🚀 go-pugleaf

I have a question for the non-5v advocates: Where did all the extra charge
come from? Q = C*V. If the final voltage is 7.07, you just made 8.28 coulombs
for free.

Maybe this will reassure some people. Look at the case where there
is a resistor R connecting the two capacitors, each with capacitance C. The
current through this resistor starts out at 10/R and decays toward zero at
infinite time by an exponential function with time constant RC/2. (See any
circuit theory book if you don't believe me.) In other "words"...

I = (10/R) * exp (-2*t/(R*C))

P = R*I^2 = (100/R) * exp (-4*t/(R*C))

To get the energy dissipated in the resistor, integrate power over time from
0 to infinity...

               /oo                                                       oo
E = (100/R) * | exp(-4*t/(R*C)) dt  =  (100/R) * (-R*C/4)exp(-4*t/(R*C)) ]
             / 0                                                         0

  = (100/R) * ( 0 - (-R*C/4)) which is, (voila!)

  = 25*C

2 Farads?   There's your missing 50 joules - in the resistor.

(Note that the factor of 25 is, more generally, 1/4 of V^2, where V is the
initial voltage difference between the capacitors.)


It's not at all hard for me to believe that in the resistanceless case, the
energy is carred off by radiation - anybody want to volunteer to do that
analysis? My Maxwell's eqs are a bit rusty.

						Jeff Winslow