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Dialogue: 0,0:00:01.96,0:00:04.94,Default,,0000,0000,0000,,Welcome to an Introduction to\NElectrical & Computer Engineering.
Dialogue: 0,0:00:04.94,0:00:06.30,Default,,0000,0000,0000,,My name is Lee Brinton,
Dialogue: 0,0:00:06.30,0:00:11.15,Default,,0000,0000,0000,,I'm an electrical engineering instructor\Nat Salt Lake Community College.
Dialogue: 0,0:00:11.15,0:00:14.73,Default,,0000,0000,0000,,In this video we'll be talking about\Nways to analyze circuits using the node
Dialogue: 0,0:00:14.73,0:00:16.22,Default,,0000,0000,0000,,voltage method.
Dialogue: 0,0:00:16.22,0:00:20.81,Default,,0000,0000,0000,,We're gonna start by making a, looking at\Nthe similarities between elevation and
Dialogue: 0,0:00:20.81,0:00:21.94,Default,,0000,0000,0000,,voltage.
Dialogue: 0,0:00:21.94,0:00:26.10,Default,,0000,0000,0000,,We'll then discuss the difference between\Nbranch voltages and node voltages.
Dialogue: 0,0:00:26.10,0:00:30.92,Default,,0000,0000,0000,,We'll introduce the concept of nodes and\Ncritical or extraordinary nodes.
Dialogue: 0,0:00:30.92,0:00:35.85,Default,,0000,0000,0000,,We'll then look at the actual process of\Nanalyzing circuits using node voltages,
Dialogue: 0,0:00:35.85,0:00:40.01,Default,,0000,0000,0000,,and we'll look at how that applies\Nwhen they are dependent sources and
Dialogue: 0,0:00:40.01,0:00:42.21,Default,,0000,0000,0000,,supernodes present in the circuit.
Dialogue: 0,0:00:43.69,0:00:46.71,Default,,0000,0000,0000,,First of all,\Nthe similarities between elevation.
Dialogue: 0,0:00:46.71,0:00:49.42,Default,,0000,0000,0000,,If we wanna talk about the elevation of,\Nsay Mount Olympus,
Dialogue: 0,0:00:49.42,0:00:53.18,Default,,0000,0000,0000,,we've got to define first of all\Nwhat we're measuring relative to.
Dialogue: 0,0:00:53.18,0:00:55.97,Default,,0000,0000,0000,,In other words,\Nwe establish our reference.
Dialogue: 0,0:00:55.97,0:01:00.41,Default,,0000,0000,0000,,Typically sea level is our\Nreference at zero feet.
Dialogue: 0,0:01:00.41,0:01:03.66,Default,,0000,0000,0000,,Mount Olympus then is about\N9,500 feet above sea level,
Dialogue: 0,0:01:03.66,0:01:08.03,Default,,0000,0000,0000,,and the Salt Lake Valley Floor is\Naround 4,100 feet above sea level.
Dialogue: 0,0:01:09.55,0:01:12.63,Default,,0000,0000,0000,,On the other hand,\Nif we're standing on the Valley Floor, and
Dialogue: 0,0:01:12.63,0:01:16.29,Default,,0000,0000,0000,,looking up to the mountains to the east,\Nwe'll see that the elevation,
Dialogue: 0,0:01:16.29,0:01:20.06,Default,,0000,0000,0000,,the amount that they rise, can be\Ncalculated by taking the higher elevation,
Dialogue: 0,0:01:20.06,0:01:27.19,Default,,0000,0000,0000,,9,500 feet less the lower elevation\Nof 4,100 feet, and we've got a 5,400
Dialogue: 0,0:01:27.19,0:01:31.40,Default,,0000,0000,0000,,foot change in elevation going from the\NValley Floor to the top of Mount Olympus.
Dialogue: 0,0:01:32.55,0:01:35.95,Default,,0000,0000,0000,,On the other hand, if instead of\Ncalling sea level our reference,
Dialogue: 0,0:01:35.95,0:01:44.25,Default,,0000,0000,0000,,we made the Valley Floor our reference and\Nsaid then the elevation here = 0 feet,
Dialogue: 0,0:01:44.25,0:01:49.24,Default,,0000,0000,0000,,then Mount Olympus would be 5,400 feet,
Dialogue: 0,0:01:49.24,0:01:54.20,Default,,0000,0000,0000,,and sea level would be -4,100 feet.
Dialogue: 0,0:01:54.20,0:01:57.78,Default,,0000,0000,0000,,Elevations are all relative\Nto a reference, and
Dialogue: 0,0:01:57.78,0:02:01.09,Default,,0000,0000,0000,,similarly, that's true with voltages.
Dialogue: 0,0:02:02.36,0:02:06.77,Default,,0000,0000,0000,,To understand that, let's make sure we\Nunderstand the difference between a branch
Dialogue: 0,0:02:06.77,0:02:08.97,Default,,0000,0000,0000,,voltage, and a no voltage.
Dialogue: 0,0:02:08.97,0:02:13.65,Default,,0000,0000,0000,,A branch voltage is the voltage\Nacross a branch within a circuit.
Dialogue: 0,0:02:13.65,0:02:17.63,Default,,0000,0000,0000,,In this case here,\Nwe've got a 3V drop across this resistor.
Dialogue: 0,0:02:18.79,0:02:22.70,Default,,0000,0000,0000,,Over here, we've got a 10V drop\Nacross that resistor, similarly a 2V,
Dialogue: 0,0:02:22.70,0:02:24.84,Default,,0000,0000,0000,,and a 12V drop across those resistors.
Dialogue: 0,0:02:26.01,0:02:29.98,Default,,0000,0000,0000,,On the other hand,\Nif we wanted to talk about the voltage,
Dialogue: 0,0:02:29.98,0:02:36.57,Default,,0000,0000,0000,,at a point in the circuit, we would need\Nto specify what we were referring it to.
Dialogue: 0,0:02:36.57,0:02:43.54,Default,,0000,0000,0000,,So let's create,\Nwe'll call this node here 0V,
Dialogue: 0,0:02:43.54,0:02:49.17,Default,,0000,0000,0000,,and then as we come along this\Nbranch we go up from 0, up 15V,
Dialogue: 0,0:02:49.17,0:02:56.02,Default,,0000,0000,0000,,so at this point, the voltage there is\N15V above 0, or above our reference.
Dialogue: 0,0:02:56.02,0:03:01.11,Default,,0000,0000,0000,,Now as we traverse this branch here,\Nwe drop 3V,
Dialogue: 0,0:03:01.11,0:03:08.03,Default,,0000,0000,0000,,getting to a voltage here of 15-\N3 = 12V above our reference.
Dialogue: 0,0:03:08.03,0:03:10.72,Default,,0000,0000,0000,,Continuing on along here,
Dialogue: 0,0:03:10.72,0:03:15.97,Default,,0000,0000,0000,,we drop another 10V to 2V\Nabove our reference, and
Dialogue: 0,0:03:15.97,0:03:23.79,Default,,0000,0000,0000,,then continuing on down across these\N2V to back to our zero reference.
Dialogue: 0,0:03:23.79,0:03:28.39,Default,,0000,0000,0000,,Thus the distinction a node voltage where\Nthe voltage add a node is a voltage at
Dialogue: 0,0:03:28.39,0:03:33.06,Default,,0000,0000,0000,,the node relative to some reference,\Nwhereas the voltage across the branch is
Dialogue: 0,0:03:33.06,0:03:36.50,Default,,0000,0000,0000,,just a drop across the single\Nelement within the branch.
Dialogue: 0,0:03:39.74,0:03:44.66,Default,,0000,0000,0000,,Up until now, as we've been analyzing\Ncircuits, we've identified branch
Dialogue: 0,0:03:44.66,0:03:49.02,Default,,0000,0000,0000,,currents and voltages, and\Nworked with those as our variables.
Dialogue: 0,0:03:49.02,0:03:53.83,Default,,0000,0000,0000,,For example, we have a branch current\Nhere, call it i1, we have another branch
Dialogue: 0,0:03:53.83,0:03:59.58,Default,,0000,0000,0000,,current here, call it i2,\Nanother here, i3, i4,
Dialogue: 0,0:04:01.33,0:04:06.54,Default,,0000,0000,0000,,and perhaps referencing like that, i5,\Nand of course we know that i5 in this
Dialogue: 0,0:04:06.54,0:04:11.86,Default,,0000,0000,0000,,case equals i0, but in order to analyse\Nthis circuit using branch currence,
Dialogue: 0,0:04:11.86,0:04:16.35,Default,,0000,0000,0000,,we would have five different variables.
Dialogue: 0,0:04:16.35,0:04:20.41,Default,,0000,0000,0000,,And when using those variables we could\Nthen write KDL and KCL equations, and
Dialogue: 0,0:04:20.41,0:04:24.54,Default,,0000,0000,0000,,solve for any branch voltage or branch\Ncurrent in that circuit we wanted to do.
Dialogue: 0,0:04:27.05,0:04:32.60,Default,,0000,0000,0000,,The node voltage method involves,\Nrather than branch currents,
Dialogue: 0,0:04:32.60,0:04:36.08,Default,,0000,0000,0000,,it involves our defining node voltages.
Dialogue: 0,0:04:37.49,0:04:41.85,Default,,0000,0000,0000,,In order to do that, we need to specify or\Nto make the distinction between nodes and
Dialogue: 0,0:04:41.85,0:04:43.47,Default,,0000,0000,0000,,extraordinary nodes.
Dialogue: 0,0:04:43.47,0:04:47.52,Default,,0000,0000,0000,,A node is a point where two or\Nmore branches are joined.
Dialogue: 0,0:04:47.52,0:04:52.16,Default,,0000,0000,0000,,Here we've got a node, there's another\Nnode here, there's another node there, and
Dialogue: 0,0:04:52.16,0:04:56.35,Default,,0000,0000,0000,,here's a node, here's a node, and then\Nall the way along the bottom here is yet
Dialogue: 0,0:04:56.35,0:04:57.18,Default,,0000,0000,0000,,another node.
Dialogue: 0,0:04:57.18,0:05:00.97,Default,,0000,0000,0000,,So we have one, two, three, four.
Dialogue: 0,0:05:00.97,0:05:02.91,Default,,0000,0000,0000,,I said that was, I made a mistake there.
Dialogue: 0,0:05:02.91,0:05:06.03,Default,,0000,0000,0000,,This is all one node, five nodes.
Dialogue: 0,0:05:06.03,0:05:11.68,Default,,0000,0000,0000,,So one, two, three, four,\Nfive regular nodes,
Dialogue: 0,0:05:11.68,0:05:17.27,Default,,0000,0000,0000,,and now we have this node here\Nwhere we have more than two,
Dialogue: 0,0:05:17.27,0:05:21.61,Default,,0000,0000,0000,,we have three or\Nmore branches coming to these.
Dialogue: 0,0:05:21.61,0:05:26.34,Default,,0000,0000,0000,,Those types of nodes are referred to as\Nextraordinary nodes, or critical nodes.
Dialogue: 0,0:05:26.34,0:05:29.61,Default,,0000,0000,0000,,In other words,\Na node is a place where two or
Dialogue: 0,0:05:29.61,0:05:32.88,Default,,0000,0000,0000,,more branches come together,\Nand extraordinary or
Dialogue: 0,0:05:32.88,0:05:37.11,Default,,0000,0000,0000,,critical node is a place where three or\Nmore branches come together.
Dialogue: 0,0:05:37.11,0:05:45.22,Default,,0000,0000,0000,,In this case we have one, two,\Nthree extraordinary or critical nodes.
Dialogue: 0,0:05:47.51,0:05:52.71,Default,,0000,0000,0000,,Our approach is going to be then\Nto identify the critical nodes,
Dialogue: 0,0:05:52.71,0:05:56.44,Default,,0000,0000,0000,,choose one of them as our reference, and
Dialogue: 0,0:05:56.44,0:06:01.48,Default,,0000,0000,0000,,define variables at the other
Dialogue: 0,0:06:01.48,0:06:05.87,Default,,0000,0000,0000,,two nodes, and
Dialogue: 0,0:06:05.87,0:06:11.16,Default,,0000,0000,0000,,then with those voltages equation
Dialogue: 0,0:06:11.16,0:06:17.78,Default,,0000,0000,0000,,at each of those critical nodes,\Nin terms of V1 and V2.
Dialogue: 0,0:06:17.78,0:06:23.07,Default,,0000,0000,0000,,For example, let's start by summing\Nthe currents leaving this node right here.
Dialogue: 0,0:06:24.46,0:06:29.34,Default,,0000,0000,0000,,In terms of V1 and this,\Nour voltage source,
Dialogue: 0,0:06:29.34,0:06:33.73,Default,,0000,0000,0000,,we can first of all note that,\Nthe voltage at that node
Dialogue: 0,0:06:33.73,0:06:38.85,Default,,0000,0000,0000,,is V0 volts above our reference.
Dialogue: 0,0:06:39.88,0:06:41.68,Default,,0000,0000,0000,,Now, we can write an expression for
Dialogue: 0,0:06:41.68,0:06:46.71,Default,,0000,0000,0000,,the current leaving this node by\Ntaking the voltage of this side
Dialogue: 0,0:06:46.71,0:06:52.12,Default,,0000,0000,0000,,minus the voltage at this side of those\Nresistors and dividing by the resistence.
Dialogue: 0,0:06:52.12,0:06:54.90,Default,,0000,0000,0000,,In other words, we're specifying\Nthe branch voltage across those two
Dialogue: 0,0:06:54.90,0:06:57.77,Default,,0000,0000,0000,,resistors in terms of our node voltages,\Nor
Dialogue: 0,0:06:57.77,0:07:05.23,Default,,0000,0000,0000,,thus we would write (V1-V0)/(R2+R3)
Dialogue: 0,0:07:05.23,0:07:10.60,Default,,0000,0000,0000,,will be the current leaving our\Nfirst node going to the left.
Dialogue: 0,0:07:12.27,0:07:15.49,Default,,0000,0000,0000,,Now let's do similarly for\Nthe other two branches, and
Dialogue: 0,0:07:15.49,0:07:19.80,Default,,0000,0000,0000,,sum those three currents together and\Nacknowledge that the sum of the three
Dialogue: 0,0:07:19.80,0:07:24.46,Default,,0000,0000,0000,,currents leaving that node must\Nequal zero, just cuz current law.
Dialogue: 0,0:07:24.46,0:07:29.69,Default,,0000,0000,0000,,So the current now leaving that node\Ncoming downs through our one would be,
Dialogue: 0,0:07:29.69,0:07:34.00,Default,,0000,0000,0000,,V1, the voltage at the top,\Nminus the voltage at this side,
Dialogue: 0,0:07:34.00,0:07:37.26,Default,,0000,0000,0000,,which in this case is just 0,\Ndivided by R1.
Dialogue: 0,0:07:37.26,0:07:43.02,Default,,0000,0000,0000,,And, finally, the current leaving\Nnode 1 going to the ride would be,
Dialogue: 0,0:07:43.02,0:07:48.11,Default,,0000,0000,0000,,the voltage of the left hand side\Nof that resistor would be V1,
Dialogue: 0,0:07:48.11,0:07:52.14,Default,,0000,0000,0000,,the voltage on the right\Nhand side would be V2, so
Dialogue: 0,0:07:52.14,0:07:58.57,Default,,0000,0000,0000,,(V1-V2)/R4 represent the sum of\Nthe three currents leaving that node,
Dialogue: 0,0:07:58.57,0:08:02.43,Default,,0000,0000,0000,,and the sum of those three\Nthings must equal 0.
Dialogue: 0,0:08:02.43,0:08:05.42,Default,,0000,0000,0000,,Similarly, we write another KCl at node 2.
Dialogue: 0,0:08:05.42,0:08:10.14,Default,,0000,0000,0000,,So this is node 1, and\Nthen at node 2 we have,
Dialogue: 0,0:08:10.14,0:08:14.63,Default,,0000,0000,0000,,the current leaving node\N2 going this way is
Dialogue: 0,0:08:14.63,0:08:19.60,Default,,0000,0000,0000,,going to be the voltage\Nat the right-hand side,
Dialogue: 0,0:08:19.60,0:08:26.17,Default,,0000,0000,0000,,V2 minus the voltage of\Nthe left-hand side V1 divided by R4.
Dialogue: 0,0:08:26.17,0:08:30.81,Default,,0000,0000,0000,,Note right now, that the current\Nleaving node 2 going to the left
Dialogue: 0,0:08:32.14,0:08:36.94,Default,,0000,0000,0000,,is equal, but of opposite sign to\Nthe current leaving node 1, and
Dialogue: 0,0:08:36.94,0:08:41.54,Default,,0000,0000,0000,,going to the right, and you'll notice\Nthose two terms are the same in each of
Dialogue: 0,0:08:41.54,0:08:43.74,Default,,0000,0000,0000,,these equations other than\Nthe different bias sign.
Dialogue: 0,0:08:44.79,0:08:49.53,Default,,0000,0000,0000,,This term in the first equation,\N(V1-V2)/R4,
Dialogue: 0,0:08:49.53,0:08:54.47,Default,,0000,0000,0000,,and that term in the second\Nequation (V2-V1)/R4.
Dialogue: 0,0:08:54.47,0:09:01.32,Default,,0000,0000,0000,,All right continuing on, we now add\Nthe current leaving node 2 going down.
Dialogue: 0,0:09:02.35,0:09:07.37,Default,,0000,0000,0000,,that current will be (V2/R5) + the current
Dialogue: 0,0:09:07.37,0:09:11.78,Default,,0000,0000,0000,,leaving the node going in this direction,
Dialogue: 0,0:09:11.78,0:09:18.89,Default,,0000,0000,0000,,well actually the current is going in so\Nwe'll subtract -I0 = 0,
Dialogue: 0,0:09:18.89,0:09:23.19,Default,,0000,0000,0000,,the sum of those three currents equal 0.
Dialogue: 0,0:09:23.19,0:09:27.77,Default,,0000,0000,0000,,So here we have two\Nequations with two unknowns,
Dialogue: 0,0:09:27.77,0:09:33.57,Default,,0000,0000,0000,,it becomes simply a matter of\Nalgebra at this point to solve for
Dialogue: 0,0:09:33.57,0:09:36.04,Default,,0000,0000,0000,,those two node voltages.