1 00:00:02,760 --> 00:00:07,030 Welcome back to Linear Circuits, last time we talked about RC circuits, so this 2 00:00:07,030 --> 00:00:11,200 time we get to do RL circuits. And the aim is basically the same as last 3 00:00:11,200 --> 00:00:14,620 time, except now we're using these differential equations to show the 4 00:00:14,620 --> 00:00:16,810 behavior of RL circuits instead of RC circuits. 5 00:00:18,840 --> 00:00:21,990 And as I mentioned before, the previous lesson covered the RC ciircuits With 6 00:00:21,990 --> 00:00:26,928 resistors and capacitors. Now we're doing the exact same thing but 7 00:00:26,928 --> 00:00:30,120 using inductors. The lesson objectives also are the same. 8 00:00:30,120 --> 00:00:33,230 We're generating the differential equation, identifying initial and final 9 00:00:33,230 --> 00:00:38,600 conditions, solving the differential equations, and graphing the result. 10 00:00:38,600 --> 00:00:40,800 So let's look at the behavior of RL circuits. 11 00:00:42,520 --> 00:00:45,715 Again we have a source. This time it's current source. 12 00:00:45,715 --> 00:00:50,320 It's going to go through here, and as this current source comes through, 13 00:00:50,320 --> 00:00:56,720 there's going to be a current divider. This current here, iL, is going to be 14 00:00:56,720 --> 00:01:00,979 dependent upon what the existing magnetic field happens to be. 15 00:01:02,490 --> 00:01:07,150 Through this inductor here. Since this inductor doesn't want to, it's 16 00:01:07,150 --> 00:01:10,900 current to change very much. So, if I make a sudden change to this 17 00:01:10,900 --> 00:01:15,760 source, or for example, I have a switch and I close a switch suddenly. 18 00:01:15,760 --> 00:01:18,140 It's going to try and send current through it a different rate. 19 00:01:20,070 --> 00:01:23,550 And so, in order to deal with that We can kind of start looking at this. 20 00:01:23,550 --> 00:01:28,820 This is closed so what's going to happen? Well initially, most of the current is 21 00:01:28,820 --> 00:01:33,540 going to go through this resistor because, the inductor doesn't want to 22 00:01:33,540 --> 00:01:37,360 change its current rate. But over time, the voltage that's 23 00:01:37,360 --> 00:01:42,460 applied, because using Kircchoff's, using Ohm's Law we see a current going through 24 00:01:42,460 --> 00:01:49,980 this resistor is going to say, V is going to be equal to whatever this resistor 25 00:01:49,980 --> 00:01:59,310 current is times the resistance. This voltage is going to drive The 26 00:01:59,310 --> 00:02:02,770 current and start to change the magnetic field in the inductor. 27 00:02:03,950 --> 00:02:07,030 But as that happens, this voltage is going to decrease. 28 00:02:07,030 --> 00:02:15,940 And notice the L and v R, because they are parallel, these two values have to be 29 00:02:15,940 --> 00:02:18,290 equal. So as this voltage decreases, so does 30 00:02:18,290 --> 00:02:20,680 this one. And the current decreases, and this 31 00:02:20,680 --> 00:02:23,626 current increases. Until such time that it reaches a 32 00:02:23,626 --> 00:02:26,160 steady-state. And that steady-state is when all of the 33 00:02:26,160 --> 00:02:31,140 current is going through the inductor and none of the current is going through the 34 00:02:31,140 --> 00:02:33,470 resistor. But this current is going to start by 35 00:02:33,470 --> 00:02:37,880 changing somewhat quickly and then over time will start to get slower and slower. 36 00:02:37,880 --> 00:02:47,540 And so again were going to see that exponential behavior In this case, we're 37 00:02:47,540 --> 00:02:51,920 going to i sub s as our maximal value, and we'll assume that the starting 38 00:02:51,920 --> 00:02:56,620 current is zero because right now time, before time equals zero, this is actually 39 00:02:56,620 --> 00:03:00,820 a gap. Now that we have some intuition in what's 40 00:03:00,820 --> 00:03:05,300 going on, let's look at some examples. Again, we'll start by looking at initial 41 00:03:05,300 --> 00:03:09,850 and final conditions. So, before time t equals 0, we just have 42 00:03:09,850 --> 00:03:13,920 an inductor and a resistor put together. So, any current going through this 43 00:03:13,920 --> 00:03:17,186 inductor is eventually going to be used up. 44 00:03:17,186 --> 00:03:22,470 As it loses energy, or loses power in this resistor. 45 00:03:22,470 --> 00:03:26,995 So, we can say this has been. It's initial position for a long time. 46 00:03:26,995 --> 00:03:37,820 The initial voltage is equal to zero. As is the current. 47 00:03:37,820 --> 00:03:41,350 After we close this switch and we let it sit for a long time. 48 00:03:41,350 --> 00:03:44,132 Again, remember that this is going to start to look like a wire, which that 49 00:03:44,132 --> 00:03:46,745 means that all of the current is going to go this direction. 50 00:03:46,745 --> 00:03:53,010 So, I look at my current after a long time, this is going to be equal to i sub 51 00:03:53,010 --> 00:03:59,080 s. We also know that the voltage after a 52 00:03:59,080 --> 00:04:06,130 long time Is also going to be equal to 0. Now, this problem's a little bit 53 00:04:06,130 --> 00:04:07,480 trickier. Because what we're actually going to 54 00:04:07,480 --> 00:04:12,260 solve for is vL. We're not going to solve for iL. 55 00:04:12,260 --> 00:04:17,480 And so, the way that we're going to do that is we're going to find iL. 56 00:04:18,880 --> 00:04:26,517 And then we will use the fact that V equals L d i l, d t to find the voltage. 57 00:04:26,517 --> 00:04:32,710 It can be an interesting exercise to go through this derivation again and solving 58 00:04:32,710 --> 00:04:35,510 using the Vs directly instead of solving for the is. 59 00:04:35,510 --> 00:04:39,850 And this will probably become a little bit more clear as we do the example. 60 00:04:40,920 --> 00:04:44,200 So, in working the example, we need to generate our differential equation. 61 00:04:44,200 --> 00:04:47,972 So, we start with from what we know. First of all, that vL equals L diL dt, 62 00:04:47,972 --> 00:04:49,466 and then iR is equal to v over R, using Ohm's law. 63 00:04:49,466 --> 00:05:00,520 And then Kirchhoff's current law tells us that iS has to equal iR plus iL. 64 00:05:00,520 --> 00:05:06,750 Putting all of that together, we see that is equals L over R, diL dt plus iL, and 65 00:05:06,750 --> 00:05:11,350 we're going to then rearrange this a little bit, so it's in the same format as 66 00:05:11,350 --> 00:05:17,570 we were using before. So now, R over L is A, R over Lis is our 67 00:05:17,570 --> 00:05:28,612 K, and that tal is going to be L over R. So if we take L over R, we're going to 68 00:05:28,612 --> 00:05:37,780 take 10 million, divided by 5 k, it's going to give us 2 time 10 to the 69 00:05:37,780 --> 00:05:43,292 negative 6, which are microseconds. So, that one time constant is going to be 70 00:05:43,292 --> 00:05:46,432 2 microseconds. Putting it all together, we get this for 71 00:05:46,432 --> 00:05:54,554 our current, we get this for our voltage. So, the voltage is i sub s times R times 72 00:05:54,554 --> 00:06:00,920 e to the negative t divided by tau. This is a little bit trickier than 73 00:06:00,920 --> 00:06:07,570 before. Because we noticed that initially, our 74 00:06:07,570 --> 00:06:13,130 voltage across our inductor is zero, because all of the current's been 75 00:06:13,130 --> 00:06:16,230 consumed by the resistor over time, so it's going to be zero. 76 00:06:16,230 --> 00:06:19,705 But we also know that this equation is going to start us off at five volts. 77 00:06:19,705 --> 00:06:24,117 And after one time constant go down to 2 3rds. 78 00:06:24,117 --> 00:06:28,768 After two time constant another 2 3rds. After three cons, time constant another 2 79 00:06:28,768 --> 00:06:29,342 3rds and so on. So, we're going to expect it to look 80 00:06:29,342 --> 00:06:41,360 something like that. A little soft at the beginning, but 81 00:06:41,360 --> 00:06:42,650 [INAUDIBLE]. So it's interesting to notice that here 82 00:06:42,650 --> 00:06:45,960 the ohm doesn't jump. Where it was at zero and suddenly it's 83 00:06:45,960 --> 00:06:48,840 times equal zero and you close the switch and it jumps up to five volts. 84 00:06:48,840 --> 00:06:54,040 But remember voltage, voltages jumping in conductors is just fine. 85 00:06:54,040 --> 00:06:56,780 We just can't have the currents instantaneously change. 86 00:06:56,780 --> 00:07:00,987 Let's see how we did. Pretty close. 87 00:07:00,987 --> 00:07:03,590 A little bit sloppy on the curve this time, it's a little hard to free hand 88 00:07:03,590 --> 00:07:10,260 draw sometimes but we see That, we see Two thirds drop, So after one time 89 00:07:10,260 --> 00:07:19,080 constants, Which is to microsecond, we've dropped 62.3% of the way and then 86.5 90 00:07:19,080 --> 00:07:24,420 after two time constants, and so on. Now we're going to look at something a 91 00:07:24,420 --> 00:07:27,815 little bit new. Instead of having just a switch that we 92 00:07:27,815 --> 00:07:31,450 close or open, now we're going to have a switch that we toggle. 93 00:07:33,030 --> 00:07:36,639 In this example, we're only going to assume that the switch was in its first 94 00:07:36,639 --> 00:07:42,280 position, here, for a very long time before we switch it into our second 95 00:07:42,280 --> 00:07:47,810 position here. It can be an interesting thing, if you do 96 00:07:47,810 --> 00:07:51,730 this analysis, to see what would happen if you toggle this switch back and forth, 97 00:07:51,730 --> 00:07:53,790 so it might be something good to discuss on the forums. 98 00:07:55,580 --> 00:07:58,435 Again, we're going to start by looking at our initial and our final conditions. 99 00:07:58,435 --> 00:08:02,360 The initial condition is with the switch in this position. 100 00:08:03,440 --> 00:08:07,540 Voltage here going through here, this is going to effect a short circuit or a 101 00:08:07,540 --> 00:08:10,900 wire. So, this voltage is going to be zero or 102 00:08:10,900 --> 00:08:12,930 solving for iL. We're going to solve for a current this 103 00:08:12,930 --> 00:08:18,259 time and since it's in that configuration after awhile we have 12 volts. 104 00:08:18,259 --> 00:08:26,210 Going across 3 kg ohms, so we see that that is going to result in a current of 4 105 00:08:26,210 --> 00:08:29,780 mA after a long time. When I put in the second configuration, 106 00:08:29,780 --> 00:08:35,880 it's in the same basic situation, you have a voltage source, a resistance and 107 00:08:35,880 --> 00:08:44,840 so in that circumstance this current is going to be equal to Minus one half of a 108 00:08:44,840 --> 00:08:47,510 million. So, we have our initial current and our 109 00:08:47,510 --> 00:08:50,170 final current. And remember that the current cannot 110 00:08:50,170 --> 00:08:53,740 instantaneously change in an inductor. So, we'll be looking to make sure that 111 00:08:53,740 --> 00:08:57,150 that holds in our solution. In generating the difference of 112 00:08:57,150 --> 00:08:58,979 equations, we'll start with The L di dt and here, we see this. 113 00:08:58,979 --> 00:09:02,821 Il in the second position is going to be equal to this, so we're using the V2 114 00:09:02,821 --> 00:09:15,710 minus Vl divided by R2. And putting all of that together, you get 115 00:09:15,710 --> 00:09:19,609 this equation, and then we're going to rearrange it into our canonical form. 116 00:09:21,080 --> 00:09:31,170 So this is now our a, this is now our k, and now tau is equal to L over R 2. 117 00:09:31,170 --> 00:09:35,283 So using the differential equation solution that we have, i L is equal to v 118 00:09:35,283 --> 00:09:41,680 2 over R 2 times quantity 1 minus e to the minus t over tau, plus V1 over R1 e 119 00:09:41,680 --> 00:09:47,550 to the minus t over tap /g. So this V2 over R2 is going to become 120 00:09:47,550 --> 00:09:58,336 minus one half mA and the V1 over R1 is going become 4 mA. 121 00:09:58,336 --> 00:10:09,290 So that's what we see here. So what we're going to expect to see is 122 00:10:09,290 --> 00:10:11,672 we're going to start up here at 4 milliamps. 123 00:10:11,672 --> 00:10:19,860 Now, this, down here at minus 1 milliamp, is where we're headed. 124 00:10:19,860 --> 00:10:23,920 So one time constant will go 2 steps away from the 4 milliamps to the minus one 125 00:10:23,920 --> 00:10:29,642 half milliamp. So, approximately here. 126 00:10:29,642 --> 00:10:37,690 2 time constants about there, 3 time constants. 127 00:10:37,690 --> 00:10:43,000 Somewhere around there. And then we can make our exponential 128 00:10:43,000 --> 00:10:45,875 curve, but we'll just draw it out. We're pretty close. 129 00:10:45,875 --> 00:10:53,350 Again, the same quantities, 63%, 86% after two time clusters, and you see the 130 00:10:53,350 --> 00:10:57,680 same behavior, the same exact behavior. So once you find your time constant, it's 131 00:10:57,680 --> 00:11:00,630 pretty easy the kind of guess what the curve is going to look like, but we've 132 00:11:00,630 --> 00:11:03,594 also seen that you can derive it mathematically directly. 133 00:11:03,594 --> 00:11:06,550 And this matches all of the things were expecting to see here. 134 00:11:08,630 --> 00:11:12,920 It can also be interesting to see what would happen, if instead of just using a 135 00:11:12,920 --> 00:11:16,870 constant source here, we changed into a function, and if you know something of 136 00:11:16,870 --> 00:11:20,570 differential equations You can probably see how you can go about doing that and 137 00:11:20,570 --> 00:11:23,090 solving that. But, that kind of goes beyond the scope 138 00:11:23,090 --> 00:11:25,780 of what we'll be covering in this class. If you do have questions about it though, 139 00:11:25,780 --> 00:11:28,160 I encourage you to go to the forums and ask them there. 140 00:11:28,160 --> 00:11:32,250 As we can discuss some things that, perhaps, go beyond the material that was 141 00:11:32,250 --> 00:11:35,520 covered in the lectures. So, to summarize. 142 00:11:35,520 --> 00:11:38,320 We got some intuition about how our L circuits behave. 143 00:11:38,320 --> 00:11:41,520 We identified their initial and final conditions and used differential 144 00:11:41,520 --> 00:11:45,270 equations to see what happens while it transitions. 145 00:11:45,270 --> 00:11:49,990 Then we also graphed the results. In the next lesson, we'll start talking 146 00:11:49,990 --> 00:11:51,830 about second order differential equations. 147 00:11:51,830 --> 00:11:55,080 And this is what's going to result when we have RLC circuits. 148 00:11:55,080 --> 00:11:58,240 Circuits that have resistors, inductors, and capacitors all together. 149 00:11:58,240 --> 00:12:00,982 And it's a little bit more interesting than just the first order. 150 00:12:00,982 --> 00:12:06,650 And these concepts with then be carried on into module three, where we'll start 151 00:12:06,650 --> 00:12:09,008 talking a little bit about frequency response and filters. 152 00:12:09,008 --> 00:12:11,715 Until next time.