1 00:00:00,012 --> 00:00:05,588 Welcome back, and today we're going to be talking about a very important property 2 00:00:05,588 --> 00:00:11,130 of Linear Circuits and actually use it for analysis. 3 00:00:11,130 --> 00:00:14,413 It's called Superposition and so we're going to describe what it is and how we 4 00:00:14,413 --> 00:00:19,818 can use it to analyze our circuits. In the previous lesson we talked about 5 00:00:19,818 --> 00:00:22,976 what it meant for something to be linear. And why this class is called Linear 6 00:00:22,976 --> 00:00:26,095 Circuits. And in order for superposition the thing 7 00:00:26,095 --> 00:00:31,280 we're using today to work, we need to have everything to be linear. 8 00:00:31,280 --> 00:00:34,360 If it's not things start to break down. But we'll talk a little more about that. 9 00:00:37,670 --> 00:00:41,576 After we've finished discussing Superposition, Doctor Fairly will be 10 00:00:41,576 --> 00:00:47,038 presenting some systematic ways of going about finding systems of equations. 11 00:00:47,038 --> 00:00:51,069 To solve for these types of circuits. A lot of them actually make use of 12 00:00:51,069 --> 00:00:54,054 superposition. But superposition itself is a tool you 13 00:00:54,054 --> 00:00:58,195 can use to solve some of these circuits and do some analysis. 14 00:00:58,195 --> 00:01:03,883 The executives for this particular lesson are, given a complicated system, generate 15 00:01:03,883 --> 00:01:09,490 a set of simpler systems. Each with a single, independent source. 16 00:01:09,490 --> 00:01:13,009 And then, use those individual solutions and combine them together to come up with 17 00:01:13,009 --> 00:01:19,238 a complete solution for a system. In order to use superposition, the first 18 00:01:19,238 --> 00:01:21,825 step is you need to be able to zero out your sources. 19 00:01:21,825 --> 00:01:26,690 And what do we mean by zeroing them out? Well first of all, if I have a voltage 20 00:01:26,690 --> 00:01:30,533 source, and I want to replace it by some type of device that will have a zero 21 00:01:30,533 --> 00:01:34,785 voltage. What's the device that I want to choose 22 00:01:34,785 --> 00:01:38,810 to replace it with? It turns out that the wire works best 23 00:01:38,810 --> 00:01:45,740 because if I measure the voltage between here and here, because there's this wire. 24 00:01:45,740 --> 00:01:48,540 We know that the wire is believed to be equipotential. 25 00:01:48,540 --> 00:01:51,891 There's no voltage drop across it. So, that means that if I want to zero out 26 00:01:51,891 --> 00:01:57,064 the voltage, I turn it into a wire. For a current source, if I want zero 27 00:01:57,064 --> 00:02:00,760 current to be flowing across a pathway, the best circuit element to use is going 28 00:02:00,760 --> 00:02:05,774 to be an open circuit. Since this is basically defined as 29 00:02:05,774 --> 00:02:11,260 something that's disconnected, so no current is able to flow. 30 00:02:16,350 --> 00:02:19,472 The Steps For Superposition. So first of all, begin by following this, 31 00:02:19,472 --> 00:02:23,050 zeroing out all of the independent sources. 32 00:02:23,050 --> 00:02:26,100 After they've all been zeroed out, you return each of the sources, one at a 33 00:02:26,100 --> 00:02:29,472 time, and then you do the analysis of the circuit. 34 00:02:29,472 --> 00:02:32,670 And solve the quantity that you're searching for. 35 00:02:32,670 --> 00:02:36,020 After you've completed that, you take the arithmetic sum of these values for the 36 00:02:36,020 --> 00:02:39,092 final quantity. Now you might wonder, why would we use 37 00:02:39,092 --> 00:02:42,420 superposition, if we took a problem that we had to solve once and now we have to 38 00:02:42,420 --> 00:02:47,270 solve it multiple times. It turns out that sometimes it can be 39 00:02:47,270 --> 00:02:50,790 useful if by using Superposition, you're able to make the individual simple 40 00:02:50,790 --> 00:02:55,667 circuits very easy to analyze. so it does, sometimes, find practical 41 00:02:55,667 --> 00:03:01,075 applications in those circumstances. But superposition is often an underlying 42 00:03:01,075 --> 00:03:05,233 method, of some of the more systematic methods, that are, generally, used more 43 00:03:05,233 --> 00:03:08,210 frequently. That will be covered next time. 44 00:03:09,930 --> 00:03:13,870 Okay, so let's do an example of solving a system using superposition. 45 00:03:13,870 --> 00:03:17,019 In this example we have three sources and we have four resistors, so it's a fairly 46 00:03:17,019 --> 00:03:21,040 complicated system. What we're going to do is start by 47 00:03:21,040 --> 00:03:24,760 removing each of the sources and then putting them in one at a time, solving 48 00:03:24,760 --> 00:03:28,450 for v. Which is the value that we're looking to, 49 00:03:28,450 --> 00:03:32,786 to find in the circuit. And after we've done this three times, 50 00:03:32,786 --> 00:03:35,714 then we take all of the individual solutions, add them together to get our 51 00:03:35,714 --> 00:03:44,840 final solution. Right, so in this example, we have a 3 52 00:03:44,840 --> 00:03:49,061 volt source here, and we're going to be zeroing out this voltage source, and this 53 00:03:49,061 --> 00:03:59,506 current source. So doing that, I get something like this, 54 00:03:59,506 --> 00:04:08,099 3 volt source, 4 ohm resistor. the current source are being removed, so 55 00:04:08,099 --> 00:04:13,461 we can just take this 36 ohm resistor and move it over here. 56 00:04:13,461 --> 00:04:26,453 There's a 45 ohm resistor over here. The 30 ohm resistor stays. 57 00:04:26,453 --> 00:04:29,810 And now this is a wire, because we've removed that other source. 58 00:04:36,370 --> 00:04:41,503 Within this we see that the 30 ohm resistor and the 45 ohm resistor are in 59 00:04:41,503 --> 00:04:48,940 parallel with each other. So you can combine them, that then allows 60 00:04:48,940 --> 00:04:53,035 us to take this combination and this 36 ohm resistors and this just because a 61 00:04:53,035 --> 00:04:57,580 voltage divider. So, we have 3 volts, split across this, 62 00:04:57,580 --> 00:05:01,320 where this is the voltage we're interested in finding. 63 00:05:01,320 --> 00:05:09,318 When you do that, you discover that the voltage is equal to 18 over 18 plus 36 64 00:05:09,318 --> 00:05:16,805 times 3 is 1 volt. So that means that from part a, we've 65 00:05:16,805 --> 00:05:24,509 discovered the voltage drop contributed by this 3 volt source is 1 volt. 66 00:05:24,509 --> 00:05:29,705 Now let's do it again, but with a different source. 67 00:05:29,705 --> 00:05:33,030 Here I've noted that our first result was 1 volt. 68 00:05:33,030 --> 00:05:36,726 Now we're going to remove this voltage source and this voltage source and leave 69 00:05:36,726 --> 00:05:40,482 the current source. So again we will draw a circuit. 70 00:05:40,482 --> 00:05:52,190 This is 30 ohms, this is 45 ohms and this is the voltage that we're trying to find. 71 00:05:53,210 --> 00:05:57,698 Notice though, that because we've zeroed out this voltage source, there's a wire 72 00:05:57,698 --> 00:06:01,881 connecting the two sides of this 54 ohm resistor. 73 00:06:01,881 --> 00:06:05,510 Consequently this resistor isn't going to be interesting. 74 00:06:05,510 --> 00:06:07,785 Any current that flows is going to circumvent it. 75 00:06:07,785 --> 00:06:10,815 And so we'll see no voltage drop across it, no current going through it. 76 00:06:10,815 --> 00:06:22,824 So we can remove it from our circuit. So that the current source remains at one 77 00:06:22,824 --> 00:06:28,124 fourth amp. And then, finally, a 36 ohm resistor. 78 00:06:28,124 --> 00:06:35,860 At this point, we see that we have 3 resistors. 79 00:06:35,860 --> 00:06:39,370 Which are all in parallel, which we can combine, because they connect the same 80 00:06:39,370 --> 00:06:43,716 two nodes, as a single resistor. And then by multiplying the current that 81 00:06:43,716 --> 00:06:47,226 is going through that resistor, which is one quarter of an amp, we can discover 82 00:06:47,226 --> 00:06:53,254 what the voltage, v happens to be. In doing that, we see that the v is equal 83 00:06:53,254 --> 00:06:58,285 to 3 volts. Simply making use of combining resistors 84 00:06:58,285 --> 00:07:05,670 and then Ohm's Law. Still one more to go. 85 00:07:06,720 --> 00:07:12,810 Now, we put the five volt source back and we remove the current source again. 86 00:07:14,110 --> 00:07:19,245 Still we see that this, voltage source, because we've zeroed it out, gives us a 87 00:07:19,245 --> 00:07:24,444 wire across this 54 ohm resistor. So again, the 54 ohm resistor's not going 88 00:07:24,444 --> 00:07:29,145 to be interesting to us. We're drawing the circuit again. 89 00:07:29,145 --> 00:07:34,359 The five volts is going to then go across the 30 ohm resistor, and then we have our 90 00:07:34,359 --> 00:07:39,930 45 ohm resistor here. Cross for measuring a voltage. 91 00:07:44,250 --> 00:07:50,515 That will also be going across a 36 ohm resistor. 92 00:07:50,515 --> 00:07:55,275 So to do the analysis for this problem. We can take these two resistors and 93 00:07:55,275 --> 00:07:58,595 combine them, because they're in parallel, into a single resistor. 94 00:07:58,595 --> 00:08:02,380 And at that point, the whole problem simplifies down to a voltage divider. 95 00:08:02,380 --> 00:08:08,396 The equivalent resistance here is 20 ohms, which means that we have 20 over 20 96 00:08:08,396 --> 00:08:15,980 plus 30 or 50, so two fifths times 5 volts is equal to two volts. 97 00:08:15,980 --> 00:08:21,951 So to get our final solution of v, I don't know if this is vc. 98 00:08:21,951 --> 00:08:28,405 We simply take v from part a. Add a v from part b, add a v from part c. 99 00:08:28,405 --> 00:08:34,380 Getting a result of one plus two plus three, Six volts. 100 00:08:35,990 --> 00:08:39,290 Now, it might be easier to use a different technique, something not like 101 00:08:39,290 --> 00:08:44,087 superposition, for a circuit like this. But, this is a good way of seeing how 102 00:08:44,087 --> 00:08:46,290 superposition can be applied to a circuit. 103 00:08:46,290 --> 00:08:49,214 And sometimes, being able to zero out the sources can simplify your system enough 104 00:08:49,214 --> 00:08:52,600 that it's worth using. Additionally, a lot of the systematic 105 00:08:52,600 --> 00:08:56,618 methods that will be presented for solving these types of circuits. 106 00:08:56,618 --> 00:09:00,523 Have underlying them the requirement that the system follows superposition or that 107 00:09:00,523 --> 00:09:04,570 everything be linear. But because this is a linear circuit 108 00:09:04,570 --> 00:09:07,670 class, you're going to pretty much be able to use the techniques that we're 109 00:09:07,670 --> 00:09:11,985 presenting here, in all of the circuits that you're analyzing. 110 00:09:11,985 --> 00:09:19,050 When you're working with dependent sources, it becomes a little trickier. 111 00:09:19,050 --> 00:09:22,178 You can't just zero out your dependent sources because the dependent sources are 112 00:09:22,178 --> 00:09:25,500 functions of something else that's going on in the system. 113 00:09:26,790 --> 00:09:29,052 In order for them to work, there's a couple of caveats. 114 00:09:29,052 --> 00:09:32,872 First of all you have to leave them in in every single implementation. 115 00:09:32,872 --> 00:09:36,274 Every time you remove your independent sources you leave in your dependent 116 00:09:36,274 --> 00:09:40,240 sources when you put back in Individual independent sources. 117 00:09:40,240 --> 00:09:42,705 And analyze the whole thing with your dependent sources. 118 00:09:42,705 --> 00:09:47,700 And in order for this to work, the dependent sources must be linear. 119 00:09:47,700 --> 00:09:51,665 And typically, what that means is that your dependent source must be some 120 00:09:51,665 --> 00:09:55,780 constant times some measured value in the system. 121 00:09:57,210 --> 00:10:00,992 So if I look at this example, I have K1 times v1, where K1 is a constant 2 volts 122 00:10:00,992 --> 00:10:04,560 per volt. That's just fine. 123 00:10:05,750 --> 00:10:09,874 And that will work with no problem. But in the other example here I do K2, 2 124 00:10:09,874 --> 00:10:15,115 milliamps per volt squared and I'm multiplying it by this V i squared. 125 00:10:15,115 --> 00:10:18,193 And because we're multiplying by something that's kind of a derived 126 00:10:18,193 --> 00:10:22,040 quantity, and not just something that was measured. 127 00:10:22,040 --> 00:10:25,450 This is no longer linear. Which means that if you have something 128 00:10:25,450 --> 00:10:28,100 like this in your circuit, super position will stop, stop working. 129 00:10:28,100 --> 00:10:31,672 And you won't be able to use it anymore. So that one's out. 130 00:10:31,672 --> 00:10:37,442 We will make available, some examples. To show you how you can analyze a circuit 131 00:10:37,442 --> 00:10:40,569 using dependent sources. But typically we're not going to use 132 00:10:40,569 --> 00:10:43,980 superposition to analyze a circuit when there are dependent sources present. 133 00:10:45,930 --> 00:10:49,258 In summary, we used superposition to solve circuits, and we presented how to 134 00:10:49,258 --> 00:10:53,448 do that with independent sources only. And the technique for doing it with 135 00:10:53,448 --> 00:10:57,380 dependent sources as well. In the next lesson we will talk about 136 00:10:57,380 --> 00:11:00,728 systematic application of using Ohm's Law, Kirchhoff's Current Law, and 137 00:11:00,728 --> 00:11:04,860 Kirchhoff's Voltage Law to generate a system of equations. 138 00:11:04,860 --> 00:11:08,252 As well as, what's called, the Thevenin and Norton equivalent circuit, that 139 00:11:08,252 --> 00:11:11,132 allows you take a very complicated system. 140 00:11:11,132 --> 00:11:14,370 And replace it by a very simple equivalent system. 141 00:11:14,370 --> 00:11:16,800 Which is going to be used in some of our later analysis.