[MUSIC] Welcome back to linear circuits, this is Dr. Ferri. This lesson is on AC circuits solution method. This particular lesson is the most important one in this module. Because we are going to finally be able to introduce the systematic way of solving circuit problems with AC inputs. Now I've introduced this particular circuit several times, giving you motivation for why we want to go through phasors and impedances and define them. So we're going to finally show you how to actually solve a problem like this. We're going to be building upon phasors and impedances, but we're also going to go back to module two, when we looked at resistive circuit elements. In that particular module, we covered foundational methods and looking at how to analyze and combine series and parallel resistors, using Kirchhoff's voltage laws, voltage divider, current divider. And we also introduced several systematic solution methods, Mesh analysis, Node analysis, Thevenin and Norton Equivalents, and Superposition. Now, it's important because when we did all of these methods, we were looking at purely resistive circuits. And when we introduce impedances, we said an impedance can be treated like a complex resistor. So in other words, we can apply all of these methods for AC analysis as long as we're willing to work with complex resistors. There's three steps to this method. The first step is to redraw the circuit. Replacing all sources with their phasors and all components with their impedances. These are the impedances, equivalent impedances, a resistor, a capacitor, and an inductor. The next step is to use the circuit analysis methods that I just mentioned, KVL, node analysis, mesh analysis, to solve the circuit. Treating the impedances like complex resistors. And the last step is to convert the output phasor to its sinusoidal equivalent. Now let's take a look at a simple example. This is an RC circuit, step one of my method is to redraw this circuit, replacing the sources with their phasors. So this would be 2 at an angle 30. And this resistor becomes 2,000. So resistors, remember, stay the same, and this is Z sub C, And my output is the same. Z sub C = 1 over j omega C, which is 1 over j, omega is right here, is 2,000, And C is 0.22 times 10 to the -6. And if I calculate that, I get 2,272.7 over j. And if I multiply through by j over j, I can clear the j in the denominator. I'm going to have a j squared in the denominator, which is -1, so this is equivalent to -2,272.7 j. So just as trick there, a j in the denominator can be replaced by a -j in the numerator. So now I go back to this circuit. And I look at this circuit and say, well, how do I analyze this? Well the method that I am going to use is a voltage divider law, so this is really step one of my process. Step two of my process was to analyze the circuit. So I've got V out = Z sub C over 2,000 + Z sub C times the input, which is 2 at an angle of 30 degrees. Okay, so let me go ahead and substitute in Z sub C in there, I get -2,272.7 j, all over 2,000- 2,272.7 j, and an angle of 2 times at an angle of 30, that's V out. So now it's just a matter of working with complex math and your calculators might do this directly. So I won't go through too much of the details in here because I'd really like for you to be able to do complex algebra with your calculator and different calculators do it different ways. So it's best to learn your particular calculator. But let me go through and just say to convert from rectangular to polar, I would have this term in the denominator. And in the numerator, I've got a magnitude of 2272.7, and this is purely imaginary. And I've gotta a -j in there. A -j corresponds to an angle of -90. And then I've got 2 at an angle of 30 degrees. And the reason I converted all this to polar because it turns out, as I've said before, it's easier to work with polar notation if I'm trying to multiply and divide these. So if I work through all of this, I'm going to get 1.5. And an angle of, let me write as -41.3 + 30 degrees, which reduces down to, 1.5 at -11.3 degrees. So that's my final solution. If I go back to the last step, that is V out = 1.5 times the cosine of the same frequency, t -11.3 degrees. So that was the three step process. Now I built this circuit and I took the measurements of it and here is the oscilloscope trace here. Now in our analysis, what we found is that with this input we calculated, this is the phasor of the output. Or in terms of the cosine form, it looks like this. So the phase difference between the input and the output was this one minus this one, was -41.3 degrees. Difference between the input phasor and the output phasor. So going over here and looking at the plot. In green is V sub S and blue is V out. And looking at this, let's say, this shows voltage peak to peak that is twice the amplitude, two times the amplitude. And this again is two times the amplitude, this is the input and this is the output amplitude. Because remember, blue is the output signal. So, the input amplitude was 2 and the peak to peak was 4, so this is at 2 right here. So the peak to peak is from the bottom of the trough to the top. So this then is 1.5. Just looking at this plot, and also verifying it from the peak to peak voltage. So, when I built the circuit, I'm getting what I predicted I would get, which is very good, in terms of the amplitudes. What about in terms of the phase? Well, from here to here, remember, is 360 degrees and half that distance is would be a delay of 180. And half of that is 90, half of that is about 45, and the peak is just to the left of 45. In other words, it's not quite 45 degrees and if I measure this more carefully, I can figure out what that phase delay is. But from this plot, we're seeing this is -45, it's not quite 45 and that matches with what we calculated with the math, that it was -41. So in other words, you build this circuit and the experimental result matches what you predicted by doing the analysis. So this particular lesson covered an impedance method for solving AC circuit problems. This particular lesson is the most important one in this entire module because it tells you how to solve circuit problems. We gave a three step process, first to redraw the circuit with its impedance equivalent. So we replace all circuit elements with their impedances, all sources with their phasors. We use standard circuit methods to solve, node analysis, KVL, KCL, any of those methods. Just treat our impedance as if it was a complex resistor. And then finally, once we solve for the output phasor, we converted it to sinusoidal form. And what we also showed is that it matches with experimental results. In our next module, in our next lesson, we will cover a more detailed example of doing a circuit problem like this. All right, thank you. [MUSIC]