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Wye-Delta Three Phase Circuit

  1. Aug 18, 2007 #1

    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.

    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.

    Am I doing something wrong?

    (I posted this in Homework Help 3 days ago, but recieved no response:frown:)

    Attached Files:

  2. jcsd
  3. Aug 21, 2007 #2


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    Are you sure the answers from the mesh analysis are correct? The line currents, IaA, IbB, IcC, should all have the same magniture and be 120 apart. Your Zwye has problems also.
  4. Aug 22, 2007 #3
    Looking at your figure you will notice you have at least one voltage supply labeled incorrectly. You show both B&C phases as 100volts at an angle of 120 degrees. Although it appears that you corrected for it in your mesh 2 equations, if mesh two is correct, then the same voltage in mesh one is shown incorrectly. I get a different set of equations for mesh one than you have. Take yor time and look at where the current for each loop flows. Ask yourself, is mesh one current the only current that flows through this impedance? Another item is I do not see the impedance (12+j12) as having any affect on your current loops. Do you think that is correct?

    Hope this helps.
  5. Mar 15, 2009 #4
    Can anybody please help me to find a book where I can find more examples of this kind of problem.
  6. Mar 15, 2009 #5


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    Last edited by a moderator: May 4, 2017
  7. Aug 7, 2010 #6
    oh u can't solve for the line current with the phase voltage divided by the total line impedance.
    that's perfectly wrong. u can only do that if that is a four-wire circuit but since it's a three wire circuit, that's wrong...there are ways to solve that
    1) calculate for the total impedance in delta connection, then convert the supply source into delta connection, dividing the voltage over the impedance gives u the total line current.
    2) use mesh equations just like in the solution.
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