1 00:00:03,100 --> 00:00:03,660 Welcome back. 2 00:00:03,660 --> 00:00:09,130 This is a lesson on Systematic Solution Methods. 3 00:00:09,130 --> 00:00:11,290 I've actually broken this lesson into two 4 00:00:11,290 --> 00:00:13,760 parts, so they'll be on two different videos. 5 00:00:15,410 --> 00:00:17,890 Looking at this module, we've covered a lot of 6 00:00:17,890 --> 00:00:21,440 background material so far, and in these two lessons. 7 00:00:21,440 --> 00:00:23,760 We will be doing the Systematic Solution Methods. 8 00:00:24,770 --> 00:00:28,290 From the previous lesson, we demonstrated the superposition principle, 9 00:00:28,290 --> 00:00:30,750 for circuits with multiple sources. 10 00:00:30,750 --> 00:00:32,778 Now, the lesson that we're covering today, 11 00:00:32,778 --> 00:00:35,780 we're going to be introducing several different methods. 12 00:00:35,780 --> 00:00:39,140 And it doesn't matter if we've got one source, or many sources, 13 00:00:39,140 --> 00:00:43,510 there are systematic ways of solving the problems of finding voltages and currents. 14 00:00:45,230 --> 00:00:46,726 They are all based on a physical behavior 15 00:00:46,726 --> 00:00:49,220 and the physical laws that we've already covered. 16 00:00:49,220 --> 00:00:53,180 Ohm's Law, the Kirchhoff's Voltage Law, or KVL, 17 00:00:53,180 --> 00:00:56,050 the Kirchhoff's Current Law or KCL. 18 00:00:56,050 --> 00:01:01,970 Let's give you an overview of these different methods. 19 00:01:01,970 --> 00:01:04,978 There's some Mesh analysis and Node analysis. 20 00:01:04,978 --> 00:01:08,030 Now, these two are both algebraic in nature. 21 00:01:08,030 --> 00:01:12,184 In one case, the Mesh analysis, we're going to be looking at Systematic 22 00:01:12,184 --> 00:01:15,199 KVLs, systematic applications of the Kirchhoff's 23 00:01:15,199 --> 00:01:18,348 Volt, Voltage Law, to obtain simultaneous equations 24 00:01:18,348 --> 00:01:19,940 for currents. 25 00:01:19,940 --> 00:01:23,490 And in the Node analysis, we'll be applying the KCL. 26 00:01:23,490 --> 00:01:27,190 Systematically to obtain simultaneous equations for voltages. 27 00:01:27,190 --> 00:01:29,430 So in both cases, they're algebraic. 28 00:01:29,430 --> 00:01:33,730 The Thevenin and Norton equivalent circuits are more graphical in nature. 29 00:01:33,730 --> 00:01:36,315 We look at a circuit, and we look at the schematic of that 30 00:01:36,315 --> 00:01:39,950 circuit, and we try to reduce down that schematic to a simpler form. 31 00:01:39,950 --> 00:01:42,370 So it's a little bit more graphical in nature. 32 00:01:42,370 --> 00:01:43,407 So in this part of the 33 00:01:43,407 --> 00:01:46,640 lesson, part one, we'll be doing the mode, the node and Mesh 34 00:01:46,640 --> 00:01:51,090 Analysis and then, the Thevenin and Norton equivalent circuits in the second part. 35 00:01:52,500 --> 00:01:54,210 Let's take a look at Mesh Analysis. 36 00:01:55,240 --> 00:01:57,385 There are several steps to do, first is 37 00:01:57,385 --> 00:02:01,070 to define mesh currents, one for each noninclusive loop. 38 00:02:01,070 --> 00:02:05,220 By non-inclusive, I mean a loop that doesn't contain other loops. 39 00:02:05,220 --> 00:02:08,993 For example, here we have three non-inclusive loops. 40 00:02:08,993 --> 00:02:12,570 And then we're going to do a KVL around each loop. 41 00:02:12,570 --> 00:02:18,150 And these, I've defined these currents as what I'm going to call mesh currents. 42 00:02:18,150 --> 00:02:22,680 Now these are the unknown parameters here. 43 00:02:22,680 --> 00:02:26,680 I've introduced some of this intermediate variables that I need to solve for. 44 00:02:26,680 --> 00:02:30,330 So I do a KVL around each loop and then I solve for those quantities. 45 00:02:30,330 --> 00:02:34,580 Now let me go back and remind you of a couple of things because 46 00:02:34,580 --> 00:02:40,170 this is something that some students have trouble with and that's plarity. 47 00:02:40,170 --> 00:02:42,010 Just going back to Ohm's law. 48 00:02:42,010 --> 00:02:50,030 If I have i current going into a resister, the corresponding voltage is here. 49 00:02:50,030 --> 00:02:52,060 V is equal to iR. 50 00:02:52,060 --> 00:02:54,465 In other words, the plus side of the voltage 51 00:02:54,465 --> 00:02:58,560 is the same direction as the current enters that element. 52 00:02:58,560 --> 00:03:00,032 Now, what happens 53 00:03:00,032 --> 00:03:01,320 when I've got. 54 00:03:05,450 --> 00:03:11,204 Two components to my current, an i sub one and i sub 55 00:03:11,204 --> 00:03:17,030 two and they're going in different directions. 56 00:03:19,260 --> 00:03:23,480 Well, I can equate that to something with like the equivalent 57 00:03:25,830 --> 00:03:32,345 Current which is the difference. So this is i sub 1 minus i sub 2 in 58 00:03:32,345 --> 00:03:39,380 this direction and then the corresponding voltage. 59 00:03:41,760 --> 00:03:43,019 Well, that V. 60 00:03:45,130 --> 00:03:51,466 Is equal to R times i1 61 00:03:51,466 --> 00:03:57,599 minus i2. Now going back to this problem, 62 00:03:57,599 --> 00:04:04,260 I want to do a KVL around each loop just keeping this, these concepts in mind. 63 00:04:04,260 --> 00:04:06,112 So as I do a KVL. 64 00:04:06,112 --> 00:04:10,216 I look at the loop and they and I'm doing it in the direction of the 65 00:04:10,216 --> 00:04:13,712 current, and I determine if I use a plus or a minus based 66 00:04:13,712 --> 00:04:18,180 on what the sign is of the voltage as I approach that element. 67 00:04:18,180 --> 00:04:21,530 So, as I approach this from this direction, I see a minus first. 68 00:04:21,530 --> 00:04:28,128 So I'm going to put a minus. So looking at I1, net nash I put a minus 69 00:04:28,128 --> 00:04:35,428 V1 first, plus coming around I want, well, the currents 70 00:04:35,428 --> 00:04:40,897 going into this resistor. So that's a plus 71 00:04:40,897 --> 00:04:46,425 voltage. R1, I1 plus R2, in this case, 72 00:04:46,425 --> 00:04:51,500 we've got I1 going in this direction and 73 00:04:51,500 --> 00:04:56,500 then I3 going in this direction. So the 74 00:04:56,500 --> 00:05:01,550 net going this direction is I1-I3. 75 00:05:01,550 --> 00:05:07,300 And that has to sum to zero. Now, I2 is a pretty easy mesh. 76 00:05:10,540 --> 00:05:13,240 This current I2 has to equal to source current. 77 00:05:20,660 --> 00:05:21,560 So now we go to I3. 78 00:05:24,880 --> 00:05:25,920 Okay. 79 00:05:25,920 --> 00:05:31,700 So, if I start right here, I hit that V2 first, so the plus sign, so it's going to 80 00:05:31,700 --> 00:05:34,675 be a plus V2, plus, and then the voltage 81 00:05:34,675 --> 00:05:38,830 plus R2, will then, I'm going in this direction. 82 00:05:38,830 --> 00:05:42,090 The current, total current will be I3, minus I1. 83 00:05:51,940 --> 00:05:52,180 Okay. 84 00:05:52,180 --> 00:05:57,288 Coming around this way, I get this voltage, which is R3 times that current. 85 00:05:57,288 --> 00:06:01,466 Again, if I'm going in this direction, it's I3 minus I2. 86 00:06:01,466 --> 00:06:07,709 [BLANK_AUDIO]. 87 00:06:07,709 --> 00:06:11,224 But I already know what I2 is, I've solved 88 00:06:11,224 --> 00:06:15,268 for that and I'll make that substitution I sub s. 89 00:06:15,268 --> 00:06:17,881 And then I've got one more element and that's 90 00:06:17,881 --> 00:06:20,771 going around this way, so that's R0 times I3. 91 00:06:20,771 --> 00:06:27,497 [BLANK_AUDIO]. 92 00:06:27,497 --> 00:06:29,110 And that has to equal 0. 93 00:06:29,110 --> 00:06:32,635 So, I really only have two unknowns now, I1 and I3 and 94 00:06:32,635 --> 00:06:36,160 I want to write all of my equations of terms of I1 and I3. 95 00:06:37,230 --> 00:06:40,420 So, let me take this, this one right up here. 96 00:06:45,160 --> 00:06:48,210 And I'm going to rewrite it. And I'm going to pull out these unknowns. 97 00:06:48,210 --> 00:06:54,714 So it's I1 times R1, plus 98 00:06:54,714 --> 00:07:00,947 R2, plus I3 times a minus 99 00:07:00,947 --> 00:07:07,220 R2, has to equal V sub 1. 100 00:07:07,220 --> 00:07:10,379 So I put all my known quantities on right and all 101 00:07:10,379 --> 00:07:12,647 my unknown quantities on the left 102 00:07:12,647 --> 00:07:16,560 multiplied by whatever their coefficients are. 103 00:07:16,560 --> 00:07:18,774 In this case, we are assuming if you actually had 104 00:07:18,774 --> 00:07:21,860 to solve for this you would know what these resistors are. 105 00:07:21,860 --> 00:07:23,480 And you would know what this voltage source is. 106 00:07:23,480 --> 00:07:27,470 So this becomes one of the equations you want to solve. 107 00:07:27,470 --> 00:07:35,840 Now let's do the same thing with this other equation. 108 00:07:35,840 --> 00:07:38,280 I'm going to pull out my I1 and I3. 109 00:07:41,720 --> 00:07:43,610 Or. Other I1 here. 110 00:07:49,360 --> 00:07:54,350 The only term involving I1 is the R2 plus I3. 111 00:07:54,350 --> 00:07:59,678 And I get an R2 plus 112 00:07:59,678 --> 00:08:04,130 R3. Plus R0. 113 00:08:04,130 --> 00:08:09,026 And I put all of my known quantities on the 114 00:08:09,026 --> 00:08:14,807 right, which means that minus V2 plus R3 i s. 115 00:08:18,310 --> 00:08:19,779 Now this is my second equation. 116 00:08:23,220 --> 00:08:26,356 Again, assuming that you knew all these resistors and you 117 00:08:26,356 --> 00:08:29,390 knew these sources, the only unknowns are I1 and I3. 118 00:08:29,390 --> 00:08:31,522 So you want to solve those using 119 00:08:31,522 --> 00:08:35,710 a simultaneous equation solver, or by back substituting. 120 00:08:35,710 --> 00:08:40,290 For example, solving for I sub 3 in this equation, in terms of. 121 00:08:40,290 --> 00:08:41,810 All the rest of the quantities. 122 00:08:41,810 --> 00:08:44,520 And then plugging that back in to I3 right here. 123 00:08:46,840 --> 00:08:50,872 There's a, a link to an online equation solver in 124 00:08:50,872 --> 00:08:56,000 the resources part of this course on the course site. 125 00:08:56,000 --> 00:09:00,190 So once I solve for I1 and I3, I can go back and solve for V0. 126 00:09:00,190 --> 00:09:02,628 V0 was the ultimate thing I might have been interested 127 00:09:02,628 --> 00:09:05,330 in here, but that's the last thing I solve for. 128 00:09:05,330 --> 00:09:08,710 I have to write all my equations in terms of mesh currents, and 129 00:09:08,710 --> 00:09:12,200 that at the very end I can go back and solve for V0. 130 00:09:12,200 --> 00:09:22,120 V0 is equal to I3 times R0. So to summarize the mesh method. 131 00:09:22,120 --> 00:09:25,310 I define the mesh currents. One for each non-inclusive loop. 132 00:09:25,310 --> 00:09:27,480 I do a KVL around each loop. 133 00:09:27,480 --> 00:09:29,774 Always trying to solve for the only unknowns 134 00:09:29,774 --> 00:09:32,120 that I need which are they mesh currents. 135 00:09:32,120 --> 00:09:36,080 So I have to write everything in terms of mesh currents as my only unknowns. 136 00:09:36,080 --> 00:09:37,242 And then I reduce 137 00:09:37,242 --> 00:09:41,690 this down to simultaneous equations with these unknowns. 138 00:09:41,690 --> 00:09:43,594 I solve for them using some sort of 139 00:09:43,594 --> 00:09:46,654 linear equation solver or back substituting and then 140 00:09:46,654 --> 00:09:50,727 at the very end I can go back and solve for any other quantities I need. 141 00:09:50,727 --> 00:09:52,229 So that's Mesh Analysis. 142 00:09:52,229 --> 00:09:58,204 [BLANK_AUDIO] 143 00:09:58,204 --> 00:10:00,270 Now let's look at node analysis. 144 00:10:03,370 --> 00:10:06,902 In this particular thing the, the steps are first defined at ground node. 145 00:10:06,902 --> 00:10:10,550 Well, a ground node here, I usually pick a ground node to 146 00:10:10,550 --> 00:10:14,730 be one that I know that can, is connected to several elements like 147 00:10:14,730 --> 00:10:18,910 this one is connected to several elements and the other thing that 148 00:10:18,910 --> 00:10:22,650 makes my life a little bit easier is if I pick a node. 149 00:10:22,650 --> 00:10:25,570 That is next to a voltage source. 150 00:10:25,570 --> 00:10:28,568 So I'm going to pick this node. It is actually 151 00:10:28,568 --> 00:10:33,630 next to two voltage sources. And I want to pick that node as my ground. 152 00:10:33,630 --> 00:10:35,390 And then I defined node voltages. 153 00:10:35,390 --> 00:10:38,100 So I look at my nodes and say, what are the voltages? 154 00:10:38,100 --> 00:10:42,260 With respect to that ground. So I'm going to call this V sub A. 155 00:10:42,260 --> 00:10:48,392 That's one node V sub B is another node V sub C is another node and I'll 156 00:10:48,392 --> 00:10:53,606 define this as V sub D as another node. And I want to solve for these 157 00:10:53,606 --> 00:10:55,260 node voltages. 158 00:10:55,260 --> 00:10:58,257 Now in this particular one, this one is easy 159 00:10:58,257 --> 00:11:02,520 to solve for because the only thing between that node. 160 00:11:02,520 --> 00:11:05,390 And the ground is this source. So that's V sub 1. 161 00:11:05,390 --> 00:11:09,800 And similarly, V sub d is equal to V sub 2. 162 00:11:09,800 --> 00:11:16,840 So that means the only other unknowns I have are V sub c and V sub b. 163 00:11:17,960 --> 00:11:18,780 To solve for 164 00:11:18,780 --> 00:11:21,850 those, I have to do a KCL at those two nodes. 165 00:11:23,190 --> 00:11:26,466 Now, I'm going to give you a little background here to figure 166 00:11:26,466 --> 00:11:31,110 out how to find the currents leaving those or, or entering those nodes. 167 00:11:31,110 --> 00:11:37,558 So for example, suppose I've got a resistor between two node 168 00:11:37,558 --> 00:11:43,990 voltages, V sub a and V sub b If I wanted to find 169 00:11:43,990 --> 00:11:50,374 the current in this direction, well that current 170 00:11:50,374 --> 00:11:56,920 i is equal to V sub a minus V sub b, all over R. 171 00:11:56,920 --> 00:12:01,910 So this current leaving V sub a is equal to V sub a minus V sub b over R. 172 00:12:03,030 --> 00:12:05,637 And the second fact that I'm going to be 173 00:12:05,637 --> 00:12:09,192 using, in solving this, is that, I'm going to be, 174 00:12:09,192 --> 00:12:14,732 looking at the currents, leaving the node. So I'm going to be summing up 175 00:12:14,732 --> 00:12:20,453 all the currents leaving the node, and that has to equal 0. 176 00:12:22,090 --> 00:12:25,320 So the two unknowns I'm solving for V sub b and V sub c. 177 00:12:25,320 --> 00:12:28,060 Let's do a KCL at those two nodes. 178 00:12:30,880 --> 00:12:37,199 At V sub b, at node b. let's look at the 179 00:12:37,199 --> 00:12:43,260 current leaving the node. So, the current in this 180 00:12:43,260 --> 00:12:49,872 direction is a simple one that's i sub s and then the current leaving 181 00:12:49,872 --> 00:12:55,749 in this direction is going to be V sub b minus V sub c over R3. 182 00:12:55,749 --> 00:13:00,929 And then the current going in this direction 183 00:13:00,929 --> 00:13:06,109 was V sub b, well and then this is ground here 184 00:13:06,109 --> 00:13:10,869 so this voltage is just V sub b over R, and 185 00:13:10,869 --> 00:13:16,650 then the current is over R2 plus V sub b. 186 00:13:16,650 --> 00:13:20,768 Over R2. And then the last one is the current in 187 00:13:20,768 --> 00:13:26,612 this direction. That's V sub b minus V sub 1, because 188 00:13:26,612 --> 00:13:34,833 we've already solved for that node, all over R1, has got to equal 0. 189 00:13:38,070 --> 00:13:40,610 And I'm trying to solve for Vb and Vc. 190 00:13:40,610 --> 00:13:43,925 So I'm going to write a, an equation that pulls out those variables. 191 00:13:43,925 --> 00:13:50,260 So V sub b, so I'm just rewriting this, V 192 00:13:50,260 --> 00:13:57,138 sub b is 1 over R3 plus 1over R2 plus 1 over 193 00:13:57,138 --> 00:14:03,473 R1, plus V sub c minus 1 over R3 and then 194 00:14:03,473 --> 00:14:09,570 everything else is known quantities. 195 00:14:09,570 --> 00:14:14,678 I sub S, we are assuming we know that and we are assuming we know V sub 1 So 196 00:14:14,678 --> 00:14:20,060 I've got a minus i sub 197 00:14:20,060 --> 00:14:25,170 s plus V1 over R1. 198 00:14:25,170 --> 00:14:28,394 And this becomes one of the equations I need to solve using, for example, 199 00:14:28,394 --> 00:14:32,149 linear equation solver. Now let's look at V sub c. 200 00:14:34,340 --> 00:14:37,440 And I wanted some of the currents leaving that node. 201 00:14:37,440 --> 00:14:41,930 So this, starting in this direction, this one's easy. 202 00:14:41,930 --> 00:14:43,770 That's, i of s comes in. 203 00:14:43,770 --> 00:14:46,210 That means what's leaving is negative i sub s. 204 00:14:50,670 --> 00:14:51,550 Okay. 205 00:14:51,550 --> 00:14:56,220 And then the current going down and through this link through this branch is. 206 00:14:56,220 --> 00:15:02,066 V sub c minus V sub d which I've already found 207 00:15:02,066 --> 00:15:08,228 to be V2 over R0 plus the current going in this 208 00:15:08,228 --> 00:15:13,840 direction is V sub c minus V sub b over R3. 209 00:15:16,640 --> 00:15:21,656 All of that has to equal 0. Well, I want to rewrite this equation, 210 00:15:21,656 --> 00:15:27,041 pulling out my unknown node voltages. So V sub b times, 211 00:15:27,041 --> 00:15:31,773 let's see, I gotta minus 1 over 212 00:15:31,773 --> 00:15:36,644 R3, plus V sub c. 1 over 213 00:15:36,644 --> 00:15:41,980 R0 plus 1 over R3 and everything 214 00:15:41,980 --> 00:15:46,764 else in this equation is a known 215 00:15:46,764 --> 00:15:51,530 quantity. I'm going to put that on 216 00:15:51,530 --> 00:15:56,170 the right, and that's equal to i sub s 217 00:15:56,170 --> 00:16:01,700 plus. V two over R0, that's 218 00:16:01,700 --> 00:16:07,250 my second equation. So now, I have 219 00:16:07,250 --> 00:16:11,950 two equations into a node and I can solve for example for V 220 00:16:11,950 --> 00:16:16,510 sub c in terms of Vb and substitute that in for Vc here. 221 00:16:16,510 --> 00:16:18,830 Or I can use, once I have numbers in 222 00:16:18,830 --> 00:16:22,460 here, actual values, I can use an equation solver. 223 00:16:24,560 --> 00:16:29,988 Once I solve for Vb and Vc, I can solve for V0 as the difference between them. 224 00:16:29,988 --> 00:16:37,180 V0 is equal to Vc minus V2. So to summarize. 225 00:16:37,180 --> 00:16:40,050 In node analysis, I first select the ground node. 226 00:16:41,060 --> 00:16:45,620 And again I want to find something that connects a lot of elements, plus 227 00:16:45,620 --> 00:16:50,620 is right next to voltage source if I can do that makes it simpler. 228 00:16:50,620 --> 00:16:54,085 And then I define my node voltages at every 229 00:16:54,085 --> 00:16:58,450 node and I solve for those that are unknown. 230 00:16:58,450 --> 00:17:02,180 By going back and do a KCL at every one of those nodes. 231 00:17:02,180 --> 00:17:06,392 And then I solve for those node voltages, in this case Vb and Vc, 232 00:17:06,392 --> 00:17:11,180 and at the end I go back and solve for whatever quantity I was interested. 233 00:17:11,180 --> 00:17:15,911 So the first thing I have to solve for, is my node voltages, and then 234 00:17:15,911 --> 00:17:19,231 once I do that, I solve for any other parameters 235 00:17:19,231 --> 00:17:23,160 or values in the in the system that I'm interested in. 236 00:17:23,160 --> 00:17:25,200 The next lesson will cover the details of 237 00:17:25,200 --> 00:17:29,060 Thevenin equivalent circuits and Norton equivalent circuit methods. 238 00:17:29,060 --> 00:17:30,700 And gives some examples. 239 00:17:30,700 --> 00:17:33,100 I do want to mention that there's extra 240 00:17:33,100 --> 00:17:36,840 work to problems available for these two lessons. 241 00:17:36,840 --> 00:17:38,780 And I do also want to mention. 242 00:17:38,780 --> 00:17:41,377 That we're introducing these different technologies, 243 00:17:41,377 --> 00:17:44,020 these different methodologies, for you. 244 00:17:44,020 --> 00:17:45,700 And I don't care, in working the 245 00:17:45,700 --> 00:17:48,570 solutions, which ones of these that you use. 246 00:17:48,570 --> 00:17:51,174 So work some of these problems, feel comfortable 247 00:17:51,174 --> 00:17:53,470 with one of them, and then use that. 248 00:17:53,470 --> 00:17:56,630 And as long as you're able to solve the problem, how you do it is fine.