Question

can I have the correct solutions, answers are given

impedance of (2 + j4)n per phase, feeds two balanced 2.7 A three-phase line, which has an three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + (60 - impedance of j40)N per phase, and the other is A-connected with an j45) per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120V3 V (rms, line-to-line). Determine (a) the current, real power and reactive power delivered by the sending end source; (b) the line-to-line voltage (c) the current per phase in each load; and (d) the total three-phase real and reactive powers absorbed by each load and by the line. Check that the total three-phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads 1S at the load (40 Рoints) F.A І — 520°A, Р%3D 1800 И, ) — 0 VAR (a) (b) (c) (d) = VL.-Load = 193.64 -19.7°V I1 2.236-63.4°A, I2 = 2.582 256.56°A P = 1800 W, Q = 0 VAR еее

Answer 1

-24 7 349 160 tivo) B per phase equivalent ckt Cek (by converting a convertinga Anconnected lead into its equivalent star connected! ..leed) - 2tsp 222 Conveisting A-load to its egn #rabad its impedence per phase becomes yod of. A loed in pederxe va : Iph= Fune 2tjy Co-jis) It II (Borj yo) Ito 20-Jis)4 You = Vote (live) - 1203 - 120V :73 Iph. Ilmer - Yeh (a) Ich live as = at Touris) || (30+;4°)). 72= 20-jis)||(304we) = 60-jis) (20 4340) 20-515 +30+340 252-36-8* x 50 1313 55.902 /26-565°

ze = 22-36 Z-10.305 = 29-34 -Zstzq = 245$ +22-24 = 24 presies In = Slime : Vfhca) Zgt ze 1201 24 I 5 * pour le real power deliveredbyn source = 3 *{porfyre pered) = 3x Yax Ixcos (44, 5s). 3X120X5K Colo P 1800 W * total reactive power delivered by Souke Q = 3x (terphone e 'n source) = 3* You Is Sin (yhi Is) 5., 3X (20X5 X sino © = O VAR b). peophese load voltage = = : Line to line lead voltage = | Vu-l= Is x zq. 5ox (le=j4) 540 x 22.36 (10.305 111.8 [-10-305 Volks -(i) 3 x peophese lead voltage 3 x ): 8/10.39% so (tzó phese angle and shoul) he added since, overal lead za inoy-Constad, so, line village leads the .' Phase voltage Hf. 30.).

Viz-lead = 19364 219.095 v c) cureat imper phase in dar conected load (5) I Vph (lead) ZA (volela 111.8 2-10.305 14.8410.305 . 50 253-130 2.236 4-631435 30+140 I, : Current per phase in A- Connected lead . Vel-load . I, Zalar (ph) 113.64 211.695 60-545 193.64 49.695° 75 236.87 2.581 56.565 A B : load . Y P= Total 3-4 pwes absorbed = = 3x ( Voth klond) * I, xcocyth. II) 3x (1118 x2.236 XC6$ 53'138) C: '-10.305--(-63-435 s. 53.138 Be 449.97 w Q = 3 * 41.842-236x56 531136 - :'599.96 VAR

_load 3 ( Vectoad * IXC8 (19:595-56-565) ahsashed = .3x193164 x 2.581 x COS(-86-87) 1199.48 W Othere 3 (Vil= load XI, X sin 636-87.)) = - 899.28 VAR actually a-tead. since Qabsoiled in negative, al soled absorbs reactive power: line : Patlatented = 3x Isx () (real past of 35=2) 5 345 x2 = 150 W Qtutelattabet = 3x R$ * (9) (imeliness gestof z=4 = 3x25x4 300 VAR - ce 449.97 therefore, and head over absorbed by Iliad, a lead & life. + 1199.88 + 150 = 1799.85 18o7 =f al p' delivered by source Total reactive power absorbed by Y-load, a.lod a line = 599.96-599.28 +-307 - = - 0.68 -0 VAR Que tal delivered by source