Lesson Video: Electromagnetic Induction in Transformers | Nagwa Lesson Video: Electromagnetic Induction in Transformers | Nagwa

Lesson Video: Electromagnetic Induction in Transformers Physics • Third Year of Secondary School

In this video, we will learn how to calculate the change in the potential difference and current produced by a transformer.

12:14

Video Transcript

In this video, we’re learning about electromagnetic induction in transformers. We’ll learn what transformers are, how they work, and how they use this phenomenon of electromagnetic induction.

As we start out, let’s recall what electromagnetic induction is in the first place. The idea is this. If we take a loop of conducting material, then if we change the magnetic field experienced by the cross-sectional area of this loop, then we’ll induce current to flow in it.

One way to do this to change the magnetic field experienced by the loop’s area is to pass a magnet through the loop. Another way is to keep the magnet stationary but change the size of the loop, say, by making it bigger or smaller or even keeping it the same overall size but rotating it so that the total area exposed to the magnetic field changes. Any and all of these methods will have the overall impact of changing the magnetic field experienced by the loop and therefore inducing current in it. And that process is known as electromagnetic induction.

One of the most useful applications of electromagnetic induction is transforming electrical power. Now the basic idea behind electrical power transformation is this. When electricity is generated at a power plant, it’s at a voltage, a potential difference, which is much higher than we could safely or reasonably use in a residential context. But, in order for the electricity to get from where it’s generated to where it’s used, the most efficient way is to keep it at a very high potential difference. That way, the least amount of power possible is lost in the transmission process. That means that shortly before we use it, we’d like to be able to transform the electricity we receive from the power plant. This transformation, as we’ll see, relies on electromagnetic induction.

An electrical transformer looks like this. There are three basic parts to it. First, there’s a coil of wire known as the primary coil. This is the wire through which electricity is introduced to the transformer. Then, across from the primary coil is what’s called the secondary coil. This is the coil of wire which will have voltage induced across it and therefore current induced in it. And then, connecting these two coils is what’s called the core. As we’ll see, the type of material the core is made out of has a significant effect on the performance of the transformer overall.

So here’s how the transformation process works. First, current flows in through the primary coil. We’ll call this current 𝐼 sub p to show that it’s in the primary coil. This current travels around every single one of the loops of the primary coil wrapped around the core. And then finally, it comes back out the other side. Now, if this was all that happened in a transformer, that would be pretty boring. And it wouldn’t really accomplish much. But at this point, we can recall that a loop of wire, when it carries a current, generates a magnetic field. And in particular, if we have a loop of wire where the current is travelling in this direction as shown, then based on what’s called a right-hand rule, the magnetic field created by this loop at the center of the loop points straight up.

Knowing that, if we go back over to our primary coil, which we see is wrapped around the core some number of times, we realize that every single one of these individual loops, these windings, is itself a loop of current. And it creates a magnetic field that points up. The combined effect of the magnetic field from all these individual loops is fairly strong. And overall, we have a fairly powerful magnetic field pointing from bottom to top. Now, this is where the core material comes in. One of the main purposes of the core of a transformer is to channel the magnetic field lines around the core as though they’re moving in a circuit.

This means that the field produced within the windings of the primary coil then travels all throughout the rest of the core and goes through the windings of the secondary coil as well. And this we can see is where electromagnetic induction comes in. Let’s look for a moment at a single one of these loops in the secondary coil. And we’ll look at it as though we’re looking straight down from above onto that loop. In that case, the loop would look to our eye like this, like a circle. And what we would see, if we could see them, is this magnetic field line going into the screen, from our perspective, through the center of the loop. So, whereas before there was no magnetic field moving through this loop, now there is a field.

In other words, there’s a change in magnetic field experienced by the area of this loop. That’s exactly the kind of effect that through electromagnetic induction will induce the flow of current in this loop. For reasons that we won’t get into in this lesson, that current flows in this direction from our perspective, counterclockwise. Of course, what we’re looking at here is just a single loop in the many loops of the secondary coil. So this is going on for all of those individual loops. So then, finally, this induced current is output through the secondary coil. And it goes on to whatever its application might be, perhaps in a residential neighborhood.

Let’s go back to 𝐼 p, the current through the primary coil, for a moment. If this current was constant in time, then still a magnetic field would be formed through the loops of the coil. And that field will be carried through the core. But, after the loops of the secondary coil had been initially exposed to this change, after that, there will be no more change. The field lines would stay the same. And if there is no more change in the total magnetic field through these loops, then there’ll be no more current induced in them. In order for a transformer to work properly, the current in the primary coil must be alternating current, AC.

When that’s the case, it means that the magnetic field lines in the core are constantly changing in magnitude. And when that’s happening, it means that every single one of the windings in the secondary coil is always seeing a different magnetic field move through it. That is, there’s a constant change in magnetic field through the windings of the secondary coil and, therefore, will constantly induce voltage and, therefore, current in that coil. All that to say, for a transformer, it’s very important to work on alternating current.

We’ve said that the current going through the primary coil is 𝐼 sub p. And let’s imagine further that we know the voltage of that current. We’ll call that voltage 𝑉 sub p. And let’s say it’s 500 volts. And then, if we go over to the secondary coil, we can say the current output there is 𝐼 sub s and that the voltage output there is 𝑉 sub s. But the question is, what is that voltage? What is the potential difference induced in the secondary coil? Believe it or not, we can solve for 𝑉 sub s by knowing 𝑉 sub p and also knowing the number of turns that each of the two coils, primary and secondary, make around the core.

In general, if we call the number of windings of the primary coil 𝑁 sub p and we call the number of windings or turns around the core of the secondary coil 𝑁 sub s, then we can write this very neat equation. 𝑉 sub s divided by 𝑉 sub p is equal to 𝑁 sub s divided by 𝑁 sub p. In other words, the ratio of the potential differences is equal to the ratio of the turns. Now, in a way, this is quite fascinating. But as we think about it, it makes sense. The more turns that a coil makes around the core, the more it will contribute to the magnetic field strength in the core. And the more that magnetic field on the core changes, the more voltage will be induced in the secondary coil.

This ratio equation tells us that since we know 𝑉 sub p, if we would count up 𝑁 sub p and 𝑁 sub s, then we could use all that information to solve for 𝑉 sub s. Let’s do that; let’s count up 𝑁 sub p and 𝑁 sub s. Starting out with 𝑁 sub p, we can count these turns as one, two, three, four, five, six, seven, eight, nine, total. So applying this equation, we have 𝑉 sub s, what we want to solve for, divided by 𝑉 sub p, 500 volts, is equal to 𝑁 sub s, what we’ll find out in a moment, divided by nine, the number of turns in the primary coil.

Moving on to counting 𝑁 sub s, that’s one, two, three, four, five, six, seven, eight, nine, 10, 11, 12 turns. So we fill in 12 for 𝑁 sub s in our equation. Then if we multiply both sides by 500 volts, we find that the voltage in the secondary coil is equal to 12 divided by nine times 500 volts. That’s the same as four-thirds times 500 volts, which is approximately 667 volts. This change from 𝑉 sub p to 𝑉 sub s is why this is called a transformer.

Now we might wonder, what about the currents 𝐼 sub p and 𝐼 sub s? Can we solve for those based on the number of turns of each coil? The answer is that we can, but the relationship is flipped from the relationship for voltage. What we mean by that is if we take the ratio of the number of turns in the secondary coil to that in the primary, that’s equal to the primary current divided by the secondary current. So we have to be careful about our subscripts here, ss and ps, and to keep them straight.

Going back to our discussion of voltage, we can see that the primary voltage is less than the secondary voltage. When this happens, it means the function of the transformer is to raise the voltage from the primary to the secondary coil. When that happens, when 𝑉 sub s is greater than 𝑉 sub p, the transformer is described as a step-up transformer. And the opposite can happen as well. When the secondary voltage is less than the primary voltage, the transformer is called a step-down transformer.

Before we get a bit of practice with an example, let’s talk about the core of the transformer. From an efficiency perspective, the core is very important. In particular, the material we choose to make the core out of will affect just how well energy is transferred from the primary to the secondary coil. Of all the materials we could choose to make the core out of, we like to choose a material which is magnetizable. That is, one which becomes a magnet when it’s exposed to a magnetic field. And we’d also like a material that can respond quickly to rapidly changing magnetic fields in the windings of these coils.

A leading material for doing this is iron. And it’s very common to make transformer cores out of this metal. It’s a material that helps to channel as well as amplify the magnetic field within the core. Now that we know a bit about transformers, let’s get some practice solving a question about one.

A step-down transformer changes the potential difference of an alternating current from 10000 volts to 250 volts. If it has 25 turns on its secondary coil, how many turns does it have on its primary coil?

Okay, let’s say that this is our transformer. This is our primary coil and here is our secondary coil. We’re told that the potential difference in the primary coil, what we we’ll call 𝑉 sub p, is equal to 10000 volts. And the potential difference in the secondary coil, what we’ll call 𝑉 sub s, is 250 volts. We’re also told that the secondary coil of our transformer has 25 turns. We’ll call that number 𝑁 sub s. And if we call the number of turns in the primary coil 𝑁 sub p, it’s that value that we want to solve for. In order to do it, we can recall a relationship between primary and secondary voltage and number of turns. This relationship says that the ratio of the turns primary to secondary is equal to the ratio of potential differences primary to secondary.

In this relationship, we want to solve for 𝑁 sub p, the number of turns in the primary coil. So to do that, we can multiply both sides of the equation by the number of turns in the secondary coil. That means that that term, 𝑁 sub s, cancels out on the left-hand side of our equation. We find that 𝑁 sub p is equal to 𝑉 sub p divided by 𝑉 sub s all multiplied by 𝑁 sub s. And since we know 𝑁 sub s, 𝑉 sub p, and 𝑉 sub s, we can substitute those values into this equation now. 𝑉 sub p is 10000 volts, 𝑉 sub s is 250 volts, and 𝑁 sub s is 25. Calculating this result, we find an answer of 1000. That’s the number of turns that are in the primary coil of this transformer.

Let’s take a moment now to summarize what we’ve learned about electromagnetic induction in transformers.

In this lesson, we’ve seen that transformers change voltage and current through the process of electromagnetic induction. We saw that in general, transformers have three basic components. There’s a primary coil that current is input through. There’s a secondary coil that current is output from. And there’s a core, typically a solid metal material, that connects these two coils.

We saw that when it comes to the effect a transformer has on voltage, the ratio of the primary voltage to the secondary voltage is equal to the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. And we furthermore saw that a transformer affects the current, that the ratio of secondary coil current to primary coil current equals the ratio of the number of turns in the primary to the secondary coil. Finally, we learned that a step-up transformer increases voltage across the transformer while a step-down transformer decreases it.

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