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如图所示电路,t
人气:390 ℃ 时间:2020-05-20 02:18:13
解答
t<0时, u(t)=25÷(4+6)×6=15V
t≥0时,
i1=Cdu(t)/dt
i2=[3Cdu(t)/dt+u(t)]/6
i={25-[3Cdu(t)/dt+u(t)]}/(2+4)=i1+i2=Cdu(t)/dt+[3Cdu(t)/dt+u(t)]/6
即25/6-1/2Cdu(t)/dt-u(t)/6=3/2Cdu(t)/dt+u(t)/6
12Cdu(t)/dt+2u(t)=25
u(0)=15
示意图自己画吧
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