《电路》（英文版）5-5 The driver RC circuit

85-5 The driver RC circuit The complete response of any RC circuit may also be obtained as the sum of the natural and the forced response 了f cn Ae tt=r =vtae t vu(t) 2(0t)=020)=0:0=V+A-→A= V-Ve Ro (r>0) i=cduc ke rc(t>o) dt R f(t)=[f(0+)-f(∞)e5+∫(∞) initial-value forced-value time-cons tant

The complete response of any RC circuit may also be obtained as the sum of the natural and the forced response. §5-5 The driver RC circuit (  0) − = = RC t t e R V dt c d i C  V cf c cn c f  = +  =      t V Ae c RC t Ae cn − = = + − = c c =  =V + A→ A= −V − = +  (0 )  (0 ) 0 0 initial−value forced − value time −constant ( ) [ (0 ) ( )] ( ) / = −  +  + − f t f f e f t  (  0) − = − RC t t V Ve c

Example: Findo(t) andi(t) Solution: =0 50×200 10 Ucf60450×20050+20~20 0+200 Ch-4 120v 50v T=RC= ×0.05=12s vc(t)=20+ Ae Oc(0)=0c0)=120x50=100V 10+50 ∵100=20+A→4=80 U(r)=20+8eV(t>0 Uc()=100(t0) )=60+200=0.1924r<0)

Example: Find (t) and . c  i(t) i(t ) . A(t ) A(t ) . t . . e (t ) c i(t ) 0 192 0 60 200 50 0 1 0 4 1 2 0 200 =  + =  − = = +  Solution:   t Ae cn − = c f 20V 50 200 50 200 50 200 50 200 60 50 = +   +  +  = 1.2 ( ) 20 t t Ae c −  = + ( 0) ( ) 100 ( 0) 1.2 ( ) 20 80  −−−− =  −  = + t V t c V t t t e c   C s Req 0.05 1.2 50 1 200 1 60 1 1  = + +  = = 100= 20+ A→ A= 80 c c 50 100V 10 50 120 (0 ) (0 )  = + = − = +  

orf()=[f(0+)-f(∞)ex+f(∞) Uc(∞)=20 U2() U(0+)=D2(0)=100 100 z=12s i(t 2()=20+80c12v(>0 2 i(t)=01+0e124(t>0) i(t)=0.1924(t<0) D Example: Find v(t)and i(t), 20k jfi,=:(a)25u(t)m4; BK 12k (b)10+15u(tm

( ) [ (0 ) ( )] + () − = + −  f t f t f f e  or s c c V c V 1.2 (0 ) (0 ) 100 ( ) 20 = = − = +  =     i(t) (t) c t 100V 20V ( b ) u(t )mA i f i : ( a ) u(t )mA; Example : Find (t ) and i (t ), s c c 10 15 25 + =  ( 0) 1.2 ( ) 20 80  − = + V t t t e c  i(t ) . A(t ) A(t ) . t i(t ) . . e 0 192 0 0 1 0 4 1 2 0 =   − = +

C SuF 00I 20k 12k Solution (a):i=25()m4 8 U(∞)=25×103× 8k+324×20k=100U U2(0+)=U2(0)=0 20 5×10-6=0.05s 2 ∴U(t)=100-100e0v(t>0) (t)=C du(t) 0.01e-24(t>0)

0.01 ( 0) ( ) ( ) 20 = =  − e A t dt d t i t C c t c  a i u t mA Solution s ( ): 25 ( ) : = k V k k k c 20 100 8 32 8 ( ) 25 10 3  = +  =   −  (0 ) = (0 ) = 0 + − c c s k 5 10 0.05 2 20 6 =   = −  ( ) 100 100 ( 0) 2 0  = −  − t e V t t c  c c i 100V 0.01A t

cc 十 SuF 100 De 6m4 20k (b):is=10+15u()m c(∞)=25×10-3 8 8+324×20k=100V U(0)=U(0)=10x1O-3xk 8k+32k×20k=40V ∴Uc(t)=100+(40-100)e-20t=100-60e~20tv(>0) =6e-20r duc(t mA(t>0

20k 40V 8k 32k 8k (0 ) (0 ) 10 10 3 c c  = + = =   + − −   ( ) 100 (40 100) 100 60 ( 0) 2 0 2 0  = + − = −  − − t e e t t t  c V  c c 40V i 100V 6mA (b) : i s = 10 + 15u(t)mA 20 100V 8 32 8 ( ) 25 10 3  = +  =   − k k k k  c 6 ( 0) ( ) ( ) 20 = =  − e t dt d t i t C c t c m A 