所示电路换路前已达稳态,在t=0时将开关S断开,试求S断开后的 ic(t)

在 t = 0 时，Uc0 = 4V。
Uc = 6 - 2* Ic
Ic = dQ/dt = C*dUc/dt
Uc = 6 -2C *dUc/dt
特解为：Uc = -2C * dUc/dt，dUc/Uc = -dt/(2C)
所以，两边同时积分，可以得到：lnUc = -t/2C，Uc = U*e[-t/(2C)]
则，dUc/dt = dU/dt*e^[-t/(2C)] + U*[-1/(2C)]*e^[-t/(2C)]，代入原式，得到：
Uc = 6 -2C * dU/dt * e^[(-t/(2C)] + U*e^[-t/(2C)] = U * e^[-t/(2C)]
dU/dt = 6* [1/(2C)] * e^[t/(2C)]
dU = 6 * e^[t/(2C)] d[t/(2C)]
所以，U = 6 * e^[t/(2C)] + U0
Uc = U * e^[-t/(2C)] = 6 + U0 * e^[-t/(2C)]
因为当 t = 0 时，Uc = 6 + U0 * e^0 = 6 + U0 = 4V
所以，U0 = -2V，Uc = 6 - 2 * e^[-t/(2C)]
因此，Ic = C*{-2 * e^[-t/(2C)] * [-1/(2C)]}
= e^[-t/(2C)]