Lunat1c
Apr15-10, 04:39 PM
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:
\overline{I_a}, \overline{I_b}, \overline{I_c}
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:
E_{AB} = I_{ao}Z_A - I_{bo}Z_B, ... eqn (1)
E_{BC} = I_{bo}Z_b - I_{co}Z_c ... eqn (2)
E_{CA} = I_{co}Z_c - I_{ao}Z_a ... eqn (3)
Applied kirchoff's current law at node O and got I_{bo} = -I_{ao} - I_{co}
Substituted this result in equations 1 and 2 and got:
E_{AB} = I_{ao}(Z_a + Z_b) + I_{co}Z_{b}
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 E_{BC} + E_{BC} = -E_{CA}
and got
I_{ao} = \frac{E_{AB}Z_C - E_{CA}Z_B}{Z_{A}Z_B + Z_{A}Z_C + Z_{B}Z_C}
I found the resistances by using
P = \frac{V^2}{R}, \therefore R=P*V^2
Substituting 415V, phase 120 degrees for E_{CA} and 415V phase 0 for E_{AB} , 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?
\overline{I_a}, \overline{I_b}, \overline{I_c}
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:
E_{AB} = I_{ao}Z_A - I_{bo}Z_B, ... eqn (1)
E_{BC} = I_{bo}Z_b - I_{co}Z_c ... eqn (2)
E_{CA} = I_{co}Z_c - I_{ao}Z_a ... eqn (3)
Applied kirchoff's current law at node O and got I_{bo} = -I_{ao} - I_{co}
Substituted this result in equations 1 and 2 and got:
E_{AB} = I_{ao}(Z_a + Z_b) + I_{co}Z_{b}
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 E_{BC} + E_{BC} = -E_{CA}
and got
I_{ao} = \frac{E_{AB}Z_C - E_{CA}Z_B}{Z_{A}Z_B + Z_{A}Z_C + Z_{B}Z_C}
I found the resistances by using
P = \frac{V^2}{R}, \therefore R=P*V^2
Substituting 415V, phase 120 degrees for E_{CA} and 415V phase 0 for E_{AB} , 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?