Welcome back to our linear circuits class. And today we're going to be covering the linear model for transformers. We'll use the linear transformers to analyze a transformer circuit. In the previous lesson we introduced transformers and how they work, and then we identified some analysis models that exist for transformers, namely the linear and the ideal models. And the objectives for this lesson are to identify the linear model of transformers, so to be able to express that linear model. To use the circuit analysis to analyze the behavior of a transformer system, and then apply this analysis to solving a transformer circuit problem. The linear model of transformers treats each of the two inductors as inductances, and because we're operating in AC, you can put this as j omega L1 and j omega L2. There's also this mutual inductance, j omega M, that relates the one to the other. And so the way that we do the analysis here is we can make a current here, I1, and current here, I2, to express the two currents that are going through each of the portions of the system. And then we can use Kirchhoff's laws to come up with the systems of equations to find the relationships between these. So here using Kirchhoff's Voltage Law, I find the V Thevenin is going to be equal to I1 times Z Thevenin. Plus I1 times j omega L1. But the thing is, there's still this mutual inductance term that we need to account for. Now remember that these dots are representative of the reference directions. Here, my current is going in the dot and out the naught dot. Here, I'm going in the naught dot and out the dot. So they're referenced in opposite directions. So this will be minus I2 times j omega L2. Because this mutual inductance is based upon the current that's moving through this inductor. Sorry, this should not be an L2, this should be M. That's this based on this j omega M, the mutual inductance based on this current. The other equation for this side, again using Kirchhoff's Voltage Law is 0 = I2 times j omega L2 + zl. But again, we have to account for the mutual inductance term, minus i1 j omega M. Again, because of the opposite reference directions of the dots. If this dot were down at the bottom instead of at the top, these then would become pluses. So, using this and some linear algebra, we can actually find and solve for the values of I1 and I2. So these are types of equations that we were using, and then here's an expression of I1. It's a very lengthy and convoluted equation, but it turns out that it might sometimes be a little bit easier to work with an equivalent impedance. So if I want to know the impedance that this source sees in al of this, that is going to be equal to the Thevenin voltage divided by I1. Since the impedance is defined to be that ratio between the voltage and the current. And putting all of that together, we can find that Zeq = Z Thevenin + j omega L1 + this term, the omega squared M squared / ZL + j omega L2. We notice that this Z Thevenin and j omega L1 correspond to this and this piece. And this final portion is sometimes referred to as the reflective impedance. This is the impedance that results by the impedance of this side being intracted through the mutual inductance twice, and so this side does actually impact this side even though it's not directly connected. And so that gives us an idea of how this type of circuit is analyzed. Now let's actually use these calculations to solve an example problem. So the first thing we'll do is we'll define a couple of things. We'll identify this mesh current I1 and this mesh current I2. Now because this current is going into the dot, and this current, or hits the dot first, this current hits the dot second, they're working against each other. Which means that it's exactly the same as the example that was used previously. And so we can actually use the results from before. Then we'll also define two voltages, V1 and v2. So we want to see the relationship of these things. The results that we got from the previous section for calculating I1 and I2, well, I1 we calculated and I2, I'm just putting it here so we can save a little bit of time. They are here on the right. So all we need to do is plug in some of these values to get our solution. So first we calculate I1. So that's equal to zeta plus j omega L2. So that's 50 minus J100 + J100. So that just gives us 50. Divided by, and then on the bottom, we have Z Thevenin plus j omega L1. So that's 4 plus j2, plus j1 is 4 plus j3. Multiplying that by zl plus j omega L2. Which is the same thing as what we have here on the numerator. Just 50. And then we're going to have to add omega squared M squared. And we can see this omega squared M squared that this quantity here is j omega M. So if you get rid of the j, you get omega M. Square that is 100, 10 times 10. Putting all of that to, and of course, we have to multiply that whole quantity by 150, the V Thevenin. And putting all of that together, we get a result of 20- j 10, and that would be 22.36 with a phase angle of -26.6 degrees. Then your current, those are in amperes. Doing I2, again, just plugging into the equation here, j omega M, it's this J10 omega. Divided by, and we'll notice that the denominator here matches the denominator for the previous calculation. So, 4 + j3 x 50 + 100. And again multiplied by that V Thevenin. And doing all of that calculation gives 2 + j4 amperes, which is equal to 4.72 with a phase angle of 63.4 degrees. Now for calculating voltages. If I wanted to take this current and multiply it by this impedance to find this voltage, we would be making a mistake. Because the impedance of this side of the transformer is not just the impedance here. But the impedance here tied with whatever reflected impedance comes from the other side. So to be able to deal with that, what we can instead do is say that V1 is going to be equal to V Thevenin minus I1 times Z Thevenin. Since I can find this voltage by taking that voltage, subtracting off that voltage. So that's going to be 150- (20- j10) times 4 + j2. And that is going to be equal to 50 angle 0, or 50 volts. Doing the same basic idea for V2, this time again, I can't just use this current and the impedance here. But because these two devices are in parallel, I can use the I2 and multiply it by this ZL impedance to find what the voltage would be across here. So multiplying those two things together, we get 2 + j4 times (50- j100) and that is equal to 500 volts with zero phase angle again. Now if I want to look at the power to see what was being consumed here, we know that power is going to be equal to one-half VI star. But I want to take the real part of this to find the actual power. So this is actually our complex power. So if I find the complex power for the first section, multiplying that one-half times VI star, we find it to be 500- j250. So the real power consumed is 500 watts and the reactive power is -250 volts amps reactive. S2, again using this one half VI star, is going to be equal to 500 + j1000. So we see here, again, 500 watts is consumed with a reactive 1000 volts amps reactive. We'll notice here that the power here is equal. It's actually possible, based upon this value here, to be consuming more power than you're generating, but clearly in a physical system that's impossible. But the basic idea is that even here, this is, albeit somewhat ideal, we see that the power consumed is equal. And we can still have some extra reactive components that are taking place here. But remember, this reactive power is not actually being consumed, it's just being temporarily stored, and being used at a later time. So there's no violation here. To summarize we've presented the linear model, and derived the phenomenon of reflected impedance. And used circuit analysis to analyze an example transformer circuit. In the next lesson we will look at the ideal transformer model and contrast it with the linear transformer model. Until then.