Question

:W10+ E A 8 ALTERNATO 6 DITES UNVALATOED Voltes a 100 lol-7' AND Es = 10 12V, T ONNECTED 10 AUD ALANCE Y-LOAD. THE ALTERRA EURAAND T LOAD 50101 tund. CALCULEt + Port WE 3 SENE CU DEMI Lmca. WLEX VOM LINE WEBC ) rozar @ I Bio Lugo le Lawy Til

Answer 1

solution - To join -- - Ib josla 122 - 66=10l-goov am gia E o 10/120 v 1 Ic L jelst J3 mm Transmission line inductance - jir As both Connected to can basis supply spole & to ground, so be calculated load side neutral is phase cument Ia, Io, by per phase evaluation Ia jin mn mo n 1060 v Ca @ EQ Amp ...Io .. (31. +31) 10 LO. да е . Ampi De 5l-ge") Amps

AL VAN . - Similarly phase b current is Ib - Jo Logo - Amp. j3 Io - Current 14.909 Amps 3.333 L-180. Amp. phase c is Is in nhase fc a 10/120 Amp j4 Ic = 25 (30' Amp. As we know the relation between the, ave & zero sequence cuprent and phase currents. - - 1907 ini I 1 Urlep 3 Il x? a Ja, L 1. where az 111201 82. 14-120° or 12.240

So the Dag sequence curent of phase á is fa, = + ( Ia t&Tb + X?Ic ) Amp. To, ( 5 (-90 +(![120)(3:394-186) + (12-18 (3510) Ia, = $( 51.909 + 3.336-60 +2.5L-90') Amp. Ja, 2 Ś ( 10.511-8489") Amps Ia, - 3.50L-80-89° Amp. 0 cu So 1 positive sequence current of phase al is Ia, = 3.50/-80-89° Amp. I . positive sequence In balance system only current is present. In unbalance system, all and zero sequence currents - positive, negative are present.