Solution 2.9.6.2 R i(t) Switch closes at t = 0 e(t) L i L (0) Figure 1: Series R;L circuit For the circuit of Figure 1, the de ning di erential equation is e(t)=i(t)R+ L di(t) dt : Applying the Laplace transform weobtain E(s) = I(s)R+ L[sI(s);i L (0)] This can be rearranged as I(s)= 1 L  E(s);Li L (0) s + R=L  For e(t)=Acos!t;; E(s)= As s 2 + ! 2 : Then for i L (0) = 0, wehave I(s)=  As=L (s 2 + ! 2 )(s + R=L)  ;; In partial fraction form I(s)= M s;j! + M  s + j! + B s + R=L : 1 Then M = (s;j!)I(s) s=j! =  As=L (s + j!)(s+ R=L)  s=j! = jA!=L (j2!)(j!+ R=L) = A=2L (j!+ R=L) ;j!+ R=L ;j!+ R=L = A(R=L;j!) 2L(! 2 +(R=L) 2 ) = A 2L p ! 2 +(R=L) 2 6 ;;  = ;tan ;1 (!L=R). B = (s + R=L)I(s) s=;R=L =  As=L (s;j!)(s+ j!)  s=;R=L = ;AR=L 2 (R=L) 2 + ! 2 = ;AR R 2 + L 2 ! 2 Then i(t) = " A L p ! 2 +(R=L) 2 ! cos(!t+ );  AR R 2 + L 2 ! 2  e ;Rt=L # 1(t): 2