1 00:00:02,540 --> 00:00:06,190 Hello, this is Dr. Cynthia Furse, from the University of Utah. 2 00:00:06,190 --> 00:00:08,470 Today, we're going to talk about Kirchhoff's Laws. 3 00:00:11,333 --> 00:00:16,068 Kirchhoff's Current Law says that the sum of the currents entering a node is = 0. 4 00:00:16,068 --> 00:00:19,604 Or equivalently, the sum of currents leaving a node is 0, or 5 00:00:19,604 --> 00:00:23,892 the sum of currents leaving is equal to the sum of currents entering a node. 6 00:00:23,892 --> 00:00:26,388 So we can write the equation four different ways. 7 00:00:26,388 --> 00:00:30,458 When we're talking about a node, we're talking about something like this right 8 00:00:30,458 --> 00:00:33,126 here, that's in the center of all four of the wires. 9 00:00:33,126 --> 00:00:37,326 Now, when we say the sum of the currents entering the node, 10 00:00:37,326 --> 00:00:42,799 we can see that i1 and i4 are entering the node while i2 and i3 are leaving. 11 00:00:42,799 --> 00:00:47,222 So we're going to say i1 + i4 those are the currents entering. 12 00:00:47,222 --> 00:00:51,277 And i2 which is leaving, we're gonna put it there with the minus sign and 13 00:00:51,277 --> 00:00:53,579 i3 we're going to put with a minus sign. 14 00:00:53,579 --> 00:00:57,453 When we write the equation where we have the sum of the currents leaving the node 15 00:00:57,453 --> 00:01:00,876 is equal to 0, what we're doing is just switching the minus signs. 16 00:01:00,876 --> 00:01:05,754 Minus and plus signs here, as you can see. 17 00:01:05,754 --> 00:01:08,795 When we talk about the sum of the currents leaving is equal to the sum of 18 00:01:08,795 --> 00:01:09,909 the currents entering. 19 00:01:09,909 --> 00:01:12,648 Then you could say, here's the sum of the currents entering, 20 00:01:12,648 --> 00:01:15,343 here's the sum of the currents leaving, and they are equal. 21 00:01:15,343 --> 00:01:18,920 These four different ways of writing the node equation are all equivalent. 22 00:01:18,920 --> 00:01:22,625 This is Kirchoff's Current Law, or KCL. 23 00:01:22,625 --> 00:01:26,944 Kirchhoff's Voltage Law or KVL is the other law that we will use. 24 00:01:26,944 --> 00:01:31,293 This says that the sum of the voltage around a closed path is = 0. 25 00:01:31,293 --> 00:01:34,869 The way you do this is you run a loop like this, and 26 00:01:34,869 --> 00:01:39,766 you just add up all of the voltage as you're going around the loop. 27 00:01:39,766 --> 00:01:45,053 So for example, as we're going through here, 28 00:01:45,053 --> 00:01:50,336 we say -VDC +, see there's the + V1 + V2, 29 00:01:50,336 --> 00:01:53,128 and that's equal to 0. 30 00:01:53,128 --> 00:01:58,904 So -VDC + V1 + V2 = 0. 31 00:01:58,904 --> 00:02:03,231 Alternatively, we could say that the sum of all the voltage drops is equal to 32 00:02:03,231 --> 00:02:05,136 the sum of all the voltage rises. 33 00:02:05,136 --> 00:02:11,746 So the voltage rise is +VDC and the voltage drop is V1 + V2. 34 00:02:11,746 --> 00:02:14,329 So there's just a convention that typically, 35 00:02:14,329 --> 00:02:17,309 we use a systematic clockwise motion around the loop. 36 00:02:17,309 --> 00:02:19,230 We could go counter clockwise, but 37 00:02:19,230 --> 00:02:21,995 convention is just that we normally go clockwise. 38 00:02:21,995 --> 00:02:25,451 We also assign a positive sign to the voltage across an element, 39 00:02:25,451 --> 00:02:28,651 if the positive sign of that voltage is encountered first. 40 00:02:28,651 --> 00:02:32,025 So that's why I was kinda circling this thing as I went to your + V1. 41 00:02:34,845 --> 00:02:39,203 Now, let's talk about a simple circuit where we can apply KCL and KVL. 42 00:02:39,203 --> 00:02:44,061 So here's our simple circuit, let's first apply the loop equation. 43 00:02:44,061 --> 00:02:46,618 Right here's our loop and let's apply the equation. 44 00:02:46,618 --> 00:02:52,676 So we can see minus, see we count the minus sign first, 45 00:02:52,676 --> 00:02:59,286 -VDC + V1 + V2 = 0, that's the equation right here. 46 00:02:59,286 --> 00:03:06,282 That is the KVL equation or the loop equation. 47 00:03:06,282 --> 00:03:09,559 Now, let's examine the circuit, what do we know? 48 00:03:09,559 --> 00:03:13,006 Well, VDC is defined that's 120 volts, 49 00:03:13,006 --> 00:03:18,280 R1 is defined that's 10 ohms, R2 is defined that's 50 ohms. 50 00:03:18,280 --> 00:03:19,863 What's unknown? 51 00:03:19,863 --> 00:03:22,541 V1 and V2 are both unknown. 52 00:03:22,541 --> 00:03:25,821 So if you remember from math, if you have two things that are unknown, 53 00:03:25,821 --> 00:03:27,837 then you're going to need two equations. 54 00:03:27,837 --> 00:03:31,666 But we only have one equation, so what else can we do? 55 00:03:31,666 --> 00:03:35,471 Well Ohm's Law will give us two more equations, so why don't we try that? 56 00:03:35,471 --> 00:03:41,114 So we can see here for example that if we have a current I1 that's running through 57 00:03:41,114 --> 00:03:47,840 R1, we know that V1 = (I1) times (R1), we also know that V2 is (I2) times (R2). 58 00:03:47,840 --> 00:03:50,567 Well, look, we got two more equations. 59 00:03:50,567 --> 00:03:55,224 But we also have two more unknowns, I1 and I2. 60 00:03:55,224 --> 00:03:58,501 So now, we have four unknowns and three equations. 61 00:03:58,501 --> 00:04:01,064 So what we always need is one more equation and 62 00:04:01,064 --> 00:04:03,229 this one's going to come from a node. 63 00:04:03,229 --> 00:04:07,393 So here's our node equation right here, that's going to be our KCL. 64 00:04:07,393 --> 00:04:11,310 And remember, that's the sum of the currents coming in to the node is equal to 65 00:04:11,310 --> 00:04:13,616 the sum of the currents going out of the node. 66 00:04:13,616 --> 00:04:18,072 So I1, which is all the currents coming into that node is equal to I2, 67 00:04:18,072 --> 00:04:20,245 where the currents are going out. 68 00:04:20,245 --> 00:04:26,161 So we now have four unknowns, V1, V2, I1, and I2, but we have four equations. 69 00:04:26,161 --> 00:04:27,426 So now, we can solve it. 70 00:04:27,426 --> 00:04:29,625 You're going to say, heavens. 71 00:04:29,625 --> 00:04:32,916 It looks like we could have done this equation a whole lot simpler and 72 00:04:32,916 --> 00:04:34,430 absolutely, you could have. 73 00:04:34,430 --> 00:04:41,033 But what I'm showing you is the systematic way of handling the KCL and the KVL. 74 00:04:41,033 --> 00:04:44,004 Okay, so now, let's actually solve this equation. 75 00:04:44,004 --> 00:04:47,802 So here's our first equation where we have VDC. 76 00:04:47,802 --> 00:04:48,708 Well, that's know. 77 00:04:48,708 --> 00:04:52,037 We could substitute that in there and V1 and V2. 78 00:04:52,037 --> 00:04:56,466 We know R1, we know R2, let's substitute those in there. 79 00:04:56,466 --> 00:05:01,685 And we know that I1 = I2 so let's just call it a value I. 80 00:05:01,685 --> 00:05:06,610 So just substituting in, what I'm gonna do is substitute I in there, 81 00:05:06,610 --> 00:05:10,126 and I in here and of course, all of our constants. 82 00:05:10,126 --> 00:05:16,965 This equation now becomes this right here, and I can solve if for I. 83 00:05:16,965 --> 00:05:22,444 I = 120 volts, let's just bring in the 120 over to that side, 84 00:05:22,444 --> 00:05:26,515 divided by the sum of 10 and 50, so I = 2 amps. 85 00:05:26,515 --> 00:05:32,706 Then if we wanted to know V1, we would just bring our 2 amps back up in here. 86 00:05:32,706 --> 00:05:35,977 Let's substitute it into the equations for V1 and V2, 87 00:05:35,977 --> 00:05:40,631 and you can see that we have 20 volts across 1 and 100 volt across the other. 88 00:05:40,631 --> 00:05:43,959 Let's just do a quick check, let's just check our loop equation. 89 00:05:43,959 --> 00:05:48,378 We have 120 volts, so that's -120 90 00:05:48,378 --> 00:05:52,811 volts + 20 + 100 is that equal to 0? 91 00:05:52,811 --> 00:05:55,799 Yes, it is, so we've been able to check our equation. 92 00:05:58,056 --> 00:06:00,834 Now, what I showed you was an extremely simple circuit, and 93 00:06:00,834 --> 00:06:03,191 yes, there's an easier way to solve the circuit. 94 00:06:03,191 --> 00:06:08,499 But let's use it to define a cookbook that will let us do more complicated problems. 95 00:06:08,499 --> 00:06:11,330 So if I were to set a cookbook for this I would say, 96 00:06:11,330 --> 00:06:15,076 the first thing I would want to do is write down all known values. 97 00:06:15,076 --> 00:06:16,374 What was known? 98 00:06:16,374 --> 00:06:19,126 The 120 volts, the 10 ohms, and the 50 ohms. 99 00:06:19,126 --> 00:06:23,771 Then I would keep track of all the unknown values as I went throughout my 100 00:06:23,771 --> 00:06:28,521 circuit because I know that I need as many equations as I have unknowns. 101 00:06:28,521 --> 00:06:32,263 Now, we would write all of the KVL or loop equations. 102 00:06:32,263 --> 00:06:37,435 An important thing is that each new loop must pick up at least one new element, 103 00:06:37,435 --> 00:06:40,850 like a new resistor or a new voltage element. 104 00:06:40,850 --> 00:06:42,530 Also current sources, 105 00:06:42,530 --> 00:06:46,390 which we'll talk about in a minute, can't be included in any loops. 106 00:06:46,390 --> 00:06:48,370 Fourth, I'd write Ohm's Law. 107 00:06:48,370 --> 00:06:51,549 In fact, I usually prefer to write Ohm's Law as I go just for 108 00:06:51,549 --> 00:06:55,061 convenience because I know I'm going to have to apply it anyway. 109 00:06:55,061 --> 00:06:57,694 Ohm's Law is just applying the current and 110 00:06:57,694 --> 00:07:01,947 the resistor together to get the voltage any time we have a resistor. 111 00:07:01,947 --> 00:07:04,027 Then I would apply as many KCL or 112 00:07:04,027 --> 00:07:07,558 node equations as are needed to fill in my unknowns. 113 00:07:07,558 --> 00:07:11,982 Each KCL equation must also pick up at least one new current, 114 00:07:11,982 --> 00:07:14,777 then I'd solve for all the unknowns. 115 00:07:14,777 --> 00:07:18,582 Okay, let's take this cookbook now and see how we can apply it. 116 00:07:18,582 --> 00:07:21,698 So the first thing we do, and let's apply it to a problem, 117 00:07:21,698 --> 00:07:25,152 that's just a little bit more complicated than one we just did. 118 00:07:25,152 --> 00:07:29,790 I'm going to add a current source I3 here which is 6 amps. 119 00:07:29,790 --> 00:07:34,224 So let's apply this, and as you can see, we're going to have another element. 120 00:07:34,224 --> 00:07:36,537 So let's just apply each of the steps. 121 00:07:36,537 --> 00:07:39,314 First of all, let's write all of the known values. 122 00:07:39,314 --> 00:07:46,058 Well, VDC is known, R1, R2 and I3 are all known values. 123 00:07:46,058 --> 00:07:48,128 Now, let's keep track of our unknowns. 124 00:07:48,128 --> 00:07:51,889 I've already drawn two unknowns here, I1 and I2. 125 00:07:51,889 --> 00:07:56,518 So we have two unknowns, and we need to have at least two equations. 126 00:07:56,518 --> 00:07:58,910 So let's write, The loop equation's next. 127 00:07:58,910 --> 00:08:02,665 Well, this same loop equation is the one that we just wrote and it hasn't 128 00:08:02,665 --> 00:08:06,932 changed at all, even though we've added another element over here on the right. 129 00:08:06,932 --> 00:08:13,611 And so for the loop equation, it's -VDC + (I1)(R1), 130 00:08:13,611 --> 00:08:20,047 so here is -VDC + (I1)(R1) + (I2)(R2) = 0. 131 00:08:20,047 --> 00:08:23,823 This is what I mean by applying Ohm's Law as I go. 132 00:08:23,823 --> 00:08:28,306 I'm just doing + (I1)(R1) instead of saying + V1, and 133 00:08:28,306 --> 00:08:31,305 later having to go back in and substitute. 134 00:08:31,305 --> 00:08:35,658 Well, we could say each loop needed to pick up at least one new element. 135 00:08:35,658 --> 00:08:37,750 But which elements did this loop pick up? 136 00:08:37,750 --> 00:08:40,712 It picked up the VDC, the R1 and the R2. 137 00:08:41,902 --> 00:08:45,243 Now, current sources cannot be counted as new elements, and 138 00:08:45,243 --> 00:08:47,631 in fact, I can't include them in my loops. 139 00:08:47,631 --> 00:08:50,857 So there's no other loop that I could do in this picture, 140 00:08:50,857 --> 00:08:54,980 because any other loop would have to go through the the current source. 141 00:08:54,980 --> 00:09:00,050 This is a current source right here, and I cannot include that in a KVL loop. 142 00:09:00,050 --> 00:09:03,325 Now, let's apply Ohm's Law, as I went, that's right there. 143 00:09:03,325 --> 00:09:06,797 And then let's apply as many node equations as needed to fill in 144 00:09:06,797 --> 00:09:07,658 the unknowns. 145 00:09:07,658 --> 00:09:09,430 Well how many unknowns do I have? 146 00:09:09,430 --> 00:09:11,620 Two, how many equations do I have? 147 00:09:11,620 --> 00:09:15,110 One, so I need at least one node equation. 148 00:09:15,110 --> 00:09:19,120 Here's my node right there, and let's apply the node equation. 149 00:09:19,120 --> 00:09:22,794 Let's say that every current that's coming in, I1 + I3. 150 00:09:22,794 --> 00:09:25,376 So here's I3, it's coming into that node. 151 00:09:25,376 --> 00:09:28,484 I1 + I3 = I2. 152 00:09:28,484 --> 00:09:30,902 Okay, that's my second equation. 153 00:09:30,902 --> 00:09:35,443 So now, I can solve for all of my unknowns. 154 00:09:35,443 --> 00:09:38,602 Let's just do the solution, just one time, for fun. 155 00:09:38,602 --> 00:09:44,089 So here was my VDC, I1, R1, I2, 156 00:09:44,089 --> 00:09:49,050 R2 = 0 and I1 + 6 amps = I2. 157 00:09:49,050 --> 00:09:53,323 So what I'm going to do is substitute in order to remove the variables. 158 00:09:53,323 --> 00:09:57,073 What I'll do here is I'll substitute in for I2, 159 00:09:57,073 --> 00:10:00,565 so this equation now becomes this right here. 160 00:10:00,565 --> 00:10:03,683 Let's solve for the remaining variable, that's I1. 161 00:10:03,683 --> 00:10:08,724 That gives us the equation that you see here, and now let's go back and 162 00:10:08,724 --> 00:10:12,748 actually do the math that tells us that I1 is -3 amps. 163 00:10:12,748 --> 00:10:15,917 And then, if I go back to my original equation, 164 00:10:15,917 --> 00:10:18,466 in order to find the other variables. 165 00:10:18,466 --> 00:10:24,020 Let's say I want to find I2, I2 = 3 amps. 166 00:10:24,020 --> 00:10:27,463 Now, let's talk about what it means to have -3 amps and 167 00:10:27,463 --> 00:10:29,414 what it means to have +3 amps. 168 00:10:29,414 --> 00:10:36,384 So right here, when I did my analysis, I said that this value I1 was a variable. 169 00:10:36,384 --> 00:10:40,100 And when I solved for it, I discovered that it is -3 amps. 170 00:10:40,100 --> 00:10:41,116 What does that mean? 171 00:10:41,116 --> 00:10:44,720 Well, it means I've guessed the direction of the current incorrectly. 172 00:10:44,720 --> 00:10:48,660 It means that the current is actually going from right to left instead of from 173 00:10:48,660 --> 00:10:53,630 left to right, so that I have three amps going from left to right. 174 00:10:53,630 --> 00:10:56,943 Now, if I had actually put my arrow this way in the first 175 00:10:56,943 --> 00:11:01,855 place when I did my solution, I would have found out that I1 was equal to 3 amps. 176 00:11:01,855 --> 00:11:03,820 And I would have had the correct direction. 177 00:11:03,820 --> 00:11:06,160 It doesn't matter which way you guess your current. 178 00:11:06,160 --> 00:11:09,600 If you guess it wrong, it's gonna come up with a negative number for 179 00:11:09,600 --> 00:11:11,430 the current, and that's just fine. 180 00:11:11,430 --> 00:11:16,761 So go ahead and leave it like this, and just tell me that you have -3 amps for I1. 181 00:11:19,581 --> 00:11:22,310 Okay, let's go one step further. 182 00:11:22,310 --> 00:11:23,873 Let's add one more loop. 183 00:11:23,873 --> 00:11:29,124 So let's apply our cookbook to this system right here. 184 00:11:29,124 --> 00:11:32,064 First of all, we need to write all known values. 185 00:11:32,064 --> 00:11:33,460 V1 will be known. 186 00:11:33,460 --> 00:11:34,940 I1 will be known. 187 00:11:34,940 --> 00:11:37,836 R1, R2, and R3. 188 00:11:37,836 --> 00:11:40,963 Then we're going to keep track of all of our unknown values. 189 00:11:40,963 --> 00:11:47,870 I already have an I1, an I2, and an I3, that I can see are going to be unknown. 190 00:11:47,870 --> 00:11:49,900 Then we're going to write all the loop equations. 191 00:11:49,900 --> 00:11:51,820 Let's see how many loops we could do. 192 00:11:51,820 --> 00:11:54,170 Well, here's our first loop. 193 00:11:54,170 --> 00:11:56,280 That's a loop we've been doing all along. 194 00:11:56,280 --> 00:11:57,900 Here is a second loop. 195 00:11:59,260 --> 00:12:00,920 Is there room for another loop? 196 00:12:00,920 --> 00:12:03,190 Can I do a third loop here? 197 00:12:03,190 --> 00:12:07,480 No, I can't, because I have this current source right there. 198 00:12:07,480 --> 00:12:09,888 Can I do a third loop around this way? 199 00:12:11,077 --> 00:12:14,685 No, I can't, because that's including a current loop. 200 00:12:14,685 --> 00:12:17,029 Could I do a third loop here? 201 00:12:18,835 --> 00:12:21,300 Well, that's a legitimate loop. 202 00:12:21,300 --> 00:12:25,325 But I can't do that, because it doesn't pick up any new elements. 203 00:12:25,325 --> 00:12:28,046 My first loop picked up R1 and R2. 204 00:12:28,046 --> 00:12:30,294 My second loop picked up R3. 205 00:12:30,294 --> 00:12:32,418 There isn't anything else to be picked up. 206 00:12:32,418 --> 00:12:37,655 So I am only going to have the two blue loops that are shown here. 207 00:12:37,655 --> 00:12:39,875 Then I would apply as many KCL or 208 00:12:39,875 --> 00:12:42,745 node equations as are needed to fill in the unknowns. 209 00:12:42,745 --> 00:12:47,891 So I'm going to have two loop equations, the blue lines that are shown. 210 00:12:47,891 --> 00:12:52,478 And I'm going to have one node equation here, where things are orange, and 211 00:12:52,478 --> 00:12:54,950 that's going to fill in my unknowns. 212 00:12:54,950 --> 00:12:57,607 And then we would solve for the unknowns. 213 00:12:57,607 --> 00:13:00,786 Okay, now, let's take a look at the special node. 214 00:13:00,786 --> 00:13:04,687 So the node remember was right there, but there was a line or 215 00:13:04,687 --> 00:13:07,090 a short circuit right here. 216 00:13:07,090 --> 00:13:12,475 Any time you have a short circuit, it just means that you have a really big node. 217 00:13:12,475 --> 00:13:15,060 You don't need to include two separate nodes. 218 00:13:15,060 --> 00:13:17,540 This is just one really big node. 219 00:13:19,610 --> 00:13:23,992 Okay, now, let's imagine even adding another node, sorry, 220 00:13:23,992 --> 00:13:25,868 even adding another loop. 221 00:13:25,868 --> 00:13:28,746 So let's talk about how we might do this problem. 222 00:13:28,746 --> 00:13:33,425 Okay, so write down all the known values, V1, 223 00:13:33,425 --> 00:13:36,630 R1, R2, R3, R4, and I1. 224 00:13:36,630 --> 00:13:37,798 Those are all known. 225 00:13:37,798 --> 00:13:39,455 Keep track of the unknowns. 226 00:13:39,455 --> 00:13:46,669 They would be I1, I2, and I3, and let's call this IDC. 227 00:13:46,669 --> 00:13:48,810 I know it says I1, let's call it IDC. 228 00:13:48,810 --> 00:13:50,473 All right, so those are the unknowns. 229 00:13:50,473 --> 00:13:56,111 So we know that we're going to need, and there's also going to be an I4. 230 00:13:56,111 --> 00:13:59,410 So we're going to have four equations and four unknowns. 231 00:14:01,300 --> 00:14:03,790 Now, let's see how many loops we can do. 232 00:14:03,790 --> 00:14:06,419 First of all, we've got our first loop right here. 233 00:14:06,419 --> 00:14:09,808 That pick up V1, R1, and R2. 234 00:14:09,808 --> 00:14:12,030 That's one loop equation. 235 00:14:12,030 --> 00:14:13,950 Here's a second loop equation. 236 00:14:13,950 --> 00:14:16,168 Picking up R1 and R3. 237 00:14:16,168 --> 00:14:19,030 There's at least one new element, the R3. 238 00:14:19,030 --> 00:14:21,400 Can we do a loop here for I4? 239 00:14:21,400 --> 00:14:25,353 No, we can't, because that's going to include the current source, so 240 00:14:25,353 --> 00:14:26,835 we cannot do that as a loop. 241 00:14:26,835 --> 00:14:33,896 All right, so we have two loop equations, here's one node equation. 242 00:14:33,896 --> 00:14:38,795 And in fact, this node is actually going to encompass all of this area where you 243 00:14:38,795 --> 00:14:41,182 can see it short circuited together. 244 00:14:41,182 --> 00:14:43,663 I need at least one more equation and 245 00:14:43,663 --> 00:14:47,610 somehow I need to pick up R4 in that equation. 246 00:14:47,610 --> 00:14:48,885 So let's see what we would do. 247 00:14:50,791 --> 00:14:53,813 Okay, here is a very good choice for handling that. 248 00:14:53,813 --> 00:15:00,334 We would do a loop all the way around the outside like this picking up R4. 249 00:15:00,334 --> 00:15:04,987 So that would give me three loop equations, 250 00:15:04,987 --> 00:15:09,779 and one node equation for the four unknowns. 251 00:15:09,779 --> 00:15:12,489 So my problem would end up looking like this. 252 00:15:12,489 --> 00:15:19,845 3 loops, 1 node would give me 4 equations, sorry, 4 unknowns. 253 00:15:19,845 --> 00:15:28,438 3 loop equations, 1 node equation would be 4 unknowns and 4 equations. 254 00:15:28,438 --> 00:15:34,000 Okay, now let's take a look and just kind of expand our understanding of the nodes. 255 00:15:34,000 --> 00:15:37,930 Could I have used this as a node equation instead? 256 00:15:37,930 --> 00:15:41,415 Well, the answer is, no, I can't, because I don't know how much 257 00:15:41,415 --> 00:15:44,365 current is going to be coming out of this voltage source. 258 00:15:44,365 --> 00:15:48,925 So just like the fact that I cannot include loops when I have a current 259 00:15:48,925 --> 00:15:53,965 source, I cannot include nodes that are attaching voltage sources. 260 00:15:53,965 --> 00:15:56,775 So I'm gonna add that information to my cookbook as well. 261 00:15:59,099 --> 00:16:00,584 Okay, so let's do this problem. 262 00:16:00,584 --> 00:16:03,270 Maybe it looks a little complicated, but it really isn't. 263 00:16:03,270 --> 00:16:05,491 So let's do each of our loops. 264 00:16:05,491 --> 00:16:11,077 My first loop is the VDC loop right here which 265 00:16:11,077 --> 00:16:16,967 is going to say minus, this is the minus plus, 266 00:16:16,967 --> 00:16:21,051 -VDC + I1R1 + I2R2 = 0. 267 00:16:21,051 --> 00:16:23,577 This is my loop 1. 268 00:16:23,577 --> 00:16:26,274 Now let's do loop 2. 269 00:16:26,274 --> 00:16:32,697 And remember, I've got the plus minus going from my current like this. 270 00:16:32,697 --> 00:16:38,447 So +I3R3, this is I3 + I3R3, 271 00:16:38,447 --> 00:16:42,525 you hear that's all short 272 00:16:42,525 --> 00:16:47,357 circuited minus I1R1 = 0. 273 00:16:47,357 --> 00:16:50,931 Again, let's call this IDC, okay? 274 00:16:50,931 --> 00:16:55,816 And let's do this outside loop right there. 275 00:16:55,816 --> 00:17:01,187 So I'm gonna say -VDC 276 00:17:01,187 --> 00:17:07,416 + I3R3 + I4R4 = 0. 277 00:17:07,416 --> 00:17:10,440 Now, let's do the node. 278 00:17:10,440 --> 00:17:15,381 So right here is the big node that I'm gonna be interested in. 279 00:17:15,381 --> 00:17:17,582 So the node equation is going to say, 280 00:17:17,582 --> 00:17:22,065 all of the currents that come in has to equal all of the currents that go out. 281 00:17:22,065 --> 00:17:24,079 Or all the currents coming in are positive, 282 00:17:24,079 --> 00:17:27,027 all the currents going out are negative and that's equal to 0. 283 00:17:27,027 --> 00:17:31,597 So I3 is coming in, that's positive. 284 00:17:31,597 --> 00:17:34,567 I4 is going out, that's negative. 285 00:17:34,567 --> 00:17:37,247 I1 is coming in, that's positive. 286 00:17:37,247 --> 00:17:39,927 I2 is going out, that's negative. 287 00:17:39,927 --> 00:17:46,969 I also have to include plus IDC, and that whole thing is equal to 0. 288 00:17:46,969 --> 00:17:50,217 Okay, now let's keep track of how many unknowns I have. 289 00:17:50,217 --> 00:17:51,367 What are my unknowns? 290 00:17:52,650 --> 00:17:59,000 My unknowns are I1, I2, I3, and I4. 291 00:17:59,000 --> 00:18:00,930 How many equations do I have? 292 00:18:00,930 --> 00:18:05,726 I have th3 ee loop equations and 1 node equation. 293 00:18:05,726 --> 00:18:08,449 So I have four equations and four unknowns. 294 00:18:11,439 --> 00:18:14,536 Okay let's talk about the math for this for just a minute. 295 00:18:14,536 --> 00:18:16,463 We can solve this using algebra, and 296 00:18:16,463 --> 00:18:19,086 I know that you've had some practice doing that. 297 00:18:19,086 --> 00:18:23,984 But I want to show you one specific method of being able to solve the KCL and 298 00:18:23,984 --> 00:18:24,720 KVL math. 299 00:18:24,720 --> 00:18:27,892 It's gonna be very handy when we apply it in MATLAB. 300 00:18:27,892 --> 00:18:30,722 Any time we have four equations like this, and 301 00:18:30,722 --> 00:18:35,312 four unknowns, we can write them as a matrix, as I'm gonna do right here. 302 00:18:39,892 --> 00:18:42,043 So let's take our first equation. 303 00:18:42,043 --> 00:18:47,938 Here's our first equation right there. 304 00:18:47,938 --> 00:18:53,385 So the unknowns are gonna be I1, I2, I3, and I4. 305 00:18:53,385 --> 00:18:57,438 Anything that does not have an unknown moves over to the right-hand side. 306 00:18:57,438 --> 00:19:00,562 So we have +VDC on the right-hand side. 307 00:19:00,562 --> 00:19:06,182 Anything that does have an unknown, I've put the multiplier, like R1 times I1. 308 00:19:06,182 --> 00:19:12,138 See here's R1 times I1 and R2 times I2. 309 00:19:12,138 --> 00:19:14,191 That's right here, R2 times I2. 310 00:19:14,191 --> 00:19:17,747 Okay, and then there are no I3's and there are no I4's in that equation. 311 00:19:17,747 --> 00:19:20,741 So they are 0 and 0. 312 00:19:20,741 --> 00:19:22,689 Let's go to our next equation. 313 00:19:22,689 --> 00:19:27,895 Our next equation right here says, there is nothing that doesn't have an unknown. 314 00:19:27,895 --> 00:19:33,001 So I have 0 over here on the right hand side, 315 00:19:33,001 --> 00:19:36,324 then I have -R1 times I1. 316 00:19:36,324 --> 00:19:39,314 I have nothing times R2 and 317 00:19:39,314 --> 00:19:43,995 I have R3 times I3 and nothing times R4. 318 00:19:43,995 --> 00:19:46,324 Okay, let's go down to the third equation. 319 00:19:46,324 --> 00:19:50,501 The third equation has VDC, which has no unknowns. 320 00:19:50,501 --> 00:19:53,619 So let's bring it over thee there's +VDC. 321 00:19:53,619 --> 00:19:55,970 Do I have anything times I1? 322 00:19:55,970 --> 00:19:57,891 No, I2? 323 00:19:57,891 --> 00:19:59,944 No, I3? 324 00:19:59,944 --> 00:20:01,349 Yes, R3? 325 00:20:01,349 --> 00:20:02,437 Yes, I4? 326 00:20:02,437 --> 00:20:04,589 Yes, R4? 327 00:20:04,589 --> 00:20:08,056 So there's the matrix, and then this very last one, 328 00:20:08,056 --> 00:20:12,270 what is over here that doesn't have a unknown in front of it? 329 00:20:12,270 --> 00:20:17,088 Take that over, that's gonna become -IDC, 330 00:20:17,088 --> 00:20:24,013 what do we have times I1, a +1, I2-1, I3 + 1, I4-1. 331 00:20:24,013 --> 00:20:25,780 So here's my matrix. 332 00:20:27,660 --> 00:20:29,180 Here's my vector of unknowns. 333 00:20:29,180 --> 00:20:30,920 Here's my vector of constants. 334 00:20:30,920 --> 00:20:34,866 Then so this is writing the matrix equation for 335 00:20:34,866 --> 00:20:38,300 this particular circuit element. 336 00:20:38,300 --> 00:20:42,258 Next, you can use Gaussian elimination or MATLAB matrix solution. 337 00:20:42,258 --> 00:20:43,706 So I'm going to leave this here. 338 00:20:43,706 --> 00:20:46,082 We're not actually going to do the algebra to finish it. 339 00:20:46,082 --> 00:20:49,527 Because in general, I'm going to assume, you'd use MATLAB to solve this problem. 340 00:20:52,092 --> 00:20:55,300 So in short, we've talked about two important laws. 341 00:20:55,300 --> 00:20:59,260 One is Kirchhoff's Current Law that says the sum of all the currents 342 00:20:59,260 --> 00:21:00,730 entering the node = 0. 343 00:21:00,730 --> 00:21:04,411 The second law is the Kirchhoff's Voltage Law or KVL, 344 00:21:04,411 --> 00:21:09,144 that says the sum of the voltage is around a closed path or loop is = 0. 345 00:21:11,310 --> 00:21:15,330 Finally, we wrote down a cookbook for the KCl and kvl. 346 00:21:15,330 --> 00:21:16,371 The cookbook is this. 347 00:21:16,371 --> 00:21:18,840 First, write down all the known values. 348 00:21:18,840 --> 00:21:21,458 Keep track of how many unknown values you have, 349 00:21:21,458 --> 00:21:24,150 write all the loop equations that you can write. 350 00:21:24,150 --> 00:21:28,616 Remember that each loop needs to pick up at least one new element and 351 00:21:28,616 --> 00:21:31,419 that current sources can't be in loops. 352 00:21:31,419 --> 00:21:35,714 Then apply Ohm's law which I usually choose to do as I go for convenience. 353 00:21:35,714 --> 00:21:40,100 And that is equal to the voltage is equal to the current times the resistance. 354 00:21:40,100 --> 00:21:44,790 And then apply as many nodes as you need in order to fill in the unknowns. 355 00:21:44,790 --> 00:21:48,906 Each KCL equation must pick up at least one new current. 356 00:21:48,906 --> 00:21:54,079 Shorts as they are here can be used to combine nodes into a single node, 357 00:21:54,079 --> 00:21:57,161 and nodes can't touch voltage sources. 358 00:21:57,161 --> 00:22:02,562 Then you solve for the unknowns using matrix math. 359 00:22:02,562 --> 00:22:05,914 So the picture that's on the front that's in Red Canyon. 360 00:22:05,914 --> 00:22:09,966 That's near the bottom of Thunder Mountain, near Bryce Canyon, Utah.