WEBVTT 00:00:00.260 --> 00:00:03.225 >> In this video, we're going to demonstrate 00:00:03.225 --> 00:00:07.845 the node analysis method of analyzing circuits in the Phasor Domain. 00:00:07.845 --> 00:00:09.810 You'll recall from our previous discussions of 00:00:09.810 --> 00:00:13.020 node analysis that the first step in analyzing circuits 00:00:13.020 --> 00:00:14.985 using the node voltage approach 00:00:14.985 --> 00:00:18.075 involves identifying the critical or essential nodes. 00:00:18.075 --> 00:00:20.120 In this circuit we have got one, 00:00:20.120 --> 00:00:23.760 two, three, four essential nodes. 00:00:23.760 --> 00:00:29.070 Again, an essential node is a node where three or more branches are connected. 00:00:29.070 --> 00:00:32.280 We choose one of these nodes to then serve as 00:00:32.280 --> 00:00:34.020 our reference node where we by 00:00:34.020 --> 00:00:36.405 definition we're going to say voltage is equal to zero. 00:00:36.405 --> 00:00:41.510 In this case, I've selected this node here to be our reference node. 00:00:41.510 --> 00:00:44.455 So we'll set that voltage equal to zero. 00:00:44.455 --> 00:00:46.670 The other three nodes are 00:00:46.670 --> 00:00:50.130 the three essential nodes we will then identify or assign variables to them. 00:00:50.130 --> 00:00:52.880 Variable names. This one we're going to call V1. 00:00:52.880 --> 00:00:54.920 The voltage at this node will refer to as 00:00:54.920 --> 00:00:58.895 V2 and the voltage at this node here will refer to as V3. 00:00:58.895 --> 00:01:01.640 Node voltage analysis then has this writing 00:01:01.640 --> 00:01:06.635 Kirchhoff's Current Law or node equations at each of these essential nodes. 00:01:06.635 --> 00:01:11.285 Summing the currents leaving each of those nodes in terms of 00:01:11.285 --> 00:01:16.400 the node voltage variables and the circuit parameters in the circuit. 00:01:16.400 --> 00:01:17.690 So let's go ahead and do this. 00:01:17.690 --> 00:01:18.935 On this first node, 00:01:18.935 --> 00:01:24.565 let's write KCL at node one. 00:01:24.565 --> 00:01:29.540 The current leaving this node coming down in this direction here will be 00:01:29.540 --> 00:01:35.195 V1 minus the voltage at this node minus the voltage at this point. 00:01:35.195 --> 00:01:37.430 The voltage here by definition is zero, 00:01:37.430 --> 00:01:41.000 so V1 minus 0. We won't even bother. 00:01:41.000 --> 00:01:45.260 Well, start let's write it V1 minus 0 divided by 00:01:45.260 --> 00:01:50.245 the impedance there which is 3 minus J4. 00:01:50.245 --> 00:01:54.394 Plus the current leaving this node going in this direction, 00:01:54.394 --> 00:02:00.710 which is V1 minus V3 divided by 00:02:00.710 --> 00:02:04.250 the impedance here of two ohms plus 00:02:04.250 --> 00:02:07.910 the current leaving this node going in this direction here. 00:02:07.910 --> 00:02:09.350 We need to be a little bit careful here. 00:02:09.350 --> 00:02:12.410 It's going to be the current leaving this node going in 00:02:12.410 --> 00:02:16.235 that direction will be the current through that resistor. 00:02:16.235 --> 00:02:19.220 The current through that resistor is equal to the voltage drop 00:02:19.220 --> 00:02:22.955 across that resistor divided by the resistance of two ohms. 00:02:22.955 --> 00:02:28.565 The voltage drop across that resistor is the voltage on this side which is V1 00:02:28.565 --> 00:02:30.500 minus the voltage on 00:02:30.500 --> 00:02:35.570 this side of the resistor and here's where we need to be careful. 00:02:35.570 --> 00:02:39.860 The voltage here in terms of node voltages can be 00:02:39.860 --> 00:02:43.775 thought of as being starting here at V2. 00:02:43.775 --> 00:02:49.820 V2 we drop going from here to here crossing this voltage source of 12 volts. 00:02:49.820 --> 00:02:51.260 We drop 12 volts.. 00:02:51.260 --> 00:02:58.115 So the voltage at this point then is going to be V2 minus 12 volts. 00:02:58.115 --> 00:03:00.890 So then the current leaving this branch going in 00:03:00.890 --> 00:03:02.150 that direction will be the voltage of 00:03:02.150 --> 00:03:04.130 the left-hand side of that resistor which 00:03:04.130 --> 00:03:07.610 is V1 minus the voltage of the right-hand side of 00:03:07.610 --> 00:03:12.390 the resistor which is V2 minus 12. 00:03:12.390 --> 00:03:15.350 We throw in the parentheses there to remind ourselves that we're 00:03:15.350 --> 00:03:20.030 subtracting the voltage there which is V2 minus 12. 00:03:20.030 --> 00:03:24.365 Finally, the current in that branch then will be that voltage drop divided by 00:03:24.365 --> 00:03:29.255 two and that equals zero. 00:03:29.255 --> 00:03:32.920 Now, KCL at node two. 00:03:32.920 --> 00:03:37.400 The current leaving this node going to the left, again, 00:03:37.400 --> 00:03:39.080 is going to be the current through this resistor only 00:03:39.080 --> 00:03:40.760 this time going from right to 00:03:40.760 --> 00:03:44.430 left and that voltage will be the voltage here, 00:03:44.430 --> 00:03:49.790 which we've already established to be V2 minus 12 volts minus 00:03:49.790 --> 00:03:56.000 the voltage over here which is V1 minus V1 divided by the two ohms. 00:03:56.000 --> 00:03:59.300 The current coming down to the node or leaving this node coming 00:03:59.300 --> 00:04:02.980 down is going to be then added in and that's 00:04:02.980 --> 00:04:11.100 Simply V2 divided by the impedance of two plus one J or simply J. 00:04:11.120 --> 00:04:16.760 Then the current leaving node to going to the right will be equal 00:04:16.760 --> 00:04:22.985 to V2 minus V3 divided by two ohms. 00:04:22.985 --> 00:04:25.240 The sum of those three currents equal zero. 00:04:25.240 --> 00:04:30.380 Now, at the third node we've got KCL at node three. 00:04:30.380 --> 00:04:34.160 The current leaving node three and come into the left is going to be 00:04:34.160 --> 00:04:39.380 V3 minus V2 divided by two ohms. 00:04:39.380 --> 00:04:44.270 Plus the current leaving this node going up and over to the left is going to 00:04:44.270 --> 00:04:50.575 be V3 minus V1 divided by two ohms. 00:04:50.575 --> 00:04:53.990 Finally, the current leaving node three coming down this 00:04:53.990 --> 00:04:57.185 way is going to be the voltage drop across that 2-ohm resistor, 00:04:57.185 --> 00:05:03.010 which will be V3 minus the voltage right there. 00:05:03.010 --> 00:05:05.660 Now, the voltage at that point is equal to starting here 00:05:05.660 --> 00:05:08.270 the reference node zero minus to plus 00:05:08.270 --> 00:05:14.570 we go up a value of negative J6 volts. 00:05:14.570 --> 00:05:18.245 So the voltage right here then is 00:05:18.245 --> 00:05:24.664 zero minus J6 and then the current through here will be V3 00:05:24.664 --> 00:05:29.850 minus a minus J6 00:05:30.680 --> 00:05:37.370 divided by the two ohms and the sum of those three currents equal zero. 00:05:37.370 --> 00:05:40.085 Those are the three note note equations. 00:05:40.085 --> 00:05:42.780 All that's left now is to calculate the node voltages V1, 00:05:42.780 --> 00:05:44.825 V2 and V3 is a little bit of algebra. 00:05:44.825 --> 00:05:47.705 Let's start by combining like terms. 00:05:47.705 --> 00:05:49.180 Let's do that on up here. 00:05:49.180 --> 00:05:51.420 The V1 terms in the first equation. 00:05:51.420 --> 00:05:53.850 I've got V1 over. 00:05:53.850 --> 00:05:57.925 Let's see 3 minus J4. 00:05:57.925 --> 00:06:01.070 I've got V1 over two and I've got another V1 over two. 00:06:01.070 --> 00:06:09.245 So plus one over two plus one over two gives us the V1 terms. 00:06:09.245 --> 00:06:13.745 Plus the V2 terms here we've 00:06:13.745 --> 00:06:19.700 got only one right here and that's a minus V2 over two. 00:06:19.700 --> 00:06:21.650 So if we pull the V2 out we'll have 00:06:21.650 --> 00:06:28.980 a minus one half there and then our V3 terms there. 00:06:28.980 --> 00:06:30.150 Again, is only one of them. 00:06:30.150 --> 00:06:31.910 It's a minus V3 over two. 00:06:31.910 --> 00:06:37.320 So we have V3 times a negative one half. 00:06:37.720 --> 00:06:39.860 That then will equal. 00:06:39.860 --> 00:06:42.530 Let's see, we've got a 12 volts here. 00:06:42.530 --> 00:06:44.680 12 divided by two is six. 00:06:44.680 --> 00:06:48.620 A minus times a minus is a plus. 00:06:48.620 --> 00:06:50.660 I've got plus six on the left-hand side. 00:06:50.660 --> 00:06:51.920 We subtract six from 00:06:51.920 --> 00:06:56.845 both sides to bring our constant over there on the right-hand side. 00:06:56.845 --> 00:06:59.690 Now, combining like terms in the second equation, 00:06:59.690 --> 00:07:03.895 we have V1 times. 00:07:03.895 --> 00:07:07.940 This one we've only got one term that's going to be a negative V1 over two. 00:07:07.940 --> 00:07:10.700 So we pull the V out or the V1 out or left with 00:07:10.700 --> 00:07:15.930 the negative one half plus V2 times. 00:07:15.930 --> 00:07:17.985 I've got V2 over two, 00:07:17.985 --> 00:07:21.270 V2 over two plus J and V2 over two. 00:07:21.270 --> 00:07:30.645 So V2 or let's see V2 would be one half 00:07:30.645 --> 00:07:41.385 plus one over two plus J plus another one half and then the V3 terms, 00:07:41.385 --> 00:07:47.105 again, there's only one and that will be a negative one half. 00:07:47.105 --> 00:07:50.570 Now, here we've got a minus 12 over two. 00:07:50.570 --> 00:07:53.480 So I've got a negative six volts on the left-hand side. 00:07:53.480 --> 00:07:57.350 We add six volts to plus to our both sides and that 00:07:57.350 --> 00:08:02.255 gives us then equaling positive six volts on the right-hand side. 00:08:02.255 --> 00:08:12.510 Now, equation three, we have V1 times a negative one half plus 00:08:12.510 --> 00:08:18.040 V2 times got a negative one half again 00:08:18.590 --> 00:08:26.910 plus V3 times I've got one half plus one half plus one half. 00:08:26.910 --> 00:08:30.960 That's three halves, V3. 00:08:30.960 --> 00:08:33.890 Now, here we've got a constant of negative times negative is 00:08:33.890 --> 00:08:36.755 a positive J6 over two. 00:08:36.755 --> 00:08:40.400 So on the left-hand side we have a positive J3, 00:08:40.400 --> 00:08:44.030 which we subtract J3 from both sides gives us 00:08:44.030 --> 00:08:49.265 then a negative J3 over here on this side. 00:08:49.265 --> 00:08:52.520 All right. Moving ahead now just cleaning things up 00:08:52.520 --> 00:08:55.535 so that we're ready to put it into a matrix solver. 00:08:55.535 --> 00:08:59.180 Let's combine all those fractions here and when we do 00:08:59.180 --> 00:09:04.915 that we've got V1 times 1.12 00:09:04.915 --> 00:09:13.920 plus 0.16J plus V2 times a negative 00:09:13.920 --> 00:09:18.870 0.5 plus V3 times a 00:09:18.870 --> 00:09:24.815 negative 0.5 equals negative six volts. 00:09:24.815 --> 00:09:35.270 The second equation we have V1 times a negative 0.5 plus V2 times 00:09:35.270 --> 00:09:43.440 1.4 minus 0.2J plus 00:09:43.440 --> 00:09:51.235 V3 times a negative 0.5 equals a positive six. 00:09:51.235 --> 00:09:53.740 Finally, the third equation, 00:09:53.740 --> 00:10:00.690 V1 times a negative 0.5 plus V2 times a negative 00:10:00.690 --> 00:10:05.820 0.5 plus V3 times 00:10:05.820 --> 00:10:12.160 1.5 is equal to a negative J3. 00:10:55.070 --> 00:10:58.925 When you plug this into a matrix solver of some sort, 00:10:58.925 --> 00:11:05.970 you come up with that V1 is equal to in rectangular coordinates, 00:11:05.970 --> 00:11:14.815 a negative 4.72 minus 0.88J, 00:11:14.815 --> 00:11:22.415 which in polar coordinates is 4.8 angle negative 169.5. 00:11:22.415 --> 00:11:32.685 V2 is equal to 2.46 minus 0.87J, 00:11:32.685 --> 00:11:43.050 which in polar coordinates is 2.6 angle negative 19.84 and V3 we 00:11:43.050 --> 00:11:49.320 find to be equal to negative 0.76 minus 00:11:49.320 --> 00:11:54.230 J2.59 in rectangular coordinates in polar 00:11:54.230 --> 00:12:01.880 coordinates it's 2.7 angle negative 106.29. 00:12:02.250 --> 00:12:06.730 Of course, are all degrees. 00:12:07.130 --> 00:12:10.190 Now, that we have the voltages at each of the nodes we can 00:12:10.190 --> 00:12:13.220 calculate any current or voltage that we might want to calculate. 00:12:13.220 --> 00:12:15.605 For example, let's determine 00:12:15.605 --> 00:12:19.720 the current leaving node two coming down this branch here. 00:12:19.720 --> 00:12:24.390 Well, it's called in this circuit here which has been identified 00:12:24.390 --> 00:12:29.050 as I5 or I sub L. So, I5 then, 00:12:29.050 --> 00:12:35.250 is equal to V2 divided by two plus J, 00:12:35.250 --> 00:12:40.615 which is equal to V2 is 2.6 00:12:40.615 --> 00:12:47.235 angle negative 19.84 divided by two plus J. 00:12:47.235 --> 00:12:52.865 We find then if that is equal to in polar coordinates 1.1. 00:12:52.865 --> 00:13:00.090 six angle negative 46.4 degrees.