0195136047.pdf

3.2 TRANSIENTS IN CIRCUITS 131

(c)

iL, ss(t)

= −10/10 = −1 A

4 Ω

S

5 Ω

6 Ω

10 V

5 A Short

b

a

−

+

t = 0−

vx(o−)

+

−

−

+

(d)

iL(o−)

= A

(by superposition)

37

4 Ω

S

5 Ω

6 Ω

10 V

5 A Short

b

a

50

Figure E3.2.3Continued

Solution

The Thévenin resistance seen by the inductor fort>0 is found by considering the circuit for

t>0 while setting all ideal sources to zero, as shown in Figure E3.2.3(b).

The time constant of the inductor current isτ =L/RTh= 5 / 10 = 0 .5 s. The steady-

state value of the inductor current fort>0,iL,ss(t), is found by replacing the inductor by

a short circuit (since the sources are both dc) in the circuit fort>0, as shown in Figure

E3.2.3(c).

The initial current att= 0 −,iL

(

0 −

)

, is found from the circuit fort<0 as the steady-state

value of the inductor current fort<0. Figure E3.2.3(d) is drawn fort<0 by replacing the

inductor with a short circuit (since both sources are dc).

One can solve foriL

(

0 −

)

by superposition to yieldiL

(

0 −

)

= 50 /37 A =iL

(

0 +

)

,by

the continuity of inductor current. Then the solution for the inductor current fort>0 can be

written as

iL(t)=

[

iL

(

0 +

)

−iL,ss

(

0 +

)]

e−RTht/L+iL,ss(t), fort> 0

In our example,

iL(t)=

(

50

37

+ 1

)

e−^2 t− 1 =

(

87

37

e−^2 t− 1

)

A, fort> 0

In the circuit fort= 0 −[Figure E3.2.3(d)], notice thatvx( 0 −)has a nonzero value, which can be

evaluated by superposition asvx

(

0 −

)

=− 220 /37 V.

Fort>0,vx(t) =− 6 iL(t) =−( 522 / 37 )e−^2 t+6. Note thatvx,ss = 6V,and

vx

(

0 +

)

=− 522 / 37 + 6 =− 300 /37 V, which shows an instantaneous change invxatt=0.

EXAMPLE 3.2.4

Consider the circuit of Figure E3.2.4(a) and obtain the complete solution for the voltagevC(t)
across the 5-F capacitor and the voltagevx(t) across the 5-resistor.