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3ºTEDI BLOCK 5 PARALLEL CIRCUITS EXERCISE 1 - Contenido educativo

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Subido el 28 de abril de 2024 por Beatriz T.

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This is block 5. Three resistors in parallel of 3, 9 and 18 ohms on a cell of 3 volts. First, you have to calculate the equivalent resistance. Remember the equation, the formula that you have to use. 00:00:00
The inverse of the equivalent resistance is there equal to the addition of the three inverses of the addition of the three of them. 00:00:17
So take the calculator and make this operation 1 divided by 3 plus 1 divided by 9 plus 1 divided by 18. 00:00:28
And the result is 0.5. 00:00:35
This is not the result. 00:00:37
This is nothing. 00:00:38
The equivalent resistance is the inverse of this result. 00:00:40
So take your calculator and divide. 1 divided by 0.5. The result is 2. So you can check that the equivalent resistance of three resistors in parallel, the total equivalent resistance is lower than the lowest of them. 00:00:45
So, 2, that's it, is lower than 3. It's okay. 00:01:05
Intensity of the circuit is always voltage of the cell divided by the equivalent resistance, 00:01:11
so 3 divided by 2, 1.5 amperes. 00:01:15
They are asking you to calculate intensity in the three of the resistors, 00:01:20
but first you need to know what is the voltage. 00:01:24
The voltage in a circuit in parallel is always the same 00:01:27
because the three resistors are connected to the same terminals. 00:01:30
and these are the terminals of the cell so voltage in the three of the resistors is always the same 00:01:33
the voltage of the cell three volts because it's in parallel then now you can calculate the 00:01:41
intensity in the three of the of the resistors remember that the intensity of the c-kit is 00:01:47
divided in each of the branches of the c-kit so it's going to be divided in three into three 00:01:54
In which amount? Well, we apply the Ohm's law. Remember that the Ohm's law, voltage, intensity, resistance, the intensity, it's voltage divided by resistance. 00:02:00
So, let's apply this formula. 00:02:16
Intensity in the resistor number 1, that is, well, sorry, this is one of 3 Ohms, is 3 volts divided by 3 Ohms. 3 divided by 3, 1 ampere. 00:02:20
the intensity in the number two of nine ohms is three volts divided by nine ohms so the result is 00:02:30
0.33 amperes you can reduce only to two figures decimal figures and intensity in three voltage 00:02:37
of three is in the resistance three remember it's always the same three volts divided by 18 ohms 00:02:45
and the result is 0.16 amperes and this is uh you have a infinite figure so you can put 0.17 00:02:53
it's always safe and the last question you have to calculate the power generated and the power 00:03:04
dissipated in each of the resistors so the body where regenerated is the voltage of the cell 00:03:09
by the total intensity 3 volts by 1.5 amperes you have 4.5 5 watts this is the power regenerated by 00:03:15
the cell and in each of the resistors to dissipate a certain amount of of power you have to calculate 00:03:29
with the this product voltage by intensity in each of the resistors check 00:03:39
that the addition of the power dissipated 3 plus 1 plus 0.5 it's equal 00:03:45
to the power generated and one other thing that I forgot is to say that the 00:03:52
intensity in the in a resistor in a signaling part in parallel these 00:03:57
intensities are directly inversely proportional to the resistors so the 00:04:04
bigger is the resistance the smaller is the intensity you can check this out 00:04:09
Autor/es:
beatriz torrejon
Subido por:
Beatriz T.
Licencia:
Reconocimiento - No comercial - Sin obra derivada
Visualizaciones:
68
Fecha:
28 de abril de 2024 - 13:36
Visibilidad:
Público
Centro:
IES CERVANTES
Duración:
04′ 18″
Relación de aspecto:
1.78:1
Resolución:
1280x720 píxeles
Tamaño:
284.79 MBytes

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