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N 에 3 5 who 4 ulu m 11 10 Table: Branch Information X(pu) B{pu) Branch Bus No-Bus No 1-2 2-5 2-8 4-5 4.11 5-6 6-7 7-10 7-11 8
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Answer #1

Given line diagram Be II Is 5 9 4 G 7 10 convect Given impedances of branches them into admitances. Z 2 = 0.01 +jo.03 = 4,2=1 Y 2 = 10-30j 0.014 j0:03 295 = 0.02+0.06] =) YaS z 1 225 425 = - 5-153 0.02 +0.061 228 = 01025 +30.075 = 428 = 228 4 -12 Y22411 = 0102+j006 A 4 41 5-15j 0102 1 jo-06 100 ys6 56 = 004 tj012 26 Ysis = 2.5-75) OO4 jo12 = 03tj0109 = y67 = 1 26 e - 3,33289 = 0.017 10:03 =) Y89 289 489 = 10-30) ooitjo.03 ZS1o = 0.01+0.073 - =) 7810= 2910 4810= 2-14) 001+0.079 Zq1o = 0.02 +01433 buses then if power system consists of matrix of admitance is given below The -Y12 41* 412 +413 -Y21 - Y31 Y21 + 7227423 -Y

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