1 23:59:59,500 --> 00:00:08,792 [BLANK_AUDIO]. 2 00:00:08,792 --> 00:00:16,128 Okay, let's take a look at this circuit. Now this circuit has a volt, a current 3 00:00:16,128 --> 00:00:23,120 source in there and it's got some dependent sources. 4 00:00:23,120 --> 00:00:27,060 This dependent sources is, this is a dependent voltage source. 5 00:00:27,060 --> 00:00:29,320 And I wanted you to be able to see a problem like that. 6 00:00:29,320 --> 00:00:34,996 I might mention that dependent sources sometimes come into use in, in models of 7 00:00:34,996 --> 00:00:39,865 non linear systems. For example, a non linear system with 8 00:00:39,865 --> 00:00:43,895 transistors might have a dependent source in it if you look at a model in the 9 00:00:43,895 --> 00:00:47,716 linear range. So, it's good for you to see some 10 00:00:47,716 --> 00:00:52,460 problems with dependent sources and become familiar with working with them. 11 00:00:53,630 --> 00:01:01,910 So, this is ah, [SOUND] a node analysis approach that we're using here. 12 00:01:01,910 --> 00:01:05,290 So node is, analysis we need to pick a ground node. 13 00:01:05,290 --> 00:01:09,190 I'm going to pick this as my ground node only because I've got a voltage coming 14 00:01:09,190 --> 00:01:13,295 off of it. So, I know this voltage is pretty easy. 15 00:01:13,295 --> 00:01:17,430 That's going to be at 2 volts. So I picked this one. 16 00:01:17,430 --> 00:01:19,528 Its not unique. I could have picked another one. 17 00:01:19,528 --> 00:01:23,066 I typically don't want to pick a node with a current source coming off of it 18 00:01:23,066 --> 00:01:26,728 though, because that's a little hard to, to handle. 19 00:01:26,728 --> 00:01:30,860 I've got two other nodes that I need to identify. 20 00:01:30,860 --> 00:01:35,276 Let me call this one node A and I'll call that voltage V sub a, referenced to the 21 00:01:35,276 --> 00:01:39,610 ground. And this is V sub b, referenced to the 22 00:01:39,610 --> 00:01:44,290 ground. So the two voltages I have, node voltages 23 00:01:44,290 --> 00:01:48,630 V sub a and V sub b, means I'm going to want to write equations with unknowns, V 24 00:01:48,630 --> 00:01:53,080 sub a and V sub b. So I should have two unknowns and I 25 00:01:53,080 --> 00:01:56,200 should come up with two equations for those unknowns. 26 00:02:00,525 --> 00:02:06,978 I need to do a KCL at each of these nodes. 27 00:02:06,978 --> 00:02:08,515 So node A [SOUND]. Let me do a K KCL. 28 00:02:08,515 --> 00:02:13,512 And I always do it where I have, I sum up the currents leaving the node. 29 00:02:13,512 --> 00:02:17,830 So, in this direction, this direction, this direction. 30 00:02:17,830 --> 00:02:22,587 So, leaving the node over here, it's going to be VA minus this voltage. 31 00:02:22,587 --> 00:02:27,211 What this voltage is written a little bit strangely because it says I over 2, where 32 00:02:27,211 --> 00:02:32,471 I is this current right there. But that's okay because it's still a 33 00:02:32,471 --> 00:02:38,580 voltage source, so this is in voltages. So this voltage right here is I over 2 34 00:02:38,580 --> 00:02:44,374 referenced to ground. So V sub a minus [SOUND] I over 2, 35 00:02:44,374 --> 00:02:51,708 divided by that resistance, plus the current going in this direction, V sub a 36 00:02:51,708 --> 00:03:00,580 [SOUND] over 5, plus the current going in this direction. 37 00:03:00,580 --> 00:03:10,050 Well, the voltage drop is V sub a minus V sub b [SOUND] and the resistance is 2. 38 00:03:10,050 --> 00:03:13,490 That has to equal 0. Now as, as I said, I want to only, have 39 00:03:13,490 --> 00:03:19,093 the variables of V sub a and V sub b in my equation. 40 00:03:19,093 --> 00:03:22,960 So I need to get rid of I. How do I get ride of I? 41 00:03:22,960 --> 00:03:26,150 Well, I've got to be able to write I in terms of V sub a or V sub b or some 42 00:03:26,150 --> 00:03:30,855 combination of them. If I look at I as its defined here I can 43 00:03:30,855 --> 00:03:37,018 see that it is related to V sub a. In fact, if I look at this expression 44 00:03:37,018 --> 00:03:42,002 right here, I find that I [SOUND] is equal to V sub a over 5 and then I'm 45 00:03:42,002 --> 00:03:47,609 going to negate that, because I is referenced in this direction, where V sub 46 00:03:47,609 --> 00:03:57,240 A over 5 by itself is the current going in this direction. 47 00:03:57,240 --> 00:03:59,760 So I have to throw the minus sign in because of the direction that I is 48 00:03:59,760 --> 00:04:05,730 defined. Now, let me group my terms here. 49 00:04:05,730 --> 00:04:14,498 I've got V sub a [SOUND], 1 over 10 plus 1 over 100 plus 1 over 5 plus one half, 50 00:04:14,498 --> 00:04:23,890 and then a minus one half V sub b is equal to 0. 51 00:04:23,890 --> 00:04:31,702 And if I add this all together, this is point 71, 0.71 V sub a minus 0.5 [SOUND] 52 00:04:31,702 --> 00:04:38,382 V sub b equals 0. So now I have an equation just in terms 53 00:04:38,382 --> 00:04:43,307 of the Va and Vb. Let's go ahead and do the same thing with 54 00:04:43,307 --> 00:04:45,706 node b. [SOUND]. 55 00:04:45,706 --> 00:04:51,626 Node b, I'm going to look at the currents leaving the node. 56 00:04:51,626 --> 00:05:00,266 So that is, this voltage drop is now Vb minus Va [SOUND] divided by 2 ohms, plus 57 00:05:00,266 --> 00:05:10,044 this voltage drop is Vb minus 2 [SOUND] divided by 1 ohm. 58 00:05:10,044 --> 00:05:18,428 And then this current leaving in this direction is minus 2 [SOUND] is equal to 59 00:05:18,428 --> 00:05:24,830 zero. If I combine my terms, I will find that I 60 00:05:24,830 --> 00:05:36,320 have [SOUND] the following equation. Minus 0.5 V sub a plus 1.5 V sub b equal 61 00:05:36,320 --> 00:05:43,936 4. And again, I've got an equation in terms 62 00:05:43,936 --> 00:05:49,058 of Va and Vb. I can now write this as a linear 63 00:05:49,058 --> 00:05:55,178 equation, a matrix equation. Looking at this, [SOUND] I want to solve 64 00:05:55,178 --> 00:05:57,678 for Va and Vb. [SOUND]. 65 00:06:03,120 --> 00:06:08,698 I want to set up the equation like this. To write this first equation into the 66 00:06:08,698 --> 00:06:18,021 matrix form, I've got a 0.71 times Va. So that's going to be a 0.71 times Va. 67 00:06:18,021 --> 00:06:27,108 And then, minus 0.5 times Vb. So minus 0.71 time Va minus 0.5 times Vb 68 00:06:27,108 --> 00:06:33,173 is equal to zero. Now the other equation can be written in 69 00:06:33,173 --> 00:06:38,003 the same way. I've got a minus 0.5 times Va, so the 70 00:06:38,003 --> 00:06:46,366 coefficient minus 0.5 times Va and 1.5 times Vb is equal to 4. 71 00:06:46,366 --> 00:06:51,064 So the first row is this first equation and the second row is the second 72 00:06:51,064 --> 00:06:57,596 equation. Many of you might have an equation solver 73 00:06:57,596 --> 00:07:02,417 on your calculator. Or if you don't, there are there are 74 00:07:02,417 --> 00:07:08,124 places that you can go on the internet. In fact, I just did that before I, I 75 00:07:08,124 --> 00:07:12,927 started this. I went on the internet and I just, I just 76 00:07:12,927 --> 00:07:19,416 searched for a linear equation solver and I found one where I can just plug in 77 00:07:19,416 --> 00:07:26,111 these values, these matrix values, and plug in what it's equal to and I was able 78 00:07:26,111 --> 00:07:36,397 to solve for the variable here. And when I did that, I found that Va 79 00:07:36,397 --> 00:07:46,162 [SOUND] and Vb are equal to 2.454, 3.485. And ultimately what I wanted to solve for 80 00:07:46,162 --> 00:07:50,360 was V0, and I just found that right there. 81 00:07:53,990 --> 00:08:03,610 So V0 is equal to Va, which is 2.454. And that solves this problem. 82 00:08:04,980 --> 00:08:08,500 So, just to summarize how we solved it, the first thing we had to do was select a 83 00:08:08,500 --> 00:08:12,990 ground node. And then we identified our other node 84 00:08:12,990 --> 00:08:16,390 voltages. We had two additional nodes. 85 00:08:16,390 --> 00:08:21,415 I wanted to find equations involved that involve only those unknowns, the V sub a 86 00:08:21,415 --> 00:08:27,740 and V sub b as my only unknowns, which are my node voltages. 87 00:08:27,740 --> 00:08:32,770 I use a KCL at each node and then I just solve it. 88 00:08:32,770 --> 00:08:34,930 I came up with two equations and two unknowns. 89 00:08:34,930 --> 00:08:37,850 I chose to solve it using a matrix method. 90 00:08:37,850 --> 00:08:42,580 You don't need to use a matrix method, you can just solve these by substitution. 91 00:08:42,580 --> 00:08:46,580 I should mention that there's a separate tutorial video that I'm going to shoot 92 00:08:46,580 --> 00:08:50,562 [INAUDIBLE] that shows how to solve this by hand. 93 00:08:50,562 --> 00:08:52,320 matrix inversion of two variables by hand. 94 00:08:52,320 --> 00:08:57,380 Otherwise, you don't need to solve it by hand. 95 00:08:57,380 --> 00:09:02,642 If you have a calculator that does it, then by all means use that. 96 00:09:02,642 --> 00:09:05,790 Thank you.