Intro to three phase systems - 3 -

Since the three-phase system is symmetrical, we need only consider one phase. The phase voltage across the load is given by:

$\begin{aligned} V_{\text{load}} & = \frac {Z_{\text{load}}}{Z_{\text{line}}+Z_{\text{load}}}\times V_{an} \\ & = \frac {10-\frac 1{2\pi \times 60 \times 531 \times 10^{-6}}j}{0.5+2\pi \times 60 \times 5.3 \times 10^{-3}j + 10 -\frac 1{2\pi \times 60 \times 531 \times 10^{-6}}j} \times 120 \\ & = \frac {10-5j}{0.5+2j+10-5j} \times 120 \\ & = \frac {600(2-j)}{10.5-3j} \\ & = \frac {1341.64\angle -26.56^\circ}{10.92\angle -15.94^\circ} \\ & \approx 122.86 \angle -10.6^\circ \end{aligned}$

Since the clockwise sequence is $\begin{bmatrix} 1 \\ \alpha^2 \\ \alpha \end{bmatrix}$ , the phasor is $\boxed{122.86 \angle -10.6^\circ \begin{bmatrix} 1 \\ \alpha^2 \\ \alpha \end{bmatrix}}$ .