Example 2: Find i, and i, for t>0. solution 120 2×3 eq +1=2.2mH 2+3 ImH 18V 2mwmv2=50(120+60)×90 120+60+90 110 ea 2.2×10 0.36 =2×10s 0.2 R 11o ea i(0)=0.364→i(0+)=i(0-)=0.364 180 0.24 i1(0)=18/90=0.2 A and i1(0)=-i2(0) =-0.24A 180+90 i1(t)=0.36e4andi1(t)=-0.24e04Example 2: Find iL and i1 for t>0. 110 120 60 90 (120 60) 90 50 = + + +  Req = + i (t ) . e A and i (t ) . e A t t L 50000 1 50000 0 36 0 24 − −  = = − i L(0 ) = 0.36A → i L(0 ) = i L(0 ) = 0.36A − + − i ( ) / . A and i ( ) i L ( ) 0.24A 180 90 180 0 18 90 0 2 0 0 1 1 = − + = = = − − + + solution: s R L eq eq 5 3 2 10 110 2 2 10 − − =   = = .  L i R i 0.36 −0.24 0.2 Leq 1 2.2mH 2 3 2 3 + = +  =