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132 TIME-DEPENDENT CIRCUIT ANALYSIS


S t = 0

(a)

iC(t)

vC(t)

1 Ω
5 Ω vx(t)

10 V 4 Ω

5 F
b

+ − a

+

+

Figure E3.2.4

(b)

Short 1 Ω

5 Ω

4 Ω

S RTh = 5 + 1 = 6 Ω
b

a

S

+10 V− +


(c)

vC, ss(t) = −10 V

1 Ω
5 Ω

4 Ω
Open

b

a

S

+10 V− +


+


vx(o−) vC(o−) = −40/9 V
t = o−

1 Ω
5 Ω

4 Ω
Open

b

a

(d)

Solution

(Note that the procedure is similar to that of Example 3.2.3.) The Thévenin resistance seen by the
capacitor fort>0 is found by considering the circuit fort>0 while setting all ideal sources to
zero, as shown in Figure E3.2.4(b).
The time constant of the capacitor voltage isτ=RThC= 6 × 5 =30 s. The steady-state
value of the capacitor voltage fort>0,vC,ss(t), is found by replacing the capacitor by an open
circuit (since the source is dc in the circuit fort>0), as shown in Figure E3.2.4(c).
The initial capacitor voltage att= 0 −,vC(0−), is found from the circuit fort<0 as the
steady-state value of the capacitor voltage fort<0. Figure E3.2.4(d) is drawn fort<0by
replacing the capacitor with an open circuit (since the source is dc).
One can solve forvC( 0 −)to yieldvC

(
0 −

)
=− 40 /9V=vC

(
0 +

)
, by the continuity of the
capacitor voltage. Then the solution for the capacitor voltage fort>0 can be written as
vC(t)=

[
vC

(
0 +

)
−vC,ss

(
0 +

)]
e−t/(RThC)+vC,ss(t), fort> 0
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