0:00:01.620,0:00:05.340 In this video we're gonna demonstrate[br]the technique for determining the Thevenin 0:00:05.340,0:00:08.560 equivalent circuit by looking at[br]a relatively simple circuit and 0:00:08.560,0:00:12.440 going through the different steps[br]you will be using to do this. 0:00:13.570,0:00:16.270 The first thing we need to determine[br]is the open circuit voltage. 0:00:16.270,0:00:20.682 The Thevenin voltage or[br]the voltage in our Thevenin model. 0:00:23.708,0:00:28.222 Consists of a Thevenin voltage which[br]we pointed out is just the open 0:00:28.222,0:00:30.210 circuit voltage. 0:00:30.210,0:00:33.170 So in this case we need to know[br]what the voltage between a and 0:00:33.170,0:00:36.740 b is with nothing connecting a and b. 0:00:38.090,0:00:39.490 As you look at this, you can see then, 0:00:39.490,0:00:44.110 that the open circuit voltage[br]is the voltage across R2. 0:00:44.110,0:00:48.320 Now, we can use any method we want[br]to determine the voltage across R2. 0:00:48.320,0:00:51.950 We could calculate the current[br]running through these. 0:00:51.950,0:00:54.740 Which, because this open circuit, 0:00:54.740,0:01:00.414 we should point out there is no current-[br][SOUND] Going through that branch. 0:01:00.414,0:01:05.080 That means that these two[br]resistors are in series. 0:01:05.080,0:01:07.430 And by definition the current is[br][INAUDIBLE] the same through both of them. 0:01:07.430,0:01:12.460 So we could calculate the current[br]through those two resistors and 0:01:12.460,0:01:15.160 we multiply R2 by that current and 0:01:15.160,0:01:18.180 that would give us the Open circuit[br]volts through the volts across R2, 0:01:18.180,0:01:23.810 we're gonna also use our voltage divider[br]which was developed, we derive a voltage 0:01:23.810,0:01:27.910 formula by taking advantage of the fact[br]that those two cells are in series. 0:01:27.910,0:01:28.710 So let's just do that. 0:01:28.710,0:01:34.264 V open circuit, is equal to then,[br]V sub s, times. 0:01:34.264,0:01:39.159 R2 over R1 + R2. 0:01:39.159,0:01:42.741 For this circuit, that is our V Thevenin. 0:01:45.714,0:01:48.717 So to get the open circuit[br]voltage using any of the voltage, 0:01:48.717,0:01:52.622 any of the certain analysis techniques[br]that you have to determine the voltage 0:01:52.622,0:01:54.080 across the open terminals. 0:01:55.550,0:01:58.390 Next, we need to determine[br]the thevenin resistance. 0:02:00.521,0:02:05.650 And at that point, we will have[br]our thevenin equivalent circuit. 0:02:05.650,0:02:08.650 Now we're gonna see there are at[br]least three different methods for 0:02:08.650,0:02:11.650 determining the thevenin resistance. 0:02:11.650,0:02:16.195 And we're going to Demonstrate each of[br]those models on the simple circuit. 0:02:18.008,0:02:22.532 The first method takes advantage[br]of the definition of R Thevenin, 0:02:22.532,0:02:27.541 you'll recall from our previous video[br]we said that R Thevenin was defined 0:02:27.541,0:02:32.330 as the ratio of the open circuit[br]voltage to the short circuit current. 0:02:33.410,0:02:36.780 Now in the previous slide we determine[br]that the open circuit voltage 0:02:38.360,0:02:42.100 which is the voltage across our[br]to in this case which was vs so 0:02:42.100,0:02:47.130 times R two over R one[br]plus R two.Now we need 0:02:47.130,0:02:52.480 to determine what the short circuit[br]current is by short circuit currently mean 0:02:52.480,0:02:57.330 if we short The wire between[br]the two terminals A and 0:02:57.330,0:03:03.180 B and then determine what that current is[br]that flows through that short circuit. 0:03:03.180,0:03:05.310 That's the short circuit current. 0:03:05.310,0:03:09.640 Well, in this case the short circuit[br]pulls the voltage here to zero. 0:03:09.640,0:03:11.660 So, there would be no[br]current going through R2. 0:03:13.790,0:03:18.390 And the short circuit current then is[br]simply the current that would be flowing 0:03:18.390,0:03:24.760 through R 1 or I short circuit is[br]equal to V sub S divided by R 1. 0:03:24.760,0:03:29.690 So you've been using this[br]first method method 1 we say 0:03:29.690,0:03:33.440 then that open the R thevenin is equal[br]to the ratio of the open circuit voltage 0:03:33.440,0:03:38.490 short circuit current, we form that And[br]say that R 7 is equal to V 0:03:38.490,0:03:43.740 open circuit divided by I short[br]circuit which is equal to V sub 0:03:43.740,0:03:48.020 S times R 2 over R 1 plus R 2. 0:03:48.020,0:03:51.550 That's our open circuit voltage 0:03:51.550,0:03:55.248 divided by the short circuit[br]current which is V sub S. 0:03:55.248,0:03:58.650 Over R one. 0:03:58.650,0:04:00.550 Well, the v of s is cancelled. 0:04:00.550,0:04:04.560 We take this one over our one in[br]the denominator, invert and multiply, and 0:04:04.560,0:04:10.784 we get then that our Is equal 0:04:10.784,0:04:17.420 to R 1 R 2 over R 1 +[br]R 2 At least it works 0:04:17.420,0:04:21.750 if you've got an independent source in the[br]circuit that you're attempting to model. 0:04:23.110,0:04:25.850 Method one then is determine[br]the open circuit voltage and 0:04:25.850,0:04:31.080 the short circuit current using any[br]circuit analysis method you'd like. 0:04:31.080,0:04:33.850 And then forming the ratio of[br]the open circuit voltage to the short 0:04:33.850,0:04:34.680 circuit current. 0:04:36.690,0:04:41.250 The second method works[br]when all you have or 0:04:41.250,0:04:45.680 the only types of sources you[br]have are independent sources. 0:04:45.680,0:04:50.357 It works only with independent 0:04:51.762,0:04:58.950 Sources and in this method,[br]or to apply this method, 0:04:58.950,0:05:04.720 we deactivate any of the sources present. 0:05:04.720,0:05:06.510 And by deactivate, 0:05:06.510,0:05:09.730 we mean the same thing that we did back[br]when we were talking about super position. 0:05:09.730,0:05:13.930 When you deactivate a voltage source,[br]you turn the voltage source to zero. 0:05:15.320,0:05:17.780 Or in this case, or 0:05:17.780,0:05:20.660 a voltage source turned to 0 is[br]effectively a short circuit. 0:05:21.870,0:05:27.010 So having deactivated the independent[br]sources present, we then calculate 0:05:28.170,0:05:33.050 the resistance seen looking[br]back into the circuit. 0:05:34.070,0:05:38.450 Well, looking back into it with this[br]point pulled to ground by deactivating 0:05:38.450,0:05:44.270 the voltage source, we see then that R1[br]and R2 are in parallel with each other. 0:05:44.270,0:05:48.845 And the resistance that we see[br]going back into the ab terminals is 0:05:48.845,0:05:54.300 R1R2 over R1 + R2. 0:05:54.300,0:05:57.420 Which is the same result[br]we got on the previews. 0:05:57.420,0:05:58.500 Using the previews method. 0:06:00.990,0:06:08.470 The third method, involves applying[br]a test voltage at the terminals, and 0:06:08.470,0:06:13.210 pushing the current back into the circuit,[br]pushing the current back into the circuit. 0:06:13.210,0:06:15.120 After deactivating the sources. 0:06:16.780,0:06:20.510 This is somewhat like going back[br]to our power training analogy, or 0:06:20.510,0:06:22.550 an automobile analogy. 0:06:22.550,0:06:26.410 It'd be like sort of[br]turning off the engine, and 0:06:26.410,0:06:31.140 then blowing back into the exhaust pipe,[br]and measuring 0:06:31.140,0:06:36.680 the resistance to the movement of air[br]going back into the Into the exhaust pipe. 0:06:36.680,0:06:42.640 So, Method 3,[br]deactivate the independent sources. 0:06:43.680,0:06:47.660 If you have dependent sources present,[br]don't deactivate them. 0:06:47.660,0:06:52.480 Deactivate just the independent sources,[br]and then 0:06:54.000,0:06:59.540 apply a test voltage We'll call it V Test,[br]VT for test. 0:06:59.540,0:07:04.170 That's not the Thevenin volt,[br]it's just V Test which will 0:07:04.170,0:07:10.100 force a test current to[br]flow back into the circuit. 0:07:10.100,0:07:13.470 Again, after we have[br]deactivated the source. 0:07:17.280,0:07:22.000 The resistance that we feel going[br]back into it is simply the ratio 0:07:22.000,0:07:26.360 of the test voltage divided[br]by the test current. 0:07:28.330,0:07:33.140 So our approach now requires us to 0:07:33.140,0:07:35.740 come up with an equation or[br]more than one equation. 0:07:35.740,0:07:40.570 It allows us then to by[br]just algebraic manipulation 0:07:40.570,0:07:43.880 form the ratio of Vtest over Itest. 0:07:43.880,0:07:46.120 Let's just do it. 0:07:46.120,0:07:48.060 It's easier to show that[br]it is to tell about it. 0:07:49.220,0:07:50.450 When we apply this, 0:07:50.450,0:07:55.750 after deactivating the voltage source,[br]we see that we have only one node here. 0:07:55.750,0:08:01.320 And that node is in fact,[br]at least in this circuit, The voltage, 0:08:01.320,0:08:03.710 our test voltage being Test. 0:08:03.710,0:08:07.920 So, let's write a node equation here at[br]this node in terms of the node voltage 0:08:07.920,0:08:09.480 V Test. 0:08:09.480,0:08:14.250 The current leaving this node going[br]through R1 is going to be V Test divided 0:08:14.250,0:08:19.180 by R1,[br]is the current leaving going this way. 0:08:19.180,0:08:23.740 The current coming down here[br]through R2 Is gonna be added to it 0:08:23.740,0:08:26.350 plus V test divided by R2. 0:08:26.350,0:08:31.140 Now you'll notice that I test[br]is directed into the nodes, so 0:08:31.140,0:08:35.840 that would be minus I test equals 0. 0:08:35.840,0:08:42.460 Now let's factor out the common[br]V test here Times 1 over R1 0:08:42.460,0:08:48.070 plus 1 over R2 and take I test to[br]the other side as a positive I test. 0:08:49.900,0:08:54.800 Getting a common denominator[br]we have then V test time 0:08:54.800,0:08:58.690 R1 plus R2 over R1 times R2. 0:09:00.640,0:09:03.070 Is equal to I test. 0:09:04.180,0:09:08.600 Now that we've got a factor[br]of the test on this side, 0:09:08.600,0:09:13.870 I test on that side divide both sides by[br]I test, divide both sides by this term 0:09:13.870,0:09:19.269 here and[br]we get that then v test over I test 0:09:21.330,0:09:26.041 Which is our R thevenin is = R1, 0:09:26.041,0:09:32.029 R2 over R1 + R2. 0:09:33.815,0:09:37.405 Which is the same resistance we[br]determined through the other two methods. 0:09:37.405,0:09:41.975 So, again, this approach drives home the[br]idea that what this Thevenin resistance 0:09:41.975,0:09:46.255 represents is the restriction or[br]the resistance seen 0:09:46.255,0:09:50.635 looking back into[br]the terminals of our circuit. 0:09:52.990,0:09:57.590 Let's review then, determining[br]we find the open circuit voltage 0:09:57.590,0:10:00.790 using any method of[br]circuit analysis we want 0:10:01.930,0:10:06.470 we then use one of three methods[br]to determine our thevenin. 0:10:06.470,0:10:13.700 The first method works When you have at[br]least one independent source present. 0:10:15.070,0:10:20.100 Method one then is too short[br]between the output terminals and 0:10:20.100,0:10:21.640 determine the short circuit current and 0:10:23.530,0:10:28.290 then form the ratio V open circuit[br]divided by high short circuit. 0:10:28.290,0:10:29.230 That's method one. 0:10:31.220,0:10:35.970 Method two works only if you[br]have independent sources 0:10:37.460,0:10:43.060 and with only independent sources present,[br]And deactivate the independent sources. 0:10:43.060,0:10:45.740 In this case, we had a voltage[br]source that we shorted out. 0:10:45.740,0:10:49.780 Had it been a current source, we would[br]have opened circuited that current source, 0:10:49.780,0:10:51.930 and then determined the resistance, 0:10:51.930,0:10:56.190 the equivalent resistance seen[br]looking back in from a to b. 0:10:58.410,0:11:00.900 Method two works only if[br]independent sources are present. 0:11:02.540,0:11:10.710 Method three works when you have both[br]independent and dependent sources. 0:11:12.210,0:11:17.990 If you have dependent sources, if you have[br]independent sources you deactivate Any 0:11:17.990,0:11:22.480 of the independent sources by replacing[br]voltage sources with short circuits, and 0:11:22.480,0:11:25.190 current sources with open circuits, and 0:11:25.190,0:11:28.430 then apply a test voltage[br]to the terminals. 0:11:29.440,0:11:32.420 And then using algebraic[br]manipulation techniques, 0:11:32.420,0:11:36.790 derive the ratio of V test over I test. 0:11:38.430,0:11:40.230 Three different options. 0:11:40.230,0:11:42.240 Some options work under all circumstances, 0:11:42.240,0:11:45.520 other options work only at[br]given certain circumstances. 0:11:45.520,0:11:49.890 And in the next two or three videos, we'll[br]go through and give examples demonstrating 0:11:49.890,0:11:53.390 each of these three techniques in[br]somewhat more complicated circuits.