Electronic – Energy in capacitors – loss

Your calculations are consistent, though your measurements of capacity look a bit low.

The 20F, 16.2v capacitor stores 20*16.2 = 324 Coulombs, 324 Ampere seconds, or about 0.09 ampere hours. As you correctly calculate, 2624 Joules.

The 500F, 2.7v capacitor stores 500*2.7 = 1350 Coulombs, 1350 ampere seconds, 0.375 ampere hours. As you say, 1822 Joules.

You can also get the capacitor energy by multiplying the charge stored by the average voltage. So for the 20F cap, 324C * 8.1v = 2624 Joules, and the 500F cap, 1350C * 1.35v = 1822J. This is actually the same equation as \$0.5CV^2\$, but it's easier to see how energy depends on both voltage and charge stored.

As you can see, energy is related to ampere hours through voltage. They are different voltage caps, so you would expect the higher voltage cap to have a higher energy to charge stored ratio.

As an interesting thought experiment, let's connect the series capacitors in parallel. Now it should be easier to compare what's happening. You have a 720F cap and a 500F cap, both rated at 2.7v. You would hope they store the same energy as before with the series connection.

\$0.5CV^2\$ for each gives 0.5*2.7*2.7*720 = 2624J, and 0.5*2.7*2.7*500 = 1824J, as before. Yes, the energy sum comes out exactly as before.

However, now the charge in the 720F is 2.7v * 720F = 1944C. The energy is the same, but as the voltage is lower, the charge is higher.

When capacitors are in series, the same charge passes through each. The total charge in the whole series string is the same as for one capacitor. When capacitors are in parallel, the charges add, just like current does.

The same thing confuses people with batteries, especially Lithiums sold as series blocks!

Then what will happen to the electrical field (energy) which was stored in the capacitor before un-rolling its electrodes?

Unrolling and separating the plates of the capacitor will require work, because the electric field between them creates an attractive force. Therefore, you will actually increase the energy stored in the field. The capacitance goes down as you increase the separation, but the voltage between the plates goes way up.

The relevant equations are

$$C = \epsilon \frac{A}{d}$$

$$V = \frac{Q}{C}$$

$$E = \frac{1}{2}C V^2$$

If area \$A\$ and permittivity \$\epsilon\$ are held constant, then capacitance \$C\$ is inversely proportional to the distance between the plates \$d\$.

If the charge \$Q\$ is also constant, then the voltage \$V\$ is inversely proportional to \$C\$, which makes it directly proportional to \$d\$.

Finally, the total energy \$E\$ turns out to be proportional to \$d\$ because the rising \$V^2\$ term overrides the falling \$C\$ term. This increase in energy is the physical work you have to put into increasing the separation.

If I now bring back together the two electrode-papers again and roll-them-up into its original shape will I again get back the original energy which was there in it when the capacitor was charged originally?

Yes. The energy will drop to its original level, assuming that you haven't allowed any of the charge to leak away in the meantime.

Note that I'm glossing over a lot of physical details associated with the construction of electrolytic capacitors specifically. The paper separator is NOT the dielectric in this case — the dielectric is actually a thin layer of aluminum oxide on one of the foil plates.

Therefore, if you keep the foils immersed in a sufficient quantity of electrolyte as you separate them, there will be no change in capacitance or energy.

However, if you try to separate them in air, you will in effect be creating two capacitors in series, one with the original oxide dielectric, and another one with air as the dielectric. The latter will have very low capacitance, and the voltage will appear across it.