3 Phase Series Equivalents (Wye & Delta)

Discussion in 'Homework Help' started by jegues, Oct 12, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    See first figure attached for my solution to the series portion of the problem.

    The answer they give to the problem is,

    Z_{series} = (1.84 - j1.38) \Omega

    I would agree if this was a Y connection but it asked for the equivalent for a delta connection.

    I first calculated the equivalent for a Y connection and converted it to a delta afterwards.

    I based my reasoning on one of the other examples given in my text. (See second figure attached)

    Is the solution they give incorrect or am I misunderstanding something?
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    My approach would be to find the parallel delta configuration first. I assumed the 208V is the line-to-line rms voltage.

    Given the power per phase is 5kW then the delta (per phase) parallel resistance will be


    Zp=Rp||-jXc [capacitive since it's a leading power factor]


    arccos (pf)=36.87^o=arctan(\frac{R_p}{X_c})




    X_c=\frac{R_p}{tan(36.87^o)}=\frac{8.6528}{0.75}=11.537 \Omega

    So Zp = 8.6528Ω||-j11.537Ω = 5.538-j4.153 Ω

    The per phase delta series equivalent of the parallel combination would therefore simply be

    Rs=5.538Ω and Xs=4.153 Ω (capacitive reactance)

    So our answers agree, although you didn't derive the parallel equivalent for part (b)
    Last edited: Oct 13, 2011