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I’m a bit confused about capacitors. I understand they store energy in a field by accumulating opposite charges on the different plates. So a 1 farad capacitor will store 1 coulomb of charge if subjected to 1 volt if I understand the math right.

1 coulomb is also 1 amp-second, so this capacitor can supply 1 amp of current for 1 second.

Now what I don’t understand is where voltage comes into this. Can this theoretical capacitor only run 1V loads? Why? Wouldn’t a .5 farad capacitor subjected to 2V also store 1 coulomb of charge? What would be the difference between the charge stored in these two capacitors?

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  • \$\begingroup\$ You've got the right of it in terms of charge storage. The reason you see voltage ratings on capacitors is at some point, if you stuff more charge in to the capacitor (and raise the voltage), the capacitor breaks. \$\endgroup\$ – pgvoorhees Sep 11 '17 at 22:29
  • \$\begingroup\$ @pgvoorhees I understand the breakdown voltage, my main confusion is how does the capacitor retain the voltage that was applied to it. Isn’t it just storing charge? Why would a .5F capacitor charged with 2V be any different than a 1F capacitor charged with 1V? They both have 1 coulomb stored. \$\endgroup\$ – Matthew de Soto Sep 11 '17 at 23:05
  • \$\begingroup\$ Going to answer in a post. See below (in a few minutes). \$\endgroup\$ – pgvoorhees Sep 11 '17 at 23:10
  • \$\begingroup\$ To confuse your mind even more, think of a cap where you can adjust the dielectric (e.g. a special rotational type). As soon you move the dielectric out of the thing, the voltage magically increases! That is because the stored charge keeps being the same but the capacitance dropped. \$\endgroup\$ – Janka Sep 11 '17 at 23:19
  • \$\begingroup\$ Higher voltages store proportionally more ENERGY. \$\endgroup\$ – analogsystemsrf Sep 12 '17 at 4:44
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Answering the second comment to the question.

Yes, that is exactly correct. They would both be storing 1C of charge. Think of a capacitor like a (perfect) balloon where the larger the capacitance, the larger the balloon volume and the more you expand the balloon, the higher the pressure inside the balloon.

Imagine one really huge balloon, and one really tiny balloon (this is only to illustrate the point.)

Imagine you wanted to fill both balloons with 5 lung fulls of air, and afterward, you pinch off the orifice. I think it is easy to imagine the really huge balloon not being very full at all after 5 lung-fulls, where the small balloon is almost full to bursting.

The pressure in both balloons corresponds to the voltage, and the amount of air in each balloon (5 lung-fulls) corresponds to the amount of charge stored in each capacitor.

Does this help illustrate the relationship between charge, capacitance, and voltage?

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It is not clear what your problem is. The main relationship of concern here is that Q=CV (clearly Q = charge, C = capacitance, and V = voltage). So any combination of C and V that results in 1 yields a capacitor with 1 coulomb of stored charge. Taken together, the capacitance and the amount of charge to store determines the voltage. A 1 Farad capacitor charged to 1 volt will have stored 1 coulomb as would a 0.5 Farad capacitor charged to 2 volts. The difference occurs when you want to transfer this stored charge to a circuit. If the circuit requires 2 volts to operate than the 1 Farad capacitor would not be suitable. If your circuit required 5 volts to operate, you would have to use a 0.2 Farad capacitor since it takes 5 volts to charge such a capacitor with 1 coulomb of charge. Of course, you can use combinations of series and parallel connected capacitors to achieve the same result.

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A simple capacitor mental model (not good for quantitative predictions but useful for "just imagining") includes gravity and water and a tank: with these provisos:

  1. The area of the tank base can be likened to the capacitance of the capacitor.
  2. The tank height is related to the maximum voltage allowed, if any, for the capacitor.
  3. The amount of water in the tank is related to the stored charge in a capacitor.
  4. It takes work to pump water into the tank and the higher the water level already in the tank, the more work it takes to pump more water into it.
  5. A tank filled higher can perform more work from the same released amount of water because of the higher pressure available.

Now let's look at one of the fundamental relationships of a capacitor, where the total charge on the capacitor has this relationship to the capacitance and voltage measured between its two terminals:

$$Q=C\: V$$

Returning to the water tank model, this is basically saying that the volume of water in the tank is equal to the area of the tank (capacitance) times the height of the water level (voltage across the two terminals) in the tank.

Which makes sense.

Let's look at another relationship of a capacitor:

$$U=\frac{1}{2}C\: V^2$$

This is pretty easy, as well. If you imagine that using a certain amount of water from a tank that has a particular water level, and therefore the ability to put a certain pressure behind that water as you use it, then it should be pretty obvious that you can do more work with a gallon of water when the tank is fuller than when it is not so full. The gallon bursts out harder with a fuller tank.

With a full tank, the pressure starts out harder working and gradually becomes weaker and weaker until at the end even with still a fair bit of water left there just isn't that much pressure behind it. If you imagine this measure of "hard work" as the y-axis, and time on the x-axis, you might draw a line that starts high at first and then slopes downward to the right until you run out of water and the line hits the x-axis at some time in the future. The total work done with the entire tank will just be the area in that triangle. If the starting pressure height is \$C\: V\$ (the total water in the tank) and the time is \$V\$ (the total height of water you have to use), then the total area here is that of a simple triangle's area, which is what the above equation of \$U\$ shows you.


A \$1\:\textrm{F}\$ capacitor is then a very large area tank. If you fill it up to the height of \$1\:\textrm{V}\$, then the total volume will be \$1\:\textrm{C}\$. But there's less "pressure" despite the large volume. So the work it can do is only \$\frac{1}{2}\:\textrm{J}\$ (optimistically assuming 100% efficiency.)

In comparison, a \$\frac{1}{2}\:\textrm{F}\$ capacitor is then a tank with half the area of base. You'd need to fill it twice as high, to a height of \$2\:\textrm{V}\$, so that the total volume will be same \$1\:\textrm{C}\$ as before. But there's more usable "pressure" overall now, despite the same volume as before. So the work it can do is twice as much, or \$1\:\textrm{J}\$!

That's because for some of the time, the water current (in this analogy) is under more pressure and therefore can do enough more work to make the difference in the end. Both have the same total water to work with. But one of them has a harder working stream for part of the time, so to speak.

None of that changes the fact that in both cases there is only so much water to work with. But some of it can do more work in one case, than in the other.

Also, notice that in both cases, the height (voltage) does decline. In the case of a garden hose hooked up to the bottom of a tank, you might find that it will spin a lawn sprinker while the tank height is above some line but not below that. Similarly, a circuit might work so long as the voltage on the capacitor is above some voltage, but not below that. It's not all that dissimilar, in this sense.

Does that help?

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  • \$\begingroup\$ Great analogy, I understand how work ties into this now, thank you! \$\endgroup\$ – Matthew de Soto Sep 12 '17 at 15:57

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