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Line Currents

  1. Apr 15, 2010 #1
    The figure below shows three 1-phase appliances connected in star to a 3-phase 3-wire system. The supply voltage is star-connected with a line-to-line voltage of 415V, 50Hz. I want to calculate:

    [tex] \overline{I_a}, \overline{I_b}, \overline{I_c} [/tex]

    The problem is i'm getting wrong phases that don't correspond to the answers that I'm given.

    This is how I tried to solve it:

    [tex] E_{AB} = I_{ao}Z_A - I_{bo}Z_B, ... eqn (1) [/tex]
    [tex] E_{BC} = I_{bo}Z_b - I_{co}Z_c ... eqn (2) [/tex]
    [tex] E_{CA} = I_{co}Z_c - I_{ao}Z_a ... eqn (3) [/tex]

    Applied kirchoff's current law at node O and got [tex] I_{bo} = -I_{ao} - I_{co} [/tex]

    Substituted this result in equations 1 and 2 and got:

    [tex] E_{AB} = I_{ao}(Z_a + Z_b) + I_{co}Z_{b} [/tex]
    [tex] E_{BC} = I_{ao}(-Z_b) + I_{co}[-(Z_b + Z_c)]

    then I used cramers rule to get:


    I_{ao} = \frac{\text{det} \left[ \begin{matrix} E_{AB} & Z_b \\ E_{BC} & (Z_b + Z_c) \end{matrix} \right]}{\text{det} \left[ \begin{matrix} Z_a + Z_c & Z_b \\ -Z_b & (Z_b + Z_c) \end{matrix} \right]}

    Finally, I substituted [tex] E_{BC} + E_{BC} = -E_{CA} [/tex]

    and got

    [tex] I_{ao} = \frac{E_{AB}Z_C - E_{CA}Z_B}{Z_{A}Z_B + Z_{A}Z_C + Z_{B}Z_C} [/tex]

    I found the resistances by using

    [tex] P = \frac{V^2}{R}, \therefore R=P*V^2 [/tex]

    Substituting 415V, phase 120 degrees for [tex] E_{CA} [/tex] and 415V phase 0 for [tex] E_{AB} [/tex], I got 5.28A, phase -54.27 degrees. My lecturer gave 5.286A, phase -24.26!

    Can someone show me what's wrong with my reasoning please?

    Attached Files:

    Last edited: Apr 15, 2010
  2. jcsd
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