Explanation Concept used: Write the expression for 4 × 4 admittance matrix. Y b u s = ( Y 11 Y 12 Y 13 Y 14 Y 21 Y 22 Y 23 Y 24 Y 31 Y 32 Y 33 Y 34 Y 41 Y 42 Y 43 Y 44 ) ...... (1) Here, Y b u s is 4 × 4 admittance matrix. Calculation: The admittance is the reciprocal value of impedance. The admittance circuit diagram is redrawn as shown in Figure 1. Y 11 is the value of sum of admittance connected to node 1 . So Y 11 = 1 S Y 12 is the negative of value of sum of admittance connected between node 1 and node 2 . So Y 12 = − 1 S Y 13 is the negative of value of sum of admittance connected between node 1 and node 3 . So Y 13 = 0 S Y 14 is the negative of value of sum of admittance connected between node 1 and node 4 . So Y 14 = 0 S Y 21 is the negative of value of sum of admittance connected between node 2 and node 1 . So Y 21 = − 1 S Y 22 is the value of sum of admittance connected to node 2 . So Y 22 = 1 S + 0 .5 S + 0 .33 S − j1 S = ( 1 .83 − j1 ) S Y 23 is the negative of value of sum of admittance connected between node 2 and node 3 . So Y 23 = − ( 0 .33 S − j1 ) S Y 24 is the negative of value of sum of admittance connected between node 2 and node 4 . So Y 24 = − 0.25 S Y 31 is the negative of value of sum of admittance connected between node 3 and node 1 . So Y 31 = 0 S Y 32 is the negative of value of sum of admittance connected between node 3 and node 2 . So Y 32 = − ( 0 .33 S − j1 ) S Y 33 is the value of sum of admittance connected to node 3 . So Y 33 = j 0.25 S + j 0 .5 S + 0 .33 S − j1 S = ( 0 .33 − j0 .25 ) S Y 34 is the negative of value of sum of admittance connected between node 3 and node 4 . So Y 34 = − j 0.25 S Y 41 is the negative of value of sum of admittance connected between node 4 and node 1 . So Y 41 = 0 S Y 42 is the negative of value of sum of admittance connected between node 4 and node 2 . So Y 42 = − 0.25 S Y 43 is the negative of value of sum of admittance connected between node 4 and node 3 . So Y 43 = − j 0.25 S Y 44 is the value of sum of admittance connected to node 4 . So Y 44 = 0.25 S + j 0 .25 S − j0 .33 S = ( 0 .25 − j0 .08 ) S Substitute 1 S for Y 11 , − 1 S for Y 12 , 0 S for Y 13 , 0 S for Y 14 , − 1 S for Y 21 , ( 1 .83 − j1 ) S for Y 22 , − ( 0 .33 S − j1 ) S for Y 23 , − 0.25 S for Y 24 , 0 S for Y 31 , − ( 0 .33 S − j1 ) S for Y 32 , ( 0 .33 − j0 .25 ) S for Y 33 , − j 0.25 S for Y 34 , 0 S for Y 41 , − 0.25 S for Y 42 , − j 0.25 S for Y 43 and ( 0 .25 − j0 .08 ) S for Y 44 in equation (1). Y b u s = ( 1 − 1 0 0 − 1 ( 1 .83 − j1 ) − ( 0

6th Edition

ISBN: 9780357700907

Author: Glover

Publisher: INTER CENG

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Question

Chapter 2, Problem 2.37P

To determine

Admittance bus matrix and nodal equation in matrix form.

Expert Solution & Answer

Chapter 2 Solutions

POWER SYS. ANALYSIS+DESIGN

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