Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!apple!bionet!ames!sun-barr!cs.utexas.edu!uunet!richsun!bold
From: bold@richsun.UUCP (Jason Bold)
Newsgroups: sci.electronics
Subject: Re: HV Cap Fun!
Message-ID: <397@richsun.UUCP>
Date: 15 Jun 89 18:51:12 GMT
References: <4924@m2c.M2C.ORG> <3806@mit-amt> <20772@quacky.mips.COM> <27119@pbhya.PacBell.COM> <4489@midas.STS.TEK.COM>
Reply-To: bold@richsun.UUCP (Jason Bold)
Organization: RICH Inc. , Franklin Park,IL
Lines: 43


If there were an inductor inserted in the circuit, no energy would be carried
off in EM radiation, because that required perpendicular electric and magnetic
fields to create a propogating wave. Anyway, an ideal inductor won't lose any
energy either, it is stored in a magnetic field. I believe the equation that
holds true is q=CV, q is a constant. I believe that NO electrons will go
flying off into space. Look at the problem from the reverse angle. In other
words, what happens if you start with two capacitors connected together at
2uF apiece, charged up to 5.0V? Does everyone agree that it will take a finite
amount of work/energy to move the charge from one capacitor to the other such
that the first capacitor is completely discharged? Then just remove it from the
circuit by disconnecting the two. Viola, you are back at your original
situation! I would guess that the amount of energy required to charge the
second capacitor to 10V. and leave the first one discharged is the amount
of energy that is "lost?" when the two are brought together. The REAL question
is "Where does the energy go?". Energy and work are the same thing, so it is
the work required to move half of the electrons from one capacitor to the
other. It is somewhat akin to the following scenario: Assume you have a 10lb.
ball suspended 1ft. above the ground by a string. There is potential energy
stored there. If you cut the string (we are assuming that 'cutting' the string
takes no energy), and the ball falls to the ground, then where did the energy
go? There is no potential energy or kinetic energy left! But WAIT, you say!!!
If this is an ideal system, the ball would bounce forever, or else the energy
is dissipated when it goes smack against the ground. That is correct. It's
the same thing everyone has been saying about the system oscillating forever.
It would with no damping factor, ie. a resistor, or damping pot, in this case,
soft ground. So, my conclusion is that if there is no damping factor, a 
resistor, the potential and kinetic energies constantly are changing forever.
the "lost" energy is the energy that is being used to keep the circuit
oscillating. It will require exactly this amount of energy to completely stop
the oscillations. I believe someone else has already posted a proof to this.
So, the case of the missing joules has been solved. The potential energy
stored in the capacitor has been (50J of it anyway) converted into kinetic
energy (the ocillations). Therefore, both q=CV and E=1/2(CV^2) still hold.
It's just that now, with an inductor (no loss), E= 1/2(CV^2) + 1/2(LI^2).

Any comments?

-- 
==============================================================================
= Jason Bold       "A trend monger is a person who dreams up a trend... and  =
= bold@richsun.UUCP spreads it throughout the land using all the frightening =
=                   little skills that science has made available." - FZ     =