1 00:00:00,720 --> 00:00:03,880 Hello my name is Sejer Garcia. I'm a UPC student. 2 00:00:03,880 --> 00:00:07,360 And we're going to work on the second part of our impetus problem. 3 00:00:07,360 --> 00:00:09,732 Please refer to the results obtained in the previous problem. 4 00:00:09,732 --> 00:00:17,170 Alright so we've already looked at the circuit, in the previous problem. 5 00:00:17,170 --> 00:00:21,096 Where we have a time varying voltage. 100 ohm resistor. 6 00:00:21,096 --> 00:00:27,060 A milifire capacitor, a 2 milifire capacitor, and a 5 milihairme inductor. 7 00:00:27,060 --> 00:00:32,268 So, in the previous problem, what we id is given a frequency of 60 hertz, and 8 00:00:32,268 --> 00:00:40,018 then taking that into radience per second, we have 377 radience per second. 9 00:00:40,018 --> 00:00:44,498 Given this information, we were able to simplify this problem into their 10 00:00:44,498 --> 00:00:50,937 independent impedances. And what we ended up having was an 11 00:00:50,937 --> 00:01:00,525 impedance for the resistor, equal to 100 ohms, and impedance for both capacitors 12 00:01:00,525 --> 00:01:15,551 in series equal to negative j, 3.98 ohms. And an impedance for the inductor equal 13 00:01:15,551 --> 00:01:22,926 to j, 1.89 ohms. So let's say We were able to simplify our 14 00:01:22,926 --> 00:01:28,906 circuit into this one with little boxes, and let's say we were giving a voltage 15 00:01:28,906 --> 00:01:34,702 source equal to 10 cosine 377 times t plus 60 degrees volts, and they ask us to 16 00:01:34,702 --> 00:01:42,615 find the current that goes through each box. 17 00:01:42,615 --> 00:01:48,900 So that would be ir, this would be ic, and this would be il. 18 00:01:51,550 --> 00:01:55,190 So what we want to take from this problem is that even though we are working with 19 00:01:55,190 --> 00:01:58,886 impedances for different elements, our laws, our kickoff laws and our Ohms law 20 00:01:58,886 --> 00:02:04,050 are still going to apply to this... To our analyses. 21 00:02:05,230 --> 00:02:08,082 So what we, the way we're going to work this is exactly the same as we would work 22 00:02:08,082 --> 00:02:11,740 any other, any other problem with, let's say resistances. 23 00:02:11,740 --> 00:02:16,886 So, in the previous problem, we saw that we were able to combine these three 24 00:02:16,886 --> 00:02:21,791 impedances to find the equivalent impedance. 25 00:02:21,791 --> 00:02:25,425 That is in parallel with the source. If we were to have resistances we could 26 00:02:25,425 --> 00:02:31,718 do exactly the same thing. So we found out that the Cerulean 27 00:02:31,718 --> 00:02:46,813 impedance between these three elements is equal to a hundred minus j 3.52 Ohms. 28 00:02:46,813 --> 00:02:50,641 And this here by combining all of these three impedances we can find that, we can 29 00:02:50,641 --> 00:02:54,005 see that that's the current that's going to flow through the equivalent 30 00:02:54,005 --> 00:02:57,775 impedance is equal to the current that's going to flow through the resistors, so 31 00:02:57,775 --> 00:03:03,238 this is IR right here, and as I mentioned earlier. 32 00:03:03,238 --> 00:03:08,510 Our Ohms law still can apply, V is equal to I times R. 33 00:03:09,530 --> 00:03:12,530 And by using this relationship we can calculate IR directly from our two 34 00:03:12,530 --> 00:03:17,472 values. So, to find IR this is going to be equal 35 00:03:17,472 --> 00:03:24,732 to our voltage source, VT. Over our impedance, Z equal. 36 00:03:24,732 --> 00:03:31,530 So substituting our values, we know that the voltage source is equal to 10 times 37 00:03:31,530 --> 00:03:41,254 cosine 377 times T plus 60 degrees. And our impedance is 100 minus J times 38 00:03:41,254 --> 00:03:45,501 3.52 ohms. So once we get to this step, we can see 39 00:03:45,501 --> 00:03:50,225 that it's kind of weird to divide these. So, because, first of all, all we have 40 00:03:50,225 --> 00:03:54,739 for the voltage is an amplitude and an an angle here, and all we have for our 41 00:03:54,739 --> 00:03:59,737 impedance is expressed as complex numbers. 42 00:03:59,737 --> 00:04:03,706 So what we want to do is, use phasers for both cases such that are equal, and we 43 00:04:03,706 --> 00:04:09,905 can perform this calculation normally. So, for the voltage source, to combine 44 00:04:09,905 --> 00:04:15,383 this into phasers, we have our magnitude which is 10, and then our angle which is 45 00:04:15,383 --> 00:04:21,234 16 degrees. And then, to combine to transform our 46 00:04:21,234 --> 00:04:26,485 impendance we can use a calculator, or you can use our, our formulas for 47 00:04:26,485 --> 00:04:32,270 converting complex numbers into polar, for polar numbers and we get 100 angle 48 00:04:32,270 --> 00:04:39,027 2.02 degrees. So by making this division you end up 49 00:04:39,027 --> 00:04:44,853 having that the current that flows through our impedance right here. 50 00:04:44,853 --> 00:04:52,533 Are equal in impedance, is going to be equal to 100 angle 57.98 degrees, and 51 00:04:52,533 --> 00:04:59,365 this is milliamps. So we've already found the current that's 52 00:04:59,365 --> 00:05:03,210 going to flow through our resistor right here. 53 00:05:03,210 --> 00:05:07,104 Now, We can go back to our original circuit knowing that we know knowing that 54 00:05:07,104 --> 00:05:11,293 we have this current right here. And as you can see these two elements are 55 00:05:11,293 --> 00:05:14,785 in parallel. So we can perform a, a current, current 56 00:05:14,785 --> 00:05:19,735 divide, current divider, sorry, current divider as we've always done with our 57 00:05:19,735 --> 00:05:23,674 resistances. So let's, let's just number our steps 58 00:05:23,674 --> 00:05:26,950 here. Let's say That this was our first step. 59 00:05:28,220 --> 00:05:32,207 Second step. Third, and the next one's going to be the 60 00:05:32,207 --> 00:05:36,074 fourth step. So next let's find the current that flows 61 00:05:36,074 --> 00:05:38,436 through inductor. IL, right here. 62 00:05:38,436 --> 00:05:40,765 And like I say, we're going to use a current divider. 63 00:05:40,765 --> 00:05:50,547 Alright? So our formula for IL is going to be our 64 00:05:50,547 --> 00:05:58,472 impedance for the capacitor, which is negative J 3.98 over the sum of both. 65 00:05:58,472 --> 00:06:04,376 1.89 times our source. And our source in this case is the 66 00:06:04,376 --> 00:06:11,746 current that's going to flow through the resistance which we already found, times 67 00:06:11,746 --> 00:06:19,860 a hundred angle 57.98 degrees. Times 10 to the minus three, because 68 00:06:19,860 --> 00:06:24,105 remember that units are important. So, once we've performed this 69 00:06:24,105 --> 00:06:27,955 calculation, remembering that in order to perform this, you can either input these 70 00:06:27,955 --> 00:06:31,805 number directly into a calculator, or you can convert these polar, I'm sorry, these 71 00:06:31,805 --> 00:06:37,373 complex. the numbers into polar coordinates, and 72 00:06:37,373 --> 00:06:41,439 then make the calculation. We end up having a value equal to 190 73 00:06:41,439 --> 00:06:46,373 angle 57.98 degrees, and this is in milliamps, too. 74 00:06:46,373 --> 00:06:52,015 Keep in mind that our angle is the same as The angle for inductors, inductor 75 00:06:52,015 --> 00:06:57,930 current is the same as the angle for the resistor current and this is because we 76 00:06:57,930 --> 00:07:06,076 can see it, we can see it right here. We can see that by adding these two 77 00:07:06,076 --> 00:07:09,508 numbers in the denominator we end up having a value equal to a constant times 78 00:07:09,508 --> 00:07:13,370 j but the j's are not going to cancel out. 79 00:07:13,370 --> 00:07:17,190 So this number ends up being just a real number, a constant. 80 00:07:17,190 --> 00:07:20,688 Which does not affect at all our angle. So both, both currents are going to have 81 00:07:20,688 --> 00:07:23,563 the same angle. Now once we have our current that flows 82 00:07:23,563 --> 00:07:27,073 through resistor and current that flows through the conductor, we can easily 83 00:07:27,073 --> 00:07:30,583 calculate the current that flows through the capacitor, like I said at the 84 00:07:30,583 --> 00:07:36,238 beginning again, just because. >> We can, our, our [INAUDIBLE] still 85 00:07:36,238 --> 00:07:42,309 apply for these type of circuits. So, our next step is to calculate IC. 86 00:07:42,309 --> 00:07:45,613 And we know that the current that goes through IR is going to be equal to the 87 00:07:45,613 --> 00:07:49,309 current that goes through the capacitor plus the current that goes through the 88 00:07:49,309 --> 00:07:54,144 inductor. And if we Solve for IC we get IC is equal 89 00:07:54,144 --> 00:07:59,604 to IR minus IL and we already know these two values because we calculated in the 90 00:07:59,604 --> 00:08:05,316 previous steps and the final values equal to minus 90 angle 57.98 degrees and it is 91 00:08:05,316 --> 00:08:10,776 also in And as I explained earlier, the angles still the same just because, by, 92 00:08:10,776 --> 00:08:21,070 when you add two numbers, the angles are not affected together. 93 00:08:22,300 --> 00:08:25,144 Are not affected, sorry, by each other. So, just to go over these steps, when we 94 00:08:25,144 --> 00:08:28,199 have elements and they're, we're not given their impedances, the first thing 95 00:08:28,199 --> 00:08:31,019 to do is calculate their individual impedances as we did in the previous 96 00:08:31,019 --> 00:08:36,040 column. Once we have this simplified circuit with 97 00:08:36,040 --> 00:08:40,330 the little boxes and their individual impedances, given a source in this case a 98 00:08:40,330 --> 00:08:44,360 voltage source, we can then start applying our Kirkoff's Law and our Ohm's 99 00:08:44,360 --> 00:08:51,320 Law as we've always done, in order to find anything that's asked for us. 100 00:08:51,320 --> 00:08:54,968 In this case, the individual currents were asked So we start by combining all 101 00:08:54,968 --> 00:08:58,720 of them just to find this first current right here. 102 00:08:58,720 --> 00:09:02,785 The current that goes to the resistor. Then we can apply a current divider as 103 00:09:02,785 --> 00:09:07,350 we've always done in order to find the current that goes to the conductor. 104 00:09:07,350 --> 00:09:09,605 We could've also done it for the capacitor first. 105 00:09:09,605 --> 00:09:13,528 In this case we did it for the conductor. And once we have two out of three values, 106 00:09:13,528 --> 00:09:18,042 we can just use Kirkoff's law in order to solve for our final current. 107 00:09:18,042 --> 00:09:21,470 Thank you.