- #36
sophiecentaur
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Of course it is. It's a definition. It also represents the energy that you could get out, which assumes an ideal capacitor, of course.Mark Harder said:When one calculates work against a field this way, dissipation is ignored.
You are introducing more 'practicalities than are appropriate in a definition. However, if you want a series internal resistance, then this only needs to be added to the series source resistance. If you look at your on line tutorial and extend it to work out the energy dissipated in the series R, you will find that the value of R cancels out and you are left with CV2/2. Charging the C slowly or fast makes no difference. Believe the Maths.Mark Harder said:hence more heating due to the finite resistance in the internal conducting path.