Path: utzoo!attcan!uunet!richsun!bold
From: bold@richsun.UUCP (Jason Bold)
Newsgroups: sci.electronics
Subject: Re: HV Cap Fun!
Keywords: capacitor,energy,paradox
Message-ID: <387@richsun.UUCP>
Date: 8 Jun 89 20:32:32 GMT
References: <4924@m2c.M2C.ORG> <3806@mit-amt> <20772@quacky.mips.COM> <10992@behemoth.phx.mcd.mot.com>
Reply-To: bold@richsun.UUCP (Jason Bold)
Organization: RICH Inc. , Franklin Park,IL
Lines: 49

>In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes:
>>A Paradox of Capacitor Energy Storage

>>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.

Let me see if I can take a crack at this one...

To start: Energy = 100 joules, c = 2uF.
To finish: Energy = 100 joules, c = 4uF.

Assume t = oo to forget about all of these infinite oscillation theories.

So, E1 = (1/2)CV^2 = 100 joules = E2, by conservation of energy.
E2 = (1/2)(4uF)(V2^2) = 100 joules
V2^2 = 100/[(1/2)(4uF)] = 50 V^2 :: V2 = 5*sqrt(2) = 7.07V.

Well, assuming a lossless circuit, noone out there can deny this law of
physics. Let's see if conservation of charge still holds. It should.

Q=CV: Q1 = C1V1 = (2uF)(10V) = Q2, based on conservation of charge.
Well, if you work it out, V2 does come out to be 5V. Gee, what went
wrong? We should have gotten the same answer both times. It must be
in the definitions of Q1,Q2,V1,V2,E1, and E2.

Q1 = the charge on C1 before the connection
Q2 = the charge on C1 and C2 after the connection
V1 = the voltage on C1 before the connection
V2 = the voltage on C1 and C2 after the connection
E1 = total energy in the system before the connection
E2 = total energy in the system after the connection
the system = capacitors C1 and C2

I think the problem lies in the fact that Q1,Q2,V1, and V2 are vectors and
E1 and E2 are scalars. For example, what would happen if you connected another
capacitor charged to 10V. in parallel with the first one (C1), but in opposite
polarity? Well, I think everyone would agree that the charges would cancel.
This would mean no energy left (E2). Where did it go. Well, E=(1/2)mv^2.
I think it is the energy required to move the electrons from one plate to the
other. If any electron moves at all, energy must be expended. That's the only
explanation I can come up with. Anyone tried this experimentally (unfortunately
intuitively, I think the voltage V2 is 7.07V)., I would like to know.
-- 
==============================================================================
= 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     =