Chapter 13, Solution 1. For coil 1, L
1
– M
12
+ M
13
= 6 – 4 + 2 = 4

For coil 2, L
2
– M
21
– M
23
= 8 – 4 – 5 = – 1

For coil 3, L
3
+ M
31
– M
32
= 10 + 2 – 5 = 7

L
T
= 4 – 1 + 7 = 10H


+ 2M
12
– 2M
23
2M
31 = 10 + 12 +8 + 2x6 – 2x6 –2x4

= 22H
Chapter 13, Solution 3. L
1
+ L
2
+ 2M = 250 mH (1)

L
1
+ L
2
– 2M = 150 mH (2)

Adding (1) and (2),

= 0.2887 Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from
its dotted terminal. Therefore, the mutually induced voltages have the same sign as the
self-induced voltages. Thus,
L
eq
= L
1
+ L
2
+ 2M L
2

L
1

L
2

L
1

(b) For the parallel coil, consider Figure (b).

I
s
= I
1
+ I
2
and Z
eq
= V
s
/I
s

Applying KVL to each branch gives,

V
s
= jωL
1
I
1
+ jωMI
2
(1)

V



2
1
2
1
s
s
I
I
LjMj
MjLj
V
V

∆ = –ω
2
L
1
L
2
+ ω
2
M
2
, ∆
1
= jωV
s
(L

1
+ L
2
– 2M)V
s
/( –ω
2
(L
1
L
2
– M))

Z
eq
= V
s
/I
s
= jω(L
1
L
2
– M)/[jω(L
1
+ L
2
– 2M)] = jωL
eq


=
−+
−
= mH
36.19x26025
36.1960x25
M2LL
MLL
L
2
21
2
21

24.31 mH
Chapter 13, Solution 6. V
1
= (R
1
+ jωL
1
)I
1
– jωMI

= I(j12 + 10 – j4) = I(10 + j8)

= 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V Chapter 13, Solution 8.

Consider the current as shown below. j 6

j 4

1
Ω

-j3

j2
10
+
–
I
2
I
1
4
Ω


+(4 + j)I
2
(2)

In matrix form,













+
+
=






2
1
I

Ω

-j1

2
Ω
For loop 1,
-j2V
+
–
j 4

j 4

8∠30
o
+
–
I
2
I

= 2I
2
= 2.074∠21.12°
Chapter 13, Solution 10. Consider the circuit below.

I
o
j
ω
L j
ω
L
1/j
ω
C
I
2
I
1
j
ω
M
I
in
∠0
o


= 0
I
o
=
j I
in
(
ω
L – 1/(
ω
C)) /(R + j
ω
L + 1/(j
ω
C))

Chapter 13, Solution 11. Consider the circuit below.

I
1
j
ω
L
1
R
1
R


For mesh 1, V
1
=
I
1
(R
1
+ 1/(j
ω
C)) – I
2
(1/j
ω
C)) –R
1
I
3

For mesh 2,
0 =
–I
1
(1/(j

1
I
1
– jω(L
1
– M)I
2
+ (R
1
+ R
2
+ jωL
1
)I
3
or V
2
=
R
1
I
1
+ j
ω
(L
1
– M)I

•
Applying KVL to the loops,

21
481 IjIj +=
(1)
21
1840 IjIj +=
(2)
Solving (1) and (2) gives I
1
= -j0.1406. Thus

H 111.7
11
11
==→==
jI
LjL
I
Z
eqeqWe can also use the equivalent T-section for the transform to find the equivalent
inductance.


j8
Ω
j10
Ω
2
Ω j10
Ω
-j6
ΩZ
j4
Ω


Ω
a
b
j10 V
+
–
j6
Ω
I
j2 Note that the two coils are connected series aiding.

ωL = ωL
1
+ ωL
2

1 A

5
Ω

j6
Ω
j8
Ω
-j3
Ω
a
j2
2
Ω
+
V
o

–


= I
1
– 1 (2)

Substituting (2) into (1), (5 + j4)I
1
+(2 + j3)(1 + I
1
) = 0

I
1
= –(2 + j3)/(7 + j7)

Now, ((5 + j6)I
1
– j2I
1
+ V
o
= 0

V
o
= –(5 + j4)I
1
= (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50°

Z
Th


+
–
20
Ω

j10
Ω

j20
Ω
a
j5
I
N
60∠30
o
+
–
I
2

For mesh 2,
0 = (j20 + j10)I
2
– j5I
1
– j10I
1 or I
1
= 2I
2
(2)

Substituting (2) into (1), 12
∠
30
°
= (8 + j3)I
2I
N
= I
2
= 12
∠
30

– j10I
1
= (4 + j2)I
2
– jI
1 or I
2
= jI
1
/(4 + j2) (4)

Substituting (4) into (3), 1 = j20I
1
+ j(j5)I
1
/(4 + j2) = (–1 + j20.5)I
1I
1
= 1/(–1 + j20.5)

Z
N
= 1/I
1

I
N80 V
I
o
0
∠
1

-
b

80)28(0)428(80
2121
=−+→=−+−+−
jIIjjIIjj
(1)
2112
606
IIjIIj
=→=−
(2)

Solving (1) and (2) leads to
A 91.126246.1362.0584.1
1148
80
2

j4
Ω
j6
Ω

I
2
2V

I
1

-
b

28
)28(0
2
121
j
jI
IjIIj
+
=→−+= (3)

0)62(2
12
=−++ jIIj
(4)
Solving (3) and (4) leads to

144
20jZ
o
+=
++−
+=

Ω−=
+−
−
= 7.9j1989.0
Z6j
xZ6j
Z
o
o

Chapter 13, Solution 18. Let
5105.0,20,5.1
2121
====== xLLkMLL
ω

Ω+=
+
+
+=++= 12.29215.2
74
)4(6
27)6//()4(27
j
j
jj
jjjjZ
ThWe find
V
Th
by looking at the circuit below. -j4 j10 j25 j2
+
-j5
+

V

a
ω25LH, 552530
C2
−=−==+=+=
MMLL
bThus, the T-section is as shown below.

j65 Ω j55 Ω
-j25
Ω
Chapter 13, Solution 20.
Transform the current source to a voltage source as shown below.


k = M/
21
LL
or M = k
21
LLωM = k
21
LL
ωω
= 0.5(10) = 5

For mesh 1, j12 = (4 + j10 – j5)I
1
+ j5I
2
+ j5I
2
= (4 + j5)I
1
+ j10I


++
++
=






2
1
I
I
5j810j
10j5j4
20
12j

∆ = 107 + j60, ∆
1
= –60 –j296, ∆
2
= 40 – j100

I
1
= ∆
1
/∆ =
i
1
= 2.462 cos(1000t + 72.18°) A

i
2
= 0.878 cos(1000t – 97.48°) A

At t = 2 ms, 1000t = 2 rad = 114.6°

i
1
= 0.9736cos(114.6° + 143.09°) = –2.445

i
2
= 2.53cos(114.6° + 153.61°) = –0.8391

The total energy stored in the coupled coils is

w = 0.5L
1
i
1
2
+ 0.5L
2
i

For mesh 1, 36∠30° = (7 + j6)I
1
– (2 + j)I
2
(1)For mesh 2, 0 = (6 + j3 – j4)I
2
– 2I
1
jI
1
= –(2 + j)I
1
+ (6 – j)I
2
(2)

Placing (1) and (2) into matrix form,








2
= (2 + j)36∠30° = 80.5∠56.56°, I
1
= ∆
1
/∆ = 3.69∠–15.56°, I
2
= ∆
2
/∆ = 1.355∠20.46°

Power absorbed fy the 4-ohm resistor, = 0.5(I
2
)
2
4 = 2(1.355)
2
= 3.672 watts
Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the
coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure
13.85 then becomes,
I
x
I

I
o
I
2
I
1
− +
−
+

+
−

100 Ω

Now all we need to do is to write the mesh equations and to solve for I
o
.

Loop # 1,

-50 + j20(I
3
– I
2
) j 40(I
1
– I
3
) + j10(I
2
– I
1
) – j30(I
3
– I
2
) + j80(I
1
– I
2
) – j10(I
1

) + j30(I
3
–I
2
) – j30(I
2
– I
1
) + j60(I
2
– I
3
) – j20(I
1
– I
3
) + 100I
2
= 0

-j60I
1
+ (100 + j80)I
2
– j20I
3
= 0 (2)
Loop # 3,

-j50I

2
+ j10I
3
= 0

Multiplying by (1/j10) yields, -4I
1
– 2I
2
+ I
3
= 0 (3)

Multiplying (2) by (1/j20) yields -3I
1
+ (4 – j5)I
2
– I
3
= 0 (4)

Multiplying (3) by (1/4) yields -I
1
– 0.5I
2
– 0.25I
3
= 0 (5)

Multiplying (4) by (-1/3) yields I

3
= (j10 – 22I
2
)/3, substituting (1) into this equation produces,

I
3
= j3.333 + (-1.2273 – j1.1623)I
3or I
3
= I
o
= 1.3040∠63
o
amp.

Chapter 13, Solution 23.

ω = 10

0.5 H converts to jωL
1
= j5 ohms

1 H converts to jωL
2
= j10 ohms

For mesh 1, 12 = (j5 – j4)I
1
+ j2I
2
– (–j4)I
2 –j2 = I
1
+ 6I
2
(1)

For mesh 2, 0 = (5 + j10)I
2
+ j2I
1
–(–j4)I
1 0 = (5 + j10)I
2
+ j6I
1
(2)

1
= 5.068cos(61.13°) = 2.446

i
2
= 2.719cos(–92.3°) = –0.1089

w = 0.5(5)(2.446)
2
+ 0.5(1)(–0.1089)
2
– (0.2)(2.446)(–0.1089) = 15.02 J

Chapter 13, Solution 24. (a) k = M/
21
LL
= 1/
2x4
= 0.3535

(b) ω = 4

1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j



12 = (2 + j16)I
1
+ j4I
2 or 6 = (1 + j8)I
1
+ j2I
2
(1)

0 = (j8 + 0.5 – j0.5)I
2
+ j4I
1
or I
1
= (0.5 + j7.5)I
2
/(–j4) (2)

Substituting (2) into (1),

24 = (–11.5 – j51.5)I
2
or I
2
= –24/(11.5 + j51.5) = –0.455∠–77.41°

2
= –0.455cos(98.37° – 77.41°) = –0.4249

w = 0.5L
1
i
1
2
+ 0.5L
2
i
2
2
+ Mi
1
i
2= 0.5(4)(0.8169)
2
+ 0.5(2)(–.4249)
2
+ (1)(0.1869)(–0.4249) = 1.168 J
Chapter 13, Solution 25. j4
12∠0° Applying the concept of reflected impedance,

Z
ab
= (2 – j)||(1 + j2 + (1)
2
/(j2 + 3 + j4))

= (2 – j)||(1 + j2 + (3/45) – j6/45)

= (2 – j)||(1 + j2 + (3/45) – j6/45)

= (2 – j)||(1.0667 + j1.8667)

=(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085∠17.9° ohms

I
o

I
2
I
1
50 Ω
j40j20

+
− 10 Ω
200∠60° For mesh 1,
200∠60° = (50 – j30 + j20)I
1
+ j17I
2
= (50 – j10)I
1
+ j17I
2



°∠
2
1
I
I
40j1017j
17j10j50
0
60200

∆ = 900 + j100, ∆
1
= 2000∠60°(1 + j4) = 8246.2∠136°, ∆
2
= 3400∠–30°

I
2
= ∆
2
/∆ = 3.755∠–36.34°

I
o
= I
2
= 3.755∠–36.34° A


by replacing the 20-ohm load with a unit source as shown below.

j10
Ω

8
-jX
Ω
••

+
j12
Ω
j15
Ω
I
2
1V
-
I
1

For mesh 1,
0

2
−−
−+
=
−
=6247275.10
8.0)1.02.1(
)5.18(12
20||
2
22
22
−+=→
+−
−+
== XX
X
X
Z
ThSolving the quadratic equation yields
X = 6.425

1
|
2
(10) = 320 leads to |I
1
|
2
= 64 or |I
1
| = 8

8 = |165(20 + j50)/(X
2
+ (10 + j30)(20 + j50))|

= |165(20 + j50)/(X
2
– 1300 + j1100)|

or 64 = 27225(400 + 2500)/((X
2
– 1300)
2
+ 1,210,000)

(X
2
– 1300)
2
+ 1,210,000 = 1,233,633


20
Ω

165∠0°

165 = (10 + j30)I
1
– j38.127I
2
(1)

0 = (20 + j50)I
2
– j38.127I
1
(2)
In matrix form,








I
1
= ∆
1
/∆ = 8∠–13.81°, I
2
= ∆
2
/∆ = 5.664∠7.97°

i
1
= 8cos(1000t – 13.83°), i
2
= 5.664cos(1000t + 7.97°)

At t = 1.5 ms, 1000t = 1.5 rad = 85.94°

i
1
= 8cos(85.94° – 13.83°) = 2.457

i
2
= 5.664cos(85.94° + 7.97°) = –0.3862

w = 0.5L
1
i

= 25 + j70 + 100/(8 + j14) =
(28.08 + j64.62) ohms(b) jωL
a
= j30 – j10 = j20, jωL
b
= j20 – j10 = j10, jωL
c
= j10

Thus the Thevenin Equivalent of the linear transformer is shown below. j40 j20 j10
25
Ω
8
Ω

Z
in

j10 –j6
L
2
– M
2
= 300 – 25 = 275

L
A
= (L
1
L
2
– M
2
)/(L
1
– M) = 275/15 = 18.33 H

L
B
= (L
1
L
2
– M
2
)/(L
1
– M) = 275/10 = 27.5 H

Z
in
’ = jωL
b
+ ω
2
M
b
2
/(R + jωL
b
) = (jωL
b
R - ω
2
L
b
2
+ ω
2
M
b
2
)/(R + jωL
b
) (1)

For the first stage, we have the circuit below.

)

= (–ω
2
L
a
2
+ ω
2
M
a
2
+ jωL
a
Z
in
)/( jωL
a
+ Z
in
) (2)

Substituting (1) into (2) gives,

=
b
2

ω+ω+ω−–Rω
2
L
a
2
+ ω
2
M
a
2
R – jω
3
L
b
L
a
+ jω
3
L
b
M
a
2
+ jωL
a
(jωL
b

ω
2
R(L
a
2
+ L
a
L
b
– M
a
2
) + jω
3
(L
a
2
L
b
+ L
a
L
b
2
– L
a
M


Chapter 13, Solution 33. Z
in
= 10 + j12 + (15)
2
/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35)

= 10 + j12 + 225(20 – j35)/(400 + 1225)

= (12.769 + j7.154) ohms
Chapter 13, Solution 34.

Insert a 1-V voltage source at the input as shown below.

j6
Ω

1
Ω
8
Ω