0:00:01.962,0:00:04.939 Welcome to an Introduction to[br]Electrical & Computer Engineering. 0:00:04.939,0:00:06.300 My name is Lee Brinton, 0:00:06.300,0:00:11.150 I'm an electrical engineering instructor[br]at Salt Lake Community College. 0:00:11.150,0:00:14.730 In this video we'll be talking about[br]ways to analyze circuits using the node 0:00:14.730,0:00:16.219 voltage method. 0:00:16.219,0:00:20.810 We're gonna start by making a, looking at[br]the similarities between elevation and 0:00:20.810,0:00:21.940 voltage. 0:00:21.940,0:00:26.100 We'll then discuss the difference between[br]branch voltages and node voltages. 0:00:26.100,0:00:30.920 We'll introduce the concept of nodes and[br]critical or extraordinary nodes. 0:00:30.920,0:00:35.854 We'll then look at the actual process of[br]analyzing circuits using node voltages, 0:00:35.854,0:00:40.011 and we'll look at how that applies[br]when they are dependent sources and 0:00:40.011,0:00:42.210 supernodes present in the circuit. 0:00:43.690,0:00:46.710 First of all,[br]the similarities between elevation. 0:00:46.710,0:00:49.420 If we wanna talk about the elevation of,[br]say Mount Olympus, 0:00:49.420,0:00:53.180 we've got to define first of all[br]what we're measuring relative to. 0:00:53.180,0:00:55.970 In other words,[br]we establish our reference. 0:00:55.970,0:01:00.410 Typically sea level is our[br]reference at zero feet. 0:01:00.410,0:01:03.660 Mount Olympus then is about[br]9,500 feet above sea level, 0:01:03.660,0:01:08.030 and the Salt Lake Valley Floor is[br]around 4,100 feet above sea level. 0:01:09.550,0:01:12.630 On the other hand,[br]if we're standing on the Valley Floor, and 0:01:12.630,0:01:16.290 looking up to the mountains to the east,[br]we'll see that the elevation, 0:01:16.290,0:01:20.055 the amount that they rise, can be[br]calculated by taking the higher elevation, 0:01:20.055,0:01:27.194 9,500 feet less the lower elevation[br]of 4,100 feet, and we've got a 5,400 0:01:27.194,0:01:31.400 foot change in elevation going from the[br]Valley Floor to the top of Mount Olympus. 0:01:32.550,0:01:35.950 On the other hand, if instead of[br]calling sea level our reference, 0:01:35.950,0:01:44.250 we made the Valley Floor our reference and[br]said then the elevation here = 0 feet, 0:01:44.250,0:01:49.240 then Mount Olympus would be 5,400 feet, 0:01:49.240,0:01:54.200 and sea level would be -4,100 feet. 0:01:54.200,0:01:57.780 Elevations are all relative[br]to a reference, and 0:01:57.780,0:02:01.090 similarly, that's true with voltages. 0:02:02.360,0:02:06.770 To understand that, let's make sure we[br]understand the difference between a branch 0:02:06.770,0:02:08.970 voltage, and a no voltage. 0:02:08.970,0:02:13.650 A branch voltage is the voltage[br]across a branch within a circuit. 0:02:13.650,0:02:17.629 In this case here,[br]we've got a 3V drop across this resistor. 0:02:18.790,0:02:22.695 Over here, we've got a 10V drop[br]across that resistor, similarly a 2V, 0:02:22.695,0:02:24.840 and a 12V drop across those resistors. 0:02:26.010,0:02:29.980 On the other hand,[br]if we wanted to talk about the voltage, 0:02:29.980,0:02:36.570 at a point in the circuit, we would need[br]to specify what we were referring it to. 0:02:36.570,0:02:43.540 So let's create,[br]we'll call this node here 0V, 0:02:43.540,0:02:49.170 and then as we come along this[br]branch we go up from 0, up 15V, 0:02:49.170,0:02:56.020 so at this point, the voltage there is[br]15V above 0, or above our reference. 0:02:56.020,0:03:01.108 Now as we traverse this branch here,[br]we drop 3V, 0:03:01.108,0:03:08.030 getting to a voltage here of 15-[br]3 = 12V above our reference. 0:03:08.030,0:03:10.715 Continuing on along here, 0:03:10.715,0:03:15.968 we drop another 10V to 2V[br]above our reference, and 0:03:15.968,0:03:23.790 then continuing on down across these[br]2V to back to our zero reference. 0:03:23.790,0:03:28.389 Thus the distinction a node voltage where[br]the voltage add a node is a voltage at 0:03:28.389,0:03:33.061 the node relative to some reference,[br]whereas the voltage across the branch is 0:03:33.061,0:03:36.498 just a drop across the single[br]element within the branch. 0:03:39.737,0:03:44.664 Up until now, as we've been analyzing[br]circuits, we've identified branch 0:03:44.664,0:03:49.020 currents and voltages, and[br]worked with those as our variables. 0:03:49.020,0:03:53.830 For example, we have a branch current[br]here, call it i1, we have another branch 0:03:53.830,0:03:59.580 current here, call it i2,[br]another here, i3, i4, 0:04:01.330,0:04:06.538 and perhaps referencing like that, i5,[br]and of course we know that i5 in this 0:04:06.538,0:04:11.860 case equals i0, but in order to analyse[br]this circuit using branch currence, 0:04:11.860,0:04:16.350 we would have five different variables. 0:04:16.350,0:04:20.406 And when using those variables we could[br]then write KDL and KCL equations, and 0:04:20.406,0:04:24.540 solve for any branch voltage or branch[br]current in that circuit we wanted to do. 0:04:27.049,0:04:32.599 The node voltage method involves,[br]rather than branch currents, 0:04:32.599,0:04:36.079 it involves our defining node voltages. 0:04:37.490,0:04:41.850 In order to do that, we need to specify or[br]to make the distinction between nodes and 0:04:41.850,0:04:43.470 extraordinary nodes. 0:04:43.470,0:04:47.516 A node is a point where two or[br]more branches are joined. 0:04:47.516,0:04:52.155 Here we've got a node, there's another[br]node here, there's another node there, and 0:04:52.155,0:04:56.353 here's a node, here's a node, and then[br]all the way along the bottom here is yet 0:04:56.353,0:04:57.180 another node. 0:04:57.180,0:05:00.972 So we have one, two, three, four. 0:05:00.972,0:05:02.907 I said that was, I made a mistake there. 0:05:02.907,0:05:06.030 This is all one node, five nodes. 0:05:06.030,0:05:11.680 So one, two, three, four,[br]five regular nodes, 0:05:11.680,0:05:17.270 and now we have this node here[br]where we have more than two, 0:05:17.270,0:05:21.610 we have three or[br]more branches coming to these. 0:05:21.610,0:05:26.340 Those types of nodes are referred to as[br]extraordinary nodes, or critical nodes. 0:05:26.340,0:05:29.610 In other words,[br]a node is a place where two or 0:05:29.610,0:05:32.880 more branches come together,[br]and extraordinary or 0:05:32.880,0:05:37.110 critical node is a place where three or[br]more branches come together. 0:05:37.110,0:05:45.220 In this case we have one, two,[br]three extraordinary or critical nodes. 0:05:47.510,0:05:52.710 Our approach is going to be then[br]to identify the critical nodes, 0:05:52.710,0:05:56.438 choose one of them as our reference, and 0:05:56.438,0:06:01.480 define variables at the other 0:06:01.480,0:06:05.870 two nodes, and 0:06:05.870,0:06:11.160 then with those voltages equation 0:06:11.160,0:06:17.780 at each of those critical nodes,[br]in terms of V1 and V2. 0:06:17.780,0:06:23.070 For example, let's start by summing[br]the currents leaving this node right here. 0:06:24.460,0:06:29.340 In terms of V1 and this,[br]our voltage source, 0:06:29.340,0:06:33.730 we can first of all note that,[br]the voltage at that node 0:06:33.730,0:06:38.850 is V0 volts above our reference. 0:06:39.880,0:06:41.680 Now, we can write an expression for 0:06:41.680,0:06:46.710 the current leaving this node by[br]taking the voltage of this side 0:06:46.710,0:06:52.120 minus the voltage at this side of those[br]resistors and dividing by the resistence. 0:06:52.120,0:06:54.900 In other words, we're specifying[br]the branch voltage across those two 0:06:54.900,0:06:57.770 resistors in terms of our node voltages,[br]or 0:06:57.770,0:07:05.226 thus we would write (V1-V0)/(R2+R3) 0:07:05.226,0:07:10.600 will be the current leaving our[br]first node going to the left. 0:07:12.270,0:07:15.487 Now let's do similarly for[br]the other two branches, and 0:07:15.487,0:07:19.799 sum those three currents together and[br]acknowledge that the sum of the three 0:07:19.799,0:07:24.460 currents leaving that node must[br]equal zero, just cuz current law. 0:07:24.460,0:07:29.689 So the current now leaving that node[br]coming downs through our one would be, 0:07:29.689,0:07:34.005 V1, the voltage at the top,[br]minus the voltage at this side, 0:07:34.005,0:07:37.260 which in this case is just 0,[br]divided by R1. 0:07:37.260,0:07:43.020 And, finally, the current leaving[br]node 1 going to the ride would be, 0:07:43.020,0:07:48.108 the voltage of the left hand side[br]of that resistor would be V1, 0:07:48.108,0:07:52.140 the voltage on the right[br]hand side would be V2, so 0:07:52.140,0:07:58.573 (V1-V2)/R4 represent the sum of[br]the three currents leaving that node, 0:07:58.573,0:08:02.428 and the sum of those three[br]things must equal 0. 0:08:02.428,0:08:05.420 Similarly, we write another KCl at node 2. 0:08:05.420,0:08:10.144 So this is node 1, and[br]then at node 2 we have, 0:08:10.144,0:08:14.628 the current leaving node[br]2 going this way is 0:08:14.628,0:08:19.596 going to be the voltage[br]at the right-hand side, 0:08:19.596,0:08:26.170 V2 minus the voltage of[br]the left-hand side V1 divided by R4. 0:08:26.170,0:08:30.809 Note right now, that the current[br]leaving node 2 going to the left 0:08:32.140,0:08:36.940 is equal, but of opposite sign to[br]the current leaving node 1, and 0:08:36.940,0:08:41.539 going to the right, and you'll notice[br]those two terms are the same in each of 0:08:41.539,0:08:43.740 these equations other than[br]the different bias sign. 0:08:44.790,0:08:49.526 This term in the first equation,[br](V1-V2)/R4, 0:08:49.526,0:08:54.474 and that term in the second[br]equation (V2-V1)/R4. 0:08:54.474,0:09:01.320 All right continuing on, we now add[br]the current leaving node 2 going down. 0:09:02.350,0:09:07.373 that current will be (V2/R5) + the current 0:09:07.373,0:09:11.783 leaving the node going in this direction, 0:09:11.783,0:09:18.887 well actually the current is going in so[br]we'll subtract -I0 = 0, 0:09:18.887,0:09:23.188 the sum of those three currents equal 0. 0:09:23.188,0:09:27.768 So here we have two[br]equations with two unknowns, 0:09:27.768,0:09:33.573 it becomes simply a matter of[br]algebra at this point to solve for 0:09:33.573,0:09:36.043 those two node voltages.