[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.62,0:00:05.34,Default,,0000,0000,0000,,In this video we're gonna demonstrate\Nthe technique for determining the Thevenin
Dialogue: 0,0:00:05.34,0:00:08.56,Default,,0000,0000,0000,,equivalent circuit by looking at\Na relatively simple circuit and
Dialogue: 0,0:00:08.56,0:00:12.44,Default,,0000,0000,0000,,going through the different steps\Nyou will be using to do this.
Dialogue: 0,0:00:13.57,0:00:16.27,Default,,0000,0000,0000,,The first thing we need to determine\Nis the open circuit voltage.
Dialogue: 0,0:00:16.27,0:00:20.68,Default,,0000,0000,0000,,The Thevenin voltage or\Nthe voltage in our Thevenin model.
Dialogue: 0,0:00:23.71,0:00:28.22,Default,,0000,0000,0000,,Consists of a Thevenin voltage which\Nwe pointed out is just the open
Dialogue: 0,0:00:28.22,0:00:30.21,Default,,0000,0000,0000,,circuit voltage.
Dialogue: 0,0:00:30.21,0:00:33.17,Default,,0000,0000,0000,,So in this case we need to know\Nwhat the voltage between a and
Dialogue: 0,0:00:33.17,0:00:36.74,Default,,0000,0000,0000,,b is with nothing connecting a and b.
Dialogue: 0,0:00:38.09,0:00:39.49,Default,,0000,0000,0000,,As you look at this, you can see then,
Dialogue: 0,0:00:39.49,0:00:44.11,Default,,0000,0000,0000,,that the open circuit voltage\Nis the voltage across R2.
Dialogue: 0,0:00:44.11,0:00:48.32,Default,,0000,0000,0000,,Now, we can use any method we want\Nto determine the voltage across R2.
Dialogue: 0,0:00:48.32,0:00:51.95,Default,,0000,0000,0000,,We could calculate the current\Nrunning through these.
Dialogue: 0,0:00:51.95,0:00:54.74,Default,,0000,0000,0000,,Which, because this open circuit,
Dialogue: 0,0:00:54.74,0:01:00.41,Default,,0000,0000,0000,,we should point out there is no current-\N[SOUND] Going through that branch.
Dialogue: 0,0:01:00.41,0:01:05.08,Default,,0000,0000,0000,,That means that these two\Nresistors are in series.
Dialogue: 0,0:01:05.08,0:01:07.43,Default,,0000,0000,0000,,And by definition the current is\N[INAUDIBLE] the same through both of them.
Dialogue: 0,0:01:07.43,0:01:12.46,Default,,0000,0000,0000,,So we could calculate the current\Nthrough those two resistors and
Dialogue: 0,0:01:12.46,0:01:15.16,Default,,0000,0000,0000,,we multiply R2 by that current and
Dialogue: 0,0:01:15.16,0:01:18.18,Default,,0000,0000,0000,,that would give us the Open circuit\Nvolts through the volts across R2,
Dialogue: 0,0:01:18.18,0:01:23.81,Default,,0000,0000,0000,,we're gonna also use our voltage divider\Nwhich was developed, we derive a voltage
Dialogue: 0,0:01:23.81,0:01:27.91,Default,,0000,0000,0000,,formula by taking advantage of the fact\Nthat those two cells are in series.
Dialogue: 0,0:01:27.91,0:01:28.71,Default,,0000,0000,0000,,So let's just do that.
Dialogue: 0,0:01:28.71,0:01:34.26,Default,,0000,0000,0000,,V open circuit, is equal to then,\NV sub s, times.
Dialogue: 0,0:01:34.26,0:01:39.16,Default,,0000,0000,0000,,R2 over R1 + R2.
Dialogue: 0,0:01:39.16,0:01:42.74,Default,,0000,0000,0000,,For this circuit, that is our V Thevenin.
Dialogue: 0,0:01:45.71,0:01:48.72,Default,,0000,0000,0000,,So to get the open circuit\Nvoltage using any of the voltage,
Dialogue: 0,0:01:48.72,0:01:52.62,Default,,0000,0000,0000,,any of the certain analysis techniques\Nthat you have to determine the voltage
Dialogue: 0,0:01:52.62,0:01:54.08,Default,,0000,0000,0000,,across the open terminals.
Dialogue: 0,0:01:55.55,0:01:58.39,Default,,0000,0000,0000,,Next, we need to determine\Nthe thevenin resistance.
Dialogue: 0,0:02:00.52,0:02:05.65,Default,,0000,0000,0000,,And at that point, we will have\Nour thevenin equivalent circuit.
Dialogue: 0,0:02:05.65,0:02:08.65,Default,,0000,0000,0000,,Now we're gonna see there are at\Nleast three different methods for
Dialogue: 0,0:02:08.65,0:02:11.65,Default,,0000,0000,0000,,determining the thevenin resistance.
Dialogue: 0,0:02:11.65,0:02:16.20,Default,,0000,0000,0000,,And we're going to Demonstrate each of\Nthose models on the simple circuit.
Dialogue: 0,0:02:18.01,0:02:22.53,Default,,0000,0000,0000,,The first method takes advantage\Nof the definition of R Thevenin,
Dialogue: 0,0:02:22.53,0:02:27.54,Default,,0000,0000,0000,,you'll recall from our previous video\Nwe said that R Thevenin was defined
Dialogue: 0,0:02:27.54,0:02:32.33,Default,,0000,0000,0000,,as the ratio of the open circuit\Nvoltage to the short circuit current.
Dialogue: 0,0:02:33.41,0:02:36.78,Default,,0000,0000,0000,,Now in the previous slide we determine\Nthat the open circuit voltage
Dialogue: 0,0:02:38.36,0:02:42.10,Default,,0000,0000,0000,,which is the voltage across our\Nto in this case which was vs so
Dialogue: 0,0:02:42.10,0:02:47.13,Default,,0000,0000,0000,,times R two over R one\Nplus R two.Now we need
Dialogue: 0,0:02:47.13,0:02:52.48,Default,,0000,0000,0000,,to determine what the short circuit\Ncurrent is by short circuit currently mean
Dialogue: 0,0:02:52.48,0:02:57.33,Default,,0000,0000,0000,,if we short The wire between\Nthe two terminals A and
Dialogue: 0,0:02:57.33,0:03:03.18,Default,,0000,0000,0000,,B and then determine what that current is\Nthat flows through that short circuit.
Dialogue: 0,0:03:03.18,0:03:05.31,Default,,0000,0000,0000,,That's the short circuit current.
Dialogue: 0,0:03:05.31,0:03:09.64,Default,,0000,0000,0000,,Well, in this case the short circuit\Npulls the voltage here to zero.
Dialogue: 0,0:03:09.64,0:03:11.66,Default,,0000,0000,0000,,So, there would be no\Ncurrent going through R2.
Dialogue: 0,0:03:13.79,0:03:18.39,Default,,0000,0000,0000,,And the short circuit current then is\Nsimply the current that would be flowing
Dialogue: 0,0:03:18.39,0:03:24.76,Default,,0000,0000,0000,,through R 1 or I short circuit is\Nequal to V sub S divided by R 1.
Dialogue: 0,0:03:24.76,0:03:29.69,Default,,0000,0000,0000,,So you've been using this\Nfirst method method 1 we say
Dialogue: 0,0:03:29.69,0:03:33.44,Default,,0000,0000,0000,,then that open the R thevenin is equal\Nto the ratio of the open circuit voltage
Dialogue: 0,0:03:33.44,0:03:38.49,Default,,0000,0000,0000,,short circuit current, we form that And\Nsay that R 7 is equal to V
Dialogue: 0,0:03:38.49,0:03:43.74,Default,,0000,0000,0000,,open circuit divided by I short\Ncircuit which is equal to V sub
Dialogue: 0,0:03:43.74,0:03:48.02,Default,,0000,0000,0000,,S times R 2 over R 1 plus R 2.
Dialogue: 0,0:03:48.02,0:03:51.55,Default,,0000,0000,0000,,That's our open circuit voltage
Dialogue: 0,0:03:51.55,0:03:55.25,Default,,0000,0000,0000,,divided by the short circuit\Ncurrent which is V sub S.
Dialogue: 0,0:03:55.25,0:03:58.65,Default,,0000,0000,0000,,Over R one.
Dialogue: 0,0:03:58.65,0:04:00.55,Default,,0000,0000,0000,,Well, the v of s is cancelled.
Dialogue: 0,0:04:00.55,0:04:04.56,Default,,0000,0000,0000,,We take this one over our one in\Nthe denominator, invert and multiply, and
Dialogue: 0,0:04:04.56,0:04:10.78,Default,,0000,0000,0000,,we get then that our Is equal
Dialogue: 0,0:04:10.78,0:04:17.42,Default,,0000,0000,0000,,to R 1 R 2 over R 1 +\NR 2 At least it works
Dialogue: 0,0:04:17.42,0:04:21.75,Default,,0000,0000,0000,,if you've got an independent source in the\Ncircuit that you're attempting to model.
Dialogue: 0,0:04:23.11,0:04:25.85,Default,,0000,0000,0000,,Method one then is determine\Nthe open circuit voltage and
Dialogue: 0,0:04:25.85,0:04:31.08,Default,,0000,0000,0000,,the short circuit current using any\Ncircuit analysis method you'd like.
Dialogue: 0,0:04:31.08,0:04:33.85,Default,,0000,0000,0000,,And then forming the ratio of\Nthe open circuit voltage to the short
Dialogue: 0,0:04:33.85,0:04:34.68,Default,,0000,0000,0000,,circuit current.
Dialogue: 0,0:04:36.69,0:04:41.25,Default,,0000,0000,0000,,The second method works\Nwhen all you have or
Dialogue: 0,0:04:41.25,0:04:45.68,Default,,0000,0000,0000,,the only types of sources you\Nhave are independent sources.
Dialogue: 0,0:04:45.68,0:04:50.36,Default,,0000,0000,0000,,It works only with independent
Dialogue: 0,0:04:51.76,0:04:58.95,Default,,0000,0000,0000,,Sources and in this method,\Nor to apply this method,
Dialogue: 0,0:04:58.95,0:05:04.72,Default,,0000,0000,0000,,we deactivate any of the sources present.
Dialogue: 0,0:05:04.72,0:05:06.51,Default,,0000,0000,0000,,And by deactivate,
Dialogue: 0,0:05:06.51,0:05:09.73,Default,,0000,0000,0000,,we mean the same thing that we did back\Nwhen we were talking about super position.
Dialogue: 0,0:05:09.73,0:05:13.93,Default,,0000,0000,0000,,When you deactivate a voltage source,\Nyou turn the voltage source to zero.
Dialogue: 0,0:05:15.32,0:05:17.78,Default,,0000,0000,0000,,Or in this case, or
Dialogue: 0,0:05:17.78,0:05:20.66,Default,,0000,0000,0000,,a voltage source turned to 0 is\Neffectively a short circuit.
Dialogue: 0,0:05:21.87,0:05:27.01,Default,,0000,0000,0000,,So having deactivated the independent\Nsources present, we then calculate
Dialogue: 0,0:05:28.17,0:05:33.05,Default,,0000,0000,0000,,the resistance seen looking\Nback into the circuit.
Dialogue: 0,0:05:34.07,0:05:38.45,Default,,0000,0000,0000,,Well, looking back into it with this\Npoint pulled to ground by deactivating
Dialogue: 0,0:05:38.45,0:05:44.27,Default,,0000,0000,0000,,the voltage source, we see then that R1\Nand R2 are in parallel with each other.
Dialogue: 0,0:05:44.27,0:05:48.84,Default,,0000,0000,0000,,And the resistance that we see\Ngoing back into the ab terminals is
Dialogue: 0,0:05:48.84,0:05:54.30,Default,,0000,0000,0000,,R1R2 over R1 + R2.
Dialogue: 0,0:05:54.30,0:05:57.42,Default,,0000,0000,0000,,Which is the same result\Nwe got on the previews.
Dialogue: 0,0:05:57.42,0:05:58.50,Default,,0000,0000,0000,,Using the previews method.
Dialogue: 0,0:06:00.99,0:06:08.47,Default,,0000,0000,0000,,The third method, involves applying\Na test voltage at the terminals, and
Dialogue: 0,0:06:08.47,0:06:13.21,Default,,0000,0000,0000,,pushing the current back into the circuit,\Npushing the current back into the circuit.
Dialogue: 0,0:06:13.21,0:06:15.12,Default,,0000,0000,0000,,After deactivating the sources.
Dialogue: 0,0:06:16.78,0:06:20.51,Default,,0000,0000,0000,,This is somewhat like going back\Nto our power training analogy, or
Dialogue: 0,0:06:20.51,0:06:22.55,Default,,0000,0000,0000,,an automobile analogy.
Dialogue: 0,0:06:22.55,0:06:26.41,Default,,0000,0000,0000,,It'd be like sort of\Nturning off the engine, and
Dialogue: 0,0:06:26.41,0:06:31.14,Default,,0000,0000,0000,,then blowing back into the exhaust pipe,\Nand measuring
Dialogue: 0,0:06:31.14,0:06:36.68,Default,,0000,0000,0000,,the resistance to the movement of air\Ngoing back into the Into the exhaust pipe.
Dialogue: 0,0:06:36.68,0:06:42.64,Default,,0000,0000,0000,,So, Method 3,\Ndeactivate the independent sources.
Dialogue: 0,0:06:43.68,0:06:47.66,Default,,0000,0000,0000,,If you have dependent sources present,\Ndon't deactivate them.
Dialogue: 0,0:06:47.66,0:06:52.48,Default,,0000,0000,0000,,Deactivate just the independent sources,\Nand then
Dialogue: 0,0:06:54.00,0:06:59.54,Default,,0000,0000,0000,,apply a test voltage We'll call it V Test,\NVT for test.
Dialogue: 0,0:06:59.54,0:07:04.17,Default,,0000,0000,0000,,That's not the Thevenin volt,\Nit's just V Test which will
Dialogue: 0,0:07:04.17,0:07:10.10,Default,,0000,0000,0000,,force a test current to\Nflow back into the circuit.
Dialogue: 0,0:07:10.10,0:07:13.47,Default,,0000,0000,0000,,Again, after we have\Ndeactivated the source.
Dialogue: 0,0:07:17.28,0:07:22.00,Default,,0000,0000,0000,,The resistance that we feel going\Nback into it is simply the ratio
Dialogue: 0,0:07:22.00,0:07:26.36,Default,,0000,0000,0000,,of the test voltage divided\Nby the test current.
Dialogue: 0,0:07:28.33,0:07:33.14,Default,,0000,0000,0000,,So our approach now requires us to
Dialogue: 0,0:07:33.14,0:07:35.74,Default,,0000,0000,0000,,come up with an equation or\Nmore than one equation.
Dialogue: 0,0:07:35.74,0:07:40.57,Default,,0000,0000,0000,,It allows us then to by\Njust algebraic manipulation
Dialogue: 0,0:07:40.57,0:07:43.88,Default,,0000,0000,0000,,form the ratio of Vtest over Itest.
Dialogue: 0,0:07:43.88,0:07:46.12,Default,,0000,0000,0000,,Let's just do it.
Dialogue: 0,0:07:46.12,0:07:48.06,Default,,0000,0000,0000,,It's easier to show that\Nit is to tell about it.
Dialogue: 0,0:07:49.22,0:07:50.45,Default,,0000,0000,0000,,When we apply this,
Dialogue: 0,0:07:50.45,0:07:55.75,Default,,0000,0000,0000,,after deactivating the voltage source,\Nwe see that we have only one node here.
Dialogue: 0,0:07:55.75,0:08:01.32,Default,,0000,0000,0000,,And that node is in fact,\Nat least in this circuit, The voltage,
Dialogue: 0,0:08:01.32,0:08:03.71,Default,,0000,0000,0000,,our test voltage being Test.
Dialogue: 0,0:08:03.71,0:08:07.92,Default,,0000,0000,0000,,So, let's write a node equation here at\Nthis node in terms of the node voltage
Dialogue: 0,0:08:07.92,0:08:09.48,Default,,0000,0000,0000,,V Test.
Dialogue: 0,0:08:09.48,0:08:14.25,Default,,0000,0000,0000,,The current leaving this node going\Nthrough R1 is going to be V Test divided
Dialogue: 0,0:08:14.25,0:08:19.18,Default,,0000,0000,0000,,by R1,\Nis the current leaving going this way.
Dialogue: 0,0:08:19.18,0:08:23.74,Default,,0000,0000,0000,,The current coming down here\Nthrough R2 Is gonna be added to it
Dialogue: 0,0:08:23.74,0:08:26.35,Default,,0000,0000,0000,,plus V test divided by R2.
Dialogue: 0,0:08:26.35,0:08:31.14,Default,,0000,0000,0000,,Now you'll notice that I test\Nis directed into the nodes, so
Dialogue: 0,0:08:31.14,0:08:35.84,Default,,0000,0000,0000,,that would be minus I test equals 0.
Dialogue: 0,0:08:35.84,0:08:42.46,Default,,0000,0000,0000,,Now let's factor out the common\NV test here Times 1 over R1
Dialogue: 0,0:08:42.46,0:08:48.07,Default,,0000,0000,0000,,plus 1 over R2 and take I test to\Nthe other side as a positive I test.
Dialogue: 0,0:08:49.90,0:08:54.80,Default,,0000,0000,0000,,Getting a common denominator\Nwe have then V test time
Dialogue: 0,0:08:54.80,0:08:58.69,Default,,0000,0000,0000,,R1 plus R2 over R1 times R2.
Dialogue: 0,0:09:00.64,0:09:03.07,Default,,0000,0000,0000,,Is equal to I test.
Dialogue: 0,0:09:04.18,0:09:08.60,Default,,0000,0000,0000,,Now that we've got a factor\Nof the test on this side,
Dialogue: 0,0:09:08.60,0:09:13.87,Default,,0000,0000,0000,,I test on that side divide both sides by\NI test, divide both sides by this term
Dialogue: 0,0:09:13.87,0:09:19.27,Default,,0000,0000,0000,,here and\Nwe get that then v test over I test
Dialogue: 0,0:09:21.33,0:09:26.04,Default,,0000,0000,0000,,Which is our R thevenin is = R1,
Dialogue: 0,0:09:26.04,0:09:32.03,Default,,0000,0000,0000,,R2 over R1 + R2.
Dialogue: 0,0:09:33.82,0:09:37.40,Default,,0000,0000,0000,,Which is the same resistance we\Ndetermined through the other two methods.
Dialogue: 0,0:09:37.40,0:09:41.98,Default,,0000,0000,0000,,So, again, this approach drives home the\Nidea that what this Thevenin resistance
Dialogue: 0,0:09:41.98,0:09:46.26,Default,,0000,0000,0000,,represents is the restriction or\Nthe resistance seen
Dialogue: 0,0:09:46.26,0:09:50.64,Default,,0000,0000,0000,,looking back into\Nthe terminals of our circuit.
Dialogue: 0,0:09:52.99,0:09:57.59,Default,,0000,0000,0000,,Let's review then, determining\Nwe find the open circuit voltage
Dialogue: 0,0:09:57.59,0:10:00.79,Default,,0000,0000,0000,,using any method of\Ncircuit analysis we want
Dialogue: 0,0:10:01.93,0:10:06.47,Default,,0000,0000,0000,,we then use one of three methods\Nto determine our thevenin.
Dialogue: 0,0:10:06.47,0:10:13.70,Default,,0000,0000,0000,,The first method works When you have at\Nleast one independent source present.
Dialogue: 0,0:10:15.07,0:10:20.10,Default,,0000,0000,0000,,Method one then is too short\Nbetween the output terminals and
Dialogue: 0,0:10:20.10,0:10:21.64,Default,,0000,0000,0000,,determine the short circuit current and
Dialogue: 0,0:10:23.53,0:10:28.29,Default,,0000,0000,0000,,then form the ratio V open circuit\Ndivided by high short circuit.
Dialogue: 0,0:10:28.29,0:10:29.23,Default,,0000,0000,0000,,That's method one.
Dialogue: 0,0:10:31.22,0:10:35.97,Default,,0000,0000,0000,,Method two works only if you\Nhave independent sources
Dialogue: 0,0:10:37.46,0:10:43.06,Default,,0000,0000,0000,,and with only independent sources present,\NAnd deactivate the independent sources.
Dialogue: 0,0:10:43.06,0:10:45.74,Default,,0000,0000,0000,,In this case, we had a voltage\Nsource that we shorted out.
Dialogue: 0,0:10:45.74,0:10:49.78,Default,,0000,0000,0000,,Had it been a current source, we would\Nhave opened circuited that current source,
Dialogue: 0,0:10:49.78,0:10:51.93,Default,,0000,0000,0000,,and then determined the resistance,
Dialogue: 0,0:10:51.93,0:10:56.19,Default,,0000,0000,0000,,the equivalent resistance seen\Nlooking back in from a to b.
Dialogue: 0,0:10:58.41,0:11:00.90,Default,,0000,0000,0000,,Method two works only if\Nindependent sources are present.
Dialogue: 0,0:11:02.54,0:11:10.71,Default,,0000,0000,0000,,Method three works when you have both\Nindependent and dependent sources.
Dialogue: 0,0:11:12.21,0:11:17.99,Default,,0000,0000,0000,,If you have dependent sources, if you have\Nindependent sources you deactivate Any
Dialogue: 0,0:11:17.99,0:11:22.48,Default,,0000,0000,0000,,of the independent sources by replacing\Nvoltage sources with short circuits, and
Dialogue: 0,0:11:22.48,0:11:25.19,Default,,0000,0000,0000,,current sources with open circuits, and
Dialogue: 0,0:11:25.19,0:11:28.43,Default,,0000,0000,0000,,then apply a test voltage\Nto the terminals.
Dialogue: 0,0:11:29.44,0:11:32.42,Default,,0000,0000,0000,,And then using algebraic\Nmanipulation techniques,
Dialogue: 0,0:11:32.42,0:11:36.79,Default,,0000,0000,0000,,derive the ratio of V test over I test.
Dialogue: 0,0:11:38.43,0:11:40.23,Default,,0000,0000,0000,,Three different options.
Dialogue: 0,0:11:40.23,0:11:42.24,Default,,0000,0000,0000,,Some options work under all circumstances,
Dialogue: 0,0:11:42.24,0:11:45.52,Default,,0000,0000,0000,,other options work only at\Ngiven certain circumstances.
Dialogue: 0,0:11:45.52,0:11:49.89,Default,,0000,0000,0000,,And in the next two or three videos, we'll\Ngo through and give examples demonstrating
Dialogue: 0,0:11:49.89,0:11:53.39,Default,,0000,0000,0000,,each of these three techniques in\Nsomewhat more complicated circuits.