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LINE 4.12 -;3.12 LOAD -wA 22020V 200 15.12 j52 Figure 1: For Question 6, 7 and 8. geoce 6. Consider the circuit shown in Figu
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line to it um 4 jam IL Isu 22016 Vreme 200 jsu lad = I 220 Lo Zeq ME 4-j3) - near 15mi jse 20. Zeq = (4-13) + 20 (15+js) 20+1Is = 11 220 Lo 13 17.0856/6-242 A, 12.67632 4-6-242 Is = 17:0856/6.242 A A) 6 Complex power of the source is is (22020) (1a-0Pline - Wi rea part of Sine) u rea part of 1167.67 - j&as.9] 11 لک W dat = Pline 11 GG wi Average power dinspatted Optim-B ReSaw Pload of load of 19 load Iload a 467. I VAR Option=b Ruey 0 it (3737-54087) VA Optione (None of the above & Plane = 1168

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