0195136047.pdf

150 TIME-DEPENDENT CIRCUIT ANALYSIS

(a)

2 F v(t)

4 Ω

1

2

10 V

2 Ω

1 H

t = 0

S

+

−

+

−

Figure E3.3.1

I(s) V(s)

(b)

10

s

2

s

2 Ω

s

+

+

−

−

Solution

With switchSin position 1 for a long time,

vC( 0 −)=vC( 0 +)=10 V

iL( 0 −)=iL( 0 +)= 0

The transformed network in the frequency domain is shown in Figure E3.3.1(b).

The KVL equation is given by

I(s)

(

2

s

+ 2 +s

)

=

10

s

or

I(s)=

10

s^2 + 2 s+ 2

and V(s)=I(s)( 2 +s)

Thus,

V(s)=

10 (s+ 2 )

s^2 + 2 s+ 2

=

K 1

s+ 1 −j 1

+

K 1 ∗

s+ 1 +j 1

where

K 1 =

10 (− 1 +j 1 + 2 )

(− 1 +j 1 )+ 1 +j 1

= 5

√

2 e−jπ/^4

By taking the inverse Laplace transform, we get

v(t)= 10

√

2 e−tcos

(

t−

π

4

)

V