Path: utzoo!attcan!uunet!cs.utexas.edu!rutgers!iuvax!watmath!egvideo!timk
From: timk@egvideo.UUCP (Tim Kuehn)
Newsgroups: sci.electronics
Subject: Re: HV Cap Fun!
Summary: Watch that conservation of energy!
Keywords: capacitor,energy,paradox
Message-ID: <2026@egvideo.UUCP>
Date: 1 Jun 89 04:00:00 GMT
Reply-To: timk@egvideo.UUCP (Tim Kuehn)
Organization: A Box in the Basement, Kitchener, ON
Lines: 67

>In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes:
>>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.
>>

I trust that C is in uF, otherwise your result would be 100E-6 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. 

Is it really? This would seem to violate the laws of conservation of energy,
since, as is further stated:

>>						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 ?

		[some stuff deleted]

>Why should the energy an electron has drop when it goes to another
>capacitor?
>
>The formulas: U=1/2 C V^2=100
>	      U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case)
>	      U=2 x .5 x 10 x 10=100 for two capacitors
>	      U=1 x .5 x 20 x 10=100 for one capacitor
>
>I'm obviously missing something....
>
>- Jeff Naiman (naiman@yale.edu)

Consider that in this ideal scenario, you have one capacitor charged to 
contain 100 J of energy. If you connect another capacitor in parallel with it
(I presume that's the configuration) then part of that charge will drain off
the charged capacitor into the uncharged capacitor until both are equally 
charged. Therefore, your system MUST end up with 100 J at the end, which 
does not necessarily mean that your final voltage will be 5V on the line. 
To wit:

In the first scenario with the 2uF cap charged to 10V we have:

1/2 C * V ^ 2 = 1/2 * 2 * 10 * 10 ^ 2 = 100 J

adding another capacitor on parallel doubles the C term, the energy 
stored must remain constant (see the law on conservation of energy!), therefore
the voltage the caps will have in the final equilibrium is the only unknown.
Solving the above equation for V with C = 4uF:

(1/2) (4) * V ^ 2 = 100 
V = (100 / (1/2 * 4)) ^ (0.5)
V =  7.07 Volts

which, as you can see, is not the same as the 5V you stated previously. No
paradox at all, when you look at it.

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