Solution 2.9.6.8 R i(t) e(t) L i L (0) Figure 1: Paralell RL circuit For the circuit shown in Figure 1 we knowfromKircho 's currentlawthat The di erential equation governing the circuit is i(t)=i R (t)+i L (t): Now i R (t)= v(t) R + Acos!t R : Applying the Laplace transform wehave I R (s)= As R(s 2 + ! 2 : For the inductor we knowthat v(t)=v L (t)=L d dt i L (t): Applying the Laplace trasform, wehave V (s)= As s 2 + ! 2 = L[sI L (s); i L (0)] = LsI L (s) ; Li L (0): Thus I L (s) = 1 Ls " As +(s 2 + ! 2 )Li L (0) s 2 + ! 2 # = (A=L)s +(s 2 + ! 2 )i L (0) s(s 2 + ! 2 ) = 1 Then the total currentis I(s) = I R (s)+I L (s) = As R(s 2 + ! 2 ) + (A=L)s+(s 2 + ! 2 )i L (0) s(s 2 + ! 2 ) = (A=R)s 2 +(s 2 + ! 2 )i L (0)+ (A=L)s s(s 2 + ! 2 ) = B s + M s ; j! + M  s + j! : All that remains is to nd B and M.Tothat end, B = " s[(A=R)s 2 +(s 2 + ! 2 )i L (0)+ (A=L)s] s(s 2 + ! 2 ) # j s=0 = i L (0)! 2 ! 2 = i L (0) M = " (s ; j!)[(A=R)s 2 +(s 2 + ! 2 )i L (0)+ (A=L)s] s(s ; j!)(s+ j!) # j s=j! = ;(A=R)! 2 +(;! 2 + ! 2 )i L (0)+ (A=L)j!] j!(j2!) = ;(A=R)! 2 +(A=L)j! ;2! 2 = A 2R ; A j2!L = (A=2)(R+ j=!L) = (A=2) p R 2 + ! 2 6 ;; where  =tan ;1 (L=R): Then i(t)=[i L (0)+ A p R 2 + ! 2 cos(!t+ )]1(t): 2