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.