1 00:00:00,012 --> 00:00:05,477 So in this video, we're going to figure out how to solve circuits in a much more 2 00:00:05,477 --> 00:00:11,552 quicker, much more efficient way, than we have by writing lots of equations and 3 00:00:11,552 --> 00:00:15,897 lots of anomalies. And, the way to see how that could be is 4 00:00:15,897 --> 00:00:21,539 by asking a rather interesting question, what does a source see? And of course 5 00:00:21,539 --> 00:00:23,981 multi-sources and current sources don't have eyes, 6 00:00:23,981 --> 00:00:27,980 but I'll show you what that means. When you go down the path, they're asking 7 00:00:27,980 --> 00:00:32,215 what the source sees, that leads to the concept of equivalent circuits, one of 8 00:00:32,215 --> 00:00:38,560 the most powerful concepts and circuit theory. And we'll explore series in 9 00:00:38,560 --> 00:00:46,402 parallel combinations and learn how to think about circuits, in terms of these 10 00:00:46,402 --> 00:00:52,128 structural components. Let's go back to our example that we did 11 00:00:52,128 --> 00:00:57,043 last time. the we have a series combination of 12 00:00:57,043 --> 00:01:03,213 resistors and a voltage source. And what we calculated last time was the 13 00:01:03,213 --> 00:01:09,114 output voltage was proportional to the input voltage with a constant 14 00:01:09,114 --> 00:01:13,382 proportionality, that's the resistor R2 / R1 + R2. 15 00:01:13,382 --> 00:01:19,929 And this as I pointed out last time is called voltage divider, and that applies 16 00:01:19,929 --> 00:01:25,622 any time, you have a series combination of resistors and you're applying a 17 00:01:25,622 --> 00:01:30,171 voltage across them. A voltage across one of them, is that 18 00:01:30,171 --> 00:01:36,514 resistance divided by the sum of the resistance as a rule that you can use to 19 00:01:36,514 --> 00:01:41,159 simplify your life. I asked you last time, what was the 20 00:01:41,159 --> 00:01:47,440 voltage across R1, and I think it's pretty clear, but if you plot a voltage 21 00:01:47,440 --> 00:01:51,288 divider, it's the resistance of the resistor. 22 00:01:51,288 --> 00:01:55,346 I'm talking about divided by sum times v in. 23 00:01:55,346 --> 00:02:02,598 very easy to see, and of course, by KVL, this and this have to add up to be v in, 24 00:02:02,598 --> 00:02:08,693 so everything checks out. Well, what does the source see? And what 25 00:02:08,693 --> 00:02:16,628 do I mean by that? The voltage source, in this case, certainly notice what it's 26 00:02:16,628 --> 00:02:20,519 voltage is. The question is, when it gets to see 27 00:02:20,519 --> 00:02:26,205 through the current. So, the idea is, I'm going to put the 28 00:02:26,205 --> 00:02:32,067 resistors in a box, so that the voltage source can't cheat. 29 00:02:32,067 --> 00:02:40,248 The only thing you can do is figure out it's current, the current going into the 30 00:02:40,248 --> 00:02:47,810 box by one and based on what that, how that's related to the voltage source, 31 00:02:47,810 --> 00:02:54,430 figure out what's inside the box. We know the V out is equal to this, 32 00:02:54,430 --> 00:02:59,055 and we want to know the current i1, but KCL says that i1 = i out, 33 00:02:59,055 --> 00:03:03,555 that's the way it always works out for a series combination, 34 00:03:03,555 --> 00:03:09,222 the current goes straight through them, and so it's easy to see if the currents 35 00:03:09,222 --> 00:03:13,041 were all the same. Well, I can figure out i out from v out 36 00:03:13,041 --> 00:03:17,513 by simply dividing by R2, and, that's what I get. 37 00:03:17,513 --> 00:03:22,676 So, what does the voltage source see? Put in a box, 38 00:03:22,676 --> 00:03:29,866 for all the world, the relationship between the source voltage and the 39 00:03:29,866 --> 00:03:37,162 current, that it's supplying to the circuit looks like that of a resistor. 40 00:03:37,162 --> 00:03:44,282 A single resistor having a value equal to the sum and that's the idea of what's 41 00:03:44,282 --> 00:03:50,759 called an equivalent circuit. So, what does the source see? It looks in 42 00:03:50,759 --> 00:03:57,037 here, through its current and it sees something that behaves just like a single 43 00:03:57,037 --> 00:04:00,568 resistor, whose value is the sum of the resistances. 44 00:04:00,568 --> 00:04:05,828 Of course, it doesn't know that there is a sum. It might see a resistor that looks 45 00:04:05,828 --> 00:04:09,804 like 1 kilo ohm. It doesn't know how many resistors or if 46 00:04:09,804 --> 00:04:13,599 there is only one resistor, that can equally be, 47 00:04:13,599 --> 00:04:19,142 well be the case. But the equivalent resistance of a series 48 00:04:19,142 --> 00:04:23,326 combination is the sum of the resistances. 49 00:04:23,326 --> 00:04:29,725 You can really use that to simplify your calculations for circuits. 50 00:04:29,725 --> 00:04:36,372 And by the way, this result all comes from the vi relations KVL and KCL. 51 00:04:36,372 --> 00:04:42,930 Putting them in a much more concise, understandable form when we think about 52 00:04:42,930 --> 00:04:47,961 equal answers. So, like I said, it's a single resistor. 53 00:04:47,961 --> 00:04:55,210 Now, let's try another simple circuit and this one has what's called a parallel 54 00:04:55,210 --> 00:04:58,906 structure. We have three circuit elements in 55 00:04:58,906 --> 00:05:03,848 parallel beside each other, connected together with wires, 56 00:05:03,848 --> 00:05:10,203 and that's called a parallel circuit. So we have two resistors in parallel and 57 00:05:10,203 --> 00:05:14,262 a current source has been placed in parallel with that. 58 00:05:14,262 --> 00:05:21,856 And what I want to do is I want to find out how the output current, the currents 59 00:05:21,856 --> 00:05:29,563 of R2 is related to the input current. Well, how many nodes, how many loops are 60 00:05:29,563 --> 00:05:35,746 there for writing KVL? And you should see, there are two nodes, 61 00:05:35,746 --> 00:05:40,631 there's a big node here, and a big node down here. 62 00:05:40,631 --> 00:05:44,896 We are only going to write KCL at that node, 63 00:05:44,896 --> 00:05:51,060 and here are two loops, so we get exactly the number of equations 64 00:05:51,060 --> 00:05:56,385 we need. We have two vi relations and the third 65 00:05:56,385 --> 00:06:02,229 one is the, where the current source. We have the KCL equation, one KCL 66 00:06:02,229 --> 00:06:10,838 equation, we have two KVL equations. One thing to point out is that when you 67 00:06:10,838 --> 00:06:17,831 have resistors in parallel, it's pretty clear that the voltage across the them, 68 00:06:17,831 --> 00:06:21,685 each of them is the same. And since the current source is in 69 00:06:21,685 --> 00:06:26,855 parallel with that, all the voltage is here, all the voltages are equal to each 70 00:06:26,855 --> 00:06:32,098 other, so that's not really a question. The question is what is this current and 71 00:06:32,098 --> 00:06:38,478 how it's related to in? Well, all we have to do is put it all together. 72 00:06:38,478 --> 00:06:45,098 it's not hard to show that this is the result and the way you do that is by 73 00:06:45,098 --> 00:06:51,661 looking at this set of equations here, the vi relations, because these two 74 00:06:51,661 --> 00:06:57,638 voltages are equal, these two quantities are equal. 75 00:06:57,638 --> 00:07:06,622 And I can substitute for i1 in this equation by solving this set for i1. 76 00:07:06,622 --> 00:07:15,212 What I get is i1 = R2 / R1 times i out. And if I substitute that in there and 77 00:07:15,212 --> 00:07:23,013 solve, I will get this result, and this is known as current divider as you might 78 00:07:23,013 --> 00:07:30,234 expect, so in a parallel circuit, if you have a current going in, it splits 79 00:07:30,234 --> 00:07:36,865 between the two resistors. So, that the, the part through this resistor is the 80 00:07:36,865 --> 00:07:41,527 other, the value of the other resistor divided by the sum, 81 00:07:41,527 --> 00:07:47,458 and so, it works in a similar way, the voltage divider, except it's not the 82 00:07:47,458 --> 00:07:54,928 resistance of the one that we're talking about, it's the other resistor. Okay. 83 00:07:54,928 --> 00:08:07,475 So what does this look like? Well, again, because we get that for the input-output 84 00:08:07,475 --> 00:08:13,812 relationship, well, the only way the, the current 85 00:08:13,812 --> 00:08:18,486 source can see what's going on is to look at its voltage. 86 00:08:18,486 --> 00:08:24,406 Well, that voltage is just that. Well, that's again the vi relationship 87 00:08:24,406 --> 00:08:28,729 for a resistor. It looks like a single resistor whose 88 00:08:28,729 --> 00:08:35,915 resistance is the product over the sum and I use this shorthand notation for 89 00:08:35,915 --> 00:08:41,476 that result. So R1 parallel R2 means R1 R2 divided by 90 00:08:41,476 --> 00:08:46,968 the sum of R1 and R2. So, again, what the source sees is a 91 00:08:46,968 --> 00:08:53,774 single resistor it does not know, there are two resistors in there. All it 92 00:08:53,774 --> 00:09:00,162 can tell it looks like one resistor. So, if the series in parallel combination 93 00:09:00,162 --> 00:09:05,802 structures are very important to recognize to help you simplify and see 94 00:09:05,802 --> 00:09:10,349 what's going on. So let's go through this little exercise 95 00:09:10,349 --> 00:09:13,732 here. I'm not going to tell you if I'm, what 96 00:09:13,732 --> 00:09:19,147 kind of a source I'm going to attach out here, It doesn't really matter, 97 00:09:19,147 --> 00:09:25,037 the question is what does the source see? Well, we have to figure out the structure 98 00:09:25,037 --> 00:09:28,432 first. It's clear that R1 is in parallel with 99 00:09:28,432 --> 00:09:35,549 all of this, or what's all of this. I see R2 in parallel with R3 at there, 100 00:09:35,549 --> 00:09:43,501 and then that's in series with R4, and that's going to allow us to figure out 101 00:09:43,501 --> 00:09:51,838 what the equivalent resistance is. So, let's start the parallel combination 102 00:09:51,838 --> 00:09:57,155 here, that looks like a single resistor whose 103 00:09:57,155 --> 00:10:04,900 value is R2 parallel R3. This set now looks like a single resistor 104 00:10:04,900 --> 00:10:13,451 which, whose value is R2 parallel R3 plus R4, because this parallel combination is 105 00:10:13,451 --> 00:10:18,167 in series with that. And then, this single resistor, is what 106 00:10:18,167 --> 00:10:21,963 that looks like there, is in parallel with R1, 107 00:10:21,963 --> 00:10:28,602 and so our final result is it's a little complicated, but very simple to write. 108 00:10:28,602 --> 00:10:35,837 And that is that R1 is in parallel with the combination R2 R2 and R3, and then, 109 00:10:35,837 --> 00:10:38,598 which is in series with R4. Well, 110 00:10:38,598 --> 00:10:45,608 I'm going to use those results to solve a circuit the easy way and this is actually 111 00:10:45,608 --> 00:10:51,377 a pretty practical circuit. So what I want to point out here is that 112 00:10:51,377 --> 00:10:56,247 this is the system we've already been talking about. 113 00:10:56,247 --> 00:11:04,067 'Kay? And what usually happens is when you try to pass that signal on to the 114 00:11:04,067 --> 00:11:11,553 sync, that it has an input resistance. In some sense, its equivalent resistance 115 00:11:11,553 --> 00:11:18,593 is something which we're going to call R sub L and L means load. And the question 116 00:11:18,593 --> 00:11:25,634 is, does the relationship between v in and v out change, because there is a load 117 00:11:25,634 --> 00:11:30,761 resistance out there? You might say, well, geez, this voltage 118 00:11:30,761 --> 00:11:35,428 here is the same as that voltage there, so it probably doesn't change. 119 00:11:35,428 --> 00:11:40,070 Well, that's not quite true, and that's because these are in parallel 120 00:11:40,070 --> 00:11:43,526 and these voltages are actually across the parallel combination. 121 00:11:43,526 --> 00:11:49,722 There's something to figure out there. I've essentially given away how you 122 00:11:49,722 --> 00:11:53,284 analyze it. What we discover is that the output 123 00:11:53,284 --> 00:11:58,226 voltage is using our series of parallel combination levels. 124 00:11:58,226 --> 00:12:04,355 This looks like a single resistor, which is R2 in parallel with RL, and then 125 00:12:04,355 --> 00:12:09,692 that's in series with R1, so we simply use voltage divider to 126 00:12:09,692 --> 00:12:14,889 figure it out. And is the concise expression, and when 127 00:12:14,889 --> 00:12:20,616 you simplify everything, you get this for a final answer. 128 00:12:20,616 --> 00:12:29,004 Well, that doesn't much look like the result that we had before, which was R2 / 129 00:12:29,004 --> 00:12:35,060 R1 + R2 * v in. So, when is, is there a situation as we 130 00:12:35,060 --> 00:12:42,325 change RL, that this result looks like that? If you think about that for a 131 00:12:42,325 --> 00:12:47,299 second, there are sort of two extremes to look at, 132 00:12:47,299 --> 00:12:52,124 RL going to zero, RL going to v in. That's the easy cases to look at, 133 00:12:52,124 --> 00:12:57,159 which is pretty clear if you had set RL to zero, the output is zero voltage, 134 00:12:57,159 --> 00:13:03,385 right? Because there's an RL [INAUDIBLE] More interesting is RL going to infinity. 135 00:13:03,385 --> 00:13:08,827 When RL goes to infinity this term down here essentially disappears, because 136 00:13:08,827 --> 00:13:13,932 these terms are getting big, and then the RLs cancel, and sure enough, 137 00:13:13,932 --> 00:13:19,942 you get this result. So, if you want to do develop the system 138 00:13:19,942 --> 00:13:26,885 such that when you attach it to the sync, it has input resistance RL. 139 00:13:28,262 --> 00:13:33,729 It better, it better be that RL is big and it's going to take until the 140 00:13:33,729 --> 00:13:39,073 succeeding video we define what being in comparison what is. 141 00:13:39,073 --> 00:13:42,086 That's in the next video. Alright. 142 00:13:42,086 --> 00:13:45,875 So, in summarize, the resistor equivalence. 143 00:13:45,875 --> 00:13:53,123 So, if you have resistors in series, the equivalent resistance looks like the sum. 144 00:13:53,123 --> 00:13:59,813 And voltage divider, what it looks like, the voltage across any one of them is 145 00:13:59,813 --> 00:14:05,130 that resistance divided by the sum of all of the resistances. 146 00:14:05,130 --> 00:14:11,216 If you have a lot of resistors in parallel, the equivalent conductance is 147 00:14:11,216 --> 00:14:17,444 the sum of the conductances and the current through any one of them is that 148 00:14:17,444 --> 00:14:21,805 conductance divided by the sum of the conductances. 149 00:14:21,805 --> 00:14:28,267 So, it turns out writing the current divider rule, which is what this is over 150 00:14:28,267 --> 00:14:31,582 here. In terms of conductances, it makes it 151 00:14:31,582 --> 00:14:34,712 look a lot like the voltage divider up here, 152 00:14:34,712 --> 00:14:39,412 is when you write in terms of resistances, the expressions get 153 00:14:39,412 --> 00:14:43,047 complicated. It's a whole lot easier to write it in 154 00:14:43,047 --> 00:14:50,406 terms of the conductances in this case. By the way, there's a, a minor error on 155 00:14:50,406 --> 00:14:54,493 this sign. It should be a voltage v here and an i 156 00:14:54,493 --> 00:14:55,682 there. Okay. 157 00:14:55,682 --> 00:15:01,204 [SOUND] So, that's what we've learned about solving circuits. 158 00:15:01,204 --> 00:15:08,385 The equivalent resistance idea comes up because you asked the question, what does 159 00:15:08,385 --> 00:15:13,600 the source see? And it turns out as you can see from all these examples, when you 160 00:15:13,600 --> 00:15:18,841 use the idea of an equivalent resistance along with the ideas of voltage divider 161 00:15:18,841 --> 00:15:22,839 and current divider, you can solve circuits rather quickly. 162 00:15:22,839 --> 00:15:27,715 And notice, we solved that circuit having a load resistor on it without ever 163 00:15:27,715 --> 00:15:32,052 writing KVL, KCL, and V averages. We just wrote down the answer. 164 00:15:32,052 --> 00:15:36,477 That's really very, very nice. So, the whole idea about being able to 165 00:15:36,477 --> 00:15:39,902 solve circuits easily means seeing the structure, 166 00:15:39,902 --> 00:15:44,852 what's in series with what, what's in parallel with what, and finding the 167 00:15:44,852 --> 00:15:47,315 equivalent resistances accordingly.