Why is inductive reactive power always represented at 90 degrees?

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Discussion Overview

The discussion centers around the representation of inductive reactive power in electrical circuits, specifically why it is depicted at 90 degrees in phasor diagrams. Participants explore the implications of this representation in the context of complex power calculations and the nature of reactive power in circuits with inductors and capacitors.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant references a professor's statement that inductive power is purely reactive and represented at 90 degrees, questioning the reasoning behind this representation.
  • Another participant points out the phase shift caused by resistors in the circuit and asks how the behavior changes when the source is applied directly to a coil or capacitor.
  • A participant discusses the formula for complex power, indicating that for purely reactive loads, the real power component is zero, leading to purely imaginary power, which is represented as reactive.
  • One participant clarifies that the term "dissipation" is typically associated with power lost as heat, suggesting that reactive power should not be described in this way.
  • Another participant expresses understanding that the overall power can be represented in complex format, identifying it as apparent power, which includes both reactive and real power components.
  • A participant revisits their calculations for reactive power, questioning the dropping of the angle in their results and receiving clarification that the magnitude squared of the current should be used, leading to a conclusion that the argument of the reactive power is 90 degrees for inductors.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the representation of inductive reactive power and its calculations. There is no consensus on the terminology used (e.g., "dissipation") and the implications of phase shifts in circuits, indicating ongoing debate and exploration of the topic.

Contextual Notes

Some participants highlight the importance of using the magnitude of current squared in calculations, which may lead to confusion regarding the angle in reactive power calculations. There are also discussions about the definitions and terminology used in describing power types, which may affect clarity.

Number2Pencil
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My professor made this statement: "the power dissipated by an inductor will always be purely reactive and represented at 90 degrees" (we label real power as a phasor at 0 degrees, inductive reactive power at 90 degrees and capacitive at -90)

Why is this?

I made a little example with a series circuit with a 20V<0 source, Rt = 310 ohms, ZL = j300 ohms, and ZC = -j50 ohms. I calculated the source current as 50.22mA<38.88 degrees. Now QL (reactive power) = I^2*ZL which gives 756.6mW<12.23. Now how come I can drop the angle off the answer? Is it because it will be purely reactive no matter what angle it's located, so you can just represent it at 90 degrees? I'm so confused.
 
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You have a phase shift due the resistor. If you apply the source straight to the coil or a capacitor what will you get?
 
I'm used to seeing complex power as

P_{complex} = VI^{\ast}

Where V and I are phasors, and the asterisk is the complex conjugate. For your load with impedance Z, this becomes

P_{complex} = |I|^2 Z^{\ast} = |I|^2 ( R - jX)

For a purely reactive load, R = 0, so the power is purely imaginary and thus purely reactive.
 
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By the way, "dissipation" is usually reserved for power lost as heat or friction, so reactive power wouldn't be called dissipation.
 
I think I understand now. If I take the overall power and put it in the complex format, this phasor is really "S" or apparent power, which is just a combination of the reative and real power.
 
Number2Pencil said:
I made a little example with a series circuit with a 20V<0 source, Rt = 310 ohms, ZL = j300 ohms, and ZC = -j50 ohms. I calculated the source current as 50.22mA<38.88 degrees. Now QL (reactive power) = I^2*ZL which gives 756.6mW<12.23. Now how come I can drop the angle off the answer? Is it because it will be purely reactive no matter what angle it's located, so you can just represent it at 90 degrees? I'm so confused.

How did you get QL = 756.6mW<12.23? In QL = I^2*ZL, note that I^2 is the square of the magnitude of I, or more aptly put as |I|^2, so arg(|I|^2*ZL) = 90 for the inductor.
 
ok well, being |I|^2 instead of I^2 does solve my problem. I guess that's kind of what I was getting at with my whole "it's that much power no matter what angle it's at" random belch of an idea
 

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