Question

Analysis of a Balanced Wye-Wye Circuit w Review Learning Goat: To yanaye a balanced thre e we wye Guit including equivalent outline currents phase voltages at the generator and at the load, and in voltages at the formats of the generator and of the load Atalanced, Tre phase, y connected generator with positive sequence has an impedance of 0.15+ and an internal voltage of 120 V / The generator ads a balanced the s e Y-connected to 34+ / The dance of the line ortage gelo e loads 0 1 . The diagra how the three-phase Y.Yasem having an impedance of Because there is and the magic of the voltage and current only by a constant angle Solving one phase, then, is s et to completely solve the circuit The phase woma wape of the specified as the

Answer 1

Sel Kaa = (0.15+30.5) /* = 7gb = Zge Van = 12010 → given it os Reference Voin = 1201-120 ; Van= 120lt 120v ZA = (34 +321) z ZB = Zc Zla = (0.8tilit) 2 = E16 = Zic The Perthase equivalent Circuit is a zlat fan Zga Lj0.15+30.shes (O.8 til they ZA &4 +54) Van GIROLO - Zgat Ela + ZA » Ran - Constjos)+(08 Phase a Current Ion = Van {2=0} 1 2010° " (0.15tjos)+(astj1.7) + (34+12) Lap =- 2.8605L-33.57° Load Voltase VLA = SAXZA { voltase division Rele} = 9.86051-33.5+ x(347324) VKA = 114-3165/60875° V PbB = 2.86051-33-577-128 = 286054153.57in VHB = 114631651-1875–128 = 114.2657–12).875° for phase "c" ICC = 2.86051-33157+120 = 2:8685/86.430A VRC = 114.31651-1-875* +128 = 114.3465118.125 V 1: Va - Vm100; Vb = Vm L=120; Vc = Um 1128 pa a Im Loo; Io = IM L. 128 ; fc = Rm11128} for phase B :