Video: The Power of Electrical Components

In this lesson, we will learn how to use the formula ๐ธ = ๐‘„๐‘‰ to find the energy dissipated to the environment by the component of an electronic circuit.


Video Transcript

In this video, weโ€™re learning about the power of electrical components. This kind of power is not always the easiest thing to notice. After all, electrical components โ€” like a computer or a lamp or a microwave โ€” are fairly stationary devices. Mechanical power involving fairly large objects in motion may be more familiar to us. But that doesnโ€™t mean that electrical power isnโ€™t a very real phenomenon.

We can start our discussion of power for electrical components by recalling the definitions of a number of terms. For starters, we can remember that energy is defined as the ability to do work. This work that weโ€™re talking about could be mechanical work such as moving a mass up a hill or it can be electrical work. An example of electrical work would be moving a charge โ€” say this one over here โ€” while itโ€™s in the presence of an electric field. The analogy would be moving a mass in the presence of a gravitational field. To do this, it takes work. And that work is the measure of the energy invested in the process.

Moving on, letโ€™s now recall the definition for power. Power is defined as the amount of energy transferred over some amount of time written as an equation. We can say that power ๐‘ is equal to energy ๐ธ divided by time ๐‘ก. And at this point, letโ€™s recall that energy โ€” as we saw earlier โ€” is the ability to do work. Going back to our positive charge in an electric field, letโ€™s say we did some amount of work on this positive charge.

Letโ€™s imagine we did work ๐‘Š in moving it towards the other positive charge given off the field. In doing that much work measured in joules, weโ€™ve exerted that much energy also measured in joules. In other words, for this process, we can rewrite the energy used in terms of the work done. We could say that in this case the power is equal to the work weโ€™ve done in moving the electric charge divided by the time it took to move it.

As weโ€™ve said, weโ€™re working with an electric charge. And letโ€™s say that this charge has a charge ๐‘„. What we can do now is come back to our equation for power and multiply both the numerator and denominator on the right-hand side by that charge ๐‘„. It will become clear in a minute why weโ€™re doing this. But for now, just notice that by multiplying by ๐‘„ divided by ๐‘„, weโ€™re effectively multiplying by one; that is, weโ€™re not changing the equation.

So we have power is equal to work divided by time multiplied by charge divided by charge. And as a last little manipulation, letโ€™s switch the denominators here. Letโ€™s switch the ๐‘„ and the ๐‘ก, which algebraically we can do. Now that we have this equation for power in this form, letโ€™s leave it alone for a moment and go on with our definitions. Next up is voltage.

Voltage โ€” also called electrical potential โ€” is equal to electrical potential energy per unit charge. What on earth does that mean? Well, letโ€™s consider it in the context of our electric charge in our electric field. As we mentioned, this is a bit analogous to a mass in a gravitational field. And if we think of it that way, it can be helpful. Right now, this electric charge has a tendency to move. Thatโ€™s because itโ€™s in an electric field.

We can see that that tendency to move โ€” how much this charge wants to move so to speak โ€” is a measure of its electric potential energy. And notice that thatโ€™s a lot like the gravitational potential energy of a mass in a gravitational field. So anyway, this charge ๐‘„ has an electric potential energy. And we can refer to it for short as EPE. If we were to take this electric potential energy that the charge has by virtue of being in an electric field and divide it by the amount of charge ๐‘„ that the charge possesses, then what our definition for voltage is saying is that this fraction is equal to electric potential or another word for that is voltage.

But now take a look at this, in the numerator of the left side, we have an energy. And as we saw earlier, energy is the ability to do work. In fact, the electric potential energy of this charge here is equal to the work it would take to bring the charge to this particular location from infinitely far away. We can say that this is the same amount of work capital ๐‘Š as we referred to in our equation for power. That just as well could be the work done to bring the charge in from infinitely far away to its current position.

If we make this substitution replacing EPE with the work done on the charge, we see something interesting in terms of the definition for voltage. We see that voltage or equivalently electrical potential is equal to the work done on a charge divided by the amount of charge it possesses. And notice this, just as in this equation, we see work divided by charge. So in our equation for power, we have a ๐‘Š divided by ๐‘„ term. That means in the equation for power, we can replace ๐‘Š divided by ๐‘„ with ๐‘‰ voltage.

Now having done that, letโ€™s move on to our last definition: the definition of current. Current is defined as the amount of electric charge passing a point over some amount of time. Writing this as an equation, we can say that ๐ผ current is equal to charge ๐‘„ divided by time ๐‘ก. And this definition is quite useful because notice up in our equation for power that we have a ๐‘„ divided by ๐‘ก term. In other words, we can replace that term with ๐ผ, the current.

With this substitution made, we now have our equation for power for an electrical component. Itโ€™s often expressed this way: ๐‘ is equal to ๐ผ current times ๐‘‰ voltage. There are a couple of helpful things to notice about this equation. And to see them, letโ€™s a bit of space on the bottom of our screen.

Okay, the first thing to notice is that according to our definition for power, power is equal to some amount of energy transferred over some amount of time. That means that energy divided by time is equal to current times voltage. And if we then multiplied both sides of the remaining equation by time, we see that term cancels on the right-hand side. And we have an equation that says that energy is equal to time times current times voltage.

And now we can remember from earlier our definition of current that current is equal to charge divided by time. This means we can substitute ๐‘„ divided by ๐‘ก in for ๐ผ in this equation. And notice what happens when we do. The factor of time ๐‘ก cancels out. And we then have an equation for energy, which says that itโ€™s equal to charge multiplied by voltage. So once we arrived at the expression power is equal to current times voltage, we were able to use that to find this expression for electrical energy that itโ€™s equal to charge times voltage.

But there is a second thing we can do with this power equation. Ohmโ€™s law says that if we have a resistor of constant resistance value then we multiply that resistance by the current running through it, then that product is equal to the potential difference across the resistor. Letโ€™s say we were to take Ohmโ€™s law and multiply both sides of the equation by the current ๐ผ.

If we did that, then the left-hand side of this expression would be equal to ๐ผ times ๐‘‰ which is equal to electrical power, which means that the right-hand side of this expression is an equivalent way to write electrical power: ๐ผ times ๐‘… times ๐ผ or ๐ผ squared ๐‘…. So then, not only is electrical power equal to ๐ผ times ๐‘‰, itโ€™s also equal to ๐ผ squared times ๐‘…. And perhaps, you can see thereโ€™s even another way to write electrical power. We could also write it as ๐‘‰ which is equal to ๐ผ times ๐‘… by Ohmโ€™s law all squared divided by the resistance ๐‘….

So what we found then is several ways to express electrical power and one way to express electrical energy. And weโ€™ve seen that these equations come from basic definitions of electrical quantities backed up by simple charge motion scenarios. Now that we know these equations, letโ€™s get a bit of practice using them through a couple of examples.

An electric motor is connected to a nine-volt battery. Over a period of time, the motor converts 450 joules of electrical energy into kinetic energy, heat, and sound. How much charge passes through the motor over this period of time?

So what we have here is an electrical motor being powered by a nine-volt battery. We want to know over the time it takes the motor to convert 450 joules of electric energy into these other kinds of energy โ€” kinetic energy, heat, and sound โ€” we want to know how much charge passes through the motor over that time. To figure this out, we can recall the relationship that connects voltage, energy, and charge.

Electrical energy ๐ธ is equal to the amount of charge ๐‘„ multiplied by the potential difference across which the charge moves ๐‘‰. In our particular case though, we donโ€™t want to solve for ๐ธ, but we do want to solve for the charge ๐‘„. So we can rearrange this equation. When we do, we see that charge ๐‘„ is equal to energy divided by voltage. And in our problem statement, weโ€™re told the energy used by the motor as well as the voltage power in it.

When we substitute in these values, we have 450 joules of energy divided by nine volts of potential difference. This fraction comes out to 50 coulombs of charge. This is the amount of charge that passes through the motor over this period of time.

Now, letโ€™s look at a second example.

The diagram shows a circuit consisting of a light bulb connected to a cell. The potential difference across the bulb is nine volts, and the current through it is four amps. How much is the power of the light bulb?

Taking a look at this circuit, we see that indeed this bulb is set up in series with a cell. With the circuit set up like this, the bulb will be on shining light and we want to know just how much power itโ€™s using up as it does so. Weโ€™re told the potential difference across the bulb as well as the current through it. And we can recall the relationship for potential difference current and power. That equation tells us that power ๐‘ is equal to current times voltage.

Applying this to our scenario, we can substitute in the given values of current four amps and voltage nine volts. Then when we go and multiply these quantities together, we find the result of 36 watts. Thatโ€™s the power of the light bulb, likely given off both in the form of light as well as heat energy.

Letโ€™s summarize what weโ€™ve learned so far about the power of electrical components.

In this section, we learned that electrical power is given by the relationship ๐‘ is equal to ๐ผ times ๐‘‰, current times voltage. We also saw that thanks to Ohmโ€™s law there are equivalent ways to write this expression. We can write it as ๐ผ squared times ๐‘… or we can write it as ๐‘‰ squared divided by ๐‘…. These are all equivalent expressions for electrical power.

We also saw that electrical energy ๐ธ is equal to charge multiplied by voltage. Specifically, this is the charge ๐‘„ that passes a certain point โ€” say in an electrical circuit โ€” multiplied by the potential difference supplying that circuit. Furthermore, we saw that mechanical processes such as lifting up a mass or climbing a set of stairs can help us understand and clarify electrical terms and phenomena, such as voltage, electrical potential energy, and power.

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