Power.System.Analysis.Design.5E-Solutions.pdf

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1、INSTRUCTOR'SSOLUTIONSMANUALTOACCOMPANYPOWERSYSTEMANALYSISANDDESIGNFIFTHEDITIONJ.DUNCANGLOVERMULUKUTLAS.SARMATHOMASJ.OVERBYEContentsChapter21Chapter327Chapter471Chapter595Chapter6137Chapter7175Chapter8195Chapter9231Chapter10303Chapter11323Chapter12339Chapte

2、r13353Chapter14379Chapter2FundamentalsANSWERSTOMULTIPLE-CHOICETYPEQUESTIONS2.1b2.19a2.2a2.20A.c2.3cB.a2.4aC.b2.5b2.21a2.6c2.22a2.7a2.23b2.8c2.24a2.9a2.25a2.10c2.26b2.11a2.27a2.12b2.28b2.13b2.29a2.14c2.30(i)c2.15a(ii)b2.16b(iii)a2.17A.a(iv)dB.b2.31aC.a2.32a

3、2.18c1©2012CengageLearning.AllRightsReserved.Maynotbescanned,copiedorduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.2.1(a)Aj=∠°=5305cos30[]°+sin30°=4.33+j2.514−°11j26.87(b)Aj=−+3491=+6t∠an=∠5126.875°=e2−3(c)Ajjj=++()4.332.5(−3+=+=∠°4)1.336

4、.56.63578.443(d)Aj=∠°∠()530(5126.87°=∠)25156.87°=−22.99+9.8214j156.87°(e)Ae=∠°∠−()530/5(126.87°=∠)1156.87°=152.2(a)Ij=∠400−°30=−346.4200(b)it()=+5sin()(ωωt15°=+5cost15°−90°)()=−5cosωt75°Ij=()52∠−°=753.536∠−°=750.9151−3.415(c)Ij=()42∠−°+∠−°=30575()2.449−1.4

5、14+()1.294−j4.83=−=∠3.743j6.2447.28−°59.062.3(a)VI==359.3V;100Amaxmax(b)VI====359.32254.1V;100270.71A(c)VI=∠254.115V;°=∠70.71−85A°−∠j66−90°2.4(a)I=∠°100=10=∠−°7.590A1866+−jj8III=−=∠°−∠−°=+1007.39010j7.512.536.87A=∠°21VIj=−=∠°∠2()(612.536.87)(6−°=∠90)75−°53

6、.13V(b)2.5(a)υω(tt)=+2772cos3()03°=+91.7cos3()ωt0°V(b)IV==∠/2013.8530A°it()19.58cos=+()ωt30A°2©2012CengageLearning.AllRightsReserved.Maynotbescanned,copiedorduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.−3(c)ZjLj==ωπ()2601010()×=∠3.77190°

7、ΩIVZ==∠()27730°()3.77190∠°=∠73.46−60A°it()=−73.462cos()ωωt60°=103.9cos()t−60A°(d)Zj=−25ΩIVZ==∠()27730°∠()25−90°=∠°11.08120Ait()11.082cos=+()()ωωt120°=15.67cost+120A°2.6(a)V=()1002∠−°30=∠−°70.730;ωdoesnotappearintheanswer.(b)υω()1002costt=+()20°;withω=377,υ

8、()141.4cos377tt=+()20°(c)AA=∠αβ;;BB=∠CAB=+=+=jωtct()at()bt()2ReCeTheresultanthasthesamefrequencyω.2.7(a)Thecircuitdiagramisshownbelow:(b)Zj=+−=+=∠°Ω38jj434553.1(c)I=∠()1000°∠°()553.1=∠20−°53.1AThecurrentl