Three Phase Y-Connection Line Voltage ?

Discussion in 'Homework Help' started by Xufyan, May 17, 2011.

  1. Xufyan

    Thread Starter Member

    Aug 3, 2010
    114
    0
    Hello,

    for 3-Phase Y-Connection the Line Voltage is,

    VL=\sqrt{3}xVp

    This can be proved using this equation,

    Vab = Van + Vnb , looking at this circuit, how this equation is derived ?

    [​IMG]

    please help
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    V_{ab} = V_{a} - V_{b}

    \text{Because, } V_{a} = V_{m} \angle 0^{o} \text{ and } V_{b} = V_{m} \angle -120^{o}, \text{ we have,}

    V_{ab} = V_{m} - V_{m}(-1.5 - j0.866)

     = V_{m}(1.5 - j0.866)

    V_{ab} = \sqrt{3}V_{m} \angle 30^{o}

    Similarly,

    V_{bc} = \sqrt{3}V_{m} \angle -90^{o}

    V_{ca} = \sqrt{3}V_{m} \angle -210^{o}

    Thus the line-to-line voltage is √(3) times the phase voltage and is displace 30° in phase.

    EDIT: After reading your post, I may have misinterpreted your question.

    If you are asking where the equation,

    Vab = Va-Vb comes from,

    see the figure attached.
     
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    Last edited: May 17, 2011
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  3. Xufyan

    Thread Starter Member

    Aug 3, 2010
    114
    0
    the attached figure solved my problem :)

    thanks alot
     
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