ejemplo superposición c a

roxana garcia
13 Dec 202010:12

Summary

TLDRThe video script discusses the method of superposition in electrical circuits, focusing on calculating the voltage across a specific point labeled 'v2'. It explains the process of analyzing the circuit with a single source at a time, calculating the current through various impedances using Ohm's law and the current divider rule. The script involves converting impedances to polar form, applying the method to alternating current, and obtaining the resultant voltage by superposing the effects of different sources. The explanation includes mathematical calculations and the use of complex numbers to find the final voltage in rectangular form.

Takeaways

  • 🔌 The method of superposition is used to solve for voltages and currents in a circuit with multiple sources.
  • 🔍 The script focuses on finding the voltage V2 at specific points in a circuit with complex impedances.
  • 🔧 The first step involves analyzing the circuit with respect to the first voltage source, treating it as an open circuit for the other sources.
  • 📡 The script describes a scenario with a current source and impedances in parallel, requiring the use of current division theorems.
  • 🔢 The current I2 is calculated using the total current from the source and the impedances involved, applying Ohm's law for AC circuits.
  • 📚 Impedance values are converted to polar form for calculation, which includes magnitude and phase angle.
  • 🔗 The voltage V2 across an impedance is found by multiplying the current through that impedance by its impedance value.
  • 🔄 The process is repeated for a second source, considering the circuit with respect to this new source and calculating the resultant voltage.
  • 📉 The equivalent impedance of the circuit is recalculated when considering different sources, which affects the current and voltage calculations.
  • 📌 The final voltage V2 is obtained by superposing the voltages calculated with respect to each individual source.
  • 📝 The script emphasizes the importance of checking calculations, especially when converting between rectangular and polar forms of complex numbers.

Q & A

  • What is the method being discussed in the script?

    -The method being discussed is the method of superposition.

  • What is the first step in applying the method of superposition according to the script?

    -The first step is to work with one source at a time, starting with the first source.

  • What is the main objective of the exercise in the script?

    -The main objective is to find the voltage V2 at certain points in the circuit.

  • What is the significance of the term 'circuito abierto' in the script?

    -'Circuito abierto' refers to an open circuit, which is a condition where the circuit is broken and no current flows.

  • What does the script mention about the impedance in the circuit?

    -The script discusses the impedance in the circuit, including the calculation of impedance in parallel and the use of impedance in the method of superposition.

  • How is the current I2 calculated in the script?

    -I2 is calculated using the current divider rule, which involves dividing the total current by the sum of the impedances.

  • What is the significance of converting impedance to polar form in the script?

    -Converting impedance to polar form helps in performing complex arithmetic and finding the magnitude and phase angle of the impedance.

  • What is the role of the current source in the second part of the exercise?

    -In the second part of the exercise, the current source is used to find the voltage V2 across the impedance when the direction of the current source is downward.

  • How is the voltage V2 related to the current I2 in the script?

    -V2 is calculated as the product of I2 and the impedance Zc, according to Ohm's law.

  • What is the final step described in the script for finding the voltage V2?

    -The final step is to sum the voltages obtained from both sources, considering their rectangular forms, to get the total voltage V2.

Outlines

00:00

🔌 Introduction to Superposition Method

The script begins with an introduction to the superposition method, a technique used to solve electrical circuits. The speaker explains that the method involves analyzing the circuit with one source at a time. The focus is on finding the voltage 'v2' across certain points in the circuit. The speaker then describes the setup of the circuit with a current source and impedances in parallel, outlining the steps to calculate the current 'i2' using the current divider rule. The current is then used to find the voltage 'v2' across the impedance 'c' by multiplying the current by the impedance in polar form, resulting in a complex number representing the voltage magnitude and phase angle.

05:00

🔍 Calculating Voltage 'v2' with Different Sources

In this paragraph, the speaker continues the exercise by considering a different source configuration. The circuit now includes a current source with a negative imaginary component, indicating a phase shift. The speaker calculates the equivalent impedance for the circuit and uses it to find the voltage 'v2' across the same points as before. The calculation involves converting the impedance and current source into polar form, multiplying them to get the voltage in polar form, and then converting it back to rectangular form. The speaker emphasizes the importance of checking the calculations and provides the resulting voltage 'v2' in both polar and rectangular forms.

10:01

📚 Conclusion of the Electrical Circuit Exercise

The final paragraph wraps up the exercise with a summary of the results obtained. The speaker combines the voltage values calculated for 'v2' from the different source configurations, emphasizing the process of adding complex numbers in rectangular form. The conclusion highlights the final result for 'v2' and expresses hope that the exercise was of interest to the audience. The speaker thanks the viewers for their attention and concludes the video script.

Mindmap

Keywords

💡Superposition Method

The superposition method is a fundamental principle in electrical engineering used to analyze linear circuits. It involves breaking down a complex circuit into simpler ones by considering each independent source separately and then combining the results. In the video's context, the superposition method is used to find the voltage across a specific point, with the speaker initially focusing on one source at a time and then moving on to the next.

💡Voltage (V2)

Voltage, denoted as V2 in the script, is the electric potential difference between two points in a circuit. It is a key parameter in understanding the behavior of the circuit. In the video, the speaker is trying to calculate the voltage across a certain point, which is a central goal of the exercise and is used to demonstrate the application of the superposition method.

💡Impedance

Impedance is a measure of the opposition that a circuit presents to the flow of alternating current (AC). It is a complex quantity that may include resistance, inductive reactance, and capacitive reactance. In the script, the speaker discusses calculating the impedance in parallel and series configurations, which is crucial for determining the current and voltage in AC circuits.

💡Current Divider

The current divider is a principle used in circuit analysis to determine the distribution of current through parallel paths. It states that the current through a branch is inversely proportional to its impedance. In the video, the speaker uses the current divider rule to find the current flowing through specific impedances when the circuit is open.

💡Polar Form

Polar form is a way of representing complex numbers, which include both magnitude and phase angle. In the context of the video, the speaker converts impedance values to polar form to perform calculations more easily, especially when dealing with AC circuits where phase angles are significant.

💡Phase Angle

The phase angle is the angle by which a sinusoidal waveform is shifted in time, and it is a critical aspect of AC circuits. In the script, the speaker calculates and uses phase angles to determine the behavior of the circuit under analysis, particularly when converting impedances to polar form.

💡Rectangular Form

Rectangular form, also known as Cartesian coordinates, is another way to represent complex numbers, using real and imaginary parts. The speaker in the video switches between polar and rectangular forms to perform different calculations related to the circuit's voltage and impedance.

💡Source of Voltage

A source of voltage is an element in a circuit that provides energy, typically in the form of an electromotive force (EMF). In the script, the speaker refers to different sources of voltage, which are essential for applying the superposition method and for understanding how the circuit will respond to different conditions.

💡Circuit Open

A circuit being 'open' refers to a break in the circuit that prevents current from flowing. In the context of the video, the speaker mentions having a circuit open to simplify the analysis by focusing on one source at a time, which is a standard approach in the superposition method.

💡AC (Alternating Current)

Alternating current is a type of electrical current that periodically reverses direction. The speaker in the video is analyzing an AC circuit, which is why concepts like impedance, phase angle, and the use of both polar and rectangular forms are essential for understanding the circuit's behavior.

💡Equivalent Impedance

Equivalent impedance is the total opposition to the flow of current in a circuit when components are combined. The speaker calculates the equivalent impedance by considering the series and parallel connections of various impedances, which is necessary for determining the overall behavior of the circuit under the superposition method.

Highlights

Introduction to the method of superposition for solving electrical circuits.

Explanation of working with a single voltage source first in the superposition method.

Description of the circuit setup with an open circuit for the first source.

Identification of the voltage V2 and its location in the circuit.

Discussion on the use of impedance in parallel in the circuit analysis.

Application of current divider theorem for AC circuits to find current I2.

Calculation of I2 using the total current and impedance values.

Conversion of impedance values to polar form for complex calculations.

Determination of the equivalent impedance for current I2.

Calculation of the voltage V2 across impedance C using current I2.

Introduction of the second part of the exercise with a different voltage source.

Setup of the circuit for the second source with an open circuit condition.

Explanation of the current flow through impedances in the second scenario.

Calculation of the equivalent impedance for the second voltage source.

Determination of the voltage V2 with respect to the second source.

Combination of results from both sources to find the final voltage V2.

Conversion of the final result into rectangular form for clarity.

Conclusion of the exercise with a summary of the findings.

Transcripts

play00:00

hola buenas tardes nuevamente vamos a

play00:02

hacer un ejercicio

play00:05

por el método de superposición

play00:07

recordemos que el método de

play00:09

superposición obtenemos bueno trabajamos

play00:13

con una fuente primeramente y luego

play00:15

trabajamos con la siguiente entonces

play00:17

primeramente bueno nos piden encontrar

play00:19

el voltaje v2 el voltaje v2 se encuentra

play00:22

en estos puntos y bueno para esto vamos

play00:26

a trabajar

play00:28

con respecto a la fuente de un camper

play00:33

entonces mi circuito va a quedar de la

play00:36

siguiente manera como esta es una nueva

play00:39

fuente de corriente entonces me dice que

play00:42

mi circuito de este lado va a quedar con

play00:44

circuito abierto entonces tengo mi

play00:47

fuente

play00:48

que es esta y luego tengo una impedancia

play00:51

que se encuentra en paralelo que sería

play00:54

impedancia

play00:58

ok y tenemos luego la impedancia d

play01:03

y tenemos la impedancia sea

play01:07

de esta manera inferencias impedancia de

play01:10

y aquí tenemos mi voltaje de 2 en estos

play01:15

puntos ok ahora

play01:19

esto es con respecto a la primera fuente

play01:22

obviamente la siguiente yo tendría aquí

play01:25

un circuito abierto entonces no lo voy a

play01:28

indicar por cuestiones esa idea de

play01:31

espacio en la hoja ok como tengo yo

play01:33

primeramente la corriente y 2 que es la

play01:36

que va a pasar por aquí bueno la

play01:39

corriente y 2 la puedo obtener porque es

play01:42

la misma que va a pasar por la

play01:43

impedancia ve por la impedancia sé si yo

play01:46

aplico divisor de corriente voy a

play01:49

obtener que la la corriente de 2 va a

play01:52

ser igual a la corriente total que en

play01:54

este caso es la corriente que se

play01:57

encuentra presente al aplicar divisor de

play01:59

corriente multiplicado

play02:02

por la impedancia o resistencia

play02:07

en donde no la quiero que en este caso

play02:10

sería la impedancia sobre la sumatoria

play02:13

de las impedancia

play02:16

recordemos que los teoremas son los

play02:18

mismos solo que ahora las estamos

play02:21

aplicando para corriente alterna ok

play02:24

entonces si yo sustituyó a esto voy a

play02:26

obtener que y 2 va a ser igual a el

play02:30

valor de la corriente total que es 1 que

play02:33

es el de la fuente

play02:35

que multiplica a la impedancia que en

play02:39

este caso la impedancia a vale 4 - j

play02:44

2 si yo esto lo convierto a polar ahí en

play02:47

sus calculadoras ustedes lo pueden

play02:49

verificar me debe de dar un valor de

play02:52

4.47 menos 26 con un ángulo de menos 26

play02:56

punto 56 grados

play03:00

eso sería mi impedancia sobre la

play03:03

sumatoria de las impedancia yo ya aquí

play03:05

traigo el resultado que es 10 con un

play03:08

ángulo de menos 53 puntos 13 ok y esto

play03:12

nos va a dar un valor de 0.40 y 447 con

play03:17

un ángulo de

play03:20

26 punto 57 grados

play03:24

aquí tengo mi corriente número 2 que es

play03:26

la corriente que va a pasar a través de

play03:27

la impedancia de ok ahora

play03:31

yo ya puedo obtener el voltaje que va a

play03:34

pasar a través de v 2 v 2 va a ser igual

play03:38

a la corriente 2 por la impedancia c

play03:43

verdad entonces el voltaje 2 va a ser

play03:46

igual a la corriente 2 que desde cero

play03:49

punto 447 con un ángulo

play03:54

con un ángulo de 26 puntos 57

play04:00

que multiplica a la impedancia hace que

play04:03

recordemos que el valor de la impedancia

play04:05

c es 2 + j4 si yo lo convierto polar ese

play04:10

valor le debe de dar 4.47 con un ángulo

play04:14

de 63 punto 43

play04:18

v2 entonces va a ser igual a 1.99 con un

play04:23

ángulo de 90 grados que esto es igual a

play04:28

jota 2 y aquí yo obtengo mi primer valor

play04:32

de v2

play04:34

recordemos que sería con respecto a la

play04:39

fuente de voltaje ahora con respecto a

play04:43

la fuente de 0.5 con un ángulo de menos

play04:47

90 entonces ahora este sería mi inciso a

play04:51

mi inciso b sería ahora mi circuito

play04:54

quedaría de la siguiente manera es

play04:58

impedancia

play05:00

impedancia be impedancia c

play05:06

y la fuente con la que voy a trabajar

play05:09

la fuente de corriente con dirección

play05:12

hacia abajo y tiene un valor en este

play05:15

caso ya lo pasó a formar forma

play05:19

rectangular y me dice que es menos j

play05:27

0.5 empresas

play05:31

ojo con esto aquí - j 0.5 empersa ok y

play05:36

aquí tengo mi voltaje v2 esto sería

play05:40

mismo impedancia se mi impedancia b y mi

play05:43

impedancia ok qué más tengo que conocer

play05:47

bueno pues que aquí voy a tener más

play05:49

menos mi voltaje v 1

play05:54

y

play05:58

creo que es lo más relevante ok vamos a

play06:00

obtener ahora nuevamente mi voltaje v2 y

play06:05

voltaje v2 en este caso es el voltaje

play06:08

que se encuentra que se encuentra en

play06:10

esta parte de quién

play06:13

yo sé que mi impedancia es igual a 2

play06:16

mascota 4 entonces mi voltaje 2 va a ser

play06:23

igual

play06:28

de acuerdo a la ley de o menos 0.5 con

play06:34

un ángulo de menos 90 que multiplica a

play06:38

mi impedancia equivalente entonces mi

play06:41

voltaje 2

play06:43

va a ser igual a menos que es este menos

play06:46

que multiplica a 0.5 con un ángulo de

play06:50

menos 90

play06:52

por mi impedancia equivalente mi

play06:56

impedancia equivalente en este caso va a

play06:59

ser igual

play07:02

5.65

play07:05

con un ángulo de 45 grados

play07:12

y esto va a ser igual

play07:15

a menos 2.82 con un ángulo de 45 grados

play07:19

menos 45 grados

play07:23

y v2 va a ser igual en forma polar a

play07:26

menos dos más j

play07:28

dos está envuelto con bolsas verdad

play07:32

ok

play07:39

por favor revisen este dato de

play07:41

impedancia equivalente si yo veo mi

play07:44

circuito a través de esta fuente pues

play07:47

voy a sacar la impedancia equivalente

play07:49

que sería la sumatoria de las tres

play07:51

verdad

play08:01

bueno no sería la sumatoria de las tres

play08:03

vamos a hacerlo

play08:09

sería mi impedancia equivalente sería

play08:11

estas dos que están en serie y en

play08:14

paralelo con zeta

play08:17

sería zeta a zp que se encuentra en

play08:23

paralelo con zeta sí sí y este valor me

play08:28

tiene que dar un valor de por favor no

play08:31

revisan ustedes de 5.65 con un ángulo de

play08:37

45 grados

play08:39

y este no es perdón por no haberme ha

play08:42

brincado esto

play08:46

pero ahí lo tenemos ya entonces mi

play08:49

voltaje 2 con respecto a la fuente que

play08:55

se encuentra

play08:57

que es de 0.5 con un ángulo de 90 es

play09:01

igual a este valor ahora sumando los los

play09:05

dos valores

play09:07

yo voy a obtener tomando la respuesta de

play09:10

ambos valores voy a obtener mi voltaje

play09:15

ahora sí

play09:18

sumando las dos respuestas tenemos que

play09:20

son en forma rectangular menos dos más

play09:26

jota cinco y en forma rectangular es j 2

play09:30

verdad entonces es j dos menos dos más j

play09:37

2 así lo tenemos y esto va a ser igual a

play09:41

menos dos más j 4

play09:45

bonito

play09:48

aquí sería mi resultado ahora eso sería

play09:53

mi resultado en forma rectangular

play09:58

ok

play10:01

pues hasta aquí chicos espero que este

play10:04

ejercicio

play10:09

de su interés muchas gracias

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