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?
