1 00:00:01,930 --> 00:00:05,380 Welcome back to Linear Circuits. Last time from Doctor Ferry, you learned 2 00:00:05,380 --> 00:00:08,500 a little about first ordered differential equations. 3 00:00:08,500 --> 00:00:12,690 So today we're actually going to use them for something real and apply them to 4 00:00:12,690 --> 00:00:15,340 something. In this case it'll be RC circuits or 5 00:00:15,340 --> 00:00:18,200 circuits with resistors and capacitors in them. 6 00:00:18,200 --> 00:00:21,920 So the aim of this class, is to allow you to use differential equations to show the 7 00:00:21,920 --> 00:00:24,750 behavior of an RC circuit, as the system changes. 8 00:00:26,140 --> 00:00:30,960 So the previous lesson you learned some basics about solving simple first order 9 00:00:30,960 --> 00:00:34,840 linear differential equations. And let's start apply them to something 10 00:00:34,840 --> 00:00:39,160 real, RC circuits, and then it turns out that RL circuits or circuits that have 11 00:00:39,160 --> 00:00:42,265 resistors and inductors in them, behave very similarly. 12 00:00:42,265 --> 00:00:45,770 And so you might notice a lot of similarity between this lesson, and the 13 00:00:45,770 --> 00:00:49,400 next lesson. The objectives for this lesson are to 14 00:00:49,400 --> 00:00:53,585 allow you to generate a differential equation from the circuit, identify 15 00:00:53,585 --> 00:00:57,240 initial and final conditions, solve the differential equation, and then graph the 16 00:00:57,240 --> 00:01:01,740 result. So we'll start by looking at the behavior 17 00:01:01,740 --> 00:01:04,010 of circuits with resistors and capacitors in them. 18 00:01:04,010 --> 00:01:05,850 So we'll take a look at this very simple example. 19 00:01:07,040 --> 00:01:09,470 Here we have a resistance and a capacitance. 20 00:01:11,150 --> 00:01:15,310 And we have labeled here a current. Now I know that this current, due to 21 00:01:15,310 --> 00:01:20,849 Ohm's law. It's going to be equal to Vs minus Vc, 22 00:01:20,849 --> 00:01:30,530 all divided by R. So Vs is this voltage, i, Vc is this 23 00:01:30,530 --> 00:01:34,760 voltage, so if I want to know this current, I can calculate it that way. 24 00:01:34,760 --> 00:01:39,210 Which, this current happens to also be equal to ic. 25 00:01:39,210 --> 00:01:42,270 So lets think of what the system will do. We've already talked about if we were to 26 00:01:42,270 --> 00:01:47,710 analyze this circuit in a steady state. That this capacitor is going to look like 27 00:01:47,710 --> 00:01:51,040 an open circuit. And this voltage is going to then be 28 00:01:51,040 --> 00:01:54,060 equal to V sub s. Since there is no where for the current 29 00:01:54,060 --> 00:01:58,350 to flow, no voltage drop cross this resistor. 30 00:01:58,350 --> 00:02:02,370 Because of that, we know what our final solutions going to be but, how does it 31 00:02:02,370 --> 00:02:04,230 get there? That's what this lesson is going to be 32 00:02:04,230 --> 00:02:07,010 about. And since we know something about this 33 00:02:07,010 --> 00:02:10,400 current. Initially, let's say that this initial 34 00:02:10,400 --> 00:02:13,420 which is zero to begin with, but times zero. 35 00:02:13,420 --> 00:02:16,540 There's a big difference in voltage between the two sides of the resistor. 36 00:02:16,540 --> 00:02:20,910 So this is a fairly strong current. And as this current is flowing, this 37 00:02:20,910 --> 00:02:27,116 capacitor is getting charged. But as this capacitor charges Vc gets 38 00:02:27,116 --> 00:02:32,540 bigger, so the difference between Vs and Vc get smaller, so this current 39 00:02:32,540 --> 00:02:36,380 decreases. And as this current decreases, this 40 00:02:36,380 --> 00:02:40,390 capacitor still continues to charge up, but at a slower rate. 41 00:02:41,860 --> 00:02:46,390 And so it turns out that if we look at the behavior of this system, our 42 00:02:46,390 --> 00:02:49,445 capacitor, voltage is going to start at zero. 43 00:02:49,445 --> 00:02:53,780 And it's going to start by increasing very rapidly, but as it starts to 44 00:02:53,780 --> 00:02:58,870 increase, the current decreases, and so it starts to tapers off. 45 00:03:01,020 --> 00:03:05,040 So it has kind of an exponential curve look to it. 46 00:03:05,040 --> 00:03:09,570 And I know that it has kind of this limit, this upper limit of V sub s. 47 00:03:09,570 --> 00:03:13,110 Cause I know the final value, if my system has been in this state for a very 48 00:03:13,110 --> 00:03:16,490 long time. Let's actually do the math to see where 49 00:03:16,490 --> 00:03:19,064 this comes from. So our first example is this one. 50 00:03:19,064 --> 00:03:23,690 We have a few different resistances, a few different capacitance's. 51 00:03:23,690 --> 00:03:27,149 But the first thing we're going to start with, is what we do already know? 52 00:03:27,149 --> 00:03:30,610 The first thing we're going to need to look at, then, is what are, where are we 53 00:03:30,610 --> 00:03:35,430 starting, and where are we getting to? So here, here you see that we have little 54 00:03:35,430 --> 00:03:40,210 switch that we're opening at time t equal 0. 55 00:03:40,210 --> 00:03:43,070 So let's consider what's going on before the switch is opened. 56 00:03:43,070 --> 00:03:46,040 We're going to assume this capacitor has been there for a very long time. 57 00:03:47,140 --> 00:03:49,910 But this is a wire, since the switch is closed. 58 00:03:51,340 --> 00:03:53,980 That means that any charge that's built up on this capacitor, is going to go 59 00:03:53,980 --> 00:04:03,670 away. So, my Vc, at 0, before the switch is 60 00:04:03,670 --> 00:04:06,490 opened is getting, is going to be equal to 0. 61 00:04:06,490 --> 00:04:09,990 Because any charge is just going to be able to go from one plate to the other. 62 00:04:09,990 --> 00:04:13,480 So we'll have no initial charge. After a long time of this switch being 63 00:04:13,480 --> 00:04:17,930 open however, it goes back to the, the previous example that we were looking at. 64 00:04:17,930 --> 00:04:25,240 And so I know that after a very long time, and then this voltage is going to 65 00:04:25,240 --> 00:04:27,666 be equal to 5 volts, because that's what the source is. 66 00:04:27,666 --> 00:04:30,602 So, we know where, where we're starting and where we're heading. 67 00:04:30,602 --> 00:04:37,725 Let's now derive a differential equation for this circuit. 68 00:04:37,725 --> 00:04:43,150 So let's start with what we know. First of all, we know that the current i, 69 00:04:43,150 --> 00:04:48,050 is equal to C times dv dt. Well, dv c dt. 70 00:04:48,050 --> 00:04:54,230 Right, the voltage across the capacitor. We also know that the current here, is 71 00:04:54,230 --> 00:04:57,620 going to be equal to V sub s minus V sub c over R. 72 00:04:57,620 --> 00:05:00,430 Just like we were looking at in the, the first problem that we showed. 73 00:05:01,620 --> 00:05:06,590 Putting these two things together, we see that Vs minus Vc over R is C dvc dt. 74 00:05:06,590 --> 00:05:09,770 We'll do a little bit of manipulation to get it into this format. 75 00:05:11,340 --> 00:05:15,270 And this is similar to the format that Doctor Ferry presented, when presenting 76 00:05:15,270 --> 00:05:19,270 differential equations. So, here we have a differential. 77 00:05:20,320 --> 00:05:23,651 The constanse times our function is equal to some constant. 78 00:05:23,651 --> 00:05:24,966 So here, 1 over RC is a, and Vs over RC is K. 79 00:05:24,966 --> 00:05:36,898 What we're going to do, is we're going to say let tau be equal to RC. 80 00:05:36,898 --> 00:05:41,990 And you already talked about tau in the previous lesson. 81 00:05:41,990 --> 00:05:45,410 And here it means the same thing that it did before, the time constant. 82 00:05:45,410 --> 00:05:50,000 Doing that, and putting this into the equation we showed here, we see that Vc 83 00:05:50,000 --> 00:05:54,810 is equal to Vs times the quantity 1 minus e to the minus t over tau. 84 00:05:54,810 --> 00:06:05,922 And we know that tau, because it's equal to R times C, Is equal to 2K times 500 85 00:06:05,922 --> 00:06:13,526 times 10 to the negative 6. So this ends up becoming 1000. 86 00:06:13,526 --> 00:06:21,020 So the time constant just being 1. The time constant, it turns out, is going 87 00:06:21,020 --> 00:06:24,560 to be in seconds. Because it's a time, time constant. 88 00:06:24,560 --> 00:06:28,070 And if you actually go through and look at the units of resistance, and the units 89 00:06:28,070 --> 00:06:31,130 of capacitance, and you multiply them together, you do get seconds. 90 00:06:32,350 --> 00:06:34,130 I'm not going to through the derivation of that, however. 91 00:06:36,260 --> 00:06:37,560 Let's see what this looks like on a graph. 92 00:06:38,610 --> 00:06:41,865 So, we already decided that we, sort of knew where we were headed. 93 00:06:41,865 --> 00:06:52,000 That we were going to start at 0. Then we're going to end up here at 5 94 00:06:52,000 --> 00:06:55,870 volts. And because of this exponential behavior, 95 00:06:55,870 --> 00:07:00,620 we know that in one time constant, we're going to make it around 2 3rds of the 96 00:07:00,620 --> 00:07:04,210 way. So we'll kind of guess at 2 3rds, maybe 97 00:07:04,210 --> 00:07:08,800 say it's about there. At 2 time constants, we're going to go 98 00:07:08,800 --> 00:07:13,803 another 2 3rds of the way. So another 2 3rds might be approximately 99 00:07:13,803 --> 00:07:18,470 here, and then 3 time constants again another 2 3rds. 100 00:07:18,470 --> 00:07:21,130 It turns out that one way to help you sketch this, is that if you look at the 101 00:07:21,130 --> 00:07:28,320 time constant where you cross this line, because of the way the exponential is 102 00:07:28,320 --> 00:07:33,430 it's own derivative, it's going to start off, kind of following along that curve. 103 00:07:33,430 --> 00:07:39,470 So, to kind of sketch it out, I believe that my curve is going to look something 104 00:07:39,470 --> 00:07:41,620 like this. So let's fill it in, and see how close we 105 00:07:41,620 --> 00:07:44,132 were. Not too bad. 106 00:07:44,132 --> 00:07:50,483 At 1 time constant we're going to be 63.2% of the way toward the final value, 107 00:07:50,483 --> 00:07:56,450 after 2 time constants 86.5%, and 3 time constants and so on, is going to follow 108 00:07:56,450 --> 00:08:00,371 the same way. Because of this 5 minus 5 over e, and 5 109 00:08:00,371 --> 00:08:05,680 minus 5 over e squared behavior. And so that's what we're going to expect, 110 00:08:05,680 --> 00:08:10,020 and we can see that our initial and our final conditions do match what we what we 111 00:08:10,020 --> 00:08:12,680 calculated initially. Now, we're going to move through another 112 00:08:12,680 --> 00:08:15,440 example, and we're going to go through this next example a little more quickly. 113 00:08:16,450 --> 00:08:21,390 In this case, we again have a switch. Now it's closing at time t equal 0, and 114 00:08:21,390 --> 00:08:24,360 we have two different resistors, but let's look at our initial and final 115 00:08:24,360 --> 00:08:27,880 conditions. Before this resist, before this switch is 116 00:08:27,880 --> 00:08:31,980 closed, we have this current source that is going to flow through here. 117 00:08:31,980 --> 00:08:36,424 This is going to look like an open circuit, so Vc is just going to be this 1 118 00:08:36,424 --> 00:08:42,540 ml amp and times 12 Kohms. So Vc, before we close the switch, is 119 00:08:42,540 --> 00:08:46,790 going to be 12 Volts. After the switch is closed, now, these 120 00:08:46,790 --> 00:08:50,260 two resistors could be seen as being in parallel. 121 00:08:50,260 --> 00:08:53,686 And their equivalent resistance would be 3 kilohms. 122 00:08:53,686 --> 00:08:57,740 And again, after a long time the capacitor will look like an open circuit. 123 00:08:57,740 --> 00:09:01,120 And vC after a very long time, is going to be equal to 3 volts. 124 00:09:03,540 --> 00:09:05,700 Let's see how we get our differential equations for this one. 125 00:09:10,040 --> 00:09:13,730 Capacitance here is 1 picofarad, 12 kilohms, 4 kilohms, and so on. 126 00:09:13,730 --> 00:09:17,840 We, again, are going to use some things we know, in this case, Kirchhoff's 127 00:09:17,840 --> 00:09:21,740 current law. I know that this source current, has to 128 00:09:21,740 --> 00:09:24,550 equal the sum of the currents going through each of these different devices. 129 00:09:26,280 --> 00:09:29,140 And I know the equations for each of these devices, for the resistors, it's 130 00:09:29,140 --> 00:09:33,015 Ohm's law. And then for this capacitors, it's C 131 00:09:33,015 --> 00:09:37,320 dvdt. So we manipulate these equations, and I 132 00:09:37,320 --> 00:09:41,550 show the derivations here. We're going to use R eq to represent 133 00:09:41,550 --> 00:09:43,590 this. Which is the same thing as when we were 134 00:09:43,590 --> 00:09:46,900 calculating equivalent resistances, in parallel. 135 00:09:46,900 --> 00:09:49,080 And finally we're going to get this equation. 136 00:09:49,080 --> 00:09:56,392 So again, we have our K, we have our a, and our tau is just going to be R eq 137 00:09:56,392 --> 00:09:59,190 times C. So it corresponds to the tau where it was 138 00:09:59,190 --> 00:10:01,740 presented before, giving us this final solution. 139 00:10:02,850 --> 00:10:15,500 And is times Req is going to be 3, and the is times R1 is going to be 12. 140 00:10:15,500 --> 00:10:24,239 Our time constant, tau, Is equal to 1 over, or sorry, it's just R eq times C, 141 00:10:24,239 --> 00:10:31,005 so 3K ohms times c, which is one picofarad. 142 00:10:31,005 --> 00:10:39,180 So it's going to give us 1 microsecond for 1 tao. 143 00:10:39,180 --> 00:10:42,050 Again, we're going to graph it. So each of these tau's now, instead of 144 00:10:42,050 --> 00:10:46,460 corresponding to 1 second, so this is 1 microsecond, and this is the equation 145 00:10:46,460 --> 00:10:50,730 we're looking to, to show. Now, what it turns out to be, is that 146 00:10:50,730 --> 00:10:56,290 initially, we're going to start at 12 volts. 147 00:10:56,290 --> 00:11:00,970 We're going to go down to 3 volts. By looking at this equation, you can kind 148 00:11:00,970 --> 00:11:05,370 of identify why. Again, 2 3rd's down, and the first time 149 00:11:05,370 --> 00:11:08,750 constant. Another 2 3rd's, and 2 time constants. 150 00:11:08,750 --> 00:11:12,500 Another 2 3rd's and 3. So, if we kind of trace a little curve 151 00:11:12,500 --> 00:11:18,830 along this line like this. We're a little, we're pretty close. 152 00:11:18,830 --> 00:11:21,760 That was a little sloppy there, especially near the end. 153 00:11:21,760 --> 00:11:26,640 But again, in the first time constant, we've gotten 63.2% of the way to our 154 00:11:26,640 --> 00:11:31,149 final value, and in 2 time constants, 86.5, and so on. 155 00:11:37,060 --> 00:11:40,250 So, we've now been able to get some intuition about how RC circuits behave. 156 00:11:40,250 --> 00:11:42,255 And we identified their initial and final conditions. 157 00:11:42,255 --> 00:11:45,820 And then we found differential equations, to find out what's going on in the 158 00:11:45,820 --> 00:11:48,870 interim. We were able to then solve these 159 00:11:48,870 --> 00:11:50,745 differential equations, and then graph the results. 160 00:11:50,745 --> 00:11:58,160 In the next lesson were applying the same methods and techniques, to circuits with 161 00:11:58,160 --> 00:12:01,190 resistors and inductors in them. And we'll see that the behavior is very 162 00:12:01,190 --> 00:12:04,640 similar, but we'll see that there's some differences as well. 163 00:12:04,640 --> 00:12:07,714 So until then, take care.