Why is inductive reactive power always represented at 90 degrees?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
6 replies · 4K views
Number2Pencil
Messages
204
Reaction score
1
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.
 
Engineering news on Phys.org
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

[tex]P_{complex} = VI^{\ast}[/tex]

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

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

For a purely reactive load, R = 0, so the power is purely imaginary and thus purely reactive.
 
Last edited:
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