Wye connected loads question

Discussion in 'Homework Help' started by matt_t, Sep 13, 2012.

  1. matt_t

    Thread Starter Member

    Aug 18, 2012
    I am stuck on a question similar to the delta question I asked:

    Three devices are connected in wye to a three-phase four-wire 240 Volt supply.

    • A-N Z = 30 ohms pf = .9 leading
    • B-N Z = 20 ohms pf = .8 lagging
    • C-N Z = 10 ohms pf = 1
    Calculate the neutral current.

    I know that the neutral current will be the vector difference of currents ABC.

    First I need to find the currents:

    Ia= 240/_0 deg / 30 /_25.84 = 8_/-25.84A

    Ib=240/_120 / 20/_36.87 = 12_/83.13A

    Ic=240/_240 / 10/_0 = 24/_0A

    Next I find the difference between the currents:

    In= 7.2-j3.49 - 1.435+j11.91 - -12-j20.78 = 17.765+j12.36 = 21.64/_34.82A

    The answer given for the question is 14.6A????
  2. matt_t

    Thread Starter Member

    Aug 18, 2012
    Any help would be appreciated. I realised the Phase to N voltage would be 138.56V instead of 240V but tried the same calculations and still did not get the correct answer. Any ideas anyone?
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    You angles are out of kilter. For instance your A phase current has a lagging phase angle when it is explicitly stated that the A phase has leading power factor.

    Referring to the attached solution - While I use different phase rotation than yours, you may be able to see how the solution would evolve using your preferred notation.
    Last edited: Sep 17, 2012
    matt_t likes this.
  4. matt_t

    Thread Starter Member

    Aug 18, 2012
    Thanks t_n_k I was so close but had my +/- on my angles all messed up. I appreciate your time and help.