Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!tut.cis.ohio-state.edu!ucbvax!hplabs!hpfcdc!hpfcdj!myers
From: myers@hpfcdj.HP.COM (Bob Myers)
Newsgroups: sci.electronics
Subject: Re: Re^2: HV Cap Fun!
Message-ID: <11170009@hpfcdj.HP.COM>
Date: 1 Jun 89 18:46:24 GMT
References: <4924@m2c.M2C.ORG>
Organization: Hewlett Packard -- Fort Collins, CO
Lines: 67

>A Paradox of Capacitor Energy Storage
>
>I've heard several competing answers to this paradox. None is entirely
>satisfactory:
>
>Consider an ideal 2uF (for computational ease) capacitor charged by a 10 volt
>source. Eventually, the energy stored is (1/2)*CV^2=100 joules.
>
>Consider the capacitor to be isolated from the voltage source, and then
>directly shorted across an identical (ideal) capacitor. Eventually, the
>voltage across each capacitor will be 5V. Now, there are two equally
>charged capacitors, each storing (1/2)*CV^2= 25 joules, for a total of
>of 50 joules.  What happened to the other 50 joules ?

This is one of the "classic" puzzles in electronics, and more than anything
else is a prime example of how you can do things with "theoretically ideal"
parts which make absolutely no sense at all in the real world.  (Here's
another example - an ideal 100 uF capacitor is instantaneously connected
across an *ideal* 10V voltage source.  How long does it take for the
voltage across the capacitor to reach 10V?)  Anyway, some possible answers
to the above:

1. Given the "ideal" components used in this problem, we are forced to 
conclude that there is zero resistance in the conductors used to connect
the two capacitors.  However, there *cannot* be zero inductance in these
conductors, if they have any physical length at all.  Therefore, what we
actually have is a resonant L-C circuit, which is excited into oscillation
at its resonant frequency when the second capacitor is connected.  With
no resistance present, the oscillation is never damped out (no loss of
energy via a resistive element), and so it continues forever.  The original
assumption, that the circuit would wind up at 5VDC across the two caps, is
incorrect.  (For that matter, consider the word "eventually" in the problem
statement; where does the "eventually" come from in the absence of resistance?)

2. Let's suppose, though, that we further "idealize" the situation by
considering that the conductors between the capacitors have no inductance.
(This is possible only if they have no physical length.)  For example,
suppose that I simply (through the typical Amazing Blackboard Magic)
double the area of the original capacitor's plates, hence doubling the
capacitance.  Now what happens?

2b.  What if there is just a little inductance?  Enough so that the
theoretical resonant frequency (given by 1/(2pi*sqrt(LC))) is EXTREMELY
high - say, that of UV light?  (See what trouble you can get into with these
"ideal" and "purely theoretical" calculations?)

3. Another go at it: this time, instead of doubling the plate area, I will
instantaneously insert a sheet of dielectric with a relative permittivity
of 2 (the original capacitor had air or vacuum between the plates).  What
happens in this case?

4. Of course, all of the above discussion ignores the fact that there will
ALWAYS be a loss of energy from this circuit; if we step back from the
realm of pure circuit theory, and look at the situation from a fields point
of view, we have all we need (time-varying electric and magnetic fields) for
this thing to lose energy via EM radiation.  Exactly how this happens, of
course, would depend on the physical structure of the apparatus.  Now the
first answrer is incorrect, and we WOULD expect a loss of energy from the
system, and so a damping of the oscillations.  We cannot, though, with the
information given so far, describe this in detail.

I certainly hope that this has further confused the issue.   :-)


Bob Myers  KC0EW   HP Graphics Tech. Div.|  Opinions expressed here are not
                   Ft. Collins, Colorado |  those of my employer or any other
myers%hpfcla@hplabs.hp.com               |  sentient life-form on this planet.