R51+1+1)?

1??=?

????

??（1+1??1?????? ????=

+ ++ -??=s4+?? - ????=??s4+解得，??s4=則

=

=1,??=-10??1 所以，L、??2并聯后與??1在ω1=100rad/s處發生串聯諧振。解得，??1=8.9×10?4F

=

=??

所以，L、??2并立案后與??1在ω1=300rad/s處發生并聯諧振。解得，??2=3.3×10?4F且

=?? =????3=3

=-1 ??1+??2 132由??1=??2+??2)=25√2,解得??3= ????=????0+????1+????3= 所以，??????=-0.5????

=??× ??3+??4?jX??

?????4

解得，X??′′由??22′=??22′′解得??′=-j????cotβ??2=-j100Ω4即cotβ??=1，則（β?? 42由β

min??=??,從而??2min= ??′′cosβ??2+????????????β ??′′=33 ,??′=-j=- ????33′??????β??+cosβ?? tanβ??2=2-

解得??2min= 可求得 Ω，τ =????1

?????? 故，開關K閉合后的等效電路為 -+-+ =0.75Ω，τ 所以??????=1????7.5??· u(t)=

·Tu(t)=A??2(??)+B??2??,且??????=解得，??2??=0.125???7.5??·由??1=1Ω，I=-0.2A

????

=??2設P網絡傳輸矩陣T= 當??1=1Ω時，????130??,??1=8??,??????=13??,??21??①則30=13A+B8=13??+ ????

|????1=0=A=2,B=ab開路，即??=02+Req+

=??1???? -當??1=4Ω，????1=30??,??2=?0.625??,??????=120.6254=-????1??1+????????2=????1??1+????????2 因??????1=??????2所以只要滿足??????1=??????2=10V此時??2=0，ab端開路，????1=2??????

=????=4.4/-故????1=????1=????1=2204.4cos53.1°=

=??????=??2=380×7.6×cos30°?83.1°=因??3為純電抗性負載故??2=0所以????=P-(??1+??2)=2333W3×22020.8=2333，解得八、解：當????=15??τ=????????=0.1s，可得??????所以??2=????????10+=1??,????30+=10??,????10+=5??,????10+=所以????0+=0.5-1=-0.5A,????20+=?所以????0?=????0+=當????=60??時，????∞=所以????=30?15???10??· =??+??????????=(30-5???10??)·??1t=3?0.5???10??·

因為??10?=3??，所以????0?=????0+=????10?=3??，且????