华北电力大学电力系统故障分析第四章答案

Uka1?ZsIka1?ZmIka2?Z?mIka0???111??Za?2Z?Ika1??Za?Z?Ika2??Za?Z?Ika0 333???1?????Za?Z??Ika1?Ika2?Ika0??ZIka13??Uka2?Z?mIka1?ZsIka2?ZmIka0???111??Za?Z?Ika1??Za?2Z?Ika2??Za?Z?Ika0 333???1?????Za?Z??Ika1?Ika2?Ika0??ZIka23??Uka0?ZmIka1?Z?mIka2?(Zs?3Zg)Ika0????111??Za?Z?Ika1??Za?Z?Ika2??Za?2Z?Ika0?3ZgIka0 333???1?????Za?Z??Ika1?Ika2?Ika0??(Z?3Zg)Ika03??所以可得复合序网图为：

（d）略 （e）

将原图等值变换为：

???U?kb?U?kc????Z由题意得：?U?ka?I?ka(Zg?)

2????I?kb??I?kc?????????????? U?ka1?U?ka2；3I?ka0?I?ka1?I?ka2?I?ka0?I?ka

???????? U?ka0?2U?ka1?3I?ka0(Zg???????Z) 2Z) 2即：?I?ka0Z?0??2U?ka1?3I?ka0(Zg??U?ka13??ZZ?0???I?ka0?(Zg?)?223????

???????ZgZZ?0?????Ika1?Ika2?I?ka0???? ?246???????????ZgZ0?Z???Ika1?I?ka2?I?ka0?????66????2U?ka1?U?ka2

??U?ka0??I?ka0Z?0?3??ZZ?0??3??ZZ?0??????Ika0?(Zg?)??Ika0?(Zg?)??I?ka0Z?0? ??223?223??????ZgZ0?Z???3???????Ika1?I?ka2?I?ka0??????I?ka0???Zg?Z0???66????2?2???复合序网如图所示：

[习题4-10] 由题意得：

???Uka?0????Ukb?Ukc????Ikb??Ikc?（2）式?Uka1?Uka2

??(1)(2) (3)（1）式?Uka0?2Uka1?0 （3）式?Ika0????????1??1???1????Ika?Ikb?Ikc??Ika??Ika1?Ika2?Ika0? 3?3??3???1?1Z0?Uka1??Uka0???(?Ika0)Z0??Ika0222??1???Z0???Ika1?Ika2?Ika0? 3?2??????Z0???Ika1?Ika2?Ika0???6Uka0??2Uka1?Uka1?3Uka1???Z0????Z0? ??Ika1?Ika2?Ika0??3Ika02??6??????Z0???3??Ika1?Ika2?Ika0??Ika0??Z0????6?2?????复合序网如图所示：

Ika1?x1??x2?????Ea1?xx?(?0?)?0?26??j10.5??j?0.25?6????3

Uka1?Ea1??Ika1jx1??j1?3?j0.25?j0.25

Ika0???Uka1j0.25???1.0 11?x0???j0.522?