Welcome back to our class on linear circuits. Today we're going to be talking about the ideal transformer model. So we're going to use the ideal transformer model to analyze basic transformer circuits. And then see the relationships between voltages and currents. So in the previous lesson we looked at the linear model. It's a fairly complicated model that makes use of phaser analysis and mutual inductance to do its calculations. In the ideal model we're going to make a few assumptions that are never completely true, but they allow us to get a decent idea of how a transformer is going to behave without having to do a more complicated analysis. So first of all we need to identify the assumptions that are used for the ideal transformer model, then use the ideal transformer model to do some basic circuit analysis. Then we will describe the importance of transformers in power transmission. First of all we need to define the Coefficient of Coupling. So the Coefficient of Coupling is going to be represented by a lower case k. And the Coefficient of Coupling is the value that gives this relationship between mutual inductance M and the self inductance is L1 and L2 in the transformer. So it's always possible to find this k but some k's are invalid in a physical sense and some k's are not. K's basically have to be somewhere between 0 and 1, 0 meaning that there's a no mutual inductance between the two coils, and they're completely independent of each other, and 1 meaning that they're very tightly coupled. But there's a physical limitation to how tightened this coupling can be. So, to do an example problem, consider that L1 is 4 millihenries, L2 is 9 millihenries, and M is 2mH, or we can simply use this equation to see that k is equal to m divided by the square root of L1 times L2. So that equals 2mH divided by 4mH times 9mH. So that's 36mH squared. Take the square root of that to give us 6 millihenries under 2 millihenries. So that means that k is going to be equal to one-third. This is quite simple to calculate the coefficient of coupling. It's going to be very important, when we start talking about ideal transformers, though, because we are going to make a certain assumption about what this coefficient of coupling happens to be. In the ideal transformer case, the coupling coefficient k is equal to 1, which means that these two coils are very tightly coupled. L2 and L1 are assumed to go to infinity, which means that so is our mutual inductance. Now, this is a limit. It's not that it's equal to the value infinity. But they're going to approach very, very large numbers. And the reason that we make this assumption is that it allows the analysis that leads to the transformer equations. We're going to skip over the actual analysis. It's available. You can find it if you're interested, but we just need to make use of it. But we clearly know that having this infinite inductance is not possible. Finally we assume that losses from coil resistances are negligible. Sometimes when we're doing analysis of transformers, we're going to stick a resistor here. And a resistor here to correspond to the resistance of the two wires that are placed here and here. It gives a better representation but in the ideal transformer case we're going to assume that both of those go to zero. Now the implications of this ideal transformer model are the following. First of all, v1/N1 = v2/N2 where v1 and v2 are the voltages across the primary and secondary coils respectively. And then N1 and N2 are the number of rotations at the coil, how many wraps of coil in the primary and the secondary coils respectively. We also have the relationship N1i1 = N2i2. Now these equations come from the Faraday's Law of Induction. So what we have is, we're making a circle like this. And then we have a changing B field through that surface. But as I wrap coils around again and again and again, that corresponds to additional surfaces. So everytime we have another wrap, it's just another one of these surfaces. And so we take the number of these wraps and multiply it by the voltage that's induced by the changing magnetic flux in each of these sections. Which is why we get this N, v behavior. But these equations are very simple to work with. They are quite easy to use for analysis. And so that's the most important that we're going to need to take out of this, so let's do an example. Here we're going to start by writing down our two equations that v1/N1 is equal to v2/n2 and that N1 times i1 is equal to N2 times i2. We have a phaser voltage here. And I'm going to do all of my analysis for this problem in terms of RMS of voltages and currents. Because it's entirely possible to do that without any kind of confusion, but I want to point that out. So we have 120 kilovolts RMS here leading to 120 volts RMS here. Now we don't know the number of coils because we don't have the number for either of these. But what we do know is the relationship between them. So frequently with ideal transformers, we'll show something like this, a ratio of the number of coils from one to the other. Because the actual number of themselves don't really matter as much as the relationship between them. And in this case we see that there's 100 coils to one over here because if I take 120 and multiply it by 100 we get 12 kilovolts over here. And then if I want to find the current that's going through over here, we have 120 volt rms, 240 ohms. So that means that i2 is going to be equal to half an amp that's phase angle is zero. On this side, i1, we're going to have the same relationship using this i2 and i1, that means that i1 is going to be equal to 5 milliamps, yeah 5 milliamps rms. And so what we can then do is we can find the power that being consumed in each of these devices. So first of all we'll look at the resistor. I2 is one-half amp rms and the voltage across this is equal to 120. And since this is a completely real device, P is going to be equal to the apparent power which is one-half times 120. The rms current and the rms voltage multiplied together. So 60 watts, so this is basically a 60 watt lightbulb. If I calculate the power that's being consumed here, we'd find that P is equal to minus 60 watts based on the rough instructions and the current and the voltage. So this transformer is generating 60 watts of power, and that's because this side, if we take this voltage and this current and multiply them together, you'll find that that P is also equal to 60 watts. So we'll call this P of the secondary, we'll call this P of the primary, this is P of the resistor. That's because the source is going to be generating 60 watts as well. And so we say that that's why we need to have this relationship between voltage and current. Because the power is being transferred from one side to the other. And that's one of the nice things about ideal transformers, is it's converting all of the power that's going over here, pushing it to the other side to be used to do some type of work. So essentially what we're doing is allowing ourselves to change the voltage to operate at a different voltage. So consequently transformers are often used in computers where you have 120 volts coming out of the wall but you need 5 volt DC to run your components. You need a transformer to change the voltage. It's also used in power applications. In this case this is what we're illustrating here. This is like the power that's being sent by the power company. This transformer is a big power transformer that's owned by the power company that changes the voltage that they're transmitting to a smaller voltage. The reason for this is that if you have high currents flowing through wires, it leads to basically heaters. And you lose a lot of power as heat lost in the wires. Power companies don't want to lose all of that heat in their wires, so they operate their transmission lines at very high voltages so that the currents can remain small. And they use transformers to go from those very high voltages to smaller voltages that are used in residential homes. So what are the implications? Well transformers allow a change from one voltage to another voltage. We take high-voltage and low-current power and transform it into, using long-distance power distribution through transformers, into something that is a low-voltage and high-current. So that you can still get the power to all of the homes. Now, it turns out that this is a very important phenomenon. Before this event, it was required for power stations to be placed very close to where the power was being consumed. By being able to use transformers, the power can now be sent over very long distances. So things like the Niagara Falls power plant can now send power all across northern the United States and Canada without having to have a whole bunch of power stations all over the place. So in summary, we showed the ideal transformer model and used this model to solve an example system. And identified that transformers are useful for power transmission because they make possible sending power using very high voltages and low currents, to be able to avoid getting high heat through high currents through transmission links. In the next lesson, we'll be looking at a sensor that makes use of this concept of mutual inductance, a linear-variable differential transformer, and see places that they can find useful applications. Until then.