I have a big 500 F 2.7 V capacitor, and a module of six 120 F 2.7 V capacitors. If I calculated the capacitance correctly, the module has 20F total capacity.

If I calculate the energy stored in them by E = 0.5 * C * V^2, I get:

- For the module: E = 0.5 * 20 * 16.2^2 = 2624 J
- For the big cap: E = 0.5 * 500 * 2.7^2 = 1822 J

So far, so good.

I bought a battery tester device and charged the capacitors to 2.7 V and 15 V, respectively (my DC source only goes to 15 V), and I've connected them like a battery to the tester device. I set 0.5 V as the lower cutoff voltage. The results I got are:

- The module yielded about 58 mAh.
- The big cap yielded about 220 mAh.

These results are in the ballbark of what an "Ah equivalence" equation, Ah = Farads * DeltaVolts / 3600 gives, but both the measurements and this equation are in conflict with the capacitor energy equation, which shows the energy stored in the module should be greater than in the big cap. The battery tester puts the current through a (constant) resistor, so it should be fairly simple as far as calculations go.

So my question is: how to reconcile the two?

## Best Answer

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!