1 00:00:01,260 --> 00:00:04,759 This is the exercise 2 in block 6. 2 00:00:05,339 --> 00:00:11,359 They are asking you to calculate the intensities through all the resistors. 3 00:00:11,900 --> 00:00:15,439 The first thing we have to do is to calculate the equivalent resistance. 4 00:00:16,179 --> 00:00:20,379 To calculate the equivalent resistance, first I have to solve this part. 5 00:00:21,100 --> 00:00:25,359 I have given names to all resistors. 6 00:00:25,539 --> 00:00:28,280 So this is 1, 2, 3, 4, 5, 6. 7 00:00:28,280 --> 00:00:44,399 Well, it doesn't matter the order, but it's important because of calculations. The first thing is to calculate this part here, because this brand that I have put these names AB, this brand, it's in parallel with this one. 8 00:00:44,399 --> 00:01:02,960 In this part I have a 60 ohms resistor and here I have two resistors of 20 and 10. So this part is 10 plus 20, 30 ohms. So this part is totally equivalent to two resistors in parallel of 60 and 30. 9 00:01:02,960 --> 00:01:24,640 I have to solve this parallel circuit, as usual, and you get a resistor that can substitute these three of them, so that's why I have put the name 2, 3 and 4, and all of them are equivalent to only one resistor of 20 ohms. 10 00:01:24,640 --> 00:01:39,120 In the next part, this part of the C-kit can be substituted with three resistors, the 100 ohms and the 150 ohms, and the third one, as I said, is 20 ohms. 11 00:01:39,120 --> 00:01:46,519 So I have three resistors in series and it's really easy to calculate the equivalent resistance that is the addition of these three. 12 00:01:47,099 --> 00:01:54,500 So finally, the equivalent resistance of the total C-kit is 270 ohms. 13 00:01:54,640 --> 00:02:14,979 Right now, I can calculate the total intensity flowing through the circuit, because it's always the voltage of the cell divided by the equivalent resistance, that's 10 volts divided by 270 ohms. 14 00:02:14,979 --> 00:02:25,050 The result is this. This is the intensity that is flowing from the cell. 15 00:02:25,050 --> 00:02:41,189 So here, in this part and also here, in this part, in the resistor 1 and in the resistor 5, I have the total intensity that is 0.037 amperes or 37 milli amperes. 16 00:02:41,189 --> 00:02:48,469 But when I get to this knot A, this intensity is divided in these two branches. 17 00:02:48,729 --> 00:02:56,330 I have to calculate the voltage between A and B, because if not, I cannot calculate the intensity through the resistor number 3. 18 00:02:56,729 --> 00:02:58,969 How do I calculate this? Really easy. 19 00:02:59,509 --> 00:03:04,110 I know the resistance and the intensity here, and I know the intensity here. 20 00:03:04,110 --> 00:03:09,789 so what I'm going to do is to calculate voltage here and voltage here and I know 21 00:03:09,789 --> 00:03:15,090 the voltage that is given by the cell is 10 so the voltage that is used by this 22 00:03:15,090 --> 00:03:20,250 resistor and this one and the one between A and B the addition of these 23 00:03:20,250 --> 00:03:26,250 three should be 10 because the voltage that is getting from the cell is 24 00:03:26,250 --> 00:03:37,689 consumed by the resistor so let's do it voltage in the resistor number one is 100 by this in 25 00:03:37,689 --> 00:03:46,889 intensity you have 3.7 volts and voltage in the number five it's 5.5 volts so the addition of 26 00:03:46,889 --> 00:03:58,969 these two it's 9.25 and the amount that rest to 10 is the voltage between a and b here i have put 27 00:03:58,969 --> 00:04:06,889 this diagram you for you to understand um i have here the 10 volts that are given by the cell and 28 00:04:06,889 --> 00:04:15,689 in the first resistor i'm consuming a 3.7 and in this resistor number five and the consumption is 29 00:04:15,689 --> 00:04:24,449 5.5 so the rest between a and b as the 3 of n has the addition of these three should be 10 30 00:04:24,449 --> 00:04:37,069 is the difference between these two amounts and 10 and it's as i said 0.275 volts so now 31 00:04:37,069 --> 00:04:43,250 i know the voltage between these two points i know the voltage i know the resistance i can 32 00:04:43,250 --> 00:04:47,970 calculate the intensity through the resistor number 3, because I applied the Ohm's law. 33 00:04:48,930 --> 00:04:57,089 The intensity is equal to the voltage divided by the resistance. So, this intensity number 3 is 0.75 34 00:04:57,089 --> 00:05:05,689 between 60. This is the intensity through the resistor number 3. And finally, if I have here 35 00:05:05,689 --> 00:05:09,850 0.037 36 00:05:09,850 --> 00:05:11,490 amperes 37 00:05:11,490 --> 00:05:13,709 and here I have 38 00:05:13,709 --> 00:05:15,129 0.0 39 00:05:15,129 --> 00:05:17,970 150 amperes 40 00:05:17,970 --> 00:05:19,870 the rest of 41 00:05:19,870 --> 00:05:21,850 the intensity is flowing 42 00:05:21,850 --> 00:05:22,970 through this brand 43 00:05:22,970 --> 00:05:25,649 imagine that I have here 44 00:05:25,649 --> 00:05:27,730 1 ampere and here 45 00:05:27,730 --> 00:05:29,689 I have the third part 46 00:05:29,689 --> 00:05:32,050 the rest of it should be flowing through this 47 00:05:32,050 --> 00:05:33,730 so with this 48 00:05:33,730 --> 00:05:35,829 this is what I said 49 00:05:35,829 --> 00:05:37,350 the intensity 50 00:05:37,350 --> 00:05:39,589 the total intensity through the c-kit 51 00:05:39,589 --> 00:05:41,430 should be equal to the 52 00:05:41,430 --> 00:05:43,769 addition of the two intensities that are 53 00:05:43,769 --> 00:05:44,149 flowing 54 00:05:44,149 --> 00:05:47,470 across the knot 55 00:05:47,470 --> 00:05:49,829 so finally I have this 56 00:05:49,829 --> 00:05:51,649 intensity that is flowing 57 00:05:51,649 --> 00:05:53,310 through this branch of the c-kit 58 00:05:53,310 --> 00:05:55,490 and here you have the totals 59 00:05:55,490 --> 00:05:57,529 and