《电路》（英文版）6-2 The source-free series circuit

86-2 The source-free series circuit i(t) di KVL: L+Ri+ idt-o(to)=0 0.+ R L L2+R+t=0---(i=Ae") dt Ls+ rs+ 0 R土√R2-4L/C R R S ata2-a=-a±jo 2L 2L 2L LC R (a 2L 9c0 VLcd vo-a) The over-damped response---- i(t)=Ae+a2e The critically-damped response----i(t)=e (A,t+A2) The under-damped response----i(t=e (B, cos@,t+B, sin@,t)

§6-2 The source-free series circuit 0 ( ) 1 2 2 s t i i Ae dt C di R dt d i L + + = − − − =  + + − = t t c idt t C Ri dt di KVL L 0 ( ) 0 1 :  0 0 2 1 + + = C Ls Rs , ) LC , L R ( d 2 2 0 0 1 2  =  =  =  − d j L LC R L R L R R L C s = −  − = −   − = −   −  − = 2 0 2 2 2 1,2 1 ) 2 ( 2 2 4 / The over-damped response---- The critically-damped response---- The under-damped response---- ( ) ( cos sin ) 1 2 i t e B t B t d d t    = + − s t s t i t A e A e 1 2 1 2 ( ) = + ( ) ( ) 1 A2 i t e A t t = + − − +  c R i(t) L C

Example 1: i(t) 0.+ Jf:L=1H,R=2000,C F 401 R (0)=2m4,U(0)=2V Find: i(t) Solution: 200 =10o0 之Z 2×1 =20025 LC 10 401 a,=√c2-a2=2000o i(t)=e(61cos20000+B2sin20000)4

Example 1: : ( ) (0) 2 , (0) 2 . , 401 1 : 1 , 2000, Find i t i mA V If L H R C F = c = = = =   20000 2 2 d = 0 − = i t e B t B t A t ( ) ( cos20000 sin20000 ) 1 2 1000  = + − 1000 2 1 2000 2 =  = = L R  20025 10 401 1 1 1 1 6 0 =   = = LC −  Solution: − +  c R i(t) L C

∴i(t)= 1000t i(0)=2×103→B1 1=2×10-sin20000) (B1 cos 20000t B 1000t (2×10c0s20000t+B2sin200001) di(t) e(-40sin20000+20000B2cos20000) 1000e-(2×103cos20000t+B2sin2000 di(t) 20000B,-0b,(0) )U2(0)-Ri(0) 2 dt t=0 .B2=0→i(t)=2×103e1cos20000t4 i(t) 2x10e-oour-exponential-envelope R 2×103e-100

3 1 3 (0) 2 10 2 10 − − i =  → B =  ( ) ( cos20000 sin20000 ) 1 2 1000 i t e B t B t t  = + − ( ) (2 10 cos 20000 sin20000 ) 2 1000 3 i t e t B t t =  + − − ( 40sin20000 20000 cos20000 ) ( ) 2 1000 e t B t dt di t t = − + − 1000 (2 10 cos20000 sin20000 ) 2 1000 3 e t B t t −  + − − 2 (0) (0) (0) 20000 2 0 ( ) 2 = − − = − = = = L Ri L B dt t di t  L  c B i t e t A t 0 ( ) 2 10 cos20000 3 1000 2 − −  = → =  e exponential envelope t  − − −3 −1000 2 10 t e 3 1000 2 10− − −  i(t) t − +  c R i(t) L C

C10.0025u (0)=0.002 v(0)=2 1H i(t)=2×103e-caos200004 0 ““ 0A 日I(C1)

iL (0)=0.002 vC(0)=2 Time 0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms I(C1) -2.0mA 0A 2.0mA i t e tA t ( ) 2 10 cos 20000 −3 −1000 =  L1 1H 1 2 C 1 0.0025u R 1 2k I 0