k1(0 已知：电感无初始储能 例4 t=0时合k1,t=0.2s时合k2 22 1H 求两次换路后的电感电流(t)。 32 k(t=0.29) 解 0<t<0.2s t>0.2s i(0.2）=2-2e50.2=1.26 i(0*)=(0)=0 1=L/R=1/5=0.2S i(0.2)=1.26A i(o)=10/5=2A 2=L/R=1/2=0.5 i(t)2-2e-A i(∞)=10/2=5A i(t)=5-3.74e2-0.2 A例4 i 10V 1H k1 (t=0) k2 (t=0.2s) 3 2 已知：电感无初始储能 t = 0 时合k1 , t =0.2s时合k2 求两次换路后的电感电流i(t)。 0 < t < 0.2s ( ) 2 2 A 5t i t e − = − t > 0.2s ( ) 10 / 5 2A / 1/ 5 0.2 s (0 ) (0 ) 0 1  = = = = = = = + − i L R i i  i A L R i A ( ) 10 / 2 5 / 1/ 2 0.5 (0.2 ) 1.26 2  = = = = = = +  (0.2 ) 2 2 1.26 5 0.2 = − = − −  i e ( ) 5 3.74 A −2( −0.2) = − t i t e 解