Question

4.3. Given a 208 V three-phase system, find the currents if a. the load is wye connected, with Zy = 8.66 +522. b. the load is delta connected, with Za = 25.98 + 152.

Answer 1

20 (0) when load is wye connected R given line voltage Vi = 208 Lo volt IR R1 & 8.66 jXL & jsn pv 30.664 8.661 JXL B where → VR = Phase voltage z = load impedance

for star connected system phase voltage will be 1/3 times of line voltage and lag 30 with line voltage." TR = L-30 - 208L-30 UR = 120-09 Z-30 voit Now 120.09 C-30 8.66 tj5 IR = 12 L-60° Amp as load is balanced hence all will be equal in magnitude and from each other phase currect 120 apart I = 12 [-180 Amp IB = 12 L60 Amp all phase current IR = 12 L-60 Amp ly = 12 2-180 Amp .. IB = 12 L60 Amp * phase current will be equal to line current

(11) complex load power 3 phase system is nothing but 3 time 1 prase system tence 530 = 3.510 334 = 3ſ(220.09 L-30). (12 Z60)] 4323.24LO p. 5 3744.04 +1 2161.62 VA When load is delta connected Y IRY PQ Z pa se on 2 25.981 | Jt1 s . where J = line voltage (fordelta connected system phase voltage is equal to line voltage) <p afety M-am gor Typ jisr la

TRY = 20820 25.98+15 = 6.93L-30 A caotias as lood is balanced hence phase current will be equal in magnitude and 120 apart from each other. IRY = 6.93L-30' A lýß = 6.93 L-150 A IBR = 6.93 L90 A Phase current ? For delta connected system line current will be 3 times of phase current and lags 30° with the phase current IR = S3 (IRY 1 /-38-30 = 12 L-60 A Iy = 13 15481 (-150-30º = 12 (-180 A Is = 1B 178RL90–30 - 12/60 A line current

(11) complex load power → 530 = 3. 31$ = 3 [vix Ipy*] 53° 3 [(208 20). (6:93 Lt30)] 530 = 43.24-32/30 530 = 3744.9.+j 2162.2 VAI Power will be same in both case" :