Chapter 16, Solution 1.

Consider the s-domain form of the circuit which is shown below.
I(s)
+
−

1
1/s
1/s
s
22
2
)23()21s(
1
1ss
1
s1s1
s1
)s(I
++
=
++
=
++
=




+
−

+
V
x

−
4

V)t(u)e2e24(v
3
8
j
3
4
s
125.0
3
8
j
3
4

2
x
x
2
x
2
x
xx
x
−−+−
++−=












−+
−
+
++
−
+−=
++








−








−
−−
Chapter 16, Solution 3.

s

5/s
1/2
+
V
o

=
+
=












++
=

v
o
(t) =
( )
V)t(ue13125.
t625.0−
−0
Chapter 16, Solution 4. The s-domain form of the circuit is shown below.
6 s

10s6s
10
1s
1
s106s
s10
)s(V
2
o10s6s
CBs
1s
A
)10s6s)(1s(
10
)s(V
22
o
++
+
+
+
=
+++
=

)1s(C)ss(B)10s6s(A10
22

−
+
=
++
+
−
+
=

=)t(v
o
V)tsin(e4)tcos(e2e2
-3t-3t-t
−− Chapter 16, Solution 5.
s
2
2s
1
+
2
I
o
s2
2s
1
2
s
2
1
s
1
1
2s
1
V
t2
t3229.1j2/t90t3229.1j2/t90t2
o
22
2
o
2
−=
++=
−+
++
+−
+
++
−−
−−
+

+
=
−
−°−−°−−
Chapter 16, Solution 6. 2 2s
5
+
I
o

10/s
s
Use current division.

t3sine

++
+
=
Chapter 16, Solution 7. The s-domain version of the circuit is shown below.

1/s
1 I
x

+ 2s

1
2
+s

–
Z


A
)5.0ss)(1s(
1s2
1s2s2
s21
x
1s
2
Z
V
I
22
2
2
2
x
++
+
+
+
=
+++
+
=
++
+
+
==
+
−
+
=

[]
A)t(ut866.0cose46)t(i
t5.0
x
−
−=
Chapter 16, Solution 8.

(a)
)1s(s
1s5.1s
s22
)s21(
s
1
)s21//(1
s
1
Z
2
+
++

+
=

Chapter 16, Solution 9.

(a)

The s-domain form of the circuit is shown in Fig. (a).

=
++
+
=+=
s1s2
)s1s(2
)s1s(||2Z
in
1s2s
)1s(2
2
2
++
+

1
1
2
s2/s
1/s
s



+
+
+






+
+
⋅
=






+
+
=
2s3
6s5
s
2s3
6s5
s


Applying KCL gives

s/12
V
V21
x
o
+
=+

But
xo
V
s
/12
2
V
+
=
. Hence

s3
)1s2(
V
s
/12
V
s/12
V4

1
2
+
s
2 V
o
2V
o
-
Applying KCL gives

)1s(3
4
V
2
V
V2
1s
2
o
o
o
+
−=→=+
+Chapter 16, Solution 11. The s-domain form of the circuit is shown below.
4/s s
I
2
I
1
+
−
+
−
2
4/(s + 2)
1/s

Write the mesh equations.

21
I2I
s
4
2
s
1
−


=






+
2
1
I
I
2s2-
2-s42
2)(s4-
s1)4s2s(
s
2
2
++=∆ ,
)2s(s
4s4s
2
1
+
+−
=∆

222
++++++=+−⋅

Equating coefficients :
2
s:
BA21 +=

1
s:
CB2A22- ++=
0
s:
C2A42 +=

Solving these equations leads to
A 2=
,
23-B =
,
-3C =22
1
)3()1s(
3s23-
2s
2
I

[]
A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2
-t-2t
−−

22
2
2
2
)3()1s(
3-
)4s2s(2
s
s
6-
I
++
=
++
⋅=
∆
∆
=

== )t3sin(e
3
3-
)t(i
t-
2

10
+=+
−
+1s
15s1510
15
1s
10
V)ss25.01(
o
2
+
++
=+
+
=++

1s25.0s
CBs
1s
A
)1s25.0s)(1s(
25s15
V
22
o
++


740A =
,
740-B =
,
7135C =4
3
2
1
s
2
3
3
2
7
155
4
3
2
1
s
2
1
s
7
40
1s




⋅+
+






+
+
−
+
=
+






+
+
+
+
=
7
40
)t(v
2t-2t-t-
o=)t(v
o
V)t866.0sin(e57.25)t866.0cos(e714.5e714.5
2-t2-t-t
+−
Chapter 16, Solution 13. Consider the following circuit.
V
o
1/(s + 2)
1/s 2s
I
o
2
1

Applying KCL at node o,


B
1s
A
)2s)(1s(
1
1s2
V
I
o
o
+
+
+
=
++
=
+
=

1A =
,
-1B =2s
1
1s
1
I
o

5 V
(a)
A5)0(i
o
=
−
, V0)0(v
c
=
−

We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
2s 5/s
I
o
V
o
+
−
1 4
15/s 4/s
(b)
At node o,
0
s44
0V
s
5
s2
V


o
2
o
22
V
)1s(s4
2s6s5
V
)1s(s4
s2s2s4s4
s
10
+
++
=
+
++++
=2s6s5
)1s(40
V
2
o
++
+
=


sCsB)4.0s2.1s(A)1s(4
s2
++++=+

Equating coefficients :
0
s: 10AA4.04 =→=
1
s
: -84-1.2ACCA2.14 =+=→+=
2
s : -10-ABBA0 ==→+=

4.0s2.1s
8s10
s
10
s
5
I
2
o
++
+
−+=

2222
o
2.0)6.0s(
)2.0(10

o
+
−

+
−

5/s
s/4
+ V
x

−
10

3V
x

2s
5
VV
2s
5
VV,But

−++−
=
+
−
+
−
+
−

We can now solve for V
x
.

)40s5.0s)(2s(
)20s(
5V
2s
)20s(
10V)40s5.0s(2
0
2s
s5
V120
2s
5
V)40ss2(
2
2
x
2

To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.
1
Ω
+
−
V
o
1 F
V
o
/2
+
−
1 H
i
o
+
−
2
Ω

3 V
(a)Hence,
A1-
3
3-
i)0(i

s
2.5/s
+
−
1
+
−
V
o
1/s
V
o
/2
+
−
I
o
−
+
2
-1 V
5/(s + 2)
(b)
For mesh 1,
0
2
V
s
5.2
I

I
s
1
2
1
I
s
1
2
21
−
+
=






−+






+
(1)

For mesh 2,

2
1
I
s
1
-
21
−=






+++ (2)

Put (1) and (2) in matrix form.











5.2
s
5.2
2s
5
I
I
s
1
s
2
1
s
1-
s
1
2
1
s
1
2
2
1s
3
2s2 ++=∆ ,
)2s(s
5

)2s(C)s2s(B)3s2s2(A132s-
222
++++++=+

Equating coefficients :
2
s: B A22- +=
1
s
:
CB2A20 ++=
0
s:
C2A313 +=

Solving these equations leads to
7143.0A =
, ,
-3.429B = 429.5C =5.1ss
714.2s7145.1
2s
7143.0
3s2s2
429.5s429.3
2s
7143.0
I

o
[ ]
A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0
-0.5t-0.5t-2t
+−
Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
I
3
+ −
1/s
I
2
I
1
4
s
1
1
For mesh 3,
0IsI
s
1
I




+−++






+ (2)

But
(3)
4II
21
−=

Substituting (3) into (1) and (2) leads to






+=





s
1
s-
32
+
−=






++






+ (5)

Adding (4) and (5) gives
1s
2
4I2
2
+
−=
s
s
−
−
−=−
1 Ω

+
V
o

−

+
−
1/s
V
s

2 Ω




+
−=−
+
=
=→=++
−
−−−
−−

+
Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below.

I
1/s
−
+
2 I
V
1
V
o
1/s
o1
Vs
s
1
V
s
1
2
1
2
2s
2
++






+=+
+
(1)

But
and
I2VV
1o
+=

Substituting (2) into (1)
oo
Vs
2s
2
V
2s
s
s
1s2
s
1
2
2s
2
+






+
−













+
+
=
+
+
+−+
+o
22
V
2s
1s4s
2s
9s2
)2s(s
s9s2
+
++
=
+
+
=

V
o
+
−
+
=Therefore,
=
)t(v
o
V)t(u)e4434.0e443.2(
-3.732t-0.2679t
− Chapter 16, Solution 20.

We incorporate the initial conditions and transform the current source to a voltage source
as shown.
1 s
+ −
1/s
2/s
V
o
+
−
1

s
1s
1s
s
++=−
+
−
+)1s2s2)(1s(
1s4s2-
V
2
2
o
+++
−−
=5.0ss
CBs
1s
A
)5.0ss)(1s(
5.0s2s-
V
22
o


222
o
)5.0()5.0s(
)5.0s(2
1s
1
5.0ss
1s2
1s
1
V
++
+
−
+
=
++
+
−
+
=

=)t(v
o
[]
V)t(u)2tcos(e2e
2-t-t
−


V
s
)1
2
()1(10
21
10
2
1
1
1
−++=→+
−
=
−
(1)

At node 2,
)1
2
(
2
2
1
1
++=→+=
−
s
s
VVsV

CBs
s
A
sss
V
oCsBsssA ++++=
22
)5.12(10
BAs +=0:
2

CAs += 20:

-40/3C -20/3,B ,3/205.110:constant ===→= AA





++
−
++
+
−=


Taking the inverse Laplace tranform finally yields

[]
V)t(ut7071.0sine414.1t7071.0cose1
3
20
)t(v
tt
o
−−
−−=

Chapter 16, Solution 22. The s-domain version of the circuit is shown below.
4s
V
1
V
2 1s +
12


+
→
−
+=
+
(1)

At node 2,






++=→+=
−
1s2s
3
4
VVV
3
s
2
V
s4
VV
2
212
221
(2)


++=






−






+






++=
+)
8
9
s

7
s(A9
22
++++++=Equating coefficients:

BA0:s
2
+=

A
4
3
CCA
4
3
CBA
4
7
0:s −=→+=++=

-18C -24,B ,24AA
8
3
CA
8
9
9:constant ===→=+=

++
+
++
+
−
+
=
++
+
−
+
=Taking the inverse of this produces:
[ ]
)t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v
t875.0t875.0t
2
−−−
+−=

Similarly,

)
8
9
s
4
7


++
=)1s(F)ss(E)
8
9
s
4
7
s(D1s2s
3
4
9
222
++++++=






++

Equating coefficients:
ED12:s
2
+=


8
7
s(
)8/7s(4
)1s(
8
)
8
9
s
4
7
s(
s4
)1s(
8
V
222
1
++
−
++
+
+
+
=
++
+
+
=


sCV
sL
V
R
V
C5
s
2
s
4
++=++







++=
+
LC
1
RC
s
s
s
CV
s
sC56

480s5
V
++
+
++
+
=
++
+
=

=)t(v
V)t2sin(e230)t2cos(e5
-4t-4t
+

)20s8s(s4
480s5
sL
V
I
2
++
+
==

20s8s
CBs
s
A

++
−
++
+
−=
++
+
−=

=)t(i
0t),t2sin(e375.11)t2cos(e6)t(u6
-4t-4t
>−−
Chapter 16, Solution 24. At t = 0
-
, the circuit is equivalent to that shown below.

+ 9A 4
Ω
5
Ω

+

36V 10A 2/s 5Ω
-
Applying KCL gives

8.12B,2.7A,
5.0s
B
s
A
)5.0s(s
s206.3
V
5
V
2
sV
10
20
V36
o
ooo
−==
−
+
−
(2s)/(s
2
+ 16)
+
−
9/s
2/s

0
s
2
I
s
9
s6
16s
s2
2
=+







16s
EDs
)3s(
C
3s
B
s
A
s
2
22
+
+
+
+
+
+
++=)s48s16s3s(B)144s96s25s6s(A288s36
2342342
++++++++=+

)s9s6s(E)s9s6s(D)s16s(C
232343
++++++++

,
D 2016.0-=
,
765.2E =