Our Energy Sources, Electricity

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]

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