Another thing I noticed is that the test works the same even if I just used a copper wire instead of the light bulb it was still around 9.35 volts in each cap.
This was a valuable observation, and it has something important to tell us.
Let's do some math with your experiment:
We will assume in our experiment that we have a 10,000 µF capacitor charged up to 24 V. There is a formula for how much charge this capacitor is holding, and it is the voltage times the capacitance. So we can say the stored charge is Q1:
Q1 = 24 V * 10,000 µF = 240,000 µC
(Don't worry what a µC is, let's just say it is a measure of electric charge.)
In our experiment, let us connect this capacitor to another identical capacitor so that the voltages equalize. We will find now that each capacitor is charged to 12 V (in a perfect world). The charge stored in each capacitor is given by the same formula, only now there are two capacitors, so the total charge is twice the charge on one capacitor. Let's call the total charge Q2:
Q2 = 2 * 12 V * 10,000 µF = 240,000 µC
It's just the same as before! So maybe we have not lost anything?
If we put a bulb in the circuit when we connect the capacitors together, the bulb will light up and give us useful light, and still we will have the same total charge left at the end. It sure seems like we have got something for nothing, does it not?
But as you can tell, there is a "but" coming. Let's rewind the experiment and look at the but.
When we look at the first capacitor, the total charge is not the important thing, the total stored energy is the important thing. The total stored energy is given by a different formula, involving the square of the voltage. The energy, E1, stored in the first capacitor, is this:
E1 = 0.5 * 10,000 µF * 24 V * 24 V = 2,880,000 µJ = 2.88 J
Here, the "J" means joules, a unit of energy. Energy is what we pay for when we buy electricity from the utility company, only they measure it by kilowatt-hours, or kWh. It's the same thing though.
Let's now see what we have after we join the capacitors together and equalize them. The new energy E2, is this:
E2 = 2 * 0.5 * 10,000 µF * 12 V * 12 V = 1.44 J
Look what happened. It is only half the energy we started with! We have "lost" half of our valuable energy.
Where did it go? It was wasted, in fact. Lost as heat due to the resistance in the wires.
If we want to recover our lost energy, we need to put that bulb in the circuit. Then at least the bulb will convert some of the lost energy into useful light.
So how can we summarize this experiment? It tells us, not how much energy can we conserve, but rather, how much energy can we avoid wasting!
There is a rule of thumb in engineering that often applies. The more complex you make a system, the more waste you are likely to get. Very often, simplest is best.
If you want light, connect your bulb to the power source directly and maximize your light.
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