Question

- There is 480V across a 3-phase load, which consists of three impedances (10<40 (2) connected in a wye configuration. Power is delivered to the load through three phase conductors (1<352) and a neutral conductor with Z-092. (a) What is the 3-phase complex power required at the source? (b) What is the 3-phase complex power required at the source if the load is connected in a delta configuration?

Answer 1

Consider the line-to-line load voltage is V, = 480 V. The load impedance is, Z1 = 10240 12. (A) The load is connected in wye configuration: Determine the phase current. ph 480 – 30° 5<-300 ="POZ400 Since Vid="2-300) O = 27.7132-70° A In a star-connection, line currents are equal to phase currents. Ij = ph = 27.7132 - 70° A The source is wye configuration. Since, the neutral conductor with Z = 0 12. The source current is, Is = 1; = 27.7132-70° A.

The per-phase source voltage is, Vph = 1,25 = (27.7132 – 70°)(1Z35°) = 27.7132–35º V Calculate the 3-phase complex power required at the source. S3 =3Vph" ph =3(27.7132 – 35°)(27.7132–70°)* =3(27.7132–350)(27.713270°) S36 =1887.35+j1321.54 VA S36 = 2304.03Z35° VA (b) The load is connected in Delta configuration: In a delta connection the line voltage is equal to phase voltage. Vph = V1 = 480 V Determine the phase current. Sph 480 10240° = 48% -40° A Calculate the line current. 11 = 131 12–30° = 13(482-40°)<-30° = 83.142 - 70° A

The source current is, Is = 1, = 83.142-70° A. The source voltage is, Vph = 1,Z = (83.142–70°)(1235°) =83.142-35° V Calculate the 3-phase complex power required at the source. S34 = 3V pul*ph = 3(83.142–35°)(83.142-70°)* = 3(83.142-35°)(83.14270°) S34 = 16.99+ jl1.89 kVA S30 = 20.737Z350 kVA