1 00:00:00,012 --> 00:00:05,000 In this video, we're going to talk about using what are called formal circuit 2 00:00:05,012 --> 00:00:10,400 methods that are needed to solve circuits that you can't use a series in parallel 3 00:00:10,412 --> 00:00:16,900 rules on. all the circuits we've seen to date, using voltage divider and current 4 00:00:16,912 --> 00:00:22,968 divider are all you need to really solve circuits and that's applies in a vast 5 00:00:22,980 --> 00:00:29,050 majority of the cases but there are a few that you can't. They had a bit different 6 00:00:29,062 --> 00:00:35,220 structure and I'll show you an example of one in just a second. So instead of going 7 00:00:35,232 --> 00:00:41,077 back to the v-i relations, KVL, KCL and solving a load of equations, we're going 8 00:00:41,077 --> 00:00:46,165 to develop something called the node method, which very efficiently uses the 9 00:00:46,177 --> 00:00:51,287 v-i relations, KVL and KCL. It turns out it's so efficient that you may want to use 10 00:00:51,299 --> 00:00:56,701 it to check your answers that you get by other techniques. It's very nice, and very 11 00:00:56,713 --> 00:01:02,499 general. One of the surprises is that you can use the node method to prove that 12 00:01:02,511 --> 00:01:08,452 circuits conserve power. the assumptions underlying that proof are pretty 13 00:01:08,464 --> 00:01:14,589 interesting, and we'll see how that works out but first, a circuit that's a little 14 00:01:14,601 --> 00:01:21,413 harder than it looks. So here's the circuit, and let's look at its structure. 15 00:01:21,563 --> 00:01:30,187 See if it fits into the mold of series and parallel circuits. Well, let's start over 16 00:01:30,199 --> 00:01:38,472 here. This resistor looks like it could be in parallel with this series combination, 17 00:01:38,622 --> 00:01:46,105 but these two aren't in series. Because there's this, this branch going off into 18 00:01:46,117 --> 00:01:53,055 that 2 ohm resistor. So, these aren't in series and because of the, that branch, 19 00:01:53,192 --> 00:01:59,930 this is not in parallel with those two either. And these two sort of look to be 20 00:01:59,942 --> 00:02:06,280 in series but they're not because of this side branch et cetera because if you go 21 00:02:06,292 --> 00:02:12,396 through it nothing is in series with anything, nothing is in parallel with 22 00:02:12,408 --> 00:02:19,120 anything, it's kind of interesting. We still want to solve circuits like this and 23 00:02:19,132 --> 00:02:25,479 we can do it, we know that given the v-i relations, KVL and KCL we can solve any 24 00:02:25,491 --> 00:02:31,850 circuit. It's just that is very tedious to solve it that way. because, lets see, this 25 00:02:31,862 --> 00:02:38,986 circuit has 1, 2, 3, 4, 5, 6 elements? That means 12 unknowns. And I don't r 26 00:02:38,986 --> 00:02:45,405 eally want to write all those equations, all 12 equations and 12 unknowns. So lets 27 00:02:45,417 --> 00:02:52,980 talk about the node method. Node method is very easy and very simple. So here's our 28 00:02:52,992 --> 00:03:00,907 circuit and what you do is you start by defining what are called node voltages. 29 00:03:01,062 --> 00:03:09,310 You pick a node as a reference and usually it's the one that you don't write KCL for 30 00:03:09,322 --> 00:03:16,885 if you use the old techniques. There's usually this big node at the bottom, and 31 00:03:16,897 --> 00:03:23,885 I'm going to pick that as being the reference. And that symbol is used for 32 00:03:23,897 --> 00:03:31,199 reference and I have defined a nide voltage is almost all the other nodes. So, 33 00:03:31,335 --> 00:03:38,605 here's a node here, a node here, and a node there. And what this, these voltages, 34 00:03:38,741 --> 00:03:45,366 e1 means, it's the voltage between that node and the reference. E2 means the 35 00:03:45,378 --> 00:03:51,744 voltage between this node and the reference. And since the voltage source is 36 00:03:51,756 --> 00:03:58,174 attached to this node in the reference, we know that this node voltage is vn, so I 37 00:03:58,186 --> 00:04:03,648 don't need to define the node voltage there, it's not necessary. However, 38 00:04:03,769 --> 00:04:10,254 something a little bit more interesting, that when we define node voltages It also 39 00:04:10,266 --> 00:04:18,520 means it doesn't matter which path we took to go between this node and the reference. 40 00:04:18,671 --> 00:04:26,127 That is the voltage there. So that saying that this voltage is equal to that 41 00:04:26,139 --> 00:04:33,690 voltage, plus that Voltage, so you can go around this way, or you can go around that 42 00:04:33,702 --> 00:04:40,765 way, you get to that reference and that voltage is e1. Well that is, that implies 43 00:04:40,777 --> 00:04:47,840 KVL, so we're using the KVL relations implicitly when we define node voltages as 44 00:04:47,852 --> 00:04:55,315 just being the voltage between some point in the circuit, a node and the reference. 45 00:04:55,457 --> 00:05:03,035 And this voltage applies no matter how what path you take to get between the two. 46 00:05:03,177 --> 00:05:10,990 So the n minus 1 here refers to the fact that we don't need to write a define a 47 00:05:11,002 --> 00:05:16,306 node voltage for the reference, it's the reference. And it turns out in this case, 48 00:05:16,416 --> 00:05:21,674 we don't need to write three node voltages. This has, this is a four node 49 00:05:21,686 --> 00:05:26,990 circuit because we already know one of the node voltage. So we only need to define 50 00:05:27,002 --> 00:05:33,760 two, E1 and E2. That is going to turn out to be our only set of unknowns. We do n't 51 00:05:33,767 --> 00:05:40,655 need anything else. So that means we only need two equations to write those two 52 00:05:40,667 --> 00:05:47,530 unknowns. We've used KVL, so the next thing we're going to do is use KCL at each 53 00:05:47,542 --> 00:05:54,118 node and we're going to write the case seal equations using the node voltages and 54 00:05:54,130 --> 00:05:59,631 the v-i relations all at once. So that uses up all the things, tools that we 55 00:05:59,643 --> 00:06:05,202 need. We know that the, defining node voltages using up KVL, we're going to 56 00:06:05,214 --> 00:06:11,679 write KCL use up the v-i relations while we do it. So how do we do that? So here's 57 00:06:11,691 --> 00:06:19,755 the set of KCL for this circuit. So let's start with e1, the node here, node 1, 58 00:06:19,898 --> 00:06:27,388 there's a current going that way, current going that way and a current going that 59 00:06:27,400 --> 00:06:33,573 way and I'm going to write the KCL equation, and all those currents sum to 60 00:06:33,585 --> 00:06:38,870 zero. But as I write the currents I'm going to go ahead and use the v-i 61 00:06:38,882 --> 00:06:45,236 relations. Now the current going that way means plus to minus that way, and so that 62 00:06:45,248 --> 00:06:52,329 is e1 minus v-n is that voltage divided by 1, which is the resistance. So that's what 63 00:06:52,341 --> 00:06:59,772 I mean by using the v-i relations at the same time. So the west going current is 64 00:06:59,784 --> 00:07:06,954 e1-, vn over 1. The software in current is simply e1 over 1. And the east going 65 00:07:08,499 --> 00:07:19,702 current is e1 minus e2 divided by 1 again. And those have to be 0. That's KCL at that 66 00:07:19,714 --> 00:07:31,336 node. So we can write KCL at the other node. , where e2 is so it's e2 minus e1 67 00:07:31,579 --> 00:07:44,246 divided by 1, e1 divided by 1, and e2 minus v in divided by 2 is the last KCL 68 00:07:44,246 --> 00:07:48,154 equations. So, we've exhaust, we've used up all the 69 00:07:48,450 --> 00:07:53,440 circuit laws. There are KVL, KCL, NDR relations. Turns out they're all 70 00:07:53,452 --> 00:07:58,626 implicitly or explicitly used. We have wound up with two equations and two 71 00:07:58,638 --> 00:08:03,781 unknowns. And all I have to do is solve them, and you, there's no tricks to 72 00:08:03,793 --> 00:08:09,966 solving them. You just solve this set of linear equations. Now I want to point out 73 00:08:09,978 --> 00:08:17,302 something that also makes the node method almost spill proof. And that is, that you 74 00:08:17,314 --> 00:08:24,840 look at the equations for node 1, since we did some of currency leaving, the node 75 00:08:24,852 --> 00:08:31,524 voltage for that node always appears with the plus sign. The node voltage at the 76 00:08:31,536 --> 00:08:37,189 other nodes always appears with a minus sign. Always, always, always, always, 77 00:08:37,307 --> 00:08:42,979 always, if you obey this convention. So, and see that happens here, too. e2 is 78 00:08:42,991 --> 00:08:48,401 always with a plus sign, and the other node voltages are always there with a 79 00:08:48,413 --> 00:08:54,319 minus sign. And, so at elast you get the signs right for the voltages, in your KCL 80 00:08:54,331 --> 00:09:00,328 equations you gotta be careful and use the right value for the resistence, other than 81 00:09:00,340 --> 00:09:06,179 that, it's foolproof, and hopefully you can solve the equations without error. So, 82 00:09:06,295 --> 00:09:12,403 2 equations, 2 unknowns and all we have to do is just solve them for the node 83 00:09:12,415 --> 00:09:19,567 voltages and I did this. You should check to make sure I got the right answers, and 84 00:09:19,579 --> 00:09:24,181 that e1 is going to be six-thirteenths of v in. It just turns out to be 85 00:09:24,193 --> 00:09:29,114 six-thirteenths. No good reason why it should be, but it is and e2 is 86 00:09:29,126 --> 00:09:34,429 five-thirteenths. So just solve, 2 equations, 2 unknowns and you get the 87 00:09:34,441 --> 00:09:41,494 result in terms of Vin. And now what you do, is you express the output in terms of 88 00:09:41,506 --> 00:09:48,686 the node voltage. And for this circuit example, I choose that current as being my 89 00:09:48,698 --> 00:09:56,721 output. And I think it's pretty easy to see that that's just e2 divided by 1 and 90 00:09:56,733 --> 00:10:03,723 so the answer is five-thirteenths v in. A little note about units here this current 91 00:10:03,735 --> 00:10:11,521 should be in amperes, and it looks like I have a voltage here, and I do but it turns 92 00:10:11,533 --> 00:10:17,440 iyt the five-thirteenths, if you look back up at this answer, it turns out to be at 93 00:10:17,452 --> 00:10:23,871 no dimensions, that 1 has a dimension in ohms. When you get numeric answers like 94 00:10:23,883 --> 00:10:30,143 this, it tends to hide the fact that the units are correct, it may look a little 95 00:10:30,155 --> 00:10:35,897 strange but if you did everything correctly the units should all work out 96 00:10:35,909 --> 00:10:42,107 and your answer would be correct. Okay, let's use the node method to prove that 97 00:10:42,119 --> 00:10:49,702 all circuits conserve power, very powerful statement. So, here is a section of a 98 00:10:49,714 --> 00:10:59,006 generic circuit, not going to tell you, and I don't even know, what these elements 99 00:10:59,018 --> 00:11:07,814 are, they're just elements. And it's a piece of a circuit, because i7 is going 100 00:11:07,826 --> 00:11:14,142 somewhere, 5, 6, and 4 They're going off. There's some elements that have those 101 00:11:14,154 --> 00:11:19,229 numbers, and, I don't need to define exactly what each circuit is, because I 102 00:11:19,241 --> 00:11:24,834 want a very general answer. I'm using this core of the circuit to illustrate the core 103 00:11:24,846 --> 00:11:29,798 of the circuit to illustrate that I want to make. Now, what I want to, what I'm 104 00:11:29,810 --> 00:11:36,961 headed for is to show. That the VK iK is equal to zero. So what I'm going to do is 105 00:11:36,973 --> 00:11:46,145 express the voltages across each element in terms of node voltages. The direction 106 00:11:46,157 --> 00:11:54,328 for that voltage is going to be defined by the direction I chose for the currents. 107 00:11:54,328 --> 00:11:58,012 So. I3 is going that way. So I'm going to 108 00:11:58,024 --> 00:12:04,800 define d3 to be that. So that I get a consistent answer that goes according to 109 00:12:04,812 --> 00:12:11,361 the rules by which we define the I relations. So, voltage across here goes 110 00:12:11,373 --> 00:12:18,372 like that. And voltage across there is going to be like that, okay? So here's 111 00:12:18,384 --> 00:12:26,189 what we're going to do, we're going to write VKIK for each element using the node 112 00:12:26,201 --> 00:12:34,193 voltages. Okay, so let's look at element 1, let's see i1 goes that way, that means 113 00:12:34,205 --> 00:12:42,432 it's plus to minus that way. So that means it's eb minus ea. And that's right up 114 00:12:42,444 --> 00:12:52,237 there. And v 2, let's see plus or minus that way, node c minus node a. Right, okay 115 00:12:52,249 --> 00:13:01,018 times i 2 and ec minus eb times 93 that looks right. And, let's see for circuit 116 00:13:01,030 --> 00:13:08,241 element five which is somewhere out here is eb minus whatever's at the end of it, 117 00:13:08,381 --> 00:13:16,575 times i5 and same thing for i6. I didn't write i4 and i7. we won't need them. Okay, 118 00:13:16,727 --> 00:13:24,130 so the next thing I want to do is because I'm considering the sum of all these 119 00:13:24,142 --> 00:13:31,775 terms, which are now expressed in terms of node voltages, I want to group them 120 00:13:31,787 --> 00:13:38,974 according to node. What I'm going to do is group these equations, according to eb. 121 00:13:39,242 --> 00:13:46,144 I'm going to find all the terms involved in eb and write them down, collect them 122 00:13:46,156 --> 00:13:53,595 together. So the eb terms we have e b times, these currents. So let's check our 123 00:13:53,772 --> 00:14:02,085 answers here we have an i1 from this equation. There's no eb in this equation. 124 00:14:02,672 --> 00:14:11,260 here's a minus i3, okay. eb times an i5, okay, and an eb times an i6, okay. Yeah, 125 00:14:11,422 --> 00:14:19,315 we got them all. Okay. Notice that this equation is the KCL 126 00:14:19,327 --> 00:14:30,680 equation for node b. So, i1, i is leaving, i3 is entering. So it gets a minus sign. 127 00:14:30,883 --> 00:14:39,452 i5 is leaving, i6 is leavin g. So that is KCL and what's going to happen when you 128 00:14:39,464 --> 00:14:46,956 group the sum of the VKs, IKs. You group them according to node voltage what you're 129 00:14:46,968 --> 00:14:53,458 going to get is that KCL because KCL applies at every node, you're going to get 130 00:14:53,470 --> 00:15:02,629 a KCL equation. So, what, since all these terms are zero, for every node voltage, 131 00:15:02,804 --> 00:15:12,329 the sum of the VKIK's is zero. Amazing. So, we have proven what me wanted. That. 132 00:15:12,332 --> 00:15:20,784 But this is summed of all elements in a circuit including sources. That the vk, 133 00:15:20,947 --> 00:15:29,827 ik's multiplying together, add 'em up you get zero. Okay, so that proves that all 134 00:15:29,839 --> 00:15:38,536 circuits conserve power. But what did we use to prove this? What we used were the, 135 00:15:38,697 --> 00:15:47,096 the node voltages and then, when we grouped terms, we used KCL. Using node 136 00:15:47,108 --> 00:15:56,149 voltages implies KVL. So turns out, the only thing that is required through, that 137 00:15:56,161 --> 00:16:05,243 circuits convert, conserve power is KVL and KCL. But that means that v-i relations 138 00:16:05,255 --> 00:16:11,880 are immaterial. So this proves that for resis-, circuits consisting of resistors, 139 00:16:11,998 --> 00:16:17,196 inductors, capacitors. Non linear elements, like diodes. All kinds of 140 00:16:17,208 --> 00:16:22,861 things. It really doesn't matter. It's just that KVL and KCL are sufficient to 141 00:16:22,873 --> 00:16:29,315 show that circuits conserve power. That's really interesting, I think. It really is 142 00:16:29,327 --> 00:16:35,660 a very interesting and important result that all circuits conserve power now 143 00:16:35,672 --> 00:16:42,291 matter what, how you build them and no matter what elements are used to construct 144 00:16:42,303 --> 00:16:48,997 them. Okay, so it turns out you can use the node method for circuits involving 145 00:16:49,009 --> 00:16:56,987 impedances. So and that's because complex amplitudes obey KVL and KCL. And, so you 146 00:16:56,999 --> 00:17:03,952 can use the node method for an RLC circuit. So I took the simple circuit we 147 00:17:03,964 --> 00:17:11,362 used in our example, and replaced an elements by some inductors and capacitors. 148 00:17:11,522 --> 00:17:18,045 So, we begin using the convention that capital letters denote complex amplitudes. 149 00:17:18,172 --> 00:17:24,270 The node voltages are now complex amplitudes. That's a capital V although it 150 00:17:24,282 --> 00:17:30,095 may not look like it. You define a reference and you do exactly what we did 151 00:17:30,107 --> 00:17:37,242 before. Here are the two, KCL equations for nodes 1 and node 2. And now you, 152 00:17:37,386 --> 00:17:45,553 instead of resistance, you're dividing by the impedance. So, let's see e1 minus Vin, 153 00:17:45,697 --> 00:17:52,373 well that's this voltage. And the impedance of the inductor is 1 times S. 154 00:17:52,517 --> 00:18:00,085 I'm using the S instead of j 2 pi f again. because it really cleans up the notation 155 00:18:00,097 --> 00:18:06,255 in the equations. And I'll substitute that back in there when we're done. E 1 is 156 00:18:06,267 --> 00:18:12,626 sitting on top of the capacitor. So is E 1 divided by the impedance of the one farad 157 00:18:12,638 --> 00:18:19,230 capacitor. And E 1 minus E 2. And divided by the resistance of the one resistor, et 158 00:18:19,242 --> 00:18:24,448 cetera. You should check the equation to make sure that I didn't make any errors, 159 00:18:24,558 --> 00:18:29,570 and of course, the final step is to express the output which is the current in 160 00:18:29,582 --> 00:18:35,494 terms of no voltages in mean and this case, it's e2 divided by the impedance of 161 00:18:35,506 --> 00:18:41,251 the inductor, which is 1s. Okay. So, everything follows. So you can solve 162 00:18:41,263 --> 00:18:47,045 any circuit using a node method. Really, just resistors, or at whether RLC 163 00:18:47,057 --> 00:18:53,704 circuits. It really doesn't matter. you can solve all of them. And similarly, 164 00:18:53,858 --> 00:19:01,592 because it works, the proof that we just used to prove conservation of power 165 00:19:01,604 --> 00:19:10,491 applies also to complex power. , the complex power. And this is the, the 166 00:19:10,503 --> 00:19:16,563 average power for the Kth element. That also applies. 167 00:19:16,566 --> 00:19:22,558 The reason is, is that if you have the KCL equations and if you just take the 168 00:19:22,570 --> 00:19:28,093 conjugates of them, that's what can be, the conjugates also, adding the 169 00:19:28,105 --> 00:19:34,842 conjugates, conjugates of complex amplitudes of current also equal zero. So 170 00:19:34,843 --> 00:19:41,899 this shows that power in all senses that we've talked about, complex amplitudes or 171 00:19:42,252 --> 00:19:48,998 power in the time domain all have to add up to be 0. All circuits from any 172 00:19:49,010 --> 00:19:55,785 viewpoint satisfy conservation of All right, so the node method we can solve any 173 00:19:55,797 --> 00:20:02,494 RL circuit using it and it's especially used when voltage and current dividers 174 00:20:00,969 --> 00:20:08,195 show currents can't be applied as we start talking about electronics, you'll see 175 00:20:08,207 --> 00:20:14,145 there's a really good reason why we need to talk about the node method. Like I 176 00:20:14,157 --> 00:20:19,693 said, it can be used to check your answers. it's so easy to use. It's so 177 00:20:19,705 --> 00:20:25,572 efficient. It's hard to write down the wrong node. I'm sorry, the wrong TCL 178 00:20:25,584 --> 00:20:32,055 equations. So it's a pretty good way of c hecking answers. And its because of the 179 00:20:32,067 --> 00:20:37,430 generality of the node method that applies to all circuits, it's the underlying 180 00:20:37,442 --> 00:20:41,455 reason we could use it to prove that circuits conserve power.