1 00:00:03,560 --> 00:00:07,710 Hi, I'm Jacob Block and I'm a ECE graduate student here at Georgia Tech. 2 00:00:09,030 --> 00:00:13,644 Today, I'll be going over an extra problem on RL circuits. 3 00:00:13,644 --> 00:00:18,731 Alright, in this extra problem, we're going to be looking at an RL circuit. 4 00:00:18,731 --> 00:00:24,451 And the RL circuit here has a current source, three resistors, an inductor and 5 00:00:24,451 --> 00:00:30,000 a switch and the switch opens at t equals 0. 6 00:00:30,000 --> 00:00:39,985 So we're interested in finding i1 of t, i2 of t and iL of t. 7 00:00:39,985 --> 00:00:48,370 This is a, a brief review. We need to know what an inductor is or 8 00:00:48,370 --> 00:00:58,674 what happens to inductor in steady state. So an inductor has some current and 9 00:00:58,674 --> 00:01:07,258 voltage across it. The voltage is related to the current by 10 00:01:07,258 --> 00:01:11,686 a derivative. So the current is proportional to the 11 00:01:11,686 --> 00:01:19,490 change in current. So in steady state, when you have no 12 00:01:19,490 --> 00:01:29,780 change in current, you have no change in voltage. 13 00:01:33,170 --> 00:01:36,828 And what that means is that we can take this inductor, and if we're in steady 14 00:01:36,828 --> 00:01:40,880 state mode, we can replace it with a short circuit. 15 00:01:43,090 --> 00:01:54,098 So knowing that, let's kind of start tackling this problem, so our first step 16 00:01:54,098 --> 00:02:05,090 is before the switch is flipped. So let's redraw the circuit with the 17 00:02:05,090 --> 00:02:12,342 switch flipped. So we have our current source, our 18 00:02:12,342 --> 00:02:22,390 resistor, a a closed switch and our inductor. 19 00:02:26,700 --> 00:02:31,055 Now, the inductor, the 100 millihenry inductor, since this circuit has been 20 00:02:31,055 --> 00:02:35,345 running for a really long time, and that switch has only been flipped at t equals 21 00:02:35,345 --> 00:02:39,637 0. This inductor is fully charged, and for 22 00:02:39,637 --> 00:02:45,720 inductors, the, that means that we have reached a steady state position. 23 00:02:45,720 --> 00:02:49,502 And we can replace the inductor with a short circuit, so let's go ahead and 24 00:02:49,502 --> 00:03:00,092 simplify this circuit. So now, we replace this with the, a short 25 00:03:00,092 --> 00:03:05,544 circuit and we can move our resistor over, and so we have a 5 ohm, a 30 ohm 26 00:03:05,544 --> 00:03:13,760 and a 30 ohm resistor here, and a 3 amp current source. 27 00:03:13,760 --> 00:03:18,982 So now, we want to solve the circuit, we want to solve for i1, i2, and iL. 28 00:03:18,982 --> 00:03:23,020 So you know, there's isn't an iL, we're just calling this branch current iL. 29 00:03:23,020 --> 00:03:28,646 So how do we go about doing that? Well, we can you know, combine all of the 30 00:03:28,646 --> 00:03:33,060 resistors together and then solve for the voltage across them. 31 00:03:33,060 --> 00:03:38,610 You know, and then do v equals i r. You can also do a current divider rule. 32 00:03:38,610 --> 00:03:46,362 So the current divider, if you remember, the current divider tells us that the 33 00:03:46,362 --> 00:03:56,789 current splits between two resistors by the weight of the second branch. 34 00:03:56,789 --> 00:04:05,779 So if this is i1 and this is i2 and this is an i input, this is R1 and R2. 35 00:04:05,779 --> 00:04:14,887 i1 is equal to i times R2 over R1 plus R2, and similarly, similarly i2 is equal 36 00:04:14,887 --> 00:04:27,555 i times R1 over R1 plus R2. So, let's go ahead and calculate the i1 37 00:04:27,555 --> 00:04:35,868 current. So, i1 is the 3 amps times the second two 38 00:04:35,868 --> 00:04:46,284 branches, the 30 and 30. Parallel over the sum of the branches, 39 00:04:46,284 --> 00:04:54,288 the 5 and the 30s, and this notation, the parallel bars, it stands for a parallel 40 00:04:54,288 --> 00:05:00,772 combination. So just as we use a sum to indicate a 41 00:05:00,772 --> 00:05:06,144 series connection, the parallel bars help us reduce notiation for a parallel 42 00:05:06,144 --> 00:05:11,948 connection. And so, when you see 30 in parallel with 43 00:05:11,948 --> 00:05:17,668 30, you should immediately think of doing the operation 1 over 30 plus 1 over 30 44 00:05:17,668 --> 00:05:30,240 all inverted, so this is 15 ohms. So i1 is 2.25 amps. 45 00:05:30,240 --> 00:05:38,304 And we can repeat the process for i2. i2 is the current in this branch. 46 00:05:38,304 --> 00:05:45,512 So the current in this branch is 3 amps times the second two branches combined, 47 00:05:45,512 --> 00:05:52,190 the 5 and the 30, divided by the sum of the branches, 30 plus 5 in parallel with 48 00:05:52,190 --> 00:05:59,999 30. And so, if you do the same operation with 49 00:05:59,999 --> 00:06:06,457 5 in parallel 30, you should get 0.375 amps. 50 00:06:06,457 --> 00:06:15,025 So we take we're looking at this branch, and we write out the same equation and we 51 00:06:15,025 --> 00:06:23,482 should get 0.375 amps. And you can check this by summing the 52 00:06:23,482 --> 00:06:32,587 currents at using a KCL equation. So, for this KSL equation we have 3 amps 53 00:06:32,587 --> 00:06:40,192 coming in, 0 2.25 amps in this first branch, and 0.375 in the second branch, 54 00:06:40,192 --> 00:06:50,780 and 0.375 in the third branch, and that all sums up to be 3. 55 00:06:50,780 --> 00:06:55,004 So we, we have our currents for this, the first part of the problem, for the 56 00:06:55,004 --> 00:06:59,802 pre-switch flip. Now we're going to be looking at the 57 00:06:59,802 --> 00:07:04,331 post-switch flip. So we are, we switch our flip we flip our 58 00:07:04,331 --> 00:07:10,540 switch, and now, we have a kind of new set a new circuit. 59 00:07:10,540 --> 00:07:12,710 So let's go ahead and draw out that circuit. 60 00:07:12,710 --> 00:07:28,300 We have a current source and our first resistor. 61 00:07:28,300 --> 00:07:34,253 And we have an open circuit at our, where our switch used to be. 62 00:07:34,253 --> 00:07:39,583 And that come the bottom comes over still, and we have our second branch 63 00:07:39,583 --> 00:07:45,747 resistor, and our third branch resistor, and our inductor. 64 00:07:45,747 --> 00:07:51,508 And this is a 5 ohm, 30 ohm, 30 ohm, and 100 millihenry conductor. 65 00:07:55,430 --> 00:08:01,496 So, what is going on at this bottom node? Is there, is there current in this 66 00:08:01,496 --> 00:08:05,700 branch? Well, we can do a KCL to figure that out. 67 00:08:07,760 --> 00:08:10,930 So, let's look at the KCL at this bottom node. 68 00:08:10,930 --> 00:08:12,448 So the KCL here well, this, this current source tells us we have 3 amps in this 69 00:08:12,448 --> 00:08:13,689 left wire. So we have 3 amps. 70 00:08:13,689 --> 00:08:15,425 And in this top top wire at the node, we have also 3 amps, right? 71 00:08:15,425 --> 00:08:28,560 The source this three amp current source tells us there's a constant 3 amp current 72 00:08:28,560 --> 00:08:40,165 on this on this branch. And by KCL, that tells us we have 0 amps 73 00:08:40,165 --> 00:08:47,778 in this connecting branch. And that means that these two circuits 74 00:08:47,778 --> 00:08:53,930 are actually separated, they're independent from each other. 75 00:08:53,930 --> 00:09:02,040 So, right away, we can say that i1 is 3 amps because of this current source. 76 00:09:02,040 --> 00:09:04,990 So now, let's take a look at this, at the right half circuit. 77 00:09:06,480 --> 00:09:17,589 The right half circuit has you know, two resistors, and an inductor and there's 78 00:09:17,589 --> 00:09:28,940 two there's two states to this circuit. There's a transient, right when the 79 00:09:28,940 --> 00:09:33,356 switch gets flipped, and a steady state, where this inductor has smoothed out, 80 00:09:33,356 --> 00:09:40,345 it's evened out. And remember for steady state, we have 81 00:09:40,345 --> 00:09:49,014 all of our time derivatives go to 0. And so, we can replace that inductor with 82 00:09:49,014 --> 00:09:53,549 a short circuit, just like in the first part of the problem. 83 00:09:54,640 --> 00:09:58,788 And if we calculate for the two branch currents, iL and i2, we'll find that 84 00:09:58,788 --> 00:10:02,706 they're 0, because there's no source here. 85 00:10:02,706 --> 00:10:10,048 So i2 is 0 and iL is 0. So now, let's look at the transient 86 00:10:10,048 --> 00:10:16,170 response, where we have, we have the full the full circuit including the inductor. 87 00:10:18,430 --> 00:10:25,570 We have our resistor, our second resistor, and our inductor, 30 ohms, 30 88 00:10:25,570 --> 00:10:35,344 ohms and 0.1 henries. What's so now, let's, to solve this 89 00:10:35,344 --> 00:10:43,830 circuit, let's do a KVL around this loop. So the KVL starting with the kind of left 90 00:10:43,830 --> 00:10:52,885 part, the KVL is 30, the resistance times the current. 91 00:10:52,885 --> 00:10:59,477 So this, the, the voltage drop across this resistor, the voltage drop across 92 00:10:59,477 --> 00:11:07,826 this resistor is the same, 30 times iL. And then the voltage drop across an 93 00:11:07,826 --> 00:11:15,720 inductor is equal to L times diL dt, and all of that equals 0. 94 00:11:15,720 --> 00:11:22,720 So, this is our, our KVL number. Now, L is 0.1 henries. 95 00:11:22,720 --> 00:11:36,120 So, you remember from lecture 3.6. We found that the derivative this, this 96 00:11:36,120 --> 00:11:45,010 differential equation, dy of t dt plus ay of t equal to k. 97 00:11:45,010 --> 00:11:51,912 It has the solution y of t equal to k over a times 1 minus e, 1 minus e to the 98 00:11:51,912 --> 00:12:03,880 negative at plus y at 0, the initial condition, times e to the negative at. 99 00:12:03,880 --> 00:12:09,592 And this is valid for t greater than, or, for, for all time after the, the switch 100 00:12:09,592 --> 00:12:17,230 flips, or basically the time after this this starting condition. 101 00:12:18,910 --> 00:12:25,670 So using that previous work, we can rewrite our KVL equation in the similar 102 00:12:25,670 --> 00:12:30,910 form. So,let's go ahead and reorganize that 103 00:12:30,910 --> 00:12:38,660 equation. diL, diL of t plus 630, and then this 0.1 104 00:12:38,660 --> 00:12:50,850 gets moved over to this constant plus 600 iL equals 0. 105 00:12:50,850 --> 00:12:54,150 And this is actually, this is in the same form as our known solution, so we can 106 00:12:54,150 --> 00:13:02,498 just plug in our numbers. So our iL of t has the solution k in this 107 00:13:02,498 --> 00:13:14,799 case, our k is 0, and our a is 600. So, iL of t is going to be 0 times, you 108 00:13:14,799 --> 00:13:22,940 know, all of this plus our initial condition. 109 00:13:22,940 --> 00:13:30,034 From the first part, we found that iL at 0 is 0.375 amps. 110 00:13:30,034 --> 00:13:43,553 So plus 0.375e to the negative 600t for t greater than or equal to 0. 111 00:13:52,490 --> 00:13:57,970 So we have our, our solution for kind of all time. 112 00:14:00,150 --> 00:14:09,157 Now, what is, what is the current in i2? So i2, remember, i2 is pointing down 113 00:14:09,157 --> 00:14:17,980 through the 30 ohm, and iL is pointing down through the inductor. 114 00:14:19,090 --> 00:14:24,265 And they're, so since they're their reference directions are opposite, we can 115 00:14:24,265 --> 00:14:32,710 say that i2 is equal to negative iL. So let's summarize kind of what we have 116 00:14:32,710 --> 00:14:38,700 now. So we have i1, where t less than 0, and 117 00:14:38,700 --> 00:14:50,500 that is zero point 2.25 amps. i2 for t less than 0 is 0.375 amps. 118 00:14:50,500 --> 00:15:05,650 And i3 for t less than 0 is 0.375 amps. And our i1, when T, f, for the, the 119 00:15:05,650 --> 00:15:14,596 switch is flipped turns into 3 amps or i2 after the switch is negative 0.375e of 120 00:15:14,596 --> 00:15:23,829 the negative 600t. And our i3 amps, and our i3 is 0.375e to 121 00:15:23,829 --> 00:15:31,045 the negative 600t amps. So, let's, let's graph that solution and 122 00:15:31,045 --> 00:15:41,691 see what it looks like. Whoops, sorry, these are L's. 123 00:15:41,691 --> 00:15:46,597 So iL of t starts right at t less than 0. We know that it's kind of a constant 124 00:15:46,597 --> 00:15:51,600 0.375, and at infinity kind of in the steady state, it approaches 0. 125 00:15:51,600 --> 00:16:18,280 So right when this switch flips, we turn from steady state to transient. 126 00:16:18,280 --> 00:16:22,765 And the transient here, the 0.375e to the negative 600t, it's connecting these two 127 00:16:22,765 --> 00:16:28,993 points together. It's connecting the steady state before 128 00:16:28,993 --> 00:16:39,310 the switch was flipped and the steady state when the inductor fully discharges. 129 00:16:39,310 --> 00:16:42,030 And that's the, this is the solution of the problem. 130 00:16:42,030 --> 00:16:53,236 So what do we, what do we do? We started with a we started with a 131 00:16:53,236 --> 00:16:58,608 circuit [NOISE ] 132 00:16:58,608 --> 00:17:05,270 [COUGH]. 133 00:17:05,270 --> 00:17:09,523 Okay. So we started with our RL circuit, and we 134 00:17:09,523 --> 00:17:15,847 solved for the circuit where, before the switch was flipped in a steady state, and 135 00:17:15,847 --> 00:17:28,125 we calculated each of those currents. And then after we looked at a new 136 00:17:28,125 --> 00:17:36,484 circuit, after the switch was flipped. So after the switch was flipped, the we, 137 00:17:36,484 --> 00:17:40,614 we had two separate independent circuits, one with a current source and one with 138 00:17:40,614 --> 00:17:46,458 just Rs and Ls. And then we proceeded to solve, set up a 139 00:17:46,458 --> 00:17:52,038 differential equation for that, for that RL circuit, and we found that the 140 00:17:52,038 --> 00:17:58,518 inductor slowly dissipates its energy and ends up fully discharged, kind of in the 141 00:17:58,518 --> 00:18:03,210 steady state.