u(c) = u(t) = 9V - i(t)*10Ω = 9 - 10i(t)
i(t) = i(c) + u(t)/6Ω = C*du(c)/dt + u(t)/6 = 2*du(t)/dt + u(t)/6
所以,
u(t) = 9 - 10i(t) = 9 - 10*2*du(t)/dt - 5u(t)/3
(1+5/3)*u(t) = 9 - 20*du(t)/dt
du(t)/dt = 1/60 * [27 - 8 * u(t)]
du(t)/[8*u(t)-27] = -dt/60
两边同时积分,可以得到:
(1/8)*ln[8*u(t) - 27] = -t/60 + c
ln[8*u(t) - 27] = -t/7.5 + c'
8*u(t) - 27 = C*e^(-t/7.5)
当 t = 0 时,u(0) = 9V.所以,C = 45V.因此,
u(t) = [45*e^(-t/7.5) + 27]/8 = 5.625*e^(-t/7.5) + 3.375
i(t) = [9 - u(t)]/10Ω = 0.9 - 0.3375 - 0.5625*e^(-t/7.5) = 0.5625 * [1 - e^(-t/7.5)]