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 V_{L} = 480 V (rms) and the phase impedance is Z_{Y} = 24 + j9 Ω per phase. Using ∠\pmb{V}_{AN} = 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 \pmb{V}_{AN} = 277∠0° and the phase A line current is found to be
\pmb{I}_{A} =\frac{ \pmb{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 \pmb{I}_{B}=10.8 ∠ – 140.6° A (rms) and \pmb{I}_{C} = 10.8 ∠ – 260.6° A (rms). In a positive phase sequence \pmb{V}_{AB} leads \pmb{V}_{AN} by 30°; hence \pmb{V}_{AB} = V_{L} ∠30° = 480 ∠30° V (rms). The other two line voltages are \pmb{V}_{BC} =480 ∠ – 90° V (rms) and \pmb{V}_{CA} = 480 ∠ – 210° V (rms).