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10-19-2010, 12:54 PM
 Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
 Originally Posted by Mario Ok, I want to calculate it anyway . We have: C1 (now in series) which is 0.0235F with a charge of 0.0235F @ 20V = 0.47J C2 which is 0.094F with a charge of 0.094F @ 10V = 0.94J To find the final voltage: Ctot.= 0.0235F + 0.094F = 0.1175F Qtot.= 0.47Q + 0.94Q = 1.41Q Vtot. = 1.41Q / 0.1175F = 12V So you were right, even if we were not talking about the same exact setup. This confirms the loss I am seeing by measurement. regards, Mario

You are not seeing a loss in measurement, unless you are not completing the cycle.

For example at this point:

C1 = .094F @ 12V
C2 = .0235F @ 12V

If we switch C2 into parallel again, we have .094F @ 6V.

C1 = .094F @ 12V
C2 = .094F @ 6V.

Which gives us a charge of
C1= 0.094c * 12v = 1.128q
C2 = 0.094c * 6v = 0.564q

Qtot = 1.128 + 0.564 = 1.844

And we have a capacitance total of .094 + .094 = 0.188

Now knowing we that Vtot = Qtot/Ctot 1.844 / 0.188 = 9.85V across C1 and C2, and they started at 10v each.

which is 98.5% of the original energy, and I would be willing to bet that the 1.5% missing is due to rounding error.

So where is this loss?
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