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11 Substituting these values in the equations determined in Prob. 1.25, compute the voltages at the nodes of Fig. 1.28. Numerically determine the corresponding Zus Matrix. Solution: Using the Ybus solution of Problem 1.25, substitute the given admittance values: —j143 8.0 j40 52.8 VY 0 48.0 —f170 340 75.0 Va} 0 340 j4.0 78.8 = 70 A 1.0 7-90° 525 75.00 j0 78.3 Vs 0.68 /— 135° Compute voltages: YouV = I Yous YbusV Yous “I 40.7187 40.6688 0.6307 70.6194 ae _ | jo.sea8 70.7045 70.6242 70.6258 where Yous = Zus = | 39.6307 30.7045 70.6840 70.5660 70.6194 70.6258 0.5660 70.6840 Vo = You I VY 50.7187 30.6688 70.6307 90.6194 0 Vy _ 0.6688 70.7045 70.6242 50.6258 0 Vs ~ 70.6307 30.7045 30.6840 70.5660 1.67—90° Va 70.6194 70.6258 70.5660 70.6840 0.68 ,—135° VY 0.9285 — j0.2978 0.9750 / -17.78° Va _ 0.9251 70.3009 | _ | 0.9728,—18.02° V3 ~ 0.9562 — 70.2721 =) 0.9941 /—15.89° Va 0.8949 — 0.3289 0.9534 / —-20.18° Chapter 2 Problem Solutions 2.1 A single-phase transformer rated 7.2 kVA, 1.2kV/120 V has a primary wind- ing of 800 turns. Determine (a) the turns ratio and the number of turns in the secondary winding, (0) the currents carried by the two windings when the transformer delivers its rated kVA at rated voltages. Hence, verify Eq. (2.7). Solution: (a} Mm YW _ 12x _ yy N° W120 Mm _ 800 _ Therefore, No = =— erefore, No 70 10 
