[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.43,0:00:04.56,Default,,0000,0000,0000,,All right, here's another example of\Nfinding the Thevenin equivalent of
Dialogue: 0,0:00:04.56,0:00:06.39,Default,,0000,0000,0000,,this circuit.
Dialogue: 0,0:00:06.39,0:00:09.15,Default,,0000,0000,0000,,Again, we have two values\Nthat we need to find.
Dialogue: 0,0:00:09.15,0:00:11.75,Default,,0000,0000,0000,,We need to determine\Nthe open circuit voltage,
Dialogue: 0,0:00:11.75,0:00:17.58,Default,,0000,0000,0000,,which is the voltage,\NAcross there with no load connected.
Dialogue: 0,0:00:17.58,0:00:21.66,Default,,0000,0000,0000,,And then we also need to determine\Nthe Thevenin equivalent resistance,
Dialogue: 0,0:00:21.66,0:00:24.34,Default,,0000,0000,0000,,which in this one we'll do\Na couple of different ways.
Dialogue: 0,0:00:24.34,0:00:27.40,Default,,0000,0000,0000,,All right, first of all, Vth open circuit.
Dialogue: 0,0:00:27.40,0:00:32.58,Default,,0000,0000,0000,,That's the voltage at this node,\Nnoting that the current
Dialogue: 0,0:00:32.58,0:00:38.30,Default,,0000,0000,0000,,leaving this node going in that direction,\Ncall it I, is = 0.
Dialogue: 0,0:00:40.00,0:00:44.68,Default,,0000,0000,0000,,So let's call this node here V sub A,\Nwe got V open circuit there and
Dialogue: 0,0:00:44.68,0:00:48.38,Default,,0000,0000,0000,,let's write the node\Nequations of these two nodes.
Dialogue: 0,0:00:49.68,0:00:54.77,Default,,0000,0000,0000,,Noting first of all, before we get\Nahead of ourselves, that this dependent
Dialogue: 0,0:00:54.77,0:01:00.34,Default,,0000,0000,0000,,current source is pushing current into the\Nnode going this direction with an amount
Dialogue: 0,0:01:00.34,0:01:06.08,Default,,0000,0000,0000,,of 1.2 times I0, where I0 is the current\Nflowing through the 3-ohm resistor.
Dialogue: 0,0:01:06.08,0:01:11.19,Default,,0000,0000,0000,,So before we get started, let's just note
Dialogue: 0,0:01:11.19,0:01:15.88,Default,,0000,0000,0000,,that I0 = 15-V sub A divided by 3.
Dialogue: 0,0:01:15.88,0:01:17.99,Default,,0000,0000,0000,,We're gonna need that\Nhere in just a second.
Dialogue: 0,0:01:17.99,0:01:21.11,Default,,0000,0000,0000,,All right, some of the current's leaving\NV sub A going in that direction.
Dialogue: 0,0:01:21.11,0:01:27.60,Default,,0000,0000,0000,,We have V sub A -15 divided by 3 +,\Nthe current coming down to
Dialogue: 0,0:01:27.60,0:01:32.65,Default,,0000,0000,0000,,the 6-ohm resistor is\Ngonna be VA divided by 6.
Dialogue: 0,0:01:32.65,0:01:37.59,Default,,0000,0000,0000,,The current leaving the sub A node\Ngoing through the 4-ohm resistor,
Dialogue: 0,0:01:37.59,0:01:42.69,Default,,0000,0000,0000,,is going to be VA, the voltage on this\Nside minus the voltage on that side,
Dialogue: 0,0:01:42.69,0:01:46.66,Default,,0000,0000,0000,,which is VOC, the open circuit voltage,\Ndivided by 4.
Dialogue: 0,0:01:46.66,0:01:53.62,Default,,0000,0000,0000,,Now we have coming into the node,\Nit's gonna be -1.2 times I0.
Dialogue: 0,0:01:53.62,0:02:01.12,Default,,0000,0000,0000,,But I0 we have already expressed\Nas 15 -VA divided by 3.
Dialogue: 0,0:02:01.12,0:02:05.99,Default,,0000,0000,0000,,So there's one, two, three, four branches,\None, two, three, four terms,
Dialogue: 0,0:02:05.99,0:02:08.68,Default,,0000,0000,0000,,the sum of those four\Nthings needs to be =0.
Dialogue: 0,0:02:08.68,0:02:14.27,Default,,0000,0000,0000,,Right in the node equation at the other\Nnode there which we're calling VOC,
Dialogue: 0,0:02:14.27,0:02:16.55,Default,,0000,0000,0000,,we have two nonzero currents.
Dialogue: 0,0:02:16.55,0:02:21.80,Default,,0000,0000,0000,,We've got the current leading\Nthis node going to the 4 ohm
Dialogue: 0,0:02:21.80,0:02:27.15,Default,,0000,0000,0000,,resistor in this direction is\NVoc-V sub A divided by 4 plus
Dialogue: 0,0:02:27.15,0:02:33.50,Default,,0000,0000,0000,,the current leaving it's not going\Nthat way which is 1.2 time I0.
Dialogue: 0,0:02:33.50,0:02:39.14,Default,,0000,0000,0000,,So +1.2 times I0 which\Nwe've already express as 15
Dialogue: 0,0:02:39.14,0:02:44.42,Default,,0000,0000,0000,,minus V sub A divided by 3,\Nas we've all ready noted,
Dialogue: 0,0:02:44.42,0:02:49.15,Default,,0000,0000,0000,,the current leaving going\Nthis direction is 0.
Dialogue: 0,0:02:49.15,0:02:53.53,Default,,0000,0000,0000,,So, the sum of those two\Ncurrents must equal 0.
Dialogue: 0,0:02:53.53,0:02:59.04,Default,,0000,0000,0000,,We're gonna skip the grimy algebra,
Dialogue: 0,0:02:59.04,0:03:05.06,Default,,0000,0000,0000,,but after you've simplified this down,
Dialogue: 0,0:03:05.06,0:03:10.05,Default,,0000,0000,0000,,you come up with VA (1.15) +
Dialogue: 0,0:03:10.05,0:03:14.37,Default,,0000,0000,0000,,VOC(-0.25) = 11.
Dialogue: 0,0:03:14.37,0:03:21.35,Default,,0000,0000,0000,,The top equation simplifies to that,
Dialogue: 0,0:03:21.35,0:03:27.90,Default,,0000,0000,0000,,and the second equation simplifies
Dialogue: 0,0:03:27.90,0:03:32.92,Default,,0000,0000,0000,,to VA(- 0.65 )+ Voc
Dialogue: 0,0:03:32.92,0:03:37.30,Default,,0000,0000,0000,,(0.25) = -6.
Dialogue: 0,0:03:37.30,0:03:41.22,Default,,0000,0000,0000,,And when you plug that in\Nyour matrix solver and
Dialogue: 0,0:03:41.22,0:03:46.36,Default,,0000,0000,0000,,however you do it, you can in\Nthat V sub A equals 10 volts and
Dialogue: 0,0:03:46.36,0:03:50.82,Default,,0000,0000,0000,,VOC which is our Thevenin\Nvoltage equals two volts.
Dialogue: 0,0:03:54.69,0:03:57.62,Default,,0000,0000,0000,,Now we have three different methods\Nof finding the Thevenin resistance.
Dialogue: 0,0:03:58.62,0:04:01.78,Default,,0000,0000,0000,,The first method would be to\Nshort out the terminals AB and
Dialogue: 0,0:04:01.78,0:04:03.96,Default,,0000,0000,0000,,determine the short circuit current, and
Dialogue: 0,0:04:03.96,0:04:07.16,Default,,0000,0000,0000,,then our Thevenin equals V open\Ncircuit divided by I short circuit.
Dialogue: 0,0:04:07.16,0:04:08.96,Default,,0000,0000,0000,,We're gonna do that here in just a minute.
Dialogue: 0,0:04:08.96,0:04:13.30,Default,,0000,0000,0000,,The second method only works when you\Nhave independent sources only, or
Dialogue: 0,0:04:13.30,0:04:15.14,Default,,0000,0000,0000,,when you only have independent sources.
Dialogue: 0,0:04:15.14,0:04:18.74,Default,,0000,0000,0000,,The fact that we have this\Ndependent source here makes it so
Dialogue: 0,0:04:18.74,0:04:20.87,Default,,0000,0000,0000,,that method 2 doesn't work.
Dialogue: 0,0:04:20.87,0:04:24.14,Default,,0000,0000,0000,,Method 3 is also gonna work,\Nwe'll do that here in just a second.
Dialogue: 0,0:04:24.14,0:04:27.31,Default,,0000,0000,0000,,Method 3 involves shorting\Nout this voltage or
Dialogue: 0,0:04:27.31,0:04:32.46,Default,,0000,0000,0000,,deactivating the voltage source and\Nthen applying a test voltage.
Dialogue: 0,0:04:32.46,0:04:36.55,Default,,0000,0000,0000,,So let's start with\Nthe short circuit method.
Dialogue: 0,0:04:36.55,0:04:42.70,Default,,0000,0000,0000,,Let's short this out,\Ncall that I short circuit and
Dialogue: 0,0:04:42.70,0:04:48.49,Default,,0000,0000,0000,,now determine the current\Nin that short circuit.
Dialogue: 0,0:04:48.49,0:04:50.17,Default,,0000,0000,0000,,We need to note a couple of things.
Dialogue: 0,0:04:50.17,0:04:54.28,Default,,0000,0000,0000,,First of all, as we short this out,
Dialogue: 0,0:04:54.28,0:04:59.49,Default,,0000,0000,0000,,we now know that\Nthe voltage there equals 0.
Dialogue: 0,0:04:59.49,0:05:03.55,Default,,0000,0000,0000,,So we only have one node where\Nwe have an unknown voltage.
Dialogue: 0,0:05:03.55,0:05:08.72,Default,,0000,0000,0000,,We're gonna calculate V sub A and\Nonce we know V sub A,
Dialogue: 0,0:05:08.72,0:05:15.30,Default,,0000,0000,0000,,we'll then write a node equation\Nat this node noting that V0 = 0.
Dialogue: 0,0:05:15.30,0:05:17.54,Default,,0000,0000,0000,,But the equation will then be\Nin terms of I short circuit,
Dialogue: 0,0:05:17.54,0:05:19.75,Default,,0000,0000,0000,,and we'll be able to\Ncalculate I short circuit.
Dialogue: 0,0:05:19.75,0:05:20.41,Default,,0000,0000,0000,,Let's go ahead and do it.
Dialogue: 0,0:05:20.41,0:05:24.98,Default,,0000,0000,0000,,It's probably easier to show it\Nthan it is to talk about it.
Dialogue: 0,0:05:24.98,0:05:28.20,Default,,0000,0000,0000,,So let's just work on over here.
Dialogue: 0,0:05:28.20,0:05:33.24,Default,,0000,0000,0000,,Write in the note equation at the A node,\Nwe have
Dialogue: 0,0:05:33.24,0:05:37.90,Default,,0000,0000,0000,,V sub A minus 15 divided by 3,
Dialogue: 0,0:05:37.90,0:05:43.44,Default,,0000,0000,0000,,plus V sub A divided by 6,\Nplus V sub A divided by four,
Dialogue: 0,0:05:43.44,0:05:49.40,Default,,0000,0000,0000,,minus 1.2 times,
Dialogue: 0,0:05:49.40,0:05:54.73,Default,,0000,0000,0000,,now notice that I zero, even though we've
Dialogue: 0,0:05:54.73,0:05:59.98,Default,,0000,0000,0000,,shorted this out I zero is still\N15 minus V sub A divided by three.
Dialogue: 0,0:05:59.98,0:06:07.58,Default,,0000,0000,0000,,So it would be minus one\Npoint two times (15-Va/3) and
Dialogue: 0,0:06:07.58,0:06:12.66,Default,,0000,0000,0000,,the sum of those terms has to equal zero.
Dialogue: 0,0:06:12.66,0:06:17.88,Default,,0000,0000,0000,,Now when you go through and\Nsimplify that and
Dialogue: 0,0:06:17.88,0:06:24.52,Default,,0000,0000,0000,,solve for V sub A you get\Nthat V sub A = 9.565 Volts.
Dialogue: 0,0:06:27.92,0:06:32.34,Default,,0000,0000,0000,,Now that we know V sub A with\Nthis output short circuited,
Dialogue: 0,0:06:32.34,0:06:36.50,Default,,0000,0000,0000,,we can go ahead and\Nwrite a node equation at this node.
Dialogue: 0,0:06:37.90,0:06:42.68,Default,,0000,0000,0000,,We'll do it in terms of V sub A,\Nbut A sub A is gonna equal 9.565.
Dialogue: 0,0:06:42.68,0:06:44.51,Default,,0000,0000,0000,,Let's just do it and\Nshow you what we get here.
Dialogue: 0,0:06:45.86,0:06:49.58,Default,,0000,0000,0000,,So we're going to have the current leaving\Nthis node going in that direction.
Dialogue: 0,0:06:49.58,0:06:54.99,Default,,0000,0000,0000,,It's going to be 0, oops.
Dialogue: 0,0:06:54.99,0:06:58.35,Default,,0000,0000,0000,,Yeah 0, the voltage there is 0.
Dialogue: 0,0:06:58.35,0:07:05.57,Default,,0000,0000,0000,,0- v sub a, which is 9.565 divided by 4.
Dialogue: 0,0:07:07.02,0:07:11.57,Default,,0000,0000,0000,,Plus the current going in this
Dialogue: 0,0:07:12.65,0:07:17.71,Default,,0000,0000,0000,,direction, which is 1.2 times I0,\NI0 is 15,
Dialogue: 0,0:07:17.71,0:07:23.25,Default,,0000,0000,0000,,minus V sub A, so that's going
Dialogue: 0,0:07:23.25,0:07:30.15,Default,,0000,0000,0000,,to be 15 minus 9.565, divided by 3.
Dialogue: 0,0:07:30.15,0:07:32.82,Default,,0000,0000,0000,,Then we have I short-circuit leaving.
Dialogue: 0,0:07:32.82,0:07:35.72,Default,,0000,0000,0000,,So it would be plus I\Nshort-circuit equals 0.
Dialogue: 0,0:07:35.72,0:07:40.16,Default,,0000,0000,0000,,Well let's just bring that short circuit\Non the other side as a equals minus I,
Dialogue: 0,0:07:40.16,0:07:42.92,Default,,0000,0000,0000,,short-circuit, running out of room here.
Dialogue: 0,0:07:42.92,0:07:47.91,Default,,0000,0000,0000,,All right,\Nlet's continue on now and solve for
Dialogue: 0,0:07:47.91,0:07:52.28,Default,,0000,0000,0000,,I short-circuit, and when you do that,
Dialogue: 0,0:07:52.28,0:07:58.04,Default,,0000,0000,0000,,you get I short circuit\Nis equal to 0.2173 Amps.
Dialogue: 0,0:07:58.04,0:08:01.71,Default,,0000,0000,0000,,We now know V open circuit,\NV open circuit is 2.
Dialogue: 0,0:08:01.71,0:08:07.33,Default,,0000,0000,0000,,I short circuit is that\NR Thevenin is equal
Dialogue: 0,0:08:07.33,0:08:13.59,Default,,0000,0000,0000,,to the ratio of V open\Ncircuit I short circuit,
Dialogue: 0,0:08:13.59,0:08:19.54,Default,,0000,0000,0000,,which equals 2 divided by 0.2173 and
Dialogue: 0,0:08:19.54,0:08:23.08,Default,,0000,0000,0000,,that equals 9.2 ohms.
Dialogue: 0,0:08:23.08,0:08:29.18,Default,,0000,0000,0000,,And we can now draw our\NThevenin equivalent circuit,
Dialogue: 0,0:08:29.18,0:08:32.97,Default,,0000,0000,0000,,it consists of a two volt source,
Dialogue: 0,0:08:32.97,0:08:37.71,Default,,0000,0000,0000,,with a 9.2 ohm resistor in series, and
Dialogue: 0,0:08:37.71,0:08:42.34,Default,,0000,0000,0000,,there are our a and b terminals.
Dialogue: 0,0:08:42.34,0:08:45.46,Default,,0000,0000,0000,,Thus, this nice concise little circuit.
Dialogue: 0,0:08:45.46,0:08:49.02,Default,,0000,0000,0000,,If all we're interested in is\Nthe terminal characteristics,
Dialogue: 0,0:08:49.02,0:08:52.74,Default,,0000,0000,0000,,this models this more complex circuit.
Dialogue: 0,0:08:54.81,0:08:58.97,Default,,0000,0000,0000,,Now lets just show that we get the same\NR Thevenin using method three.
Dialogue: 0,0:08:58.97,0:09:03.07,Default,,0000,0000,0000,,Method three requires us to\Ndeactivate the independent sources.
Dialogue: 0,0:09:03.07,0:09:07.75,Default,,0000,0000,0000,,So this is a voltage source return it to\Nzero volts, that is effectively shorting
Dialogue: 0,0:09:07.75,0:09:13.41,Default,,0000,0000,0000,,it out and we then apply a test\Nvoltage called a V test.
Dialogue: 0,0:09:16.43,0:09:24.63,Default,,0000,0000,0000,,Which then drives a current\NI test into the circuit.
Dialogue: 0,0:09:24.63,0:09:29.76,Default,,0000,0000,0000,,We have two nodes we got the node here,\Na node here we will call this one again
Dialogue: 0,0:09:29.76,0:09:34.68,Default,,0000,0000,0000,,V sub A noting that this V sub A is\Ndifferent than the V sub A we did before.
Dialogue: 0,0:09:34.68,0:09:39.20,Default,,0000,0000,0000,,The V sub A that we did before had a 15\Nvolt source connected at this point.
Dialogue: 0,0:09:39.20,0:09:42.35,Default,,0000,0000,0000,,This circuit has that 15\Nvolts source shorted out.
Dialogue: 0,0:09:42.35,0:09:44.43,Default,,0000,0000,0000,,So we gonna use the same name V sub A and
Dialogue: 0,0:09:44.43,0:09:47.11,Default,,0000,0000,0000,,over here this is no\Nlonger 0 this is now v10.
Dialogue: 0,0:09:47.11,0:09:52.38,Default,,0000,0000,0000,,Let's go ahead and write the node
Dialogue: 0,0:09:52.38,0:09:57.42,Default,,0000,0000,0000,,equation for this A node.
Dialogue: 0,0:09:57.42,0:10:04.67,Default,,0000,0000,0000,,So the current leaving, noting also\Nwe're gonna need this I0 in this case,
Dialogue: 0,0:10:04.67,0:10:09.81,Default,,0000,0000,0000,,is the current going in that\Ndirection that's going to
Dialogue: 0,0:10:09.81,0:10:15.86,Default,,0000,0000,0000,,equal 0-V sub A divided by 3, or\Njust negative V sub A over 3.
Dialogue: 0,0:10:15.86,0:10:18.41,Default,,0000,0000,0000,,All right,\Nnow let's write the node equation there.
Dialogue: 0,0:10:18.41,0:10:24.21,Default,,0000,0000,0000,,The current leaving this\Nnode going in this direction
Dialogue: 0,0:10:24.21,0:10:29.48,Default,,0000,0000,0000,,is going to be Va- 0\Nbecause that's now shorted
Dialogue: 0,0:10:29.48,0:10:34.92,Default,,0000,0000,0000,,out divided by 3 + Va\Ndivided by 6 + Va minus.
Dialogue: 0,0:10:34.92,0:10:37.64,Default,,0000,0000,0000,,We're now talking about the current\Nleaving this node going
Dialogue: 0,0:10:37.64,0:10:38.85,Default,,0000,0000,0000,,through the four ohm resistor.
Dialogue: 0,0:10:38.85,0:10:42.89,Default,,0000,0000,0000,,It's gonna be V sub a minus Vtest
Dialogue: 0,0:10:45.90,0:10:52.96,Default,,0000,0000,0000,,over 4 minus the current entering this\Nnode through the dependent source.
Dialogue: 0,0:10:52.96,0:10:58.86,Default,,0000,0000,0000,,It's going to be minus cuz it's\Nentering 1.2 times I zero and
Dialogue: 0,0:10:58.86,0:11:03.02,Default,,0000,0000,0000,,I zero is negative v sub a over three.
Dialogue: 0,0:11:04.54,0:11:09.73,Default,,0000,0000,0000,,One, two, three, four terms, the sum\Nof those four terms must equal zero.
Dialogue: 0,0:11:11.43,0:11:15.29,Default,,0000,0000,0000,,Now, let's write the node equation\Nover here at the test note.
Dialogue: 0,0:11:16.34,0:11:18.75,Default,,0000,0000,0000,,We have three branches, three currents,
Dialogue: 0,0:11:18.75,0:11:24.43,Default,,0000,0000,0000,,the current leaving the node going in this\Ndirection is going to be V test minus V
Dialogue: 0,0:11:24.43,0:11:29.33,Default,,0000,0000,0000,,sub a divided by 4 plus....
Dialogue: 0,0:11:29.33,0:11:34.62,Default,,0000,0000,0000,,Now this current here is leaving\Nthe node so it would be plus 1.2 I
Dialogue: 0,0:11:34.62,0:11:39.46,Default,,0000,0000,0000,,zero, but I zero we've determined\Nto be negative V sub a over 3e,
Dialogue: 0,0:11:39.46,0:11:44.39,Default,,0000,0000,0000,,ngative V sub a over 3.
Dialogue: 0,0:11:44.39,0:11:47.02,Default,,0000,0000,0000,,Now, IT is coming in, so\Nthat would be negative I test, so
Dialogue: 0,0:11:47.02,0:11:49.74,Default,,0000,0000,0000,,let's just take it to the other\Nside as a positive I test and
Dialogue: 0,0:11:49.74,0:11:54.11,Default,,0000,0000,0000,,say at the end of those two terms\Nwill equal positive I test.
Dialogue: 0,0:11:54.11,0:11:57.74,Default,,0000,0000,0000,,Now you notice we have three unknowns,\NV sub a,
Dialogue: 0,0:11:57.74,0:12:03.27,Default,,0000,0000,0000,,v test, and I test, but\Nhave only two equations.
Dialogue: 0,0:12:03.27,0:12:04.66,Default,,0000,0000,0000,,But that's all right,\Ncuz what we're gonna try and
Dialogue: 0,0:12:04.66,0:12:09.70,Default,,0000,0000,0000,,do is through algebraic manipulation, come\Nup with an expression here in this upper
Dialogue: 0,0:12:09.70,0:12:14.91,Default,,0000,0000,0000,,equation that gives us v\Nsub a in terms of v-test.
Dialogue: 0,0:12:14.91,0:12:18.07,Default,,0000,0000,0000,,We'll then plug in down here in\Nthe second equation for v sub a.
Dialogue: 0,0:12:19.20,0:12:24.89,Default,,0000,0000,0000,,All the terms then will have either\Nfactors of v-test or i-test.
Dialogue: 0,0:12:24.89,0:12:29.52,Default,,0000,0000,0000,,We will factor out the v-test to be able\Nto make the ratio of v-test over i-test
Dialogue: 0,0:12:29.52,0:12:36.25,Default,,0000,0000,0000,,and again that is what our Thevenin\Nis the ratio of v-test over i-test.
Dialogue: 0,0:12:36.25,0:12:41.02,Default,,0000,0000,0000,,So this upper equation cleans
Dialogue: 0,0:12:41.02,0:12:45.23,Default,,0000,0000,0000,,up to give us 1.15 V sub
Dialogue: 0,0:12:45.23,0:12:51.03,Default,,0000,0000,0000,,A Is equal
Dialogue: 0,0:12:51.03,0:12:55.82,Default,,0000,0000,0000,,to 0.25 VT.
Dialogue: 0,0:12:55.82,0:13:02.64,Default,,0000,0000,0000,,So V sub A is equal to 0.2174v test.
Dialogue: 0,0:13:02.64,0:13:08.22,Default,,0000,0000,0000,,Let's take those values in then,\Nand plug them in down here.
Dialogue: 0,0:13:08.22,0:13:10.90,Default,,0000,0000,0000,,In fact, let's just clean this\None up a little bit down here.
Dialogue: 0,0:13:10.90,0:13:18.27,Default,,0000,0000,0000,,Let's rewrite this bottom one as\Nif we were gonna have v-test times
Dialogue: 0,0:13:18.27,0:13:23.46,Default,,0000,0000,0000,,one-fourth, that's the only v-test term.
Dialogue: 0,0:13:23.46,0:13:30.70,Default,,0000,0000,0000,,Then we're gonna have plus v sub a times,
Dialogue: 0,0:13:30.70,0:13:35.67,Default,,0000,0000,0000,,we've got a negative one and
Dialogue: 0,0:13:35.67,0:13:41.26,Default,,0000,0000,0000,,a fourth, and we have a negative
Dialogue: 0,0:13:41.26,0:13:46.87,Default,,0000,0000,0000,,1.2 over 3 equals I-test.
Dialogue: 0,0:13:46.87,0:13:50.01,Default,,0000,0000,0000,,Now we'll take this expression for
Dialogue: 0,0:13:50.01,0:13:55.06,Default,,0000,0000,0000,,V sub A in terms of V test, and\Nplug it down here for V sub A.
Dialogue: 0,0:13:55.06,0:14:03.13,Default,,0000,0000,0000,,And we have 1/4Vt + .2174 v test,
Dialogue: 0,0:14:03.13,0:14:09.18,Default,,0000,0000,0000,,times,a negative one fourth,
Dialogue: 0,0:14:09.18,0:14:13.44,Default,,0000,0000,0000,,minus 1.2 over 3,
Dialogue: 0,0:14:13.44,0:14:17.72,Default,,0000,0000,0000,,that equals, i test.
Dialogue: 0,0:14:17.72,0:14:20.94,Default,,0000,0000,0000,,Now you go through and clean this up, and
Dialogue: 0,0:14:20.94,0:14:30.05,Default,,0000,0000,0000,,we get 0.10869V
Dialogue: 0,0:14:30.05,0:14:35.06,Default,,0000,0000,0000,,test equals I test.
Dialogue: 0,0:14:35.06,0:14:41.02,Default,,0000,0000,0000,,Now, for me the ratio of V test\Nover I test gives us our Thevenin.
Dialogue: 0,0:14:41.02,0:14:46.19,Default,,0000,0000,0000,,So R Thevenin equals V test\Nover I test which equals one
Dialogue: 0,0:14:46.19,0:14:50.67,Default,,0000,0000,0000,,over 0.10869,
Dialogue: 0,0:14:50.67,0:14:54.34,Default,,0000,0000,0000,,which once again equals 9.2 Ohms.
Dialogue: 0,0:14:55.91,0:14:59.74,Default,,0000,0000,0000,,Thus we get the same R Thevenin\Nfrom method three as we got for
Dialogue: 0,0:14:59.74,0:15:01.64,Default,,0000,0000,0000,,method one in the previous page,
Dialogue: 0,0:15:01.64,0:15:04.17,Default,,0000,0000,0000,,method two doesn't work because\Nthere is a depended source in there.