详细问题描述及疑问:期待您的答案,你就是当代的活雷锋,太感谢了 !
以CA支路为待接入支路,R5和R4支路设为含源两端网络。则:Uab=Us2-业德路单加怎立获[(Us2-Us3)/(R2+R4+R5)](R2+R4)-Us3=20v-[(20v-6v)/(30Ω+100Ω+10Ω)](30Ω+10Ω)-6v=10vRab=R5∥(R2+R4)=10Ω∥(30Ω+10Ω)=8Ω接入CA支路:Uca=-Uac=-Us1+[(Uab-Us1)/(Rab+R1+R6)](R1)=问答-2v+[(10v-2v)/(8Ω+5Ω+5Ω)](5Ω)=-2/9v回答完毕保证正确
已知us1=2vus2=20vus3=6vR1=5ΩR2=30ΩR3=20ΩR4=10ΩR5=10ΩR6=5Ω用戴维南定理求电压UcaWEI30
以CA支路为待接入支路,R5和R4支路设为含源两端网络。底阿卷则:Uab=Us2-[(Us2-Us3)/(R2+R4+R5)](R2+R4)比清实苗或区施来子-Us3=20v-[(20v-6v)/(30Ω+100Ω+10Ω)](30Ω+10Ω)-6v=10vRab=R5∥(R2+R4)=10Ω∥(30Ω+10Ω)=8Ω接入CA支路:Uca=-Uac=-Us1+[(Ua报红b-Us1)/(Rab+R1+R6)](口效口并信R1)=-2v+[(10v-2v)/(8Ω+5Ω+5Ω)](5Ω)=盾硫振身毛-2/9v
Uab=Us探2-[(Us2-Us3)/(R2+R4+R5)](R2+R4)-Us3=20v-[(20v-6v)/(30Ω+100Ω+10Ω拉垂年村而身曲)](30Ω+10Ω)-6v=10vRab=R5∥(R2+R4)=10Ω∥(30Ω+10Ω)=8Ω接入CA支路:谈位国述则助还日引啊Uca=-Uac=-Us1+[(Uab-Us1)/(Rab+R1+R6)](R1)=-2v+[(10v-2v)/(8Ω+5Ω+5Ω)](5Ω)=-2/9v