1 00:00:00,012 --> 00:00:04,262 In this video, we're going to connect those circuit elements together to make 2 00:00:04,262 --> 00:00:07,527 an entire circuit. Right now, all we know about the circuit 3 00:00:07,527 --> 00:00:10,812 elements are what their individual VI characteristics are. 4 00:00:10,812 --> 00:00:15,062 How the current and voltage are related to each other because of the element's 5 00:00:15,062 --> 00:00:18,572 construction. When you connect them together, we have 6 00:00:18,572 --> 00:00:23,617 to not only connect them physically, but we have to have the laws that govern the 7 00:00:23,617 --> 00:00:27,457 voltages and currents when elements are connected together. 8 00:00:27,457 --> 00:00:32,337 And these are called the Kirchhoff's Laws, and as you might expect, there's a 9 00:00:32,337 --> 00:00:36,632 current law and a voltage law. And what we're going to find is that once 10 00:00:36,632 --> 00:00:43,554 we have the VR relations KCL, Kirchhoff's Current Law, and KVL, Kirchhoff's Voltage 11 00:00:43,554 --> 00:00:48,973 Law, we can solve any circuit. So let's start with a very simple 12 00:00:48,973 --> 00:00:53,369 circuit. And here we have three elements connected 13 00:00:53,369 --> 00:00:57,507 together. I have a voltage source and two resistors 14 00:00:57,507 --> 00:01:02,487 never connecting together one after another, in what's called a series 15 00:01:02,487 --> 00:01:06,684 configuration. And what series means is that, if I was 16 00:01:06,684 --> 00:01:10,207 electron, this is the way I like to think about it. 17 00:01:10,207 --> 00:01:15,515 An electron, if I try to go out of one element, and I have no choice, but to go 18 00:01:15,515 --> 00:01:19,796 through the next one, that means you're in series. 19 00:01:19,796 --> 00:01:26,357 So we have an entire series, configuration of a voltage source and two 20 00:01:26,357 --> 00:01:28,401 resistors. All right. 21 00:01:28,401 --> 00:01:35,593 Now, I would point out that we have three circuit elements and we have a voltaging 22 00:01:35,593 --> 00:01:43,287 incurrent for each element, which means, we have six unknowns and in order to 23 00:01:43,287 --> 00:01:49,130 define six unknowns, you're going to need six equations. 24 00:01:49,130 --> 00:01:56,606 And all we have right now are the Vi relations, so I'm calling the voltage 25 00:01:56,606 --> 00:02:04,361 across R2 our output forward just because I wanted to and here's the Vi relation 26 00:02:04,361 --> 00:02:11,806 for the source and here it is for the 2 resistors and of course anything could be 27 00:02:11,806 --> 00:02:18,819 a function of time as indicated. Alright, but 3 equations isn't 6, we need 28 00:02:18,819 --> 00:02:22,710 3 more. And that's where the interconnection laws 29 00:02:22,710 --> 00:02:26,139 come in. Alright, so the first one we're going to 30 00:02:26,139 --> 00:02:31,851 talk about is Kirchhoff's Current Law, KCL, which basically says, the sum of 31 00:02:31,851 --> 00:02:36,268 currents entering a node, or leaving a node, equals zero. 32 00:02:36,268 --> 00:02:39,972 So, what's a node? So we look back at our circuit. 33 00:02:39,972 --> 00:02:45,781 A node is a connection point. That's where 2 or more elements are 34 00:02:45,781 --> 00:02:50,282 connected together and they're really attached. 35 00:02:50,282 --> 00:02:57,012 So, in this circuit, I see a node there, I see a node here where R1 and R2 are 36 00:02:57,012 --> 00:03:03,544 connected and I see a big node here where essentially R2 is connected to the other 37 00:03:03,544 --> 00:03:08,312 end of the voltage source. And this wire going between them as we 38 00:03:08,312 --> 00:03:11,395 know is just a short circuit, an ideal wire. 39 00:03:11,395 --> 00:03:14,481 And it doesn't contribute much to anything. 40 00:03:14,481 --> 00:03:19,536 There's no voltage drop, and the current just goes through it as it would 41 00:03:19,536 --> 00:03:23,087 normally. So I'm going to consider this one being 42 00:03:23,087 --> 00:03:28,266 So we have three nodes. Now what I'm going to do is I'm going to 43 00:03:28,266 --> 00:03:35,548 write KCL for each node and I'm going to do it in terms of currents entering the 44 00:03:35,548 --> 00:03:39,708 node. So let's consider them one at a time. 45 00:03:39,708 --> 00:03:45,892 So if we're the blue node, the current entering from the south, 46 00:03:45,892 --> 00:03:52,995 the current going in that way is -i. And the current entering from the east is 47 00:03:52,995 --> 00:03:56,899 -i1. Right? You just re-flip the sign to 48 00:03:56,899 --> 00:04:01,127 reverse the direction of the current arrows. 49 00:04:01,127 --> 00:04:06,592 And so, in terms of currents entering a node, -i-i1, is 0. 50 00:04:06,592 --> 00:04:12,714 Okay. Go on to the next node, and the red node, 51 00:04:12,714 --> 00:04:23,616 the current entering from the west is the is i1, the current entering from the 52 00:04:23,616 --> 00:04:31,167 south is -i out, so i1 - iout 0. And then finally, for the, green node. 53 00:04:31,167 --> 00:04:36,729 currents entering, we have I entering here, and I entering there. 54 00:04:36,729 --> 00:04:40,753 So, those, we have 3 nodes we had 3 KCL equations. 55 00:04:40,753 --> 00:04:45,795 Well, I want you to notice something about these KCL equations. 56 00:04:45,795 --> 00:04:49,472 If I were to add these two equations together, 57 00:04:49,472 --> 00:04:54,527 take a minus sign and negate it, I would get the other equation. 58 00:04:54,527 --> 00:04:58,852 And that's not good. Remember we're trying to find six 59 00:04:58,852 --> 00:05:03,922 equations and six unknowns. And if the equations are not what's 60 00:05:03,922 --> 00:05:08,652 called linearly independent, they don't do us much good. 61 00:05:08,652 --> 00:05:13,222 There's some sense one too many equations here. 62 00:05:13,222 --> 00:05:20,908 So, what's going on, and what's going on is a detail about KCL and construction of 63 00:05:20,908 --> 00:05:26,093 every circuit. So, if you think about this circuit in 2 64 00:05:26,093 --> 00:05:30,953 pieces. So we have the big yellowish kind of area 65 00:05:30,953 --> 00:05:38,571 here, and our green node at the bottom. Because KCL is satisfied at each of the 66 00:05:38,571 --> 00:05:44,670 individual nodes here. The sum of currents leaving the yellow 67 00:05:44,670 --> 00:05:51,703 surface has to be is, i + iout. Well those currents have to leave and go 68 00:05:51,703 --> 00:05:58,970 into the green node by definition. So that gives us the other KCL equation 69 00:05:58,970 --> 00:06:05,147 of currents entering. So, it turns out that in any circuit, if 70 00:06:05,147 --> 00:06:09,749 you have n nodes, you have n-1KCL equations that are 71 00:06:09,749 --> 00:06:15,704 linearly independent and the only nature writing for any set of n-1 nodes, you can 72 00:06:15,704 --> 00:06:20,282 pick anything more. As you're going to as we go through the 73 00:06:20,282 --> 00:06:27,227 course, we typically write circuits such that there's always sort of I note at the 74 00:06:27,227 --> 00:06:32,752 bottom where lots of things are connected together, and that tends to be the one we 75 00:06:32,752 --> 00:06:36,477 don't write KCL for. And so I'm going to throw out that 76 00:06:36,477 --> 00:06:38,302 equation. I don't need it. 77 00:06:38,302 --> 00:06:41,952 It can be derived from the other two. So let's move on. 78 00:06:41,952 --> 00:06:47,514 we're going to need another equation. We've got three VI relations, two KCL 79 00:06:47,514 --> 00:06:50,912 equations that are independent of each other. 80 00:06:50,912 --> 00:06:56,692 we need that last equation so we can solve our circuit, and of course that's 81 00:06:56,692 --> 00:07:01,634 going to come from the voltage law. So Kirchhoff's voltage law is that the 82 00:07:01,634 --> 00:07:06,922 sum of voltages around any closed loop in a circuit must add up to zero. 83 00:07:06,922 --> 00:07:12,385 So here's our circuit, and we have three voltages here, that are defined. 84 00:07:12,385 --> 00:07:17,840 And what we're going to do is go around a little loop, and what, the way you do 85 00:07:17,840 --> 00:07:23,361 that is if we define we're going around the loop clockwise, that's what the 86 00:07:23,361 --> 00:07:30,132 graphic says, we're going to take The voltage signs according if it goes plus 87 00:07:30,132 --> 00:07:34,342 to minus. So, I'm going to bring in the one with 88 00:07:34,342 --> 00:07:42,112 the plus sign, the out, with a plus sign. But over here, d is defined minus the 89 00:07:42,112 --> 00:07:46,432 plus, so I have to bring it in with a minus. 90 00:07:46,432 --> 00:07:52,437 And that's my, KVL equation. So, from one point of view I'm very 91 00:07:52,437 --> 00:07:59,553 happy, in terms of solving my circuit. I've got 3Vi relations, 2 KCL equations, 92 00:07:59,553 --> 00:08:05,912 and 1 KVL equation, so that's 6 equations; we should be able to solve our 93 00:08:05,912 --> 00:08:09,739 circuit. But the similar issues arises with KVL 94 00:08:09,739 --> 00:08:13,762 that does in KCL. How many K, KVL equations are there 95 00:08:13,762 --> 00:08:19,297 actually for any circuit? Here there's only one loop, but how about this 96 00:08:19,297 --> 00:08:26,055 circuit? How many loops are there? Well, I get one loop, involving On the voltage 97 00:08:26,055 --> 00:08:31,230 source; r1 and r2. And, I get another loop out here, in 98 00:08:31,230 --> 00:08:34,770 around it, that's surrounded by r2 and r3. 99 00:08:34,770 --> 00:08:38,498 However, I point out, there's another loop. 100 00:08:38,498 --> 00:08:42,583 And that is, a green loop, that's also there too. 101 00:08:42,583 --> 00:08:48,931 Well, it turns out, the number of KVL equations you need Is determined by how 102 00:08:48,931 --> 00:08:55,235 many open areas there are in the circuit. If you thought about this as a a window, 103 00:08:55,235 --> 00:09:01,265 and these were pieces of wood holding together the pieces of glass, the number 104 00:09:01,265 --> 00:09:05,107 of KVL equations is the number of pieces of glass. 105 00:09:05,107 --> 00:09:09,539 In this case, two. So, the number of KVL equations I would 106 00:09:09,539 --> 00:09:14,712 write would be two, and I'd write it for each pane of glass, if you will. 107 00:09:14,712 --> 00:09:20,826 So, that's how you find the number of KVL equations that any, that need, that you 108 00:09:20,826 --> 00:09:25,924 need to solve for any circuit. All right, so let's solve that circuit. 109 00:09:25,924 --> 00:09:33,043 So we have 3Vi relations, we had 2 KCL equations and we have 1 KVL equation. 110 00:09:33,043 --> 00:09:40,611 Aren't we happy? All we understand is finding, what the output voltage is for 111 00:09:40,611 --> 00:09:48,403 the given source voltage and we have 6 equations and 6 unknowns to plow through. 112 00:09:48,403 --> 00:09:54,415 So we want to eliminate everything. We will eliminate all the currents, we 113 00:09:54,415 --> 00:09:59,675 will eliminate the one, and we want a just relationship between Vi and V. 114 00:09:59,675 --> 00:10:05,259 And with a little bit of going through this and thinking about it, is actually 115 00:10:05,259 --> 00:10:09,628 pretty easy to do, and I won't go through all that agony. 116 00:10:09,628 --> 00:10:13,032 I'll just say that after some manipulation. 117 00:10:13,032 --> 00:10:16,841 you should be a little worried about that. 118 00:10:16,841 --> 00:10:22,964 It's easily shown, that what you get for an output, is that it is the output 119 00:10:22,964 --> 00:10:29,377 voltages proportional to the input voltage, and constant proportionality is 120 00:10:29,377 --> 00:10:32,992 given by the ratio of R2 to the sum, R1 + R2. 121 00:10:32,992 --> 00:10:38,971 It turns out, we have derived this in general, right? We've, have, don't have 122 00:10:38,971 --> 00:10:43,177 specific resistors. I usually derive this in general. 123 00:10:43,177 --> 00:10:48,945 So anytime you see a voltage source attached to two resistors in series like 124 00:10:48,945 --> 00:10:55,245 this, you'll always find that the voltage across a resistor of phases rule. 125 00:10:55,245 --> 00:11:01,050 And in fact, I'm going to ask you to tell me, what is the voltage v1? And I want 126 00:11:01,050 --> 00:11:04,707 you to do that without doing any calculations. 127 00:11:04,707 --> 00:11:09,822 I want you, you should be able to just tell me what that answer is. 128 00:11:09,822 --> 00:11:16,124 All right. Well that relationship that we, for, 129 00:11:16,124 --> 00:11:25,762 voltage across a series connection of resistors is called voltage divider, and 130 00:11:25,762 --> 00:11:34,632 the resistors the voltage, divide it. Is where the terminology comes from. 131 00:11:34,632 --> 00:11:39,417 So the bigger R2 is the closer the ratio here is to 1. 132 00:11:39,417 --> 00:11:46,382 The smaller r2 is relative to r1 the smaller this proportionality constant 133 00:11:46,382 --> 00:11:50,042 gets. So, what does our circuit do? So, 134 00:11:50,042 --> 00:11:56,966 remember what we're trying to do is build systems and we're trying to build them at 135 00:11:56,966 --> 00:12:01,852 this point out of circuits. So in terms of black diagram models, we 136 00:12:01,852 --> 00:12:07,603 have a source that's producing some voltage VN, and there's the source, it's 137 00:12:07,603 --> 00:12:12,943 literally the voltage sources. The system is the entire circuit it's 138 00:12:12,943 --> 00:12:18,647 attached to, and we've taken as an output the voltage across R2 And what our 139 00:12:18,647 --> 00:12:25,464 system, the input output relationship, we have shown is given by that expression. 140 00:12:25,464 --> 00:12:31,630 So what does our system do? Have we encountered a system like that already? 141 00:12:31,630 --> 00:12:37,332 And the answer is we certainly have. It serves as a kind of amplifier. 142 00:12:37,332 --> 00:12:43,709 The gain of our amplifier is simply R2 divided by R1 + R2 and because our gain 143 00:12:43,709 --> 00:12:50,086 has to be less than 1, resistors are always positive valued, it turns out we 144 00:12:50,086 --> 00:12:54,706 have an attenuator. And so what our little system, our 145 00:12:54,706 --> 00:13:01,496 circuit does In terms of system theory, is that it serves as an attenuator that 146 00:13:01,496 --> 00:13:07,101 makes voltages smaller. So, let's do a little power calculation. 147 00:13:07,101 --> 00:13:11,166 I'm very interested in what is consuming power. 148 00:13:11,166 --> 00:13:16,082 And what is the source of the power that's being consumed. 149 00:13:16,082 --> 00:13:26,713 So as we know power for each element is the voltage that we have * it's current 150 00:13:26,713 --> 00:13:32,957 and I want to consider that for each element. 151 00:13:32,957 --> 00:13:41,024 So we know for resistor that the voltage, the power rather, is the voltage squared 152 00:13:41,024 --> 00:13:47,490 divided by the resistance, that's v times i for a resistor, and since we solved for 153 00:13:47,490 --> 00:13:53,223 the output voltage, we might as well use this version of the formula. 154 00:13:53,223 --> 00:13:58,231 So what we get is that the voltage, output, 155 00:13:58,231 --> 00:14:06,905 the output voltage, just to remind you was R2 over R1 + R2 - VN, so you square 156 00:14:06,905 --> 00:14:14,182 that and divide by R2 you get that, and similarly the hour 157 00:14:14,182 --> 00:14:20,817 dissipated across resistor one is if you had a similar expression where R2 R2 is 158 00:14:20,817 --> 00:14:24,692 replaced by R1, makes things pretty easy to see. 159 00:14:24,692 --> 00:14:31,457 Because these are positive valued powers, right? Resistors are always positive 160 00:14:31,457 --> 00:14:35,792 value and we're the [UNKNOWN] the voltage coming in. 161 00:14:35,792 --> 00:14:42,467 The terms of that means these 2 resistors dissipate power, they consume it. 162 00:14:42,467 --> 00:14:49,317 Well it has to come from somewhere, so let's see what's going on at the source. 163 00:14:49,317 --> 00:14:52,967 Yeah. The output current is what we need. 164 00:14:52,967 --> 00:14:59,237 We need to find We have the voltage v that's just given by the source and we 165 00:14:59,237 --> 00:15:03,027 won't need to find that current i is going in. 166 00:15:03,027 --> 00:15:09,747 Well, for a series combination, I think it's pretty clear that i1 = iout and that 167 00:15:09,747 --> 00:15:13,327 i = -i1. So, if I just find the iout which is 168 00:15:13,327 --> 00:15:18,791 given by the voltage that we found divided by R2. 169 00:15:18,791 --> 00:15:27,138 We know that that is iout, and I will be equal to -iout and we get for a power 170 00:15:27,138 --> 00:15:35,433 produced by the source, is the negative of VN squared divided by R1+R2, and 171 00:15:35,433 --> 00:15:41,725 negative means it produces power. It is the one that is producing power in 172 00:15:41,725 --> 00:15:45,542 a circuit. If you add these two up, if you add 173 00:15:45,542 --> 00:15:50,767 everything up, you get zero, and that's the way it should be. 174 00:15:50,767 --> 00:15:56,685 It has to be that power is conserved. There are going to be parts of a circuit 175 00:15:56,685 --> 00:16:02,068 that consume power, and there's going to be parts that produce it and they have to 176 00:16:02,068 --> 00:16:06,389 be in balance or something is wrong with, with the calculations. 177 00:16:06,389 --> 00:16:10,686 And so, no net power is produced or consumed by a simple circuit. 178 00:16:10,686 --> 00:16:17,521 That's the way the laws of physics are. So, and this is the general formula that 179 00:16:17,521 --> 00:16:25,349 if you, for a given circuit, if you sum over all the circuit elements the sum of 180 00:16:25,349 --> 00:16:30,436 the b k, i k's have to be zero. We'll prove that later. 181 00:16:30,436 --> 00:16:37,938 So, along with the v i relations, KVL and KCL always give you exactly the number of 182 00:16:37,938 --> 00:16:44,221 equations you need, to solve any circuit. So you have to go through them all, and 183 00:16:44,221 --> 00:16:48,418 of course you don't need all the set of equations. 184 00:16:48,418 --> 00:16:54,060 What I mean by that, if our, n-node circuit, you need n-1 KCL equations. 185 00:16:54,060 --> 00:16:59,796 And for, a number of openings in a circuit are only, the number of loops you 186 00:16:59,796 --> 00:17:04,354 need for which you write KVR. Now, I do not want to spend my life, 187 00:17:04,354 --> 00:17:09,390 solving lots and lots of sets of equations just to find an input, output 188 00:17:09,390 --> 00:17:13,249 relationship. So we really need methods to streamline 189 00:17:13,249 --> 00:17:17,422 our calculations, and we're going to talk about those soon. 190 00:17:17,422 --> 00:17:23,837 And it turns out, they're very easy, to use, and, not only simplifying our life, 191 00:17:23,837 --> 00:17:27,402 but give us insight, into how circuits behave. 192 00:17:27,402 --> 00:17:32,952 Now, a little bit later I'm going to prove, that only KVL and KCL are needed, 193 00:17:32,952 --> 00:17:36,212 to show that every circuit Conserves power. 194 00:17:36,212 --> 00:17:41,014 Some of the BK, IKs is indeed zero. That's really interesting and we'll see 195 00:17:41,014 --> 00:17:42,856 of that in a succeeding video.