Solution 2.9.6.7 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 + V R : Applying the Laplace transform wehave I R (s)= V Rs : For the inductor we knowthat v(t)=v L (t)=L d dt i L (t): Applying the Laplace trasform, wehave V(s)= V s = L[sI L (s);i L (0)] = LsI L (s);Li L (0): Thus I l (s) = 1 Ls  V + Lsi L (0) s  =  V + Lsi L (0) Ls 2  = V Ls 2 + i L (0) s : 1 Then I(s) = I R (s)+I L (s) = V Rs + V Ls 2 + i L (0) s = V ;i L (0)R Rs + V Ls 2 : Finally, i(t)=  V L  t +  V ;i L (0)R R  1(t): Thus, the currentbecomes in nitely large, since the inductor becomes a short circuit at steady state. 2