Wye-Delta Three Phase Circuit

Discussion in 'Homework Help' started by bobby19, Aug 14, 2007.

  1. bobby19

    Thread Starter Member

    Jun 3, 2007
    13
    0
    Hi,

    In the attached file, the question asks to find the three line currents. I understand the solution using Mesh Analysis that was used. However, I am unsure as to why when I convert the Delta Load to a Wye load, I get the wrong answer.

    Using Zwye = (1/3)Zdelta, adding the series line impedance with the load impedance, and then finding the line current using a single phase equivalent. Am I doing something wrong?
     
  2. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    well that wud depend on your analysis,can u post the one for y eq.?
    what do u exactly mean by single phase eq?- does it mean taking Vphase and Zphase?
     
  3. bobby19

    Thread Starter Member

    Jun 3, 2007
    13
    0
    Essentially the Balanced Delta load becomes a Balance Wye load of 4 + j0.666. Then the line impedances will be in series with each individual load. So each impedance should be (4 + j0.666) + (1 + j2). Finally, the line currents will be the phase voltages (100 angle0, 100 angle-120, 100 angle120) divided by the series line and load impedance, 5 + j2.666.

    I cant seem to figure this out. :confused:
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    ok right now, cant try to solve it analytically,
    see if u can get any leads from here (assuming u got the eq resistances from the transformation figured out correctly).
    simply divide the phase voltages by impedances phase impedance values (since its balanced) this will get u phase current which is same as line current.
    does this answer not match the one by above method?
    if not post back with a little info abt your attempt , i will see if i can help (i m stuck here as i gotta submit a project report this week + some designing in m/c + something else)
     
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