《电路》（英文版）6-3 The complete response of the RLC

86-3 The complete response of the RLC circuit The complete response is written as the sum of the forced and the natural response. and the initial conditions are then determined and applied to the complete response to find the values of the constants Dr()=vU1()=4e2+A2c3 U(t)=Vr+Ae”+A1e U(0)=1+A1+A2 do(t (0+s141e+s2412e dt t=0 t=0 C

§6-3 The complete response of the RLC circuit The complete response is written as the sum of the forced and the natural response, and the initial conditions are then determined and applied to the complete response to find the values of the constants.      = = = + + = = + + C i t s A e s A e dt t d t V A A s t s t c f (0) 0 (0 ) 0 ( ) (0) 1 2 1 1 2 2 1 2   s t s t f f n t V t A e A e 1 2 1 2  ( ) =  ( ) = + s t s t f t V A e A e 1 2 1 2 ( ) = + +

Example: Find Uc and ii(t), iR(t) Solution 30×5=150 30 R M4u(tA 5A 5 3 2L 0 1/27F LO 5+√52-3 5+4 s,=-5-√52-32=-5-4=-9 Ucn =Ae+ a ∴U2=150+Ae+A2e U2(0+)=U(0)=150=150+A1+A2 0rA1+A2=0 d i(0) -A1e--9 A1-942 t=0 5+i1(0)5+(4-i1(0)5+(4-5) l08 A1=13.5mdA2=-13.5

Example: Find and i (t ), i (t ).  c L R 5 5 3 5 4 9 5 5 3 5 4 1 2 2 2 2 2 1 = − − − = − − = − = − + − = − + = − s s t t cn A e A e 9 1 2 − −  = + t t c A e A e 9 150 1 2 − −  = + + (0 ) (0 ) 150 150 0 = = = + 1 + 2 1 + 2 = + − c c A A or A A 0 [ 9 ] 0 9 1 2 = = − − = − − t A e A e dt t d c t t A1 = 13.5 and A2 = −13.5 V V Solution cf 30 5 150 : =  = 3 1 5 2 = = 0 = = LC , L R   108 5 4 0 5 4 5 = + − = + − = C ( ) C ( i ( )) L C i R 5 + (0) = C i C (0) = −A1 − 9A2 =

R 30 A1=13.5andA2=-13.5 kurtA 5A L 1/27T Dc ∴D(t)=150+13.5(e--e)(t>0) U2(t)=150(t<0) ∴2(1)=0.(-2+9e)A i(1)=9-0.5(-+9c)A R()=-5+0.5(-e+9)4

A1 = 13.5 and A2 = −13.5 ( ) 150 13.5( ) ( 0) 9  = + −  − − t e e V t t t c (t) = 150V (t  0) c i t e e A t t c ( ) 0.5( 9 ) − −9  = − + i t e e A t t L( ) 9 0.5( 9 ) − −9 = − − + i t e e A t t R( ) 5 0.5( 9 ) − −9 = − + − +

R1 i()=0.5( 9t C1 3H 个2i(0)=9-0.5(+9)A 0.037 i()=-5+0.5(--+9")A 10A ,, 4.0s 口I(L1)◇I(R1)V-I(C1

C 1 0.037 I R 1 30 L1 3H 1 2 I 1 4Adc I I 2 5Adc I 0 i t e e A t t c ( ) 0.5( 9 ) − −9  = − + i t e e A t t L( ) 9 0.5( 9 ) − −9 = − − + i t e e A t t R( ) 5 0.5( 9 ) − −9 = − + − + Time 0s 1.0s 2.0s 3.0s 4.0s I(L1) I(R1) -I(C1) -5A 0A 5A 10A