WEBVTT 00:00:01.010 --> 00:00:03.390 >> All right. In our previous video, 00:00:03.390 --> 00:00:08.750 we found that the Thevenin equivalent voltage, V_Th, 00:00:08.750 --> 00:00:16.290 was equal to 6.25 angle 51.34 degrees. 00:00:16.290 --> 00:00:19.800 Once again, I'm going to be using this shorthand notation 00:00:19.800 --> 00:00:24.750 of representing the polar coordinates of the impedances of the phasors 00:00:24.750 --> 00:00:30.300 as opposed to writing 6.25e to the j51.34. 00:00:30.300 --> 00:00:34.180 All right. With the Thevenin voltage, 00:00:35.510 --> 00:00:38.910 we need now simply to determine 00:00:38.910 --> 00:00:44.060 the Thevenin equivalent impedance 00:00:44.060 --> 00:00:47.150 and we're going to do that each of three different ways. 00:00:47.150 --> 00:00:52.070 This first method is what we'll refer to as the Equivalent Impedance method. 00:00:52.070 --> 00:00:59.640 It works when we have only independent sources or no sources. 00:00:59.640 --> 00:01:02.060 This equivalent impedance method that we're going to show first 00:01:02.060 --> 00:01:07.040 does not work if the circuit you're analyzing has dependent sources. 00:01:07.040 --> 00:01:14.630 So, this approach requires us to deactivate any independent sources present. 00:01:14.630 --> 00:01:16.670 In this case, there's an independent voltage source here 00:01:16.670 --> 00:01:18.770 which we deactivate by turning it to zero 00:01:18.770 --> 00:01:22.735 and thus replacing it with a short circuit. 00:01:22.735 --> 00:01:27.589 Then, looking back into the circuit from the terminals AB 00:01:27.589 --> 00:01:30.860 and determining what the equivalent impedance is that we would see 00:01:30.860 --> 00:01:32.735 looking back in that way. 00:01:32.735 --> 00:01:37.910 Well, hopefully you can see pretty easily that with the source deactivated- 00:01:37.910 --> 00:01:39.140 replaced with short-circuit, 00:01:39.140 --> 00:01:42.080 it brings this capacitor and this resistor in parallel, 00:01:42.080 --> 00:01:48.830 -and that parallel combination is in series with the j50-Ohm resistor 00:01:48.830 --> 00:01:53.840 such that the impedance seen looking back into these terminals, 00:01:53.840 --> 00:02:01.545 Z Thevenin, will equal Z parallel plus the j50-Ohm inductor. 00:02:01.545 --> 00:02:05.595 So, Z parallel is equal to 00:02:05.595 --> 00:02:13.425 20 times minus j25 divided by 20 minus j25, 00:02:13.425 --> 00:02:24.960 which turns out to equal 12.2 minus j9.8 Ohms. 00:02:24.960 --> 00:02:29.830 So finally, Z Thevenin then is equal to Zp; 00:02:29.830 --> 00:02:38.120 12.2 minus j9.8 plus the j50, 00:02:38.120 --> 00:02:47.200 and that then equals 12.2 plus j40.2 Ohms. 00:02:48.530 --> 00:02:51.530 Z Thevenin, bringing it up here, 00:02:51.530 --> 00:02:56.225 then rewriting it just to show what we've got to show our incomplete model. 00:02:56.225 --> 00:03:05.060 Our complete model involves the Z Thevenin equaling 12.2 plus j40.2 ohms 00:03:05.060 --> 00:03:10.730 and our claim then is that this Thevenin equivalent circuit 00:03:10.730 --> 00:03:13.910 has the same terminal characteristics at the AB terminals 00:03:13.910 --> 00:03:17.880 as our original more complicated circuit.