How did they draw root locus? P=2? And how they get the breakpoint? (b)S(s+1)G(6)Root locas Branch P>>2と=oP= 2S=0fii) Break pointヒー。CE = 1+ks(st1)4.dkdk=>(25+1)=0%3Dd6Breakpoint e2.

we will do draw root locus , we will find out break point . We will find out root locus branch .

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How did they draw root locus? P=2? And how they get the breakpoint?

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.22P

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Question

How did they draw root locus? P=2? And how they get the breakpoint?

(b)
S(s+1)
G(6)
Root locas Branch P>>2
と=o
P= 2
S=0
fii) Break point
ヒー。
CE = 1+k
s(st1)
4.
dk
dk
=>
(25+1)=0
%3D
d6
Breakpoint e
2.

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