- #1
Understanding a Concept: Seeking Deeper Answers
- Thread starter Special One
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Discussion Overview
The discussion revolves around understanding complex impedance in AC circuit analysis, particularly the role of the imaginary unit 'j' in representing reactance. Participants seek deeper explanations and clarifications on the topic, which includes theoretical concepts and practical applications.
Discussion Character
- Exploratory
- Technical explanation
- Conceptual clarification
- Debate/contested
- Homework-related
Main Points Raised
- Some participants express difficulty in understanding how 'j' appears in calculations related to reactance and impedance.
- One participant explains that 'j' indicates that the current is 90 degrees out of phase with the voltage in reactive components.
- Another participant suggests that the notation for reactance and impedance can vary, which may lead to confusion.
- A detailed explanation is provided regarding the behavior of voltage and current in resistive, inductive, and capacitive circuits, including the concept of phasors and reactive power.
- Some participants recommend video resources for a foundational understanding of AC circuit analysis before tackling complex impedance.
- There is mention of the term "complex number" as a foundational concept that may aid in understanding phasors and impedance.
Areas of Agreement / Disagreement
Participants generally agree on the complexity of the topic and the need for deeper understanding, but there remains uncertainty about the notation and the conceptual underpinnings of 'j' in the context of reactance and impedance. Multiple competing views on how to approach learning this material are present.
Contextual Notes
Some limitations include the varying definitions and notations used in different contexts, which may lead to confusion. The discussion also highlights the need for foundational knowledge in complex numbers to facilitate understanding of the concepts being discussed.
Who May Find This Useful
This discussion may be useful for students and practitioners in electrical engineering or physics who are grappling with the concepts of AC circuit analysis, complex impedance, and phasor notation.
- #2
tech99
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The LC network is in series with R, so work out its total reactance first.
As with resistors in parallel, reactances use the same formula, but with either + or - signs.
So 1/X = 1/XL - 1/Xc
1/X = 1/10 - 1/60
X = 10 x 60 / (10 -60) = 600/-50 = -12
In other words, -j12.
Now add this to R,
Z = R + X
Z = 50 - j12
As with resistors in parallel, reactances use the same formula, but with either + or - signs.
So 1/X = 1/XL - 1/Xc
1/X = 1/10 - 1/60
X = 10 x 60 / (10 -60) = 600/-50 = -12
In other words, -j12.
Now add this to R,
Z = R + X
Z = 50 - j12
- #3
Special One
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But I still didn’t get how j appeared!?
- #4
tech99
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If you have a reactance, rather than a resistance, you put j infront. It means that the current in the component is 90 degrees to the voltage.
- #5
Special One
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Alright, thanks a lot bro
- #6
berkeman
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I sympathize. ##L, X=\omega L, j\omega L, jX, \vec Z## are all notations for the same thing. Why so many? Because practitioners find it convenient. You'll just have to learn them all and live with it.Special One said:
- #8
zoki85
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There are lot of threads on PF regarding application of phasors. Just scroll down to the bottom of this page to see some them.Special One said:
- #9
DaveE
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I would look at this video series if I were you. Start at the beginning of AC circuit analysis. If you jump right to the complex impedance lecture I think you'll stay confused.
https://www.khanacademy.org/science...-topic/ee-ac-analysis/v/ee-ac-analysis-intro1
https://www.khanacademy.org/science...-topic/ee-ac-analysis/v/ee-ac-analysis-intro1
- #10
Special One
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DaveE said:
Thx mate,
- #11
Babadag
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One could measure voltage and current with an a.c.ammetter. Then you'll get an average voltage named "rms" [root mean square].
Actually the alternative instant current [a.c.] value changes all the time very fast in a cyclic way.
From maxim to maxim [peak to peak] values the time is very short T=1/60=0.0167 sec[16.7 millisecond] or 1/50=0.02 sec.
An oscilloscope can measure that[See the link] and we can represent the measurement results on a paper so we get a sinusoidal wave.
https://en.wikipedia.org/wiki/Oscilloscope
If the voltage is applied on a [pure] resistance a current will pass through the resistance and the current wave will be the same as the voltage wave only at a different scale.
Not the same phenomenon occurs if instead of a resistance we have a coil that means an inductive reactance: the current wave will lag the voltage wave with 0.0041 sec[90o or π/2 radians].
If instead of resistance we have a capacitor the current will lead the voltage by 90o.
One may represent this wave on a circle and since α=ω.t changes with t always we may think the circle rotates with ω radians/sec. Let's sit on the circle plane [in the same way we stand on the rotated Earth and we think the Earth is fixed and does not rotate].Then v(t)=√2.Vrms. Let's represent -for our convenience- on a scale 1/√2.Then v(t)=Vrms=V.
The line representing the voltage or current in the above circle we call it phasor [and some time vector].
We may use for V the angle 0 since V.cos(o)=V.
If i(t) angle will be φ<>o then the actual part of the current will be I*cos(φ) and in case of pure resistance
when φ=0 i(t)=I.
I.sin(φ) it does not represent a real part of the current but a "parasite" one.
If we use the factor j=√-1 -which does not exist actually-we may assemble an imaginary expression j.sin(φ)
So we can consider y ordinate as imaginary one while abscissa x as the real [as voltage ].
i(φ)=I.cos(φ)+j.I.sin(φ)
If φ=0 i(φ)=I
If φ=-90o i(φ)=-j.I.sin(φ)
If φ=+90o i(φ)=+j.I.sin(φ)
If we multiply I^2 by R we get an actual power which we can measured with a wattmeter.
If we divide V by XL=2.π.f.L where f=supply system frequency and L the circuit inductance, or XC=1/(2.π.f.C) we get the current [IL or IC].But IL^2.XL [or IC^2.XC] it cannot be measured on a wattmeter. It is not a real power but a "reactive power".
If we combine a resistance and an inductance or capacitor we get a mixed current where tangent(φ)=XL/R [or XC/R].
Actually the alternative instant current [a.c.] value changes all the time very fast in a cyclic way.
From maxim to maxim [peak to peak] values the time is very short T=1/60=0.0167 sec[16.7 millisecond] or 1/50=0.02 sec.
An oscilloscope can measure that[See the link] and we can represent the measurement results on a paper so we get a sinusoidal wave.
https://en.wikipedia.org/wiki/Oscilloscope
If the voltage is applied on a [pure] resistance a current will pass through the resistance and the current wave will be the same as the voltage wave only at a different scale.
Not the same phenomenon occurs if instead of a resistance we have a coil that means an inductive reactance: the current wave will lag the voltage wave with 0.0041 sec[90o or π/2 radians].
If instead of resistance we have a capacitor the current will lead the voltage by 90o.
One may represent this wave on a circle and since α=ω.t changes with t always we may think the circle rotates with ω radians/sec. Let's sit on the circle plane [in the same way we stand on the rotated Earth and we think the Earth is fixed and does not rotate].Then v(t)=√2.Vrms. Let's represent -for our convenience- on a scale 1/√2.Then v(t)=Vrms=V.
The line representing the voltage or current in the above circle we call it phasor [and some time vector].
We may use for V the angle 0 since V.cos(o)=V.
If i(t) angle will be φ<>o then the actual part of the current will be I*cos(φ) and in case of pure resistance
when φ=0 i(t)=I.
I.sin(φ) it does not represent a real part of the current but a "parasite" one.
If we use the factor j=√-1 -which does not exist actually-we may assemble an imaginary expression j.sin(φ)
So we can consider y ordinate as imaginary one while abscissa x as the real [as voltage ].
i(φ)=I.cos(φ)+j.I.sin(φ)
If φ=0 i(φ)=I
If φ=-90o i(φ)=-j.I.sin(φ)
If φ=+90o i(φ)=+j.I.sin(φ)
If we multiply I^2 by R we get an actual power which we can measured with a wattmeter.
If we divide V by XL=2.π.f.L where f=supply system frequency and L the circuit inductance, or XC=1/(2.π.f.C) we get the current [IL or IC].But IL^2.XL [or IC^2.XC] it cannot be measured on a wattmeter. It is not a real power but a "reactive power".
If we combine a resistance and an inductance or capacitor we get a mixed current where tangent(φ)=XL/R [or XC/R].
- #12
hutchphd
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Just for completeness someone should mention the term "complex number". That is what this phasor stuff really is and I think it may just be easier to learn rudimentary complex numbers and then this stuff is easy. I was taught using phasors from Sears and Zemansky and to this day do not know why. This is a general comment as well as for the OP because he/she seems at peace with it