Actually I threw you a curve there, and also erred when I wrote, A joule is a farad-volt squared. It ain't. A farad (volt squared) is half a joule. Only the volts get squared, and I omitted the 1/2. To derive that from the simple equations requires the integral calculus.
Now then, how many joules are stored in a two-farad capacitor charged to three volts?
Most people would use the calculus to solve this. Let's see if we can do it another way.
Let's charge those two farads [C] with a constant two-ampere current. The voltage e will go up at one v/s.
At start, the capacitor is at zero volts, and the current is two amperes, so the power going in is zero. But the power starts rising immediately, because the voltage [e] starts rising immediately and the current is constant.
After one second, it's charged to one volt. Our two amperes are now flowing at one volt, so the power going in at that instant is two watts.
After two seconds, it's at two volts, and still two amperes, so the power going in is four watts.
After three seconds, it's at three volts, and still two amperes, so the power going in is six watts.
Since the voltage went up linearly and the current was constant, the power went up linearly too, from zero to six watts. So the average power was three watts over the three seconds. The energy, then, must be nine joules.
Does this consist with the answer from calculus? Yes. The result from integral calculus would be that the energy = (1/2)Ce^2, using your excellent notation. (I wish I could use superscripts here.) C = 2. e = 3. e^2 = 9. Thus (1/2)Ce^2 is nine joules.
We're good to go. We did it at constant current, but actually the energy in that capacitor doesn't depend on how it got there. It depends only on the capacitance and on the voltage across it. At two farads and three volts, it's nine joules, regardless. The capacitor doesn't "remember" the history, it only remembers the end result.
__________________
-- Carl
|