Chapter 11, Solution 1. )t50cos(160)t(v =
)9018030t50cos(2)30t50sin(20-)t(i °−°+°−=°−=

)60t50cos(20)t(i °+=

)60t50cos()t50cos()20)(160()t(i)t(v)t(p °
+==
[ ]
W)60cos()60t100cos(1600)t(p °+°+=

=)t(p W)60t100cos(1600800 °
++

)60cos()20)(160(
2
1
)cos(IV
2
1
P
ivmm
°=θ−θ=

=P
W800

°
V
j150
Ω
-j100
Ω
200
Ω2.0j-902.0
150j
030
1
=°−∠=
°∠
=
I)t500sin(2.0)90t500cos(2.0)t(i
1
=°−=06.0j12.056.261342.0
j2
3.0
100j200
030

=
)351000cos()1000cos(532.5p
°−=

)935.0)(5624.0)(532.5(p
=

=p
W91.2For the inductor,
])t500sin(2.0[])t500cos(30[)t(i)t(v)t(p ×==At
,
s2t = )1000sin()1000cos(6p =
)8269.0)(5624.0)(6(p
=

=p
W79.2For the capacitor,
°
∠== 63.44-42.13)100j-(
2c
IV

Chapter 11, Solution 3. 10 , °∠→°+ 3010)30t2cos( 2
=ω
2jLjH1 =ω→

-j2
Cj
1
F25.0 =
ω
→
4 Ω
2
Ω
I I
1
I
2
+
−
10∠30° V
j2 Ω -j2 Ω

2j2
2

For the source,
)565.11-581.1)(3010(
2
1
*
°∠°∠== IVS

5.2j5.718.43905.7 +=°∠=S
The average power supplied by the source = W5.7For the 4-Ω resistor, the average power absorbed is
=== )4()581.1(
2
1
R
2
1
P
2
2
IW5
For the inductor,
5j)2j()236.2(
2
1
2
1
2
L


The average power absorbed by the capacitor =
W0
Chapter 11, Solution 4.
20 Ω 10 Ω
I
2
I
1
+
−
-j10 Ω
50 V
j5 Ω

For mesh 1,
21
10j)10j20(50 II +−=

21
j)j2(5
II
+−=
(1)
For mesh 2,
12
10j)10j5j10(0

2
1
j22j
jj2
0
5
I
I4j5 −=∆ ,
)j2(5
1
−=∆
,
-j10
2
=∆°∠=
−
−
=
∆
∆
= 1.12746.1
4j5
)j2(5
1

1
IW48.30
For the inductor and capacitor,
=P W0

For the 10-Ω resistor,
== R
2
1
P
2
2
IW2.12 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 Ω 2 Ω+
−
j6

P
3H
= P
0.25F
= 0 W097.52
2
258.2
P
258.238.256828.1
2j26j
6j
I
2
2
2
==
=°−∠
−+
=
Ω
Ω

Chapter 11, Solution 6.
20 Ω 10 Ω

2
−
°∠
=I
Hence,
j10
608j-
j10
6040
6042
o
−
°∠
=






−
°∠
−°∠=V
Substituting this into (1),







P
2
2
14
IW48.12

Chapter 11, Solution 7.
20 Ω 10 Ω
I
2
I
1
+
−
-j10 Ω
50 V
j5 Ω

Applying KVL to the left-hand side of the circuit,
oo
1.04208 VI +=°∠ (1)
Applying KCL to the right side of the circuit,
0
5j105j
8
11
o
=
−
++

025.0j VI = (2)
Substituting (2) into (1),
)j1(1.0208
o
+=°∠ V

j1
2080
o
+
°∠
=V

°∠== 25-
2
10
10
o
1
V
I

=








At node 1,
I
o
V
2
V
1
6∠0° A
0.5 I
o

j10 Ω
-j20 Ω
I
2
40 Ω
20j-10j
6
211
VVV −
+=
21
120j
VV
−=
(1)
At node 2,
40
5.0
2

=+−− VVVj6)-1(
37
360
j6
360j
2
+=
−
=V

j6)-1(
37
9
40
2
2
+==
V
I=










+=

=== mW
10
64
R
V
P
2
o
10
mW4.6

The current through the 2 -kΩ resistor is
mA1
k2
V
s
=== RIP
2
2
mW2


-1Ω°∠=°∠
+
=
k37.02-375.637.02-
)754.0(1
k10
Z
2
abmA)68t377cos(2)22t377sin(2)t(i
°−=°+=

°∠= 68-2I3
2
-3
ab
2
rms
10)37.02-375.6(
2
102
ZIS ×°∠

-
j2
Ω
(a)
882.1j471.0)4j1(
17
8
j28
(8)(-j2)
-j2||8
Th
−=−=
−
==
Z

==
*
ThL
ZZ
Ω+
882.1j471.0

We find using the circuit in Fig. (b).
Th
V
I
o
+
V

V

=






==
)471.0)(8(
68
64
R8
P
2
L
2
Th
max
V
W99.15

(b)

We obtain from the circuit in Fig. (c).
Th
Z
5
Ω
(a)

We find at the load terminals using the circuit in Fig. (a).
Th
Z
j100
Ω
-j40
Ω
Z
th
80
Ω

(a)

6.1j2.51
j6080
j100)(-j40)(80
j100)(80||-j40
Th
−=
+
+
=+=
Z

==


6j8
)203)(8(
)203(
40j100j80
80
o
+
°∠
=°∠
−+
=
I

6j8
)2024)(40j(-
40j-
oTh
+
°
∠
==
IV

=






+
−
10
∠
30
°
V
j2
Ω
4
Ω
(d)







°∠






−
−
=°∠


10
5
R8
P
2
L
2
Th
max
V
W389.1 Chapter 11, Solution 14.
I
+

V
Th

_
16
Ω
–j10
Ω
j8
Ω

Th
ThW6.456245.8
)245.8x2(
2
V
245.8IP
V12.158j53.7166.6555.173
)8j16(40j
8j1624j10
10
)8j16(IV
2
2
2
Th
2
rmsmax
Th
===
+=°∠=
+
+++
=+=
Chapter 11, Solution 15.

−
=+ (1)
At node 2,
o2
o2
o
)j2(j1
j-
21 VV
VV
V +−=→
−
=+
(2)
Substituting (1) into (2),
222
)j1()j)(j2(j1
VVV
−=+−=j1
1
2
−
=V

5.0j5.0
2
j1

12∠0° V
+
V
o

−
2 V
o

j Ω
(b)
j1
12
2
oo
o
VV
V =
−
+
j1
12-
o
+
=V

0)2j-(
Thoo
=+×− VVV

Chapter 11, Solution 16.

5
20/14
11
F20/1,4H1,4
j
xjCj
jLj
−==→=→=
ω
ωωWe find the Thevenin equivalent at the terminals of Z
L
. To find V
Th
, we use the circuit
shown below.
0.5V
o
2
Ω V
1

VV
−+=→
−
++
−
=
−
(1)

At node 2,
)25.025.0(5.00
4
25.0
4
21
2
1
21
jVV
j
V
V
VV
+−+=→=+
−
(2)

Solving (1) and (2) leads to
2
6.1947 7.0796 9.4072 48.81

(a)

10j40
-j10))(40(
20j30
)20j)(30(
)10j-(||4020j||30
Th
−
+
+
=+=
Z41.9j353.285.13j23.9
Th
−++=Z

Ω+= 44.4j583.11
Th
Z

==
*
ThL
ZZ

10j70
20j30
1
+=
+
+
=
I7.2j1.1)5j(
10j70
10j40
2
+=
+
−
=
I70j1010j30
12Th
+=+=
IIV

===
)583.11)(8(
5000
R8

Ω j20
Ω

j1080
(80)(-j10)
20j20-j10)(||8040||4020j
Th
−
++=++=
Z154.10j23.21
Th
+=Z==
*
ThL
ZZ

Th
439.0j049.2)j3(||6 +=+=Z
.

By transforming the current sources, we obtain
756.1j196.8)04(
Th
+=°∠= ZV===
608.20
258.70
R8
P
L
2
Th
max
V
W409.3
Chapter 11, Solution 20. Combine j20 Ω and -j10 Ω to get
-j20-j10||20j =


(1)

Also, , where
o21
4IVV +=
40
-
1
o
V
I =1.1
1.1
2
121
V
VVV =→=
(2)

Substituting (2) into (1),
2
2
4j
1.1
)2j1(40 V
V
+


+
−
120∠0° V
+
V
th

−
I
o
-j20 Ω
4 I
o

+ −
V
2
V
1
-j10 Ω
40 Ω
(b)
At the supernode,
j10-j20-40
120
211
VVV
+=
−


)818.5j9091.0(82.21j09.109 V+=−°∠=
+
−
== 92.43-893.18
818.5j9091.0
82.21j09.109
2Th
VV

===
)792.6)(8(
)893.18(
R8
P
2
L
2
Th
max
V
W569.6
Chapter 11, Solution 21.
)634.4j707.31(||50
Th
+
+
=+=Z

73.1j5.19
Th
+=Z==
ThL
R Z
Ω58.19
Chapter 11, Solution 22.
π
<<= ttti 0,sin4)(8)0
2
(
16

Chapter 11, Solution 23.



<<
<<
=
6t2,5
2t0,15
)t(v

[]
6
550
dt5dt15
6
1
V
6
2
2
2
0
22

2
25
dt-5)(dt5
2
1
V
2
1
2
1
0
22
rms
=+=+=
∫∫=
rms
V
V5
Chapter 11, Solution 25.
[]
266.3

++−==
∫∫∫∫
Chapter 11, Solution 26. , 4T =



<<
<<
=
4t210
2t05
)t(v

[]
5.62]20050[
4
1
dt)10(dt5
4
1
V

t
5
1
dtt
5
1
I
5
0
3
5
0
22
rms
==⋅==
∫=
rms
I
A887.2
Chapter 11, Solution 28. []
∫∫

=
rms
V
V92.2===
2
533.8
R
V
P
2
rms
W267.4
Chapter 11, Solution 29.
,
20T =



<<+
<<−
=

15
5
22
eff
dt400)t40t(dt)tt20100(
5
1
I













+−+






+−=
25

A773.5== RIP
2
eff
W400
Chapter 11, Solution 30. 


<<
<<
=
4t21-
2t0t
)t(v

[]
1667.12
3
8
4
1
dt-1)(dtt
6667.816
3
4
2
1
)4()2(
2
1
)(
2
1
2
0
1
0
2
1
222
=






+=



dt0dt)t10(
2
1
I

10
5
t
50dtt50I
1
0
5
1
0
42
rms
=⋅==
∫=
rms
I
A162.3
Chapter 11, Solution 33.

33.133)31)(100(100dt)tt44(100100I3
2
1
22
rms
=+=+−+=
∫==
3
33.133
I
rms
A667.6
Chapter 11, Solution 34.
[]
472.420f
2036
3
t9
3
1




+=
+==
∫∫∫
Chapter 11, Solution 35. []
∫∫∫∫∫
++++=
6
5
2
5
4
2
4
2
2
2
1
2
1
0
22

(b)

V 528.4
2
9
16
2
3
4
2
22
=+=→






+=
rms
rms
VV
(checked)
(c)

A 055.9
2
36
64 =+=
rms

3
2
2
2
1
2
==++=++=
rmsrmsrms
rms
IIII
Chapter 11, Solution 38. 08.157j)5.0)(50)(2(jLjH5.0 =π=ω→

08.157j30jXR
L
+=+=Z08.157j30
)210(
2
*
2
−
==


=θ==pf (lagging)1876.0
Chapter 11, Solution 39. 4j12
)8j12)(4j(
)8j12(||4j
T
−
−
=−=
Z

°∠=+= 74.7456.4)11j3(4.0
T
Z=°= )74.74cos(pf
2631.0

VVVV
++−=→
−
+=
−
(2) Solving (1) and (2) leads to V
1
= 45.045 + j66.935,
V
2
= 9.423 + j9.193