Re: energy in, energy out

[LUDWIK KOWALSKI <kowalskil@alpha.montclair.edu>]

The method used in this student project was described earlier (see below).
The anticipated "spectacular" aspect turned out to be a "false alarm", a 
small procedural error not worth taking about. Yes, the amount of energy
entering the capacitor (nominally C=3.3 F, 5.5 V) in charging (from 0.00 
to 4.64 volts) was found to be 43.6 J while the energy exiting from it 
during the discharging ( 4.64 to 0.00 volts) was only 35 joules. Nearly
20% of what was delivered did not come out as electric energy. Was it
dissipated inside the capacitor as heat or was it stored in a chemical 
form? A good question for another project. 

The subject was discussed in an earlier thread. Will the next capacitor 
(nominally 10F and 2.2 V) behave in the same way? We will see. Here are 
some other details. The charge received was 18.16 coulombs while the 
charge relesed was 16.58. Using C=Q/4.64 these charges correspond to 
C=3.91 and 3.57 F, respectively. The charging curve, I(t) was not exactly
exponential. The value of C calculated from this curve ranged from 2.4 F 
(initial slope) to about 3.3 F (final slope). The mean C from these four 
evaluations is practically the same as the nominal value of 3.3 F.
The methodology is repeated below.
                                          Ludwik Kowalski
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> Suppose a black box with two electric terminals (for example, a supercap, 
> C=10 F) is connected to a constant voltage source (for example, Vs=2.5 V) 
> through a resistor (for example, 15 ohms). How much energy is received by 
> the box during a finite time interval (for example, 500 seconds)?

> A voltage probe from Vernier was used to measure the d.o.p. on C, Vc, 
> every two seconds. There were no need to measure Vr, the d.o.p. on R, 
> because Vs+Vr=Vb. For any known Vc one has I=(Vs-Vc)/R. The total amount 
> of energy entering the box during a given time t is the sum of partial
> energies, dE, delivered during consecutive time sub-intervals, dt. 
>
>       dE = Iav * Vav * dt = 0.5*(I1+I2) * 0.5*(Vc1+Vc2) * dt
>
> where I1 and Vc1 refer to the beginning of each short time sub-interval 
> while I2 and Vc2 refer to the end of that sub-interval.
>
> And how much energy flows out from the box when it is disconnected from 
> the source and discharged through R? The same approach (measurements of
> many instant values of Vc, calculations of dE and a summation) can be 
> used to answer the second question. Do you agree? By the way, the input 
> impedance of the Vc voltmeter was enormous (many giga-ohms, as verified 
> by discharging a very good capacitor through the probe)