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3.2 TRANSIENTS IN CIRCUITS 139

steady-state value while it gradually approaches the steady-state value. Practical systems are
generally designed to yield slightly underdamped response, restricting the overshoot to be less
than 10%.
Let us next consider circuits with two energy-storage elements (LandC) and nonzero initial


conditions.


EXAMPLE 3.2.6


DetermineiL(t)andvC(t)fort>0 in the circuit given in Figure E3.2.6(a).


(a)

iL(t)

vC(t)

t = 0

S

2 Ω 4 Ω

4 V

1 H

(^13) F
10 V



  • − +



  • Figure E3.2.6
    (b)
    2 Ω 4 Ω Short
    4 V
    10 V




  • − +

    Open





  • (c)
    iL
    vC
    4 Ω
    4 V
    1 H
    (^13) F








  • Solution
    From the circuit fort= 0 −drawn in Figure E3.2.6(b), we have
    iL( 0 −)= 0 =iL( 0 +) and vC
    (
    0 −
    )


    (
    10 − 4
    6
    × 4
    )




  • 4 =8V=vC
    (
    0 +
    )
    From the circuit fort>0 drawn in Figure E3.2.6(c), we haveiL,ss= 0 ;vC,ss=4 V. With
    R= 4 , L=1H,andC=^1 / 3 F,s^2 +RLs+LC^1 =s^2 + 4 s+ 3 =0. Withs 1 =−1 and
    s 2 =− 3 ,iL,tr(t)=A 1 e−t+A 2 e−^3 t;vC,tr(t)=B 1 e−t+B 2 e−^3 t. The complete solutions are
    given by
    iL(t)=A 1 e−t+A 2 e−^3 t and vC(t)= 4 +B 1 e−t+B 2 e−^3 t
    Since
    vL
    (
    0 +
    )
    =L
    diL
    dt
    (
    0 +
    )
    =−A 1 − 3 A 2 = 4 − 4 iL
    (
    0 +
    )
    −vC
    (
    0 +
    )
    = 4 − 8 =− 4
    iC
    (
    0 +
    )
    =C
    dvC
    dt
    (
    0 +
    )


    1
    3
    (−B 1 − 3 B 2 )=iL
    (
    0 +
    )
    = 0
    it follows thatA 1 + 3 A 2 =4 andB 1 + 3 B 2 =0.



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