1 00:00:00,200 --> 00:00:03,690 >> In the last few videos, we've been developing the tools that we need 2 00:00:03,690 --> 00:00:07,105 to analyze these circuits that are being driven by sinusoidal sources. 3 00:00:07,105 --> 00:00:10,675 Specifically, we've developed the construct 4 00:00:10,675 --> 00:00:12,660 of the mathematical tool of a phasor, 5 00:00:12,660 --> 00:00:15,330 we've introduced the concepts of impedance, 6 00:00:15,330 --> 00:00:17,580 and now we're ready to move on and show how 7 00:00:17,580 --> 00:00:21,750 Kirchhoff's and Ohm's Law is can be used in terms of phasors impedances, 8 00:00:21,750 --> 00:00:23,550 to analyze these types of circuits. 9 00:00:23,550 --> 00:00:24,570 My name is Lee Brinton. 10 00:00:24,570 --> 00:00:25,830 I'm an Electrical Engineering Instructor 11 00:00:25,830 --> 00:00:28,090 in Salt Lake Community College. 12 00:00:29,780 --> 00:00:36,410 Specifically, we're going to derive or demonstrate how Kirchhoff's voltage and 13 00:00:36,410 --> 00:00:39,500 current laws apply in these circuits that are being 14 00:00:39,500 --> 00:00:43,225 driven by sinusoidal sources, 15 00:00:43,225 --> 00:00:46,070 and we'll be analyzing it in this phasor domain, 16 00:00:46,070 --> 00:00:50,570 and then we'll give an example of how these laws 17 00:00:50,570 --> 00:00:51,770 and these tools that we've developed at 18 00:00:51,770 --> 00:00:55,330 this point are used to analyze circuits. 19 00:00:55,330 --> 00:00:57,955 First of all, Kirchhoff's Voltage Law. 20 00:00:57,955 --> 00:01:00,380 We have here a circuit defined involving 21 00:01:00,380 --> 00:01:06,385 three different devices with a voltage reference here plus to minus v_1 of t, 22 00:01:06,385 --> 00:01:08,310 plus to minus v_2 of t, 23 00:01:08,310 --> 00:01:11,040 and plus to minus v_3 of t. We know from 24 00:01:11,040 --> 00:01:12,860 Kirchhoff's Voltage Law that the sum of the voltage 25 00:01:12,860 --> 00:01:15,095 drops around that loop must equal 0, 26 00:01:15,095 --> 00:01:21,195 or v_1 of t plus v_2 27 00:01:21,195 --> 00:01:27,380 of t plus v_3 of t must all add to be 0. 28 00:01:27,380 --> 00:01:32,405 Now, if we assume that we're operating in the sinusoidal steady-state, 29 00:01:32,405 --> 00:01:36,745 and we've asserted that in that type of a circuit, 30 00:01:36,745 --> 00:01:40,010 all of the voltages and currents associated with 31 00:01:40,010 --> 00:01:43,685 that circuit will be oscillating at the same frequency. 32 00:01:43,685 --> 00:01:46,580 Let's just say that the source is oscillating at 33 00:01:46,580 --> 00:01:49,995 some Omega radians per second, 34 00:01:49,995 --> 00:01:52,670 that means that v_1, v_2 and v_3 will also be 35 00:01:52,670 --> 00:01:55,130 oscillating at that same frequency, 36 00:01:55,130 --> 00:01:57,605 but they'll have different amplitudes and different phases. 37 00:01:57,605 --> 00:01:59,405 Or more specifically then, 38 00:01:59,405 --> 00:02:05,530 let's write it as v sub m1, 39 00:02:05,530 --> 00:02:10,620 cosine of Omega t plus Theta 1, 40 00:02:10,620 --> 00:02:13,790 so the amplitude of v_1 is v sub m1, 41 00:02:13,790 --> 00:02:16,085 and it has a phase of Theta 1, 42 00:02:16,085 --> 00:02:23,825 then plus v sub m2 cosine Omega t, it's the same Omega. 43 00:02:23,825 --> 00:02:29,100 Omega t plus Theta 2 plus finally 44 00:02:29,100 --> 00:02:36,840 v sub m3 cosine Omega t plus Theta sub 3, 45 00:02:36,840 --> 00:02:40,115 and the sum of those three terms has to equal 0. 46 00:02:40,115 --> 00:02:44,450 Now, let's represent these in terms of phasors, and by that, 47 00:02:44,450 --> 00:02:49,830 we mean then that the real part of whether this v sub 48 00:02:49,830 --> 00:02:57,345 m cosine Omega t plus Theta 1 can be thought of as the real part of v sub m1, 49 00:02:57,345 --> 00:03:00,135 e to the j Theta 1, 50 00:03:00,135 --> 00:03:03,120 e to the j Omega t, 51 00:03:03,120 --> 00:03:06,200 plus the second term here can be thought of as being 52 00:03:06,200 --> 00:03:09,790 the real part of v sub m2, 53 00:03:09,790 --> 00:03:13,200 e to the j Theta 2, that's a j, 54 00:03:13,200 --> 00:03:18,070 Theta 2, e to the j Omega t, 55 00:03:20,060 --> 00:03:25,695 plus the real part of v sub m3, 56 00:03:25,695 --> 00:03:28,829 e to the j Theta 3, 57 00:03:28,829 --> 00:03:34,770 e to the j Omega t, must equal 0. 58 00:03:34,770 --> 00:03:38,060 Now, realizing that we're talking about the real part of all of 59 00:03:38,060 --> 00:03:41,770 these and that this e to the j Omega t is common to all of them, 60 00:03:41,770 --> 00:03:46,435 we can rewrite this then as the real part of, 61 00:03:46,435 --> 00:03:52,790 and also pointing out that that term right there is just phasor v_1, 62 00:03:52,790 --> 00:03:57,380 and this right here is phasor v_2, 63 00:03:57,380 --> 00:04:03,330 and then of course right there is phasor v_3. 64 00:04:03,330 --> 00:04:06,850 We can now rewrite this as the real part of 65 00:04:06,850 --> 00:04:16,290 phasor v_1 plus phasor v_2 plus phasor v_3, 66 00:04:16,290 --> 00:04:20,380 e to the j Omega t, 67 00:04:21,620 --> 00:04:24,270 and that must equal 0. 68 00:04:24,270 --> 00:04:26,890 Well, we know that e to the j Omega t doesn't equal 0, 69 00:04:26,890 --> 00:04:29,110 therefore the sum of the phasors v_1, 70 00:04:29,110 --> 00:04:31,645 v_2, and v_3, must equal 0. 71 00:04:31,645 --> 00:04:36,399 Or specifically, phasor v_1 plus 72 00:04:36,399 --> 00:04:43,205 phasor v_2 plus phasor v_3 must equal 0. 73 00:04:43,205 --> 00:04:44,915 This is incredibly important. 74 00:04:44,915 --> 00:04:46,820 What that's saying is that up here, 75 00:04:46,820 --> 00:04:48,570 we have this circuit operating in the time domain, 76 00:04:48,570 --> 00:04:51,800 we know that the sum of those three voltages has to equal zero. 77 00:04:51,800 --> 00:04:55,160 What this demonstrates is that not only must the sum 78 00:04:55,160 --> 00:04:58,240 of the time signals around that closed loop equals zero, 79 00:04:58,240 --> 00:05:01,550 but the sum of the phasor representations of 80 00:05:01,550 --> 00:05:05,495 each of these voltages must also equal zero. 81 00:05:05,495 --> 00:05:08,045 In other words, and this is the final thing, 82 00:05:08,045 --> 00:05:12,780 Kirchhoff's Voltage Laws apply or Law, 83 00:05:12,780 --> 00:05:15,620 singular, the Kirchhoff's Voltage Law applies in 84 00:05:15,620 --> 00:05:21,470 this phasor domain in exactly the same way that it does in the time domain. 85 00:05:21,470 --> 00:05:24,745 Now, let's take a look at Kirchhoff's Current Law. 86 00:05:24,745 --> 00:05:27,730 We have a similar set of circumstances here 87 00:05:27,730 --> 00:05:31,360 where the sum of the currents i_1 of t 88 00:05:31,360 --> 00:05:37,690 plus i_2 of t plus i_3 of t, must equal 0. 89 00:05:37,690 --> 00:05:39,590 I'm not going to take the time right now to do it, 90 00:05:39,590 --> 00:05:41,830 suffice it to say that the same type of analysis 91 00:05:41,830 --> 00:05:44,350 that we just did for Kirchhoff's Voltage Law, 92 00:05:44,350 --> 00:05:47,230 results in the sum of 93 00:05:47,230 --> 00:05:52,120 the phasor currents associated with 94 00:05:52,120 --> 00:05:57,810 the node must also equal zero. 95 00:05:57,810 --> 00:06:01,484 So once again, in terms of impedances and phasors, 96 00:06:01,484 --> 00:06:05,120 the sum of the phasors or the sum of 97 00:06:05,120 --> 00:06:10,430 the currents represented in the phasor form must equal zero also.