接3楼，设电源电压u(t) = A sin wt，则
Uc = U*(1/jwC)/(1/jwC + R) = U/(1+jwCR),
= U * [ 1/(1+(wRC)^2 - j * wRC/(1+(wRC)^2]
取复数的幅值和角度，得到uc(t)的时域表达
uc(t) = A/(1+(wRC)^2 sin (wt - arctg(wRC) )
最后的u(t)为什么是这样得出的