1、RC电路分析 架+告0(20) c uc uc(O+)-Uo uc(O)=Uo RCduc+uc=0 方程的通解为：uc=Aet dt 1 特征方程为：RCp+1=0→p= RC uc Ae'rct % ,=Ue能 (t≥0) 0 uc 1R Uoe R=-ic (t>0) t Uo W(0,0)=0.5CU哈 0 t U0 R iC U0 uC U0 uC=U0e R iR - RC t ˄tt˅ iR = - RC t U0 R e = -iC˄t!˅ WR(0, f)=0.5CU2 0 uC ic R + iR - C uC(0)=U0 uC(0+)=U0 C =0 duC dt uC R + ˄Wı 0˅ 1ǃRC⭥䐟࠶᷀ C C du RC + u = 0 dt ᯩ〻Ⲵ䙊䀓Ѫ˖ pt u =A C e o 1 RCp +1 = 0 p = - RC ⢩ᖱᯩ〻Ѫ˖ 1 RC - t u =A C e