# Line Currents

1. Apr 15, 2010

### Lunat1c

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?

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Last edited: Apr 15, 2010