I answer and demonstrate that the yield is indeed 50%. I take again the graph of 'dirk pitt'
dirk pitt wrote:then ONCE FOR EVERYONE and this is the last one because I feel it is starting to roll. YES the 1 / 2CV² formula is valid all the time. it is like 1 / 2mV² for kinetic energy or what do I still know but NO the energy stored in a constant current charged capacitor is not half that injected. it is the same as that injected minus the low joule losses of the circuit, in particular ESR of the capacitor.
we can estimate ESR elsewhere from the deltaV at the start of charging and beginning of charging.
here is a typical curve of charge then charge at constant current. apart from the small vertical offsets, the area under the voltage curve is almost the same at load and unload, right? However, since I is constant in charge and discharge, the power is therefore also almost the same in charge and discharge AND as the charge and discharge times are the same, the energy recovered is very much the same as that injected. at least not half. QED.
and I say, follow carefully:
I take the graphic provided by 'dirk pitt'.
-1) To have a current (even constant), you need a non-zero voltage. If zero voltage, zero current, ok ...
-2) On the graph, we see the constant current supplied by an electronic circuit, this circuit must be supplied by a voltage energy supplier U (take 1 volt for simplicity). It is he who ultimately provides the energy to the capacitor, ok ...
-3) during the charging time (1 second to simplify) of the 1 amp constant current capacitor (C) (to simplify the calculations), the voltage rises linearly across its terminals. Okay.
-4) The power given by the energy supplier is U (ct) x I (ct) or P. The energy (watt hour) is therefore P x T (charging time) = W = 1v * 1a * 1s = 1 watt..ok.
-5) The power absorbed by the capacitor during this time is u (function of t) x I (ct). The power absorbed by the capacitor is also a variable as a function of time t. It is also a line which is superimposed on the line (inclined) of the voltage u across the terminals of C. I took 1amp, and Umax1volt).
-6) The energy stored by C, is: 1 Amp x average voltage of ux T (charging time). The average voltage of u is easy to calculate; it varies linearly from zero to 1 (v) or 0,5 volts!
-7) The stored energy is therefore 1amp x 1 sec x 0.5 volt or 0.5 Watt.
It is also the area of the surface of the right triangle under the line u (ft) and the abscissa) so half the rectangle under the line U (ct) x I (ct) ...
-8) to compare with the 1 Watt given by the energy supplier in §4 ...
The yield is therefore 0.5, or 50% ...!
I can not break down and simplify more to explain!