Question

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In the following three phase system Van = 120020 VRMs. Van 0.522 + ia > w j212 10000 7212 0000 M. 822 Von + 0.522 M in n 2 7212 llll 812 کې

Answer 1

0.52 Van=120020 8 8 2j8 2j Non=1200/240° Von 11 0.5.2 8 28 1200L120 0.52 Iq 0-52 Ic 0.5-12 Van 2.67 Vi ijo.67 → Van= 1200 Lov Svon 0.52 Ib Iq 1200 3.167+0.67 Iq = 370.818.-11.88 A Considering the phase sequence as abe. Von = 1200_240'v → Ib = 1200 L 240 = 370.818L-131.88 A 3.167 +0.67) Ven= 1200 L120 V → Ic = 108.05' 1200 L 120 370.818_xxarxes 3.167+0.673 a) Thus, line cuments are : ia = Iq = 370.818 L-11-88"A in = Ib= 370.818 L-131.88 A = Ic = 370.818 L 108.05'A ic by Complex power dissipation by a load , 5= 3 V. Ia CS Scanned with CamScanner

Vi = Igx (2.67+30.67) 1020.7822.206. S = = 3x1020-78L2-206 x 370.818 L-11.88 S= 1135-57 L-9.673' KVA complex powes dissipated by delta load. Phase of Ia = -11.88° phase of Ib = -131.88 • Phase of Ic 108.05 Phase of complex powes dissipated by the loads = -9.673 Magnitude of the line cuments = 370.818 A Magnitude of the complex powes 1135:57 KVA. dissipated by the loads CS Scanned with CamScanner
Thus by converting the Delta load to Star load we can solve the problem and get the desired results.