Question about reactive power

  • #1
Number2Pencil
208
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.
 

Answers and Replies

  • #2
waht
1,517
4
You have a phase shift due the resistor. If you apply the source straight to the coil or a capacitor what will you get?
 
  • #3
mdelisio
29
0
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:
  • #4
marcusl
Science Advisor
Gold Member
2,797
456
By the way, "dissipation" is usually reserved for power lost as heat or friction, so reactive power wouldn't be called dissipation.
 
  • #5
Number2Pencil
208
1
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.
 
  • #6
doodle
161
0
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.
 
  • #7
Number2Pencil
208
1
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
 

Suggested for: Question about reactive power

  • Last Post
Replies
19
Views
461
  • Last Post
Replies
5
Views
454
Replies
4
Views
444
Replies
7
Views
411
  • Last Post
Replies
4
Views
2K
Replies
6
Views
429
Replies
25
Views
1K
Replies
11
Views
349
  • Last Post
Replies
10
Views
1K
Top