Re: 50%

To: tesla-at-pupman-dot-com
Subject: Re: 50%
From: "Edward V. Phillips" <ed-at-alumni.caltech.edu>
Date: Mon, 11 Nov 1996 19:51:15 -0800 (PST)

"Total energy in caps
Joules = 1/2C(V/2)^2 + 1/2C(V/2)^2  =  1/2(1)(V/2)^2 + 1/2(1)(V/2)^2
Joules = 1/2(1)(V^2/4) + 1/2(1)(V^2/4) = V^2/8 + V^2/8 = 200/8 = 25 total
joules in the caps.
Energy is conserved

Coulombs in first cap = CV = (1)(10) 10 coulombs
Total coulombs in caps = 2CV = 2(1)(5) = 10 coulombs
Coulombs are conserved

Does anyone know how to find what the voltage would be if the second
capacitor was 1/2 farad instead of 1 farad? The voltage would be greater
than V/2."
Answer:
1 farad capacitor charged to 10 volts = 1*10 = 10 coulombs
1 farad capacitor connected to 0.5 farad capacitor = 1.5 farads
Final charge = 10 coulombs
Final voltage =10 coulombs/1.5 farads = 6.66666666 volts
Final energy = 0.5*1.5*(6.66666666)^2 = 33.333333 joules
compared to 25 joules for the two capacitor case.
	This result is reasonable.  In the limit, when an
infinitessimally (sp?) small capacitor is connected across
a large one there is no change in voltage or energy.
Ed Phillips