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BLOCK 6 EXERCISE 2 SOLVED - Contenido educativo

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Subido el 28 de abril de 2024 por Beatriz T.

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This is the exercise 2 in block 6. 00:00:01
They are asking you to calculate the intensities through all the resistors. 00:00:05
The first thing we have to do is to calculate the equivalent resistance. 00:00:11
To calculate the equivalent resistance, first I have to solve this part. 00:00:16
I have given names to all resistors. 00:00:21
So this is 1, 2, 3, 4, 5, 6. 00:00:25
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. 00:00:28
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. 00:00:44
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. 00:01:02
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. 00:01:24
So I have three resistors in series and it's really easy to calculate the equivalent resistance that is the addition of these three. 00:01:39
So finally, the equivalent resistance of the total C-kit is 270 ohms. 00:01:47
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. 00:01:54
The result is this. This is the intensity that is flowing from the cell. 00:02:14
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. 00:02:25
But when I get to this knot A, this intensity is divided in these two branches. 00:02:41
I have to calculate the voltage between A and B, because if not, I cannot calculate the intensity through the resistor number 3. 00:02:48
How do I calculate this? Really easy. 00:02:56
I know the resistance and the intensity here, and I know the intensity here. 00:02:59
so what I'm going to do is to calculate voltage here and voltage here and I know 00:03:04
the voltage that is given by the cell is 10 so the voltage that is used by this 00:03:09
resistor and this one and the one between A and B the addition of these 00:03:15
three should be 10 because the voltage that is getting from the cell is 00:03:20
consumed by the resistor so let's do it voltage in the resistor number one is 100 by this in 00:03:26
intensity you have 3.7 volts and voltage in the number five it's 5.5 volts so the addition of 00:03:37
these two it's 9.25 and the amount that rest to 10 is the voltage between a and b here i have put 00:03:46
this diagram you for you to understand um i have here the 10 volts that are given by the cell and 00:03:58
in the first resistor i'm consuming a 3.7 and in this resistor number five and the consumption is 00:04:06
5.5 so the rest between a and b as the 3 of n has the addition of these three should be 10 00:04:15
is the difference between these two amounts and 10 and it's as i said 0.275 volts so now 00:04:24
i know the voltage between these two points i know the voltage i know the resistance i can 00:04:37
calculate the intensity through the resistor number 3, because I applied the Ohm's law. 00:04:43
The intensity is equal to the voltage divided by the resistance. So, this intensity number 3 is 0.75 00:04:48
between 60. This is the intensity through the resistor number 3. And finally, if I have here 00:04:57
0.037 00:05:05
amperes 00:05:09
and here I have 00:05:11
0.0 00:05:13
150 amperes 00:05:15
the rest of 00:05:17
the intensity is flowing 00:05:19
through this brand 00:05:21
imagine that I have here 00:05:22
1 ampere and here 00:05:25
I have the third part 00:05:27
the rest of it should be flowing through this 00:05:29
so with this 00:05:32
this is what I said 00:05:33
the intensity 00:05:35
the total intensity through the c-kit 00:05:37
should be equal to the 00:05:39
addition of the two intensities that are 00:05:41
flowing 00:05:43
across the knot 00:05:44
so finally I have this 00:05:47
intensity that is flowing 00:05:49
through this branch of the c-kit 00:05:51
and here you have the totals 00:05:53
and 00:05:55
Autor/es:
BEATRIZ TORREJON
Subido por:
Beatriz T.
Licencia:
Reconocimiento - No comercial - Sin obra derivada
Visualizaciones:
93
Fecha:
28 de abril de 2024 - 22:03
Visibilidad:
Público
Centro:
IES CERVANTES
Duración:
05′ 58″
Relación de aspecto:
1.78:1
Resolución:
1280x720 píxeles
Tamaño:
393.92 MBytes

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