Solution 2.9.6.4 R i(t) Switch closes at t = 0 e(t) C v C (0) Figure 1: Series RC circuit For the circuit of Figure 1, weassume v c (0) = 0. Kircho 's law yields e(t)= 1 C Z t 0 i()d + v c (0)+ i(t)R: Applying the Laplace transform weobtain E(s) = 1 Cs I(s)+ v c (0) s + I(s)R This can be rearranged as I(s)= 1 R  sE(s) ; v c (0) s +1=RC  For e(t)=Asin!t;; E(s)= A! s 2 + ! 2 : Then for v c (0)=0,wehave I(s)= A! R  1 (s ; j!)(s+ j!)(s+1=RC)  : Then i(t)= A! R  M s ; j! + M  s + j! + B s +1=RC  : 1 Then M =  1 (s + j!)(s+1=RC)  j s=j! = 1 j2!)(j!+1=RC) ;j!+1=RC ;j!+1=RC = 1=RC ; j! j2!)((! 2 +(1=RC) 2 ) = 6  ; =2 2! p ! 2 +(1=RC) 2 ;; where  = ;tan ;1 (!RC). We next nd B =  1 (s ; j!)(s+ j!)  s=;1=RC = 1 (1=RC) 2 + ! 2 = R 2 C 2 1+R 2 C 2 ! 2 : Then i(t) = "  A! R  cos(!t; =2+) ! p ! 2 +(1=RC) 2 + R 2 C 2 e ;t=RC 1+R 2 C 2 ! 2 !# 1(t): 2