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From: naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman)
Newsgroups: sci.electronics,sci.physics
Subject: Re: HV Cap Fun!
Keywords: capacitor,energy,paradox
Message-ID: <62169@yale-celray.yale.UUCP>
Date: 31 May 89 18:06:49 GMT
References: <4924@m2c.M2C.ORG> <3806@mit-amt> <20772@quacky.mips.COM>
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Reply-To: naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman)
Organization: Yale University Computer Science Dept, New Haven CT  06520-2158
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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.
>
>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 ?

OK, here's a silly question. Why is the voltage across the two identical
capacitors 5V? Volts are joules/coulomb, or more simply, energy per
electron. Voltage drops when going through a resistor (sticking to DC current).
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)