1 00:00:00,025 --> 00:00:07,840 Hi, I'm Jacob Block, and I'm an EECE student here at Georgia Tech. 2 00:00:09,030 --> 00:00:11,690 Today I'll be going over and extra problem on RC Circuits. 3 00:00:14,660 --> 00:00:18,570 So, in this extra problem, were going to be looking at an RC circuit. 4 00:00:18,570 --> 00:00:24,366 And in this, so, for here, we have a a 5 volt source for a 300k resistor, a 10 5 00:00:24,366 --> 00:00:32,072 microfarad capacitor, a switch. and the switch closes at t equals 0, and 6 00:00:32,072 --> 00:00:37,990 a 150k ohm resistor. So, in this problem, we want to know what 7 00:00:37,990 --> 00:00:43,070 is the voltage across the capacitor for all time t. 8 00:00:45,170 --> 00:00:48,980 So, let's take this setup the circuit in two stages. 9 00:00:48,980 --> 00:00:52,820 So, the first stage we're going to look at before the switch is flipped. 10 00:01:01,680 --> 00:01:04,590 Before the switch is flipped, we have an open circuit here. 11 00:01:04,590 --> 00:01:08,239 And that means that we have simply a resistor and a capacitor. 12 00:01:17,670 --> 00:01:21,762 And this capacitor, since this circuit has been running for a really long time, 13 00:01:21,762 --> 00:01:25,811 for t less than than 0, the capacitor is filled up. 14 00:01:25,811 --> 00:01:30,410 And for a filled capacitors, we can replace those with an open circuit. 15 00:01:38,290 --> 00:01:42,688 So, really, quickly we know there is no current because of the open circuit. 16 00:01:42,688 --> 00:01:47,320 So, this top node has to have a voltage of 5 relative to this bottom node. 17 00:01:47,320 --> 00:01:52,512 So, the voltage across here, Vc or, t less than 0, is 5 volts. 18 00:01:54,330 --> 00:02:08,160 Now, we have our switch flipped. There's kind of 2 so now that our switch 19 00:02:08,160 --> 00:02:14,784 is flipped, we have our voltage source, resistor, capacitor, and our parallel 20 00:02:14,784 --> 00:02:26,330 resistor. 5 volts, 300k, 10 microfarads, and 150k. 21 00:02:26,330 --> 00:02:35,420 So, how do we analyze this circuit now? Well, we can set up a KCL at this node. 22 00:02:35,420 --> 00:02:40,028 And then, use the currents calculate the current in this top branch, the capacitor 23 00:02:40,028 --> 00:02:44,160 branch, and this resistor branch. So, let's go ahead and do that. 24 00:02:46,570 --> 00:02:52,904 So, we kind of blow up this node, here. We know that the the voltage at this node 25 00:02:52,904 --> 00:02:58,909 is Vc. And the voltage at this top node is 5. 26 00:02:58,909 --> 00:03:03,527 And then, we have our resistor in the middle. 27 00:03:03,527 --> 00:03:10,259 So, the current through this resistor is 5 minus Vc over 300k, and similarly for 28 00:03:10,259 --> 00:03:17,300 this branch. We know that the voltage is Vc here. 29 00:03:17,300 --> 00:03:25,312 And the voltage is 0 here. And so, we know the current over across 30 00:03:25,312 --> 00:03:31,243 the resistor is Vc minus 0 over 150k. And finally, for the bottom branch, the 31 00:03:31,243 --> 00:03:36,598 current through a capacitor is the capacitance times the change in voltage. 32 00:03:36,598 --> 00:03:46,704 So, we can write our, our KCL, so the sum of the currents in have to be equal to 33 00:03:46,704 --> 00:04:02,533 the sum of the currents out. That tells us that we have 5 minus Vc. 34 00:04:02,533 --> 00:04:13,846 on the left side, over 300k. And on the right side, we have Vc over 35 00:04:13,846 --> 00:04:28,480 150k, plus, this 10 micro microfarad capacitor times dVc, dt. 36 00:04:31,910 --> 00:04:36,530 So, lets and using, so now that we have our, kind of our KCL equation, we can and 37 00:04:36,530 --> 00:04:43,679 it's a first order differential equation with a constant forcing function. 38 00:04:43,679 --> 00:04:49,101 Or, a constant source on it. We can rewrite it in terms that we've 39 00:04:49,101 --> 00:04:56,070 seen in the lecture slides. So, let's, let's go ahead and do that. 40 00:04:56,070 --> 00:05:09,102 So, by moving our, our equation around, we get dVc, dVc of t, dt plus 1 times Vc 41 00:05:09,102 --> 00:05:21,953 of t. equal to 10 E to the negative oops, sorry 42 00:05:21,953 --> 00:05:31,481 5 3rds equal to five thirds. So, so this is our differential equation 43 00:05:31,481 --> 00:05:36,182 for the for that node. And it describes the currents entering 44 00:05:36,182 --> 00:05:43,150 and leaving that node. And we also know that Vc at t equal 0, 45 00:05:43,150 --> 00:05:51,376 right when the switch flips, is 5 volts. So, using the formula that we have in the 46 00:05:51,376 --> 00:05:57,638 notes, Vc of t is equal to 5 3rds, the k over a times 1 minus E to the negative a, 47 00:05:57,638 --> 00:06:06,993 which is 1 times t. plus the initial condition, 5, times E to 48 00:06:06,993 --> 00:06:15,159 the negative at, which is just t. And this is valid for all time t after 0. 49 00:06:16,940 --> 00:06:22,802 And we can kind of simplify that in one, in one more step. 50 00:06:22,802 --> 00:06:29,570 So, Vc of t is out of 5 or t less than 0. And 5 3rds plus 10 3rds, e to the 51 00:06:29,570 --> 00:06:37,730 negative t, or t greater than 0, after the switch is flipped. 52 00:06:37,730 --> 00:06:44,790 Let's go ahead and, and graph that solution. 53 00:06:44,790 --> 00:06:49,897 So, the voltage starts at 5. And this extends out to however, I mean, 54 00:06:49,897 --> 00:06:54,902 the circuit's been running for a really long time, so this gets extended out to 55 00:06:54,902 --> 00:07:01,300 the left. And we know that the circuit ends up at 5 56 00:07:01,300 --> 00:07:06,910 3rds because as t goes to infinity, this term drops out and we're just left with 5 57 00:07:06,910 --> 00:07:12,334 3rds. So, it kind of approaches this level 58 00:07:12,334 --> 00:07:16,310 here. And in between, we have this transient 59 00:07:16,310 --> 00:07:20,608 response of 5 3rds plus 10 10 3rds e to the negative t. 60 00:07:20,608 --> 00:07:25,640 So, this exp, there's an exponential curve that connects these two points. 61 00:07:28,420 --> 00:07:32,965 And we know that this so this, this, this switch flips at 0. 62 00:07:32,965 --> 00:07:40,717 And 1 second later, if we plug in 1 here, we drop significantly in our value, and 63 00:07:40,717 --> 00:07:51,013 this, kind of, value here, after one, one time step, is 2.89 volts. 64 00:07:51,013 --> 00:07:59,273 And if we actually look at our circuit, we can kind of see where this negative 1 65 00:07:59,273 --> 00:08:05,342 comes from. For our C circuits, we can calculate a 66 00:08:05,342 --> 00:08:11,865 tao constant, which is equal to RC. And this RC, the R for this C is the 67 00:08:11,865 --> 00:08:18,780 parallel combination of this 300k and 150k. 68 00:08:18,780 --> 00:08:25,475 So, when you combine those two parallel resistors of 1 over 300k plus 1 over 69 00:08:25,475 --> 00:08:31,861 150k, and multiply it by our 10 microfarad capacitor, you get a time 70 00:08:31,861 --> 00:08:38,750 constant of 1 second. And that's exactly what shows up in our 71 00:08:38,750 --> 00:08:43,980 differential equation here. So in this problem, we looked at an RC 72 00:08:43,980 --> 00:08:47,731 circuit with a switch. And we analyzed the circuit before and 73 00:08:47,731 --> 00:08:50,945 after the switch. And then looked at, and then calculated 74 00:08:50,945 --> 00:08:54,814 the transient response, and then plotted the steady state before the switch was 75 00:08:54,814 --> 00:08:58,850 flipped, the transient. And then, the steady state after the 76 00:08:58,850 --> 00:09:00,250 switch was flipped.