t=0时开关S闭合,闭合前电路处于稳态,求uc和ic
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t=0时开关S闭合,闭合前电路处于稳态,求uc和ic
![](http://img.wesiedu.com/upload/4/cb/4cb2a3e625a34f521ae67845ece1c883.jpg)
![](http://img.wesiedu.com/upload/4/cb/4cb2a3e625a34f521ae67845ece1c883.jpg)
![t=0时开关S闭合,闭合前电路处于稳态,求uc和ic](/uploads/image/z/15495682-58-2.jpg?t=t%3D0%E6%97%B6%E5%BC%80%E5%85%B3S%E9%97%AD%E5%90%88%2C%E9%97%AD%E5%90%88%E5%89%8D%E7%94%B5%E8%B7%AF%E5%A4%84%E4%BA%8E%E7%A8%B3%E6%80%81%2C%E6%B1%82uc%E5%92%8Cic)
ic(t)=Cduc(t)/dt
{[ic(t)*2k+uc(t)]/4k+ic(t)}*4k+ic(t)*2k+uc(t)=20V
=[3ic(t)/2+uc(t)/4k]*4k+ic(t)*2k+uc(t)
=8k*ic(t)+2uc(t)
=8k*Cduc(t)/dt+2uc(t)
=0.016duc(t)/dt+2uc(t)
duc(t)/dt+125uc(t)=1250
uc(t)=ae^(bt)+c,duc(t)/dt=abe^(bt)
(125+b)ae^(bt)+125c=1250
b=-125,c=10
uc(t)=ae^(-125t)+10,
t=0,uc(0-)=uc(0+)=20V=a+10,a=10,uc(t)=10e^(-125t)+10
duc(t)/dt=-1250e^(-125t)
ic(t)=Cduc(t)/dt=-2.5*10^(-3)e^(-125t)=0.0025e^(-125t)
{[ic(t)*2k+uc(t)]/4k+ic(t)}*4k+ic(t)*2k+uc(t)=20V
=[3ic(t)/2+uc(t)/4k]*4k+ic(t)*2k+uc(t)
=8k*ic(t)+2uc(t)
=8k*Cduc(t)/dt+2uc(t)
=0.016duc(t)/dt+2uc(t)
duc(t)/dt+125uc(t)=1250
uc(t)=ae^(bt)+c,duc(t)/dt=abe^(bt)
(125+b)ae^(bt)+125c=1250
b=-125,c=10
uc(t)=ae^(-125t)+10,
t=0,uc(0-)=uc(0+)=20V=a+10,a=10,uc(t)=10e^(-125t)+10
duc(t)/dt=-1250e^(-125t)
ic(t)=Cduc(t)/dt=-2.5*10^(-3)e^(-125t)=0.0025e^(-125t)
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