[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.26,0:00:03.22,Default,,0000,0000,0000,,>> In this video, we're\Ngoing to demonstrate
Dialogue: 0,0:00:03.22,0:00:07.84,Default,,0000,0000,0000,,the node analysis method of analyzing\Ncircuits in the Phasor Domain.
Dialogue: 0,0:00:07.84,0:00:09.81,Default,,0000,0000,0000,,You'll recall from\Nour previous discussions of
Dialogue: 0,0:00:09.81,0:00:13.02,Default,,0000,0000,0000,,node analysis that the first step\Nin analyzing circuits
Dialogue: 0,0:00:13.02,0:00:14.98,Default,,0000,0000,0000,,using the node voltage approach
Dialogue: 0,0:00:14.98,0:00:18.08,Default,,0000,0000,0000,,involves identifying\Nthe critical or essential nodes.
Dialogue: 0,0:00:18.08,0:00:20.12,Default,,0000,0000,0000,,In this circuit we have got one,
Dialogue: 0,0:00:20.12,0:00:23.76,Default,,0000,0000,0000,,two, three, four essential nodes.
Dialogue: 0,0:00:23.76,0:00:29.07,Default,,0000,0000,0000,,Again, an essential node is a node where\Nthree or more branches are connected.
Dialogue: 0,0:00:29.07,0:00:32.28,Default,,0000,0000,0000,,We choose one of these nodes\Nto then serve as
Dialogue: 0,0:00:32.28,0:00:34.02,Default,,0000,0000,0000,,our reference node where we by
Dialogue: 0,0:00:34.02,0:00:36.40,Default,,0000,0000,0000,,definition we're going to say\Nvoltage is equal to zero.
Dialogue: 0,0:00:36.40,0:00:41.51,Default,,0000,0000,0000,,In this case, I've selected this\Nnode here to be our reference node.
Dialogue: 0,0:00:41.51,0:00:44.46,Default,,0000,0000,0000,,So we'll set that voltage equal to zero.
Dialogue: 0,0:00:44.46,0:00:46.67,Default,,0000,0000,0000,,The other three nodes are
Dialogue: 0,0:00:46.67,0:00:50.13,Default,,0000,0000,0000,,the three essential nodes we will then\Nidentify or assign variables to them.
Dialogue: 0,0:00:50.13,0:00:52.88,Default,,0000,0000,0000,,Variable names. This one\Nwe're going to call V1.
Dialogue: 0,0:00:52.88,0:00:54.92,Default,,0000,0000,0000,,The voltage at this node will refer to as
Dialogue: 0,0:00:54.92,0:00:58.90,Default,,0000,0000,0000,,V2 and the voltage at this\Nnode here will refer to as V3.
Dialogue: 0,0:00:58.90,0:01:01.64,Default,,0000,0000,0000,,Node voltage analysis then has this writing
Dialogue: 0,0:01:01.64,0:01:06.64,Default,,0000,0000,0000,,Kirchhoff's Current Law or node equations\Nat each of these essential nodes.
Dialogue: 0,0:01:06.64,0:01:11.28,Default,,0000,0000,0000,,Summing the currents leaving\Neach of those nodes in terms of
Dialogue: 0,0:01:11.28,0:01:16.40,Default,,0000,0000,0000,,the node voltage variables and\Nthe circuit parameters in the circuit.
Dialogue: 0,0:01:16.40,0:01:17.69,Default,,0000,0000,0000,,So let's go ahead and do this.
Dialogue: 0,0:01:17.69,0:01:18.94,Default,,0000,0000,0000,,On this first node,
Dialogue: 0,0:01:18.94,0:01:24.56,Default,,0000,0000,0000,,let's write KCL at node one.
Dialogue: 0,0:01:24.56,0:01:29.54,Default,,0000,0000,0000,,The current leaving this node coming\Ndown in this direction here will be
Dialogue: 0,0:01:29.54,0:01:35.20,Default,,0000,0000,0000,,V1 minus the voltage at this node\Nminus the voltage at this point.
Dialogue: 0,0:01:35.20,0:01:37.43,Default,,0000,0000,0000,,The voltage here by definition is zero,
Dialogue: 0,0:01:37.43,0:01:41.00,Default,,0000,0000,0000,,so V1 minus 0. We won't even bother.
Dialogue: 0,0:01:41.00,0:01:45.26,Default,,0000,0000,0000,,Well, start let's write\Nit V1 minus 0 divided by
Dialogue: 0,0:01:45.26,0:01:50.24,Default,,0000,0000,0000,,the impedance there which is 3 minus J4.
Dialogue: 0,0:01:50.24,0:01:54.39,Default,,0000,0000,0000,,Plus the current leaving this\Nnode going in this direction,
Dialogue: 0,0:01:54.39,0:02:00.71,Default,,0000,0000,0000,,which is V1 minus V3 divided by
Dialogue: 0,0:02:00.71,0:02:04.25,Default,,0000,0000,0000,,the impedance here of two ohms plus
Dialogue: 0,0:02:04.25,0:02:07.91,Default,,0000,0000,0000,,the current leaving this node\Ngoing in this direction here.
Dialogue: 0,0:02:07.91,0:02:09.35,Default,,0000,0000,0000,,We need to be a little bit careful here.
Dialogue: 0,0:02:09.35,0:02:12.41,Default,,0000,0000,0000,,It's going to be the current\Nleaving this node going in
Dialogue: 0,0:02:12.41,0:02:16.24,Default,,0000,0000,0000,,that direction will be\Nthe current through that resistor.
Dialogue: 0,0:02:16.24,0:02:19.22,Default,,0000,0000,0000,,The current through that resistor\Nis equal to the voltage drop
Dialogue: 0,0:02:19.22,0:02:22.96,Default,,0000,0000,0000,,across that resistor divided\Nby the resistance of two ohms.
Dialogue: 0,0:02:22.96,0:02:28.56,Default,,0000,0000,0000,,The voltage drop across that resistor\Nis the voltage on this side which is V1
Dialogue: 0,0:02:28.56,0:02:30.50,Default,,0000,0000,0000,,minus the voltage on
Dialogue: 0,0:02:30.50,0:02:35.57,Default,,0000,0000,0000,,this side of the resistor and\Nhere's where we need to be careful.
Dialogue: 0,0:02:35.57,0:02:39.86,Default,,0000,0000,0000,,The voltage here in terms\Nof node voltages can be
Dialogue: 0,0:02:39.86,0:02:43.78,Default,,0000,0000,0000,,thought of as being starting here at V2.
Dialogue: 0,0:02:43.78,0:02:49.82,Default,,0000,0000,0000,,V2 we drop going from here to here\Ncrossing this voltage source of 12 volts.
Dialogue: 0,0:02:49.82,0:02:51.26,Default,,0000,0000,0000,,We drop 12 volts..
Dialogue: 0,0:02:51.26,0:02:58.12,Default,,0000,0000,0000,,So the voltage at this point then\Nis going to be V2 minus 12 volts.
Dialogue: 0,0:02:58.12,0:03:00.89,Default,,0000,0000,0000,,So then the current leaving\Nthis branch going in
Dialogue: 0,0:03:00.89,0:03:02.15,Default,,0000,0000,0000,,that direction will be the voltage of
Dialogue: 0,0:03:02.15,0:03:04.13,Default,,0000,0000,0000,,the left-hand side of that resistor which
Dialogue: 0,0:03:04.13,0:03:07.61,Default,,0000,0000,0000,,is V1 minus the voltage\Nof the right-hand side of
Dialogue: 0,0:03:07.61,0:03:12.39,Default,,0000,0000,0000,,the resistor which is V2 minus 12.
Dialogue: 0,0:03:12.39,0:03:15.35,Default,,0000,0000,0000,,We throw in the parentheses there\Nto remind ourselves that we're
Dialogue: 0,0:03:15.35,0:03:20.03,Default,,0000,0000,0000,,subtracting the voltage\Nthere which is V2 minus 12.
Dialogue: 0,0:03:20.03,0:03:24.36,Default,,0000,0000,0000,,Finally, the current in that branch then\Nwill be that voltage drop divided by
Dialogue: 0,0:03:24.36,0:03:29.26,Default,,0000,0000,0000,,two and that equals zero.
Dialogue: 0,0:03:29.26,0:03:32.92,Default,,0000,0000,0000,,Now, KCL at node two.
Dialogue: 0,0:03:32.92,0:03:37.40,Default,,0000,0000,0000,,The current leaving this node\Ngoing to the left, again,
Dialogue: 0,0:03:37.40,0:03:39.08,Default,,0000,0000,0000,,is going to be the current through\Nthis resistor only
Dialogue: 0,0:03:39.08,0:03:40.76,Default,,0000,0000,0000,,this time going from right to
Dialogue: 0,0:03:40.76,0:03:44.43,Default,,0000,0000,0000,,left and that voltage\Nwill be the voltage here,
Dialogue: 0,0:03:44.43,0:03:49.79,Default,,0000,0000,0000,,which we've already established\Nto be V2 minus 12 volts minus
Dialogue: 0,0:03:49.79,0:03:56.00,Default,,0000,0000,0000,,the voltage over here which is V1\Nminus V1 divided by the two ohms.
Dialogue: 0,0:03:56.00,0:03:59.30,Default,,0000,0000,0000,,The current coming down to the\Nnode or leaving this node coming
Dialogue: 0,0:03:59.30,0:04:02.98,Default,,0000,0000,0000,,down is going to be then\Nadded in and that's
Dialogue: 0,0:04:02.98,0:04:11.10,Default,,0000,0000,0000,,Simply V2 divided by the impedance\Nof two plus one J or simply J.
Dialogue: 0,0:04:11.12,0:04:16.76,Default,,0000,0000,0000,,Then the current leaving node to\Ngoing to the right will be equal
Dialogue: 0,0:04:16.76,0:04:22.98,Default,,0000,0000,0000,,to V2 minus V3 divided by two ohms.
Dialogue: 0,0:04:22.98,0:04:25.24,Default,,0000,0000,0000,,The sum of those three currents equal zero.
Dialogue: 0,0:04:25.24,0:04:30.38,Default,,0000,0000,0000,,Now, at the third node we've\Ngot KCL at node three.
Dialogue: 0,0:04:30.38,0:04:34.16,Default,,0000,0000,0000,,The current leaving node three and\Ncome into the left is going to be
Dialogue: 0,0:04:34.16,0:04:39.38,Default,,0000,0000,0000,,V3 minus V2 divided by two ohms.
Dialogue: 0,0:04:39.38,0:04:44.27,Default,,0000,0000,0000,,Plus the current leaving this node going\Nup and over to the left is going to
Dialogue: 0,0:04:44.27,0:04:50.58,Default,,0000,0000,0000,,be V3 minus V1 divided by two ohms.
Dialogue: 0,0:04:50.58,0:04:53.99,Default,,0000,0000,0000,,Finally, the current leaving node\Nthree coming down this
Dialogue: 0,0:04:53.99,0:04:57.18,Default,,0000,0000,0000,,way is going to be the voltage drop\Nacross that 2-ohm resistor,
Dialogue: 0,0:04:57.18,0:05:03.01,Default,,0000,0000,0000,,which will be V3 minus\Nthe voltage right there.
Dialogue: 0,0:05:03.01,0:05:05.66,Default,,0000,0000,0000,,Now, the voltage at that point\Nis equal to starting here
Dialogue: 0,0:05:05.66,0:05:08.27,Default,,0000,0000,0000,,the reference node zero minus to plus
Dialogue: 0,0:05:08.27,0:05:14.57,Default,,0000,0000,0000,,we go up a value of negative J6 volts.
Dialogue: 0,0:05:14.57,0:05:18.24,Default,,0000,0000,0000,,So the voltage right here then is
Dialogue: 0,0:05:18.24,0:05:24.66,Default,,0000,0000,0000,,zero minus J6 and then\Nthe current through here will be V3
Dialogue: 0,0:05:24.66,0:05:29.85,Default,,0000,0000,0000,,minus a minus J6
Dialogue: 0,0:05:30.68,0:05:37.37,Default,,0000,0000,0000,,divided by the two ohms and the sum\Nof those three currents equal zero.
Dialogue: 0,0:05:37.37,0:05:40.08,Default,,0000,0000,0000,,Those are the three note note equations.
Dialogue: 0,0:05:40.08,0:05:42.78,Default,,0000,0000,0000,,All that's left now is to\Ncalculate the node voltages V1,
Dialogue: 0,0:05:42.78,0:05:44.82,Default,,0000,0000,0000,,V2 and V3 is a little bit of algebra.
Dialogue: 0,0:05:44.82,0:05:47.70,Default,,0000,0000,0000,,Let's start by combining like terms.
Dialogue: 0,0:05:47.70,0:05:49.18,Default,,0000,0000,0000,,Let's do that on up here.
Dialogue: 0,0:05:49.18,0:05:51.42,Default,,0000,0000,0000,,The V1 terms in the first equation.
Dialogue: 0,0:05:51.42,0:05:53.85,Default,,0000,0000,0000,,I've got V1 over.
Dialogue: 0,0:05:53.85,0:05:57.92,Default,,0000,0000,0000,,Let's see 3 minus J4.
Dialogue: 0,0:05:57.92,0:06:01.07,Default,,0000,0000,0000,,I've got V1 over two and I've\Ngot another V1 over two.
Dialogue: 0,0:06:01.07,0:06:09.24,Default,,0000,0000,0000,,So plus one over two plus one\Nover two gives us the V1 terms.
Dialogue: 0,0:06:09.24,0:06:13.74,Default,,0000,0000,0000,,Plus the V2 terms here we've
Dialogue: 0,0:06:13.74,0:06:19.70,Default,,0000,0000,0000,,got only one right here and\Nthat's a minus V2 over two.
Dialogue: 0,0:06:19.70,0:06:21.65,Default,,0000,0000,0000,,So if we pull the V2 out we'll have
Dialogue: 0,0:06:21.65,0:06:28.98,Default,,0000,0000,0000,,a minus one half there and\Nthen our V3 terms there.
Dialogue: 0,0:06:28.98,0:06:30.15,Default,,0000,0000,0000,,Again, is only one of them.
Dialogue: 0,0:06:30.15,0:06:31.91,Default,,0000,0000,0000,,It's a minus V3 over two.
Dialogue: 0,0:06:31.91,0:06:37.32,Default,,0000,0000,0000,,So we have V3 times a negative one half.
Dialogue: 0,0:06:37.72,0:06:39.86,Default,,0000,0000,0000,,That then will equal.
Dialogue: 0,0:06:39.86,0:06:42.53,Default,,0000,0000,0000,,Let's see, we've got a 12 volts here.
Dialogue: 0,0:06:42.53,0:06:44.68,Default,,0000,0000,0000,,12 divided by two is six.
Dialogue: 0,0:06:44.68,0:06:48.62,Default,,0000,0000,0000,,A minus times a minus is a plus.
Dialogue: 0,0:06:48.62,0:06:50.66,Default,,0000,0000,0000,,I've got plus six on the left-hand side.
Dialogue: 0,0:06:50.66,0:06:51.92,Default,,0000,0000,0000,,We subtract six from
Dialogue: 0,0:06:51.92,0:06:56.84,Default,,0000,0000,0000,,both sides to bring our constant\Nover there on the right-hand side.
Dialogue: 0,0:06:56.84,0:06:59.69,Default,,0000,0000,0000,,Now, combining like terms\Nin the second equation,
Dialogue: 0,0:06:59.69,0:07:03.90,Default,,0000,0000,0000,,we have V1 times.
Dialogue: 0,0:07:03.90,0:07:07.94,Default,,0000,0000,0000,,This one we've only got one term that's\Ngoing to be a negative V1 over two.
Dialogue: 0,0:07:07.94,0:07:10.70,Default,,0000,0000,0000,,So we pull the V out or\Nthe V1 out or left with
Dialogue: 0,0:07:10.70,0:07:15.93,Default,,0000,0000,0000,,the negative one half plus V2 times.
Dialogue: 0,0:07:15.93,0:07:17.98,Default,,0000,0000,0000,,I've got V2 over two,
Dialogue: 0,0:07:17.98,0:07:21.27,Default,,0000,0000,0000,,V2 over two plus J and V2 over two.
Dialogue: 0,0:07:21.27,0:07:30.64,Default,,0000,0000,0000,,So V2 or let's see V2 would be one half
Dialogue: 0,0:07:30.64,0:07:41.38,Default,,0000,0000,0000,,plus one over two plus J plus another\None half and then the V3 terms,
Dialogue: 0,0:07:41.38,0:07:47.10,Default,,0000,0000,0000,,again, there's only one and that\Nwill be a negative one half.
Dialogue: 0,0:07:47.10,0:07:50.57,Default,,0000,0000,0000,,Now, here we've got a minus 12 over two.
Dialogue: 0,0:07:50.57,0:07:53.48,Default,,0000,0000,0000,,So I've got a negative six volts\Non the left-hand side.
Dialogue: 0,0:07:53.48,0:07:57.35,Default,,0000,0000,0000,,We add six volts to plus\Nto our both sides and that
Dialogue: 0,0:07:57.35,0:08:02.26,Default,,0000,0000,0000,,gives us then equaling positive six volts\Non the right-hand side.
Dialogue: 0,0:08:02.26,0:08:12.51,Default,,0000,0000,0000,,Now, equation three, we have V1\Ntimes a negative one half plus
Dialogue: 0,0:08:12.51,0:08:18.04,Default,,0000,0000,0000,,V2 times got a negative one half again
Dialogue: 0,0:08:18.59,0:08:26.91,Default,,0000,0000,0000,,plus V3 times I've got one half\Nplus one half plus one half.
Dialogue: 0,0:08:26.91,0:08:30.96,Default,,0000,0000,0000,,That's three halves, V3.
Dialogue: 0,0:08:30.96,0:08:33.89,Default,,0000,0000,0000,,Now, here we've got a constant\Nof negative times negative is
Dialogue: 0,0:08:33.89,0:08:36.76,Default,,0000,0000,0000,,a positive J6 over two.
Dialogue: 0,0:08:36.76,0:08:40.40,Default,,0000,0000,0000,,So on the left-hand side\Nwe have a positive J3,
Dialogue: 0,0:08:40.40,0:08:44.03,Default,,0000,0000,0000,,which we subtract J3\Nfrom both sides gives us
Dialogue: 0,0:08:44.03,0:08:49.26,Default,,0000,0000,0000,,then a negative J3 over here on this side.
Dialogue: 0,0:08:49.26,0:08:52.52,Default,,0000,0000,0000,,All right. Moving ahead now\Njust cleaning things up
Dialogue: 0,0:08:52.52,0:08:55.54,Default,,0000,0000,0000,,so that we're ready to put\Nit into a matrix solver.
Dialogue: 0,0:08:55.54,0:08:59.18,Default,,0000,0000,0000,,Let's combine all those fractions\Nhere and when we do
Dialogue: 0,0:08:59.18,0:09:04.92,Default,,0000,0000,0000,,that we've got V1 times 1.12
Dialogue: 0,0:09:04.92,0:09:13.92,Default,,0000,0000,0000,,plus 0.16J plus V2 times a negative
Dialogue: 0,0:09:13.92,0:09:18.87,Default,,0000,0000,0000,,0.5 plus V3 times a
Dialogue: 0,0:09:18.87,0:09:24.82,Default,,0000,0000,0000,,negative 0.5 equals negative six volts.
Dialogue: 0,0:09:24.82,0:09:35.27,Default,,0000,0000,0000,,The second equation we have V1\Ntimes a negative 0.5 plus V2 times
Dialogue: 0,0:09:35.27,0:09:43.44,Default,,0000,0000,0000,,1.4 minus 0.2J plus
Dialogue: 0,0:09:43.44,0:09:51.24,Default,,0000,0000,0000,,V3 times a negative 0.5\Nequals a positive six.
Dialogue: 0,0:09:51.24,0:09:53.74,Default,,0000,0000,0000,,Finally, the third equation,
Dialogue: 0,0:09:53.74,0:10:00.69,Default,,0000,0000,0000,,V1 times a negative 0.5\Nplus V2 times a negative
Dialogue: 0,0:10:00.69,0:10:05.82,Default,,0000,0000,0000,,0.5 plus V3 times
Dialogue: 0,0:10:05.82,0:10:12.16,Default,,0000,0000,0000,,1.5 is equal to a negative J3.
Dialogue: 0,0:10:55.07,0:10:58.92,Default,,0000,0000,0000,,When you plug this into\Na matrix solver of some sort,
Dialogue: 0,0:10:58.92,0:11:05.97,Default,,0000,0000,0000,,you come up with that V1 is equal\Nto in rectangular coordinates,
Dialogue: 0,0:11:05.97,0:11:14.82,Default,,0000,0000,0000,,a negative 4.72 minus 0.88J,
Dialogue: 0,0:11:14.82,0:11:22.42,Default,,0000,0000,0000,,which in polar coordinates\Nis 4.8 angle negative 169.5.
Dialogue: 0,0:11:22.42,0:11:32.68,Default,,0000,0000,0000,,V2 is equal to 2.46 minus 0.87J,
Dialogue: 0,0:11:32.68,0:11:43.05,Default,,0000,0000,0000,,which in polar coordinates is 2.6\Nangle negative 19.84 and V3 we
Dialogue: 0,0:11:43.05,0:11:49.32,Default,,0000,0000,0000,,find to be equal to negative 0.76 minus
Dialogue: 0,0:11:49.32,0:11:54.23,Default,,0000,0000,0000,,J2.59 in rectangular coordinates in polar
Dialogue: 0,0:11:54.23,0:12:01.88,Default,,0000,0000,0000,,coordinates it's 2.7 angle negative 106.29.
Dialogue: 0,0:12:02.25,0:12:06.73,Default,,0000,0000,0000,,Of course, are all degrees.
Dialogue: 0,0:12:07.13,0:12:10.19,Default,,0000,0000,0000,,Now, that we have the voltages\Nat each of the nodes we can
Dialogue: 0,0:12:10.19,0:12:13.22,Default,,0000,0000,0000,,calculate any current or voltage\Nthat we might want to calculate.
Dialogue: 0,0:12:13.22,0:12:15.60,Default,,0000,0000,0000,,For example, let's determine
Dialogue: 0,0:12:15.60,0:12:19.72,Default,,0000,0000,0000,,the current leaving node two\Ncoming down this branch here.
Dialogue: 0,0:12:19.72,0:12:24.39,Default,,0000,0000,0000,,Well, it's called in this circuit\Nhere which has been identified
Dialogue: 0,0:12:24.39,0:12:29.05,Default,,0000,0000,0000,,as I5 or I sub L. So, I5 then,
Dialogue: 0,0:12:29.05,0:12:35.25,Default,,0000,0000,0000,,is equal to V2 divided by two plus J,
Dialogue: 0,0:12:35.25,0:12:40.62,Default,,0000,0000,0000,,which is equal to V2 is 2.6
Dialogue: 0,0:12:40.62,0:12:47.24,Default,,0000,0000,0000,,angle negative 19.84 divided by two plus J.
Dialogue: 0,0:12:47.24,0:12:52.86,Default,,0000,0000,0000,,We find then if that is equal\Nto in polar coordinates 1.1.
Dialogue: 0,0:12:52.86,0:13:00.09,Default,,0000,0000,0000,,six angle negative 46.4 degrees.