Question 16.14: In a balanced Y-Y circuit, the line voltage is VL = 480 V(rm......
In a balanced Y-Y circuit, the line voltage is VL = 480 V(rms) and the phase impedance is ZY = 24 + j9 Ω per phase. Using ∠VAN = 0° as the phase reference, find the line current and line voltage phasors for a positive phase sequence.
The magnitude of the phase voltage is V_{P} = V_{L}/ \sqrt{3} = 277 V(rms). For the given phase reference, we have VAN = 27∠0° and the phase A line current is found to be
\textbf{I}_{A} = \frac{\textbf{V}_{AN}}{Z_{Y}} = \frac{277 ∠ 0°}{|24 + j9| ∠ 20.6°} = 10.8 ∠ -20.6° A(rms)
The other two line currents have the same magnitude and are separated in phase by 120°. For a positive phase sequence, these currents are IB = 10.8 ∠ −140.6° A(rms) and IC = 10.8 ∠ -260.6° A(rms). In a positive phase sequence VAB leads VAN by 30°; hence VAB = VL ∠ 30° = 480 ∠ 30° V(rms). The other two line voltages are VBC = 480 ∠ −90° V(rms) and VCA = 480 ∠ −210° V(rms).