>