1 00:00:01,010 --> 00:00:03,390 >> All right. In our previous video, 2 00:00:03,390 --> 00:00:08,750 we found that the Thevenin equivalent voltage, V_Th, 3 00:00:08,750 --> 00:00:16,290 was equal to 6.25 angle 51.34 degrees. 4 00:00:16,290 --> 00:00:19,800 Once again, I'm going to be using this shorthand notation 5 00:00:19,800 --> 00:00:24,750 of representing the polar coordinates of the impedances of the phasors 6 00:00:24,750 --> 00:00:30,300 as opposed to writing 6.25e to the j51.34. 7 00:00:30,300 --> 00:00:34,180 All right. With the Thevenin voltage, 8 00:00:35,510 --> 00:00:38,910 we need now simply to determine 9 00:00:38,910 --> 00:00:44,060 the Thevenin equivalent impedance 10 00:00:44,060 --> 00:00:47,150 and we're going to do that each of three different ways. 11 00:00:47,150 --> 00:00:52,070 This first method is what we'll refer to as the Equivalent Impedance method. 12 00:00:52,070 --> 00:00:59,640 It works when we have only independent sources or no sources. 13 00:00:59,640 --> 00:01:02,060 This equivalent impedance method that we're going to show first 14 00:01:02,060 --> 00:01:07,040 does not work if the circuit you're analyzing has dependent sources. 15 00:01:07,040 --> 00:01:14,630 So, this approach requires us to deactivate any independent sources present. 16 00:01:14,630 --> 00:01:16,670 In this case, there's an independent voltage source here 17 00:01:16,670 --> 00:01:18,770 which we deactivate by turning it to zero 18 00:01:18,770 --> 00:01:22,735 and thus replacing it with a short circuit. 19 00:01:22,735 --> 00:01:27,589 Then, looking back into the circuit from the terminals AB 20 00:01:27,589 --> 00:01:30,860 and determining what the equivalent impedance is that we would see 21 00:01:30,860 --> 00:01:32,735 looking back in that way. 22 00:01:32,735 --> 00:01:37,910 Well, hopefully you can see pretty easily that with the source deactivated- 23 00:01:37,910 --> 00:01:39,140 replaced with short-circuit, 24 00:01:39,140 --> 00:01:42,080 it brings this capacitor and this resistor in parallel, 25 00:01:42,080 --> 00:01:48,830 -and that parallel combination is in series with the j50-Ohm resistor 26 00:01:48,830 --> 00:01:53,840 such that the impedance seen looking back into these terminals, 27 00:01:53,840 --> 00:02:01,545 Z Thevenin, will equal Z parallel plus the j50-Ohm inductor. 28 00:02:01,545 --> 00:02:05,595 So, Z parallel is equal to 29 00:02:05,595 --> 00:02:13,425 20 times minus j25 divided by 20 minus j25, 30 00:02:13,425 --> 00:02:24,960 which turns out to equal 12.2 minus j9.8 Ohms. 31 00:02:24,960 --> 00:02:29,830 So finally, Z Thevenin then is equal to Zp; 32 00:02:29,830 --> 00:02:38,120 12.2 minus j9.8 plus the j50, 33 00:02:38,120 --> 00:02:47,200 and that then equals 12.2 plus j40.2 Ohms. 34 00:02:48,530 --> 00:02:51,530 Z Thevenin, bringing it up here, 35 00:02:51,530 --> 00:02:56,225 then rewriting it just to show what we've got to show our incomplete model. 36 00:02:56,225 --> 00:03:05,060 Our complete model involves the Z Thevenin equaling 12.2 plus j40.2 ohms 37 00:03:05,060 --> 00:03:10,730 and our claim then is that this Thevenin equivalent circuit 38 00:03:10,730 --> 00:03:13,910 has the same terminal characteristics at the AB terminals 39 00:03:13,910 --> 00:03:17,880 as our original more complicated circuit.