1 00:00:00,260 --> 00:00:03,225 >> In this video, we're going to demonstrate 2 00:00:03,225 --> 00:00:07,845 the node analysis method of analyzing circuits in the Phasor Domain. 3 00:00:07,845 --> 00:00:09,810 You'll recall from our previous discussions of 4 00:00:09,810 --> 00:00:13,020 node analysis that the first step in analyzing circuits 5 00:00:13,020 --> 00:00:14,985 using the node voltage approach 6 00:00:14,985 --> 00:00:18,075 involves identifying the critical or essential nodes. 7 00:00:18,075 --> 00:00:20,120 In this circuit we have got one, 8 00:00:20,120 --> 00:00:23,760 two, three, four essential nodes. 9 00:00:23,760 --> 00:00:29,070 Again, an essential node is a node where three or more branches are connected. 10 00:00:29,070 --> 00:00:32,280 We choose one of these nodes to then serve as 11 00:00:32,280 --> 00:00:34,020 our reference node where we by 12 00:00:34,020 --> 00:00:36,405 definition we're going to say voltage is equal to zero. 13 00:00:36,405 --> 00:00:41,510 In this case, I've selected this node here to be our reference node. 14 00:00:41,510 --> 00:00:44,455 So we'll set that voltage equal to zero. 15 00:00:44,455 --> 00:00:46,670 The other three nodes are 16 00:00:46,670 --> 00:00:50,130 the three essential nodes we will then identify or assign variables to them. 17 00:00:50,130 --> 00:00:52,880 Variable names. This one we're going to call V1. 18 00:00:52,880 --> 00:00:54,920 The voltage at this node will refer to as 19 00:00:54,920 --> 00:00:58,895 V2 and the voltage at this node here will refer to as V3. 20 00:00:58,895 --> 00:01:01,640 Node voltage analysis then has this writing 21 00:01:01,640 --> 00:01:06,635 Kirchhoff's Current Law or node equations at each of these essential nodes. 22 00:01:06,635 --> 00:01:11,285 Summing the currents leaving each of those nodes in terms of 23 00:01:11,285 --> 00:01:16,400 the node voltage variables and the circuit parameters in the circuit. 24 00:01:16,400 --> 00:01:17,690 So let's go ahead and do this. 25 00:01:17,690 --> 00:01:18,935 On this first node, 26 00:01:18,935 --> 00:01:24,565 let's write KCL at node one. 27 00:01:24,565 --> 00:01:29,540 The current leaving this node coming down in this direction here will be 28 00:01:29,540 --> 00:01:35,195 V1 minus the voltage at this node minus the voltage at this point. 29 00:01:35,195 --> 00:01:37,430 The voltage here by definition is zero, 30 00:01:37,430 --> 00:01:41,000 so V1 minus 0. We won't even bother. 31 00:01:41,000 --> 00:01:45,260 Well, start let's write it V1 minus 0 divided by 32 00:01:45,260 --> 00:01:50,245 the impedance there which is 3 minus J4. 33 00:01:50,245 --> 00:01:54,394 Plus the current leaving this node going in this direction, 34 00:01:54,394 --> 00:02:00,710 which is V1 minus V3 divided by 35 00:02:00,710 --> 00:02:04,250 the impedance here of two ohms plus 36 00:02:04,250 --> 00:02:07,910 the current leaving this node going in this direction here. 37 00:02:07,910 --> 00:02:09,350 We need to be a little bit careful here. 38 00:02:09,350 --> 00:02:12,410 It's going to be the current leaving this node going in 39 00:02:12,410 --> 00:02:16,235 that direction will be the current through that resistor. 40 00:02:16,235 --> 00:02:19,220 The current through that resistor is equal to the voltage drop 41 00:02:19,220 --> 00:02:22,955 across that resistor divided by the resistance of two ohms. 42 00:02:22,955 --> 00:02:28,565 The voltage drop across that resistor is the voltage on this side which is V1 43 00:02:28,565 --> 00:02:30,500 minus the voltage on 44 00:02:30,500 --> 00:02:35,570 this side of the resistor and here's where we need to be careful. 45 00:02:35,570 --> 00:02:39,860 The voltage here in terms of node voltages can be 46 00:02:39,860 --> 00:02:43,775 thought of as being starting here at V2. 47 00:02:43,775 --> 00:02:49,820 V2 we drop going from here to here crossing this voltage source of 12 volts. 48 00:02:49,820 --> 00:02:51,260 We drop 12 volts.. 49 00:02:51,260 --> 00:02:58,115 So the voltage at this point then is going to be V2 minus 12 volts. 50 00:02:58,115 --> 00:03:00,890 So then the current leaving this branch going in 51 00:03:00,890 --> 00:03:02,150 that direction will be the voltage of 52 00:03:02,150 --> 00:03:04,130 the left-hand side of that resistor which 53 00:03:04,130 --> 00:03:07,610 is V1 minus the voltage of the right-hand side of 54 00:03:07,610 --> 00:03:12,390 the resistor which is V2 minus 12. 55 00:03:12,390 --> 00:03:15,350 We throw in the parentheses there to remind ourselves that we're 56 00:03:15,350 --> 00:03:20,030 subtracting the voltage there which is V2 minus 12. 57 00:03:20,030 --> 00:03:24,365 Finally, the current in that branch then will be that voltage drop divided by 58 00:03:24,365 --> 00:03:29,255 two and that equals zero. 59 00:03:29,255 --> 00:03:32,920 Now, KCL at node two. 60 00:03:32,920 --> 00:03:37,400 The current leaving this node going to the left, again, 61 00:03:37,400 --> 00:03:39,080 is going to be the current through this resistor only 62 00:03:39,080 --> 00:03:40,760 this time going from right to 63 00:03:40,760 --> 00:03:44,430 left and that voltage will be the voltage here, 64 00:03:44,430 --> 00:03:49,790 which we've already established to be V2 minus 12 volts minus 65 00:03:49,790 --> 00:03:56,000 the voltage over here which is V1 minus V1 divided by the two ohms. 66 00:03:56,000 --> 00:03:59,300 The current coming down to the node or leaving this node coming 67 00:03:59,300 --> 00:04:02,980 down is going to be then added in and that's 68 00:04:02,980 --> 00:04:11,100 Simply V2 divided by the impedance of two plus one J or simply J. 69 00:04:11,120 --> 00:04:16,760 Then the current leaving node to going to the right will be equal 70 00:04:16,760 --> 00:04:22,985 to V2 minus V3 divided by two ohms. 71 00:04:22,985 --> 00:04:25,240 The sum of those three currents equal zero. 72 00:04:25,240 --> 00:04:30,380 Now, at the third node we've got KCL at node three. 73 00:04:30,380 --> 00:04:34,160 The current leaving node three and come into the left is going to be 74 00:04:34,160 --> 00:04:39,380 V3 minus V2 divided by two ohms. 75 00:04:39,380 --> 00:04:44,270 Plus the current leaving this node going up and over to the left is going to 76 00:04:44,270 --> 00:04:50,575 be V3 minus V1 divided by two ohms. 77 00:04:50,575 --> 00:04:53,990 Finally, the current leaving node three coming down this 78 00:04:53,990 --> 00:04:57,185 way is going to be the voltage drop across that 2-ohm resistor, 79 00:04:57,185 --> 00:05:03,010 which will be V3 minus the voltage right there. 80 00:05:03,010 --> 00:05:05,660 Now, the voltage at that point is equal to starting here 81 00:05:05,660 --> 00:05:08,270 the reference node zero minus to plus 82 00:05:08,270 --> 00:05:14,570 we go up a value of negative J6 volts. 83 00:05:14,570 --> 00:05:18,245 So the voltage right here then is 84 00:05:18,245 --> 00:05:24,664 zero minus J6 and then the current through here will be V3 85 00:05:24,664 --> 00:05:29,850 minus a minus J6 86 00:05:30,680 --> 00:05:37,370 divided by the two ohms and the sum of those three currents equal zero. 87 00:05:37,370 --> 00:05:40,085 Those are the three note note equations. 88 00:05:40,085 --> 00:05:42,780 All that's left now is to calculate the node voltages V1, 89 00:05:42,780 --> 00:05:44,825 V2 and V3 is a little bit of algebra. 90 00:05:44,825 --> 00:05:47,705 Let's start by combining like terms. 91 00:05:47,705 --> 00:05:49,180 Let's do that on up here. 92 00:05:49,180 --> 00:05:51,420 The V1 terms in the first equation. 93 00:05:51,420 --> 00:05:53,850 I've got V1 over. 94 00:05:53,850 --> 00:05:57,925 Let's see 3 minus J4. 95 00:05:57,925 --> 00:06:01,070 I've got V1 over two and I've got another V1 over two. 96 00:06:01,070 --> 00:06:09,245 So plus one over two plus one over two gives us the V1 terms. 97 00:06:09,245 --> 00:06:13,745 Plus the V2 terms here we've 98 00:06:13,745 --> 00:06:19,700 got only one right here and that's a minus V2 over two. 99 00:06:19,700 --> 00:06:21,650 So if we pull the V2 out we'll have 100 00:06:21,650 --> 00:06:28,980 a minus one half there and then our V3 terms there. 101 00:06:28,980 --> 00:06:30,150 Again, is only one of them. 102 00:06:30,150 --> 00:06:31,910 It's a minus V3 over two. 103 00:06:31,910 --> 00:06:37,320 So we have V3 times a negative one half. 104 00:06:37,720 --> 00:06:39,860 That then will equal. 105 00:06:39,860 --> 00:06:42,530 Let's see, we've got a 12 volts here. 106 00:06:42,530 --> 00:06:44,680 12 divided by two is six. 107 00:06:44,680 --> 00:06:48,620 A minus times a minus is a plus. 108 00:06:48,620 --> 00:06:50,660 I've got plus six on the left-hand side. 109 00:06:50,660 --> 00:06:51,920 We subtract six from 110 00:06:51,920 --> 00:06:56,845 both sides to bring our constant over there on the right-hand side. 111 00:06:56,845 --> 00:06:59,690 Now, combining like terms in the second equation, 112 00:06:59,690 --> 00:07:03,895 we have V1 times. 113 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. 114 00:07:07,940 --> 00:07:10,700 So we pull the V out or the V1 out or left with 115 00:07:10,700 --> 00:07:15,930 the negative one half plus V2 times. 116 00:07:15,930 --> 00:07:17,985 I've got V2 over two, 117 00:07:17,985 --> 00:07:21,270 V2 over two plus J and V2 over two. 118 00:07:21,270 --> 00:07:30,645 So V2 or let's see V2 would be one half 119 00:07:30,645 --> 00:07:41,385 plus one over two plus J plus another one half and then the V3 terms, 120 00:07:41,385 --> 00:07:47,105 again, there's only one and that will be a negative one half. 121 00:07:47,105 --> 00:07:50,570 Now, here we've got a minus 12 over two. 122 00:07:50,570 --> 00:07:53,480 So I've got a negative six volts on the left-hand side. 123 00:07:53,480 --> 00:07:57,350 We add six volts to plus to our both sides and that 124 00:07:57,350 --> 00:08:02,255 gives us then equaling positive six volts on the right-hand side. 125 00:08:02,255 --> 00:08:12,510 Now, equation three, we have V1 times a negative one half plus 126 00:08:12,510 --> 00:08:18,040 V2 times got a negative one half again 127 00:08:18,590 --> 00:08:26,910 plus V3 times I've got one half plus one half plus one half. 128 00:08:26,910 --> 00:08:30,960 That's three halves, V3. 129 00:08:30,960 --> 00:08:33,890 Now, here we've got a constant of negative times negative is 130 00:08:33,890 --> 00:08:36,755 a positive J6 over two. 131 00:08:36,755 --> 00:08:40,400 So on the left-hand side we have a positive J3, 132 00:08:40,400 --> 00:08:44,030 which we subtract J3 from both sides gives us 133 00:08:44,030 --> 00:08:49,265 then a negative J3 over here on this side. 134 00:08:49,265 --> 00:08:52,520 All right. Moving ahead now just cleaning things up 135 00:08:52,520 --> 00:08:55,535 so that we're ready to put it into a matrix solver. 136 00:08:55,535 --> 00:08:59,180 Let's combine all those fractions here and when we do 137 00:08:59,180 --> 00:09:04,915 that we've got V1 times 1.12 138 00:09:04,915 --> 00:09:13,920 plus 0.16J plus V2 times a negative 139 00:09:13,920 --> 00:09:18,870 0.5 plus V3 times a 140 00:09:18,870 --> 00:09:24,815 negative 0.5 equals negative six volts. 141 00:09:24,815 --> 00:09:35,270 The second equation we have V1 times a negative 0.5 plus V2 times 142 00:09:35,270 --> 00:09:43,440 1.4 minus 0.2J plus 143 00:09:43,440 --> 00:09:51,235 V3 times a negative 0.5 equals a positive six. 144 00:09:51,235 --> 00:09:53,740 Finally, the third equation, 145 00:09:53,740 --> 00:10:00,690 V1 times a negative 0.5 plus V2 times a negative 146 00:10:00,690 --> 00:10:05,820 0.5 plus V3 times 147 00:10:05,820 --> 00:10:12,160 1.5 is equal to a negative J3. 148 00:10:55,070 --> 00:10:58,925 When you plug this into a matrix solver of some sort, 149 00:10:58,925 --> 00:11:05,970 you come up with that V1 is equal to in rectangular coordinates, 150 00:11:05,970 --> 00:11:14,815 a negative 4.72 minus 0.88J, 151 00:11:14,815 --> 00:11:22,415 which in polar coordinates is 4.8 angle negative 169.5. 152 00:11:22,415 --> 00:11:32,685 V2 is equal to 2.46 minus 0.87J, 153 00:11:32,685 --> 00:11:43,050 which in polar coordinates is 2.6 angle negative 19.84 and V3 we 154 00:11:43,050 --> 00:11:49,320 find to be equal to negative 0.76 minus 155 00:11:49,320 --> 00:11:54,230 J2.59 in rectangular coordinates in polar 156 00:11:54,230 --> 00:12:01,880 coordinates it's 2.7 angle negative 106.29. 157 00:12:02,250 --> 00:12:06,730 Of course, are all degrees. 158 00:12:07,130 --> 00:12:10,190 Now, that we have the voltages at each of the nodes we can 159 00:12:10,190 --> 00:12:13,220 calculate any current or voltage that we might want to calculate. 160 00:12:13,220 --> 00:12:15,605 For example, let's determine 161 00:12:15,605 --> 00:12:19,720 the current leaving node two coming down this branch here. 162 00:12:19,720 --> 00:12:24,390 Well, it's called in this circuit here which has been identified 163 00:12:24,390 --> 00:12:29,050 as I5 or I sub L. So, I5 then, 164 00:12:29,050 --> 00:12:35,250 is equal to V2 divided by two plus J, 165 00:12:35,250 --> 00:12:40,615 which is equal to V2 is 2.6 166 00:12:40,615 --> 00:12:47,235 angle negative 19.84 divided by two plus J. 167 00:12:47,235 --> 00:12:52,865 We find then if that is equal to in polar coordinates 1.1. 168 00:12:52,865 --> 00:13:00,090 six angle negative 46.4 degrees.