1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Supplied complex power from source vs consumed by load

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-27_0-28-37.png
    upload_2016-10-27_0-28-49.png

    2. Relevant equations
    S= 3VaIa*

    3. The attempt at a solution
    After transformation:
    upload_2016-10-27_0-30-5.png
    Ia = 120<0 / (6+8j) = 12<-53.13 A
    Total complex power = 3 * Va * Ia* = 3*120<0 * 12<53.13 = 2592W + j3456 VAR
    This is the power supplied from source. What would be the power consumed by load?
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2016 #2

    cnh1995

    User Avatar
    Homework Helper

    You know the load impedance and load current. How will you determine the active and reactive power associated with the load?
     
  4. Oct 27, 2016 #3
    Load impedance = 5 + j6
    Load current = 12<-53.13A
    I*I*Z
    ?
     
  5. Oct 27, 2016 #4

    cnh1995

    User Avatar
    Homework Helper

    Yes. Take magnitudes only.
    S=I2Z.
     
    Last edited: Oct 27, 2016
  6. Oct 27, 2016 #5
    So 144 * (5+j6)
    ?

    power of (1+j2)= 144 * (1+2j)
    and
    (load) = 144*(5+j6)
    combined become 864W + j1152VAR

    but the total complex power is 2592W + j3456VAR

    Is it right for these values to not match?
     
    Last edited: Oct 27, 2016
  7. Oct 27, 2016 #6

    cnh1995

    User Avatar
    Homework Helper

    Yes.
    Yes. Can you say why?
     
  8. Oct 27, 2016 #7
    Is it because it's 1/3 of the "actual" circuit?
    Do I need to multiply by 3 to get the total complex power consumed by load?

    If no, does this look correct?
    720W + j864VAR

    Load power factor:
    S = sqrt(Q^2+P^2) = 1124VA
    PF = P/S = 720/1124 = 0.64 lagging (Q>0)
    correct?
     
  9. Oct 27, 2016 #8

    cnh1995

    User Avatar
    Homework Helper

    Well, you calculated the total power for three phases. So, you should multiply the load power by 3. But they are not equal because out of total input power, some power is associated with the line impedance 1+j2 ohm. Remaining power is fed to the load.
     
  10. Oct 27, 2016 #9
    If the prompt is:
    Calculate the total complex power consumed by the load and determine the load's power factor

    Do I use what I calculated (720W + j864VAR) or do I need to multiply that by 3?
     
  11. Oct 27, 2016 #10

    cnh1995

    User Avatar
    Homework Helper

    You should multiply by 3. You have transformed the circuit into its per phase equivalent. Total power will be thrice the per phase power.
    Here, while applying S=I2Z, you should consider the magnitude only. If you want to use the phasor form, you should take the conjugate of the current i.e. (IZ)(I*).
     
  12. Oct 27, 2016 #11
    If the prompt is:
    Calculate the total complex power consumed by the load and determine the load's power factor

    Do I use what I calculated or do I need to multiply that by 3?
    So magnitude of (5+j6) = sqrt(61)?
     
  13. Oct 27, 2016 #12

    cnh1995

    User Avatar
    Homework Helper

    For load power factor, you can use the single phase equivalent network since it is a balanced network.
    Yes.
     
  14. Oct 27, 2016 #13
    so load would be 3 * 144*sqrt(61)?
    How do i get power factor from this?
     
    Last edited: Oct 27, 2016
  15. Oct 27, 2016 #14

    cnh1995

    User Avatar
    Homework Helper

    If the load impedance is R+jX, what would its power factor?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Supplied complex power from source vs consumed by load
Loading...