This video explains the potential of a capacitor and how they function in a circuit. By David Santo Pietro. . Created by David SantoPietro.
SPEAKER 1: Check out this capacitor. Look at what happens if I hook it up to this light bulb. CHILDREN (IN UNISON): Boo! SPEAKER 1: Yeah, nothing happened because the capacitor is not charged up. But if we look at up to a battery first, to charge up the capacitor, and then hook it up to the light bulb, the light bulb lights up. CHILDREN (IN UNISON): Ooh! SPEAKER 1: The reason this happens is because when a capacitor is charged up, it not only stores charge, but it stores energy as well. When we hooked up the capacitor to the battery, the charges got separated. These separated charges want to come back together when given the chance, because opposites attract. So if you complete the circuit with some wires and a light bulb, currents going to flow. And the energy that was stored in the capacitor turns into light and heat that comes out of the light bulb. Once the capacitor discharges itself, and there's no more charges left to transfer, the process stops and the light goes out. The type of energy that's stored in capacitors is electrical potential energy. So if we want to figure out how much energy is stored in a capacitor, we need to remind ourselves what the formula is for electrical potential energy. If a charge, Q, moves through a voltage, V, the change in electrical potential energy of that charge is just Q times V. Looking at this formula, what do you think the energy would be of a capacitor that's been charged up to a charge Q, and a voltage V? CHILDREN (IN UNISON): Q times V! SPEAKER 1: Yeah, and that's what I thought it would have been, too. But it turns out the energy of a capacitor is 1/2 Q times V. CHILDREN (IN UNISON): Boo! SPEAKER 1: Where does this 1/2 come from? How come the energy is not just Q times V? Well, the energy of a capacitor would be Q times V if during discharge, all of the charges were to drop through the total initial voltage, V. But during discharge, all of the charges won't drop through the total voltage, V. In fact, only the first charge that gets transferred is going to drop through the total initial voltage, V. All of the charges that get transferred after that are going to drop through less and less voltage. The reason for this is that each time a charge gets transferred it decreases the total amount of charge stored on the capacitor. And as the charge on the capacitor keeps decreasing, the voltage of the capacitor keeps decreasing. Remember that the capacitance is defined to be the charge stored on a capacitor divided by the voltage across that capacitor. So as the charge goes down, the voltage goes down. As more and more charge gets transferred, there'll be a point where a charge only drops through 3/4 of the initial voltage. Wait longer, and there'll come a time when a charge gets transferred through only a half of the initial voltage. Wait even longer, and a charge will only get transferred through a fourth of the initial voltage. And the last charge to get transferred drops through almost no voltage at all, because there's basically no charge left that's stored on the capacitor. If you were to add up all of these drops in electrical potential energy, you'd find that the total drop in energy of the capacitor is just Q, the total charge that was initially on the capacitor, times 1/2 the initial voltage of the capacitor. So basically that 1/2 is there because not all the charge dropped through the total initial voltage, V. On average, the charges dropped through only a half the initial voltage. So if you take the charge stored on a capacitor at any moment, and multiply by the voltage across the capacitor at that same moment, divide by 2, you'll have the energy stored on the capacitor at that particular moment. There's another form of this equation that can be useful. Since capacitance is defined to be charge over voltage, we can rewrite this as charge equals capacitance times voltage. If we substitute the capacitance times voltage in for the charge, we see that the energy of a capacitor can also be written as 1/2 times the capacitance times the voltage across the capacitor squared. But now we have a problem. In one of these formulas the V is squared, and in one of these formulas the V a not squared. I used to have trouble remembering which is which. But here's how I remember now. If you use the formula with the C in it, then you can see the V squared. And if you use the formula that doesn't have the C in it, then you can't see the V squared. So these are the two formulas for the energy stored in a capacitor. But you have to be careful. The voltage, V, in these formulas refers to the voltage across the capacitor. It's not necessarily the voltage of the battery in the problem. If you're just looking at the simplest case of one battery that has fully charged up a single capacitor, then in that case, the voltage across the capacitor will be the same as the voltage of the battery. So if a 9-volt battery has charges up a capacitor to a maximum charge of four coulombs, then the energy stored by the capacitor is going to be 18 joules. Because the voltage across the capacitor is going to be the same as the voltage of the battery. But if you're looking at a case where multiple batteries are hooked up to multiple capacitors, then in order to find the energy of a single capacitor, you've got to use the voltage across that particular capacitor. In other words, if you were given this circuit with these values, you could determine the energy stored in the middle capacitor by using 1/2 Q-V. You would just have to be careful to use the voltage of that capacitor, and not the voltage of the battery. Plugging in five coulombs for the charge lets you figure out that the energy is 7.5 joules. CHILDREN (IN UNISON): Ooh! [MUSIC PLAYING]