- 18 0 Moooo Elle 26 25 J5 0 J3 75 6 25 It 0000 41 V 0000 2 5 1 4 J0 8 0 8 1 00 90 0000 0 68 135 Figure 7 1 (134.72 KiB) Viewed 37 times
-18.0 moooo elle + -26.25 -j5.0 j3.75 -6.25 it: 0000 41-V 0000 --;2.5 1,4 -j0.8 -0.8 1.00 -90° 0000 0.68/- 135 FIGURE 7.
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answerhappygod
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-18.0 moooo elle + -26.25 -j5.0 j3.75 -6.25 it: 0000 41-V 0000 --;2.5 1,4 -j0.8 -0.8 1.00 -90° 0000 0.68/- 135 FIGURE 7.
-18.0 moooo elle + -26.25 -j5.0 j3.75 -6.25 it: 0000 41-V 0000 --;2.5 1,4 -j0.8 -0.8 1.00 -90° 0000 0.68/- 135 FIGURE 7.11 Per-unit admittance diagram for Example 7.3. Consider the power system network in Example 7.3 of the textbook. (a) Find the current flow through each transmission line and the total power loss in the network (b) Given the bus admittance matrix, modify the matrix to obtain the new bus admittance matrix of the network when the line connecting buses no. 2 and 3 is tripped. (c) Calculate the changes in bus voltages. (d) Find the new current flow through each line and the power loss in the network.