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!

Homework Help: 3 Phase Systems Theory and Procedure -- Determining Average Power

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the 3 following Scenarios:

    rms value of line voltage is 208 V

    Three equal impedances, 60 +j30 W each, are delta-connected to a 230-V rms, three-phase circuit. Another three equal impedances, 40 +j10 W are wye-connected across the same circuit at the same points.

    2. Relevant equations
    S = 3|Ia|^2(Zy)
    P = Real Component (S)

    3. The attempt at a solution
    How come when analyzing circuit one the procedure for determining the average power is to
    1. find the line current Ia which is simply 110<0 divided by the impedance (Z(delta)/3 + 2)
    2. calculate S = 3|Ia|^2(Zy) where Zy =(Z(delta)/3 + 2)
    3. take the real component of S and that is your average power.

    But when analyzing circuit 2 the procedure is:
    1. Convert Van = 208 to Vp = 208/ root(3)
    2. compute S by using S = (3 * Vp^2)/ (Zpcomplex conjugate)
    3. take the real component of S and that is your average power.

    And when analyzing circuit 3 the procedure is:
    1. Find the line current
    2. compute S by using S = (3 * Vs * (complex conjugate of line current)
    3. take the real component of S and that is your average power.

    Essentially I am kind of confused about what this Vp, Zp are that is used in the second consideration. Also why for every one of these scenarios the procedure for getting average power is different and the previous methods do not work. All of these examples seem the exact same is what really confuses me. What makes these 3 analyses different? How do I know what procedure to follow? I am doing self study so any help that can be provided is greatly appreciated.

    Thank you in advanced!
    Last edited: Dec 7, 2016
  2. jcsd
  3. Dec 7, 2016 #2


    User Avatar
    Homework Helper
    Gold Member

    In this method, they have first converted the delta impedance into its star equivalent impedance and then Zstar=Zdelta/3 is used. Then of course the 2Ω line resistances come in series with the load impedances in the respective phases, so you see the term Z/3+2 ohm in the equation.
    Van=Vphase and in star, Vph=Vline/√3. Are you sure the problem says Van=208 V?
    You can use the procedure in circuit 1 for this circuit too. This is already in star configuration, so you need not divide the load impedance by 3 as you did for circuit 1. But here, you do not have any line impedance and hence, you do not have a "voltage-divider" in the line. So, instead of calculating the line current like in circuit 1, you can directly use the given voltage to find the power.
    This is the "general" procedure and can be applied to any balanced three phase network.
    Last edited: Dec 7, 2016
  4. Dec 8, 2016 #3
    Oh I see thank you very kindly for taking the time to explain this. Yes problem 2 does say that the line voltage is that value.
    Last edited: Dec 8, 2016
  5. Dec 8, 2016 #4


    User Avatar
    Homework Helper
    Gold Member

    That means Vab=208V, not Van as you have written.Van=208/root3 V.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted