Electrical Engineering (EE) Exam > Electrical Engineering (EE) Notes > Network Theory (Electric Circuits) > Detailed Notes: Network Theorems in A.C. circuits - 2
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Page 1
EXAMPLE 18.17 Using each method described for dependent sources,
find the Norton equivalent circuit for the network in Fig. 18.77.
Solution:
I
N
For each method, I
N
is determined in the same manner. From Fig.
18.78 using Kirchhoff’s current law, we have
0 I hI I
sc
or I
sc
(1 h)I
Applying Kirchhoff’s voltage law gives us
E IR
1
I
sc
R
2
0
and IR
1
I
sc
R
2
E
or
so
or
Z
N
Method 1: E
oc
is determined from the network in Fig. 18.79. By Kirch-
hoff’s current law,
0 I hI or 1(h 1) 0
For h, a positive constant I must equal zero to satisfy the above. Therefore,
I 0 and hI 0
and E
oc
E
with
Method 2: Note Fig. 18.80. By Kirchhoff’s current law,
I
g
I hI (I h)I
By Kirchhoff’s voltage law,
E
g
I
g
R
2
IR
1
0
or
Substituting, we have
and I
g
R
1
(1 h)E
g
(1 h)I
g
R
2
I
g
11 h2I 11 h2a
E
g
I
g
R
2
R
1
b
I
E
g
I
g
R
2
R
1
Z
N
E
oc
I
sc
E
11 h2E
R
1
11 h2R
2
R
1
11 h2R
2
11 h2
I
sc
11 h2E
R
1
11 h2R
2
I
N
I
sc
3R
1
11 h2R
2
4 11 h2E
R
1
I
sc
11 h2I
sc
R
2
11 h2E
I
sc
11 h2I 11 h2a
I
sc
R
2
E
R
1
b
I
I
sc
R
2
E
R
1
R
2
+
hI E
–
Norton
R
1
I
FIG. 18.77
Example 18.17.
R
2
+
hI E
–
I
sc
R
1
I + – V
R
2
I
sc
FIG. 18.78
Determining I
sc
for the network in Fig. 18.77.
+
hI E
–
E
oc
R
1
I
+
V = 0
–
+
–
FIG. 18.79
Determining E
oc
for the network in Fig. 18.77.
+
hI E
g
–
R
1
I
+ –
Z
N
I
g
R
2
+ – V
R
1
V
R
2
FIG. 18.80
Determining the Norton impedance using the
approach Z
N
E
g
/ E
g
.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 811
Page 2
EXAMPLE 18.17 Using each method described for dependent sources,
find the Norton equivalent circuit for the network in Fig. 18.77.
Solution:
I
N
For each method, I
N
is determined in the same manner. From Fig.
18.78 using Kirchhoff’s current law, we have
0 I hI I
sc
or I
sc
(1 h)I
Applying Kirchhoff’s voltage law gives us
E IR
1
I
sc
R
2
0
and IR
1
I
sc
R
2
E
or
so
or
Z
N
Method 1: E
oc
is determined from the network in Fig. 18.79. By Kirch-
hoff’s current law,
0 I hI or 1(h 1) 0
For h, a positive constant I must equal zero to satisfy the above. Therefore,
I 0 and hI 0
and E
oc
E
with
Method 2: Note Fig. 18.80. By Kirchhoff’s current law,
I
g
I hI (I h)I
By Kirchhoff’s voltage law,
E
g
I
g
R
2
IR
1
0
or
Substituting, we have
and I
g
R
1
(1 h)E
g
(1 h)I
g
R
2
I
g
11 h2I 11 h2a
E
g
I
g
R
2
R
1
b
I
E
g
I
g
R
2
R
1
Z
N
E
oc
I
sc
E
11 h2E
R
1
11 h2R
2
R
1
11 h2R
2
11 h2
I
sc
11 h2E
R
1
11 h2R
2
I
N
I
sc
3R
1
11 h2R
2
4 11 h2E
R
1
I
sc
11 h2I
sc
R
2
11 h2E
I
sc
11 h2I 11 h2a
I
sc
R
2
E
R
1
b
I
I
sc
R
2
E
R
1
R
2
+
hI E
–
Norton
R
1
I
FIG. 18.77
Example 18.17.
R
2
+
hI E
–
I
sc
R
1
I + – V
R
2
I
sc
FIG. 18.78
Determining I
sc
for the network in Fig. 18.77.
+
hI E
–
E
oc
R
1
I
+
V = 0
–
+
–
FIG. 18.79
Determining E
oc
for the network in Fig. 18.77.
+
hI E
g
–
R
1
I
+ –
Z
N
I
g
R
2
+ – V
R
1
V
R
2
FIG. 18.80
Determining the Norton impedance using the
approach Z
N
E
g
/ E
g
.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 811
so E
g
(1 h) I
g
[R
1
(1 h)R
2
]
or
which agrees with the above.
EXAMPLE 18.18 Find the Norton equivalent circuit for the network
configuration in Fig. 18.57.
Solution: By source conversion,
and (18.12)
which is I
sc
as determined in Example 18.13, and
(18.13)
For k
1
0, we have
k
1
0 (18.14)
k
1
0 (18.15)
18.5 MAXIMUM POWER TRANSFER THEOREM
When applied to ac circuits, the maximum power transfer theorem
states that
maximum power will be delivered to a load when the load impedance
is the conjugate of the Thévenin impedance across its terminals.
That is, for Fig. 18.81, for maximum power transfer to the load,
Z
N
R
2
I
N
k
2
V
i
R
1
Z
N
Z
Th
R
2
1
k
1
k
2
R
2
R
1
I
N
k
2
V
i
R
1
I
N
E
Th
Z
Th
k
2
R
2
V
i
R
1
k
1
k
2
R
2
R
1
R
2
R
1
k
1
k
2
R
2
Z
N
E
g
I
g
R
1
11 h2R
2
1 h
E
Th
= E
Th
? v
Th
s
Z
Th
Z
L
Z
Th
? v
Th
z
= Z
L
? v
L
FIG. 18.81
Defining the conditions for maximum power transfer to a load.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 812
Page 3
EXAMPLE 18.17 Using each method described for dependent sources,
find the Norton equivalent circuit for the network in Fig. 18.77.
Solution:
I
N
For each method, I
N
is determined in the same manner. From Fig.
18.78 using Kirchhoff’s current law, we have
0 I hI I
sc
or I
sc
(1 h)I
Applying Kirchhoff’s voltage law gives us
E IR
1
I
sc
R
2
0
and IR
1
I
sc
R
2
E
or
so
or
Z
N
Method 1: E
oc
is determined from the network in Fig. 18.79. By Kirch-
hoff’s current law,
0 I hI or 1(h 1) 0
For h, a positive constant I must equal zero to satisfy the above. Therefore,
I 0 and hI 0
and E
oc
E
with
Method 2: Note Fig. 18.80. By Kirchhoff’s current law,
I
g
I hI (I h)I
By Kirchhoff’s voltage law,
E
g
I
g
R
2
IR
1
0
or
Substituting, we have
and I
g
R
1
(1 h)E
g
(1 h)I
g
R
2
I
g
11 h2I 11 h2a
E
g
I
g
R
2
R
1
b
I
E
g
I
g
R
2
R
1
Z
N
E
oc
I
sc
E
11 h2E
R
1
11 h2R
2
R
1
11 h2R
2
11 h2
I
sc
11 h2E
R
1
11 h2R
2
I
N
I
sc
3R
1
11 h2R
2
4 11 h2E
R
1
I
sc
11 h2I
sc
R
2
11 h2E
I
sc
11 h2I 11 h2a
I
sc
R
2
E
R
1
b
I
I
sc
R
2
E
R
1
R
2
+
hI E
–
Norton
R
1
I
FIG. 18.77
Example 18.17.
R
2
+
hI E
–
I
sc
R
1
I + – V
R
2
I
sc
FIG. 18.78
Determining I
sc
for the network in Fig. 18.77.
+
hI E
–
E
oc
R
1
I
+
V = 0
–
+
–
FIG. 18.79
Determining E
oc
for the network in Fig. 18.77.
+
hI E
g
–
R
1
I
+ –
Z
N
I
g
R
2
+ – V
R
1
V
R
2
FIG. 18.80
Determining the Norton impedance using the
approach Z
N
E
g
/ E
g
.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 811
so E
g
(1 h) I
g
[R
1
(1 h)R
2
]
or
which agrees with the above.
EXAMPLE 18.18 Find the Norton equivalent circuit for the network
configuration in Fig. 18.57.
Solution: By source conversion,
and (18.12)
which is I
sc
as determined in Example 18.13, and
(18.13)
For k
1
0, we have
k
1
0 (18.14)
k
1
0 (18.15)
18.5 MAXIMUM POWER TRANSFER THEOREM
When applied to ac circuits, the maximum power transfer theorem
states that
maximum power will be delivered to a load when the load impedance
is the conjugate of the Thévenin impedance across its terminals.
That is, for Fig. 18.81, for maximum power transfer to the load,
Z
N
R
2
I
N
k
2
V
i
R
1
Z
N
Z
Th
R
2
1
k
1
k
2
R
2
R
1
I
N
k
2
V
i
R
1
I
N
E
Th
Z
Th
k
2
R
2
V
i
R
1
k
1
k
2
R
2
R
1
R
2
R
1
k
1
k
2
R
2
Z
N
E
g
I
g
R
1
11 h2R
2
1 h
E
Th
= E
Th
? v
Th
s
Z
Th
Z
L
Z
Th
? v
Th
z
= Z
L
? v
L
FIG. 18.81
Defining the conditions for maximum power transfer to a load.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 812
(18.16)
or, in rectangular form,
(18.17)
The conditions just mentioned will make the total impedance of the cir-
cuit appear purely resistive, as indicated in Fig. 18.82:
Z
T
(R
jX) (R j X)
and (18.18) Z
T
2R
R
L
R
Th
and
j X
load
j X
Th
Z
L
Z
Th
andu
L
u
Th
Z
Z
Th
= R
Th
± jX
Th
Z
L
E
Th
=
E
Th
?
v
Th
s
+
–
Z
T
= R
±
jX
I
FIG. 18.82
Conditions for maximum power transfer to Z
L
.
Since the circuit is purely resistive, the power factor of the circuit un-
der maximum power conditions is 1; that is,
(maximum power transfer) (18.19)
The magnitude of the current I in Fig. 18.82 is
The maximum power to the load is
and (18.20)
EXAMPLE 18.19 Find the load impedance in Fig. 18.83 for maximum
power to the load, and find the maximum power.
Solution: Determine Z
Th
[Fig. 18.84(a)]:
13.33 36.87° 10.66 j 8
Z
Th
Z
1
Z
2
Z
1
Z
2
110 53.13°218 90°2
6 j 8 j 8
80 36.87°
6 0°
Z
2
j X
L
j 8
Z
1
R j X
C
6 j 8 10 53.13°
P
max
E
Th
2
4R
P
max
I
2
R a
E
Th
2R
b
2
R
I
E
Th
Z
T
E
Th
2R
F
p
1
E = 9 V ? 0°
R
6
+
–
X
C
8
X
L
8 Z
L
FIG. 18.83
Example 18.19.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 813
Page 4
EXAMPLE 18.17 Using each method described for dependent sources,
find the Norton equivalent circuit for the network in Fig. 18.77.
Solution:
I
N
For each method, I
N
is determined in the same manner. From Fig.
18.78 using Kirchhoff’s current law, we have
0 I hI I
sc
or I
sc
(1 h)I
Applying Kirchhoff’s voltage law gives us
E IR
1
I
sc
R
2
0
and IR
1
I
sc
R
2
E
or
so
or
Z
N
Method 1: E
oc
is determined from the network in Fig. 18.79. By Kirch-
hoff’s current law,
0 I hI or 1(h 1) 0
For h, a positive constant I must equal zero to satisfy the above. Therefore,
I 0 and hI 0
and E
oc
E
with
Method 2: Note Fig. 18.80. By Kirchhoff’s current law,
I
g
I hI (I h)I
By Kirchhoff’s voltage law,
E
g
I
g
R
2
IR
1
0
or
Substituting, we have
and I
g
R
1
(1 h)E
g
(1 h)I
g
R
2
I
g
11 h2I 11 h2a
E
g
I
g
R
2
R
1
b
I
E
g
I
g
R
2
R
1
Z
N
E
oc
I
sc
E
11 h2E
R
1
11 h2R
2
R
1
11 h2R
2
11 h2
I
sc
11 h2E
R
1
11 h2R
2
I
N
I
sc
3R
1
11 h2R
2
4 11 h2E
R
1
I
sc
11 h2I
sc
R
2
11 h2E
I
sc
11 h2I 11 h2a
I
sc
R
2
E
R
1
b
I
I
sc
R
2
E
R
1
R
2
+
hI E
–
Norton
R
1
I
FIG. 18.77
Example 18.17.
R
2
+
hI E
–
I
sc
R
1
I + – V
R
2
I
sc
FIG. 18.78
Determining I
sc
for the network in Fig. 18.77.
+
hI E
–
E
oc
R
1
I
+
V = 0
–
+
–
FIG. 18.79
Determining E
oc
for the network in Fig. 18.77.
+
hI E
g
–
R
1
I
+ –
Z
N
I
g
R
2
+ – V
R
1
V
R
2
FIG. 18.80
Determining the Norton impedance using the
approach Z
N
E
g
/ E
g
.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 811
so E
g
(1 h) I
g
[R
1
(1 h)R
2
]
or
which agrees with the above.
EXAMPLE 18.18 Find the Norton equivalent circuit for the network
configuration in Fig. 18.57.
Solution: By source conversion,
and (18.12)
which is I
sc
as determined in Example 18.13, and
(18.13)
For k
1
0, we have
k
1
0 (18.14)
k
1
0 (18.15)
18.5 MAXIMUM POWER TRANSFER THEOREM
When applied to ac circuits, the maximum power transfer theorem
states that
maximum power will be delivered to a load when the load impedance
is the conjugate of the Thévenin impedance across its terminals.
That is, for Fig. 18.81, for maximum power transfer to the load,
Z
N
R
2
I
N
k
2
V
i
R
1
Z
N
Z
Th
R
2
1
k
1
k
2
R
2
R
1
I
N
k
2
V
i
R
1
I
N
E
Th
Z
Th
k
2
R
2
V
i
R
1
k
1
k
2
R
2
R
1
R
2
R
1
k
1
k
2
R
2
Z
N
E
g
I
g
R
1
11 h2R
2
1 h
E
Th
= E
Th
? v
Th
s
Z
Th
Z
L
Z
Th
? v
Th
z
= Z
L
? v
L
FIG. 18.81
Defining the conditions for maximum power transfer to a load.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 812
(18.16)
or, in rectangular form,
(18.17)
The conditions just mentioned will make the total impedance of the cir-
cuit appear purely resistive, as indicated in Fig. 18.82:
Z
T
(R
jX) (R j X)
and (18.18) Z
T
2R
R
L
R
Th
and
j X
load
j X
Th
Z
L
Z
Th
andu
L
u
Th
Z
Z
Th
= R
Th
± jX
Th
Z
L
E
Th
=
E
Th
?
v
Th
s
+
–
Z
T
= R
±
jX
I
FIG. 18.82
Conditions for maximum power transfer to Z
L
.
Since the circuit is purely resistive, the power factor of the circuit un-
der maximum power conditions is 1; that is,
(maximum power transfer) (18.19)
The magnitude of the current I in Fig. 18.82 is
The maximum power to the load is
and (18.20)
EXAMPLE 18.19 Find the load impedance in Fig. 18.83 for maximum
power to the load, and find the maximum power.
Solution: Determine Z
Th
[Fig. 18.84(a)]:
13.33 36.87° 10.66 j 8
Z
Th
Z
1
Z
2
Z
1
Z
2
110 53.13°218 90°2
6 j 8 j 8
80 36.87°
6 0°
Z
2
j X
L
j 8
Z
1
R j X
C
6 j 8 10 53.13°
P
max
E
Th
2
4R
P
max
I
2
R a
E
Th
2R
b
2
R
I
E
Th
Z
T
E
Th
2R
F
p
1
E = 9 V ? 0°
R
6
+
–
X
C
8
X
L
8 Z
L
FIG. 18.83
Example 18.19.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 813
(a)
Z
2
Z
1
Z
Th
(b)
E
+
Z
2
+
–
E
Th
Z
1
–
FIG. 18.84
Determining (a) Z
Th
and (b) E
Th
for the network external to the load in Fig. 18.83.
and Z
L
13.3 36.87° 10.66 j 8
To find the maximum power, we must first find E
Th
[Fig. 18.84(b)], as
follows:
(voltage divider rule)
Then
EXAMPLE 18.20 Find the load impedance in Fig. 18.85 for maximum
power to the load, and find the maximum power.
Solution: First we must find Z
Th
(Fig. 18.86).
Z
1
j X
L
j 9 Z
2
R 8
Converting from a to a Y (Fig. 18.87), we have
The redrawn circuit (Fig. 18.88) shows
j 3
3 90°1 j 3 8 2
j 6 8
Z
Th
Z
1
Z
1
1Z
1
Z
2
2
Z
1
1Z
1
Z
2
2
Z
1
Z
1
3
j 3 Z
2
8
P
max
E
Th
2
4R
112 V2
2
4110.66 2
144
42.64
3.38 W
18 90°219 V 0°2
j 8 6 j 8
72 V 90°
6 0°
12 V 90°
E
Th
Z
2
E
Z
2
Z
1 R
8
Z
L
E =
10 V
? 0°
+
–
X
L
9
X
L
9
9
X
L
FIG. 18.85
Example 18.20.
Z
Th
Z
1
Z
2
Z
1
Z
1 1
2
3
FIG. 18.86
Defining the subscripted impedances for the network
in Fig. 18.85.
Z
Th
Z
2
1
2
3
Z
1
Z
1 Z
1
FIG. 18.87
Substituting the Y equivalent for the upper
configuration in Fig. 18.86.
Z
Th
Z
1
Z
1
Z
2
Z
1
FIG. 18.88
Determining Z
Th
for the network in Fig. 18.85.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 814
Page 5
EXAMPLE 18.17 Using each method described for dependent sources,
find the Norton equivalent circuit for the network in Fig. 18.77.
Solution:
I
N
For each method, I
N
is determined in the same manner. From Fig.
18.78 using Kirchhoff’s current law, we have
0 I hI I
sc
or I
sc
(1 h)I
Applying Kirchhoff’s voltage law gives us
E IR
1
I
sc
R
2
0
and IR
1
I
sc
R
2
E
or
so
or
Z
N
Method 1: E
oc
is determined from the network in Fig. 18.79. By Kirch-
hoff’s current law,
0 I hI or 1(h 1) 0
For h, a positive constant I must equal zero to satisfy the above. Therefore,
I 0 and hI 0
and E
oc
E
with
Method 2: Note Fig. 18.80. By Kirchhoff’s current law,
I
g
I hI (I h)I
By Kirchhoff’s voltage law,
E
g
I
g
R
2
IR
1
0
or
Substituting, we have
and I
g
R
1
(1 h)E
g
(1 h)I
g
R
2
I
g
11 h2I 11 h2a
E
g
I
g
R
2
R
1
b
I
E
g
I
g
R
2
R
1
Z
N
E
oc
I
sc
E
11 h2E
R
1
11 h2R
2
R
1
11 h2R
2
11 h2
I
sc
11 h2E
R
1
11 h2R
2
I
N
I
sc
3R
1
11 h2R
2
4 11 h2E
R
1
I
sc
11 h2I
sc
R
2
11 h2E
I
sc
11 h2I 11 h2a
I
sc
R
2
E
R
1
b
I
I
sc
R
2
E
R
1
R
2
+
hI E
–
Norton
R
1
I
FIG. 18.77
Example 18.17.
R
2
+
hI E
–
I
sc
R
1
I + – V
R
2
I
sc
FIG. 18.78
Determining I
sc
for the network in Fig. 18.77.
+
hI E
–
E
oc
R
1
I
+
V = 0
–
+
–
FIG. 18.79
Determining E
oc
for the network in Fig. 18.77.
+
hI E
g
–
R
1
I
+ –
Z
N
I
g
R
2
+ – V
R
1
V
R
2
FIG. 18.80
Determining the Norton impedance using the
approach Z
N
E
g
/ E
g
.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 811
so E
g
(1 h) I
g
[R
1
(1 h)R
2
]
or
which agrees with the above.
EXAMPLE 18.18 Find the Norton equivalent circuit for the network
configuration in Fig. 18.57.
Solution: By source conversion,
and (18.12)
which is I
sc
as determined in Example 18.13, and
(18.13)
For k
1
0, we have
k
1
0 (18.14)
k
1
0 (18.15)
18.5 MAXIMUM POWER TRANSFER THEOREM
When applied to ac circuits, the maximum power transfer theorem
states that
maximum power will be delivered to a load when the load impedance
is the conjugate of the Thévenin impedance across its terminals.
That is, for Fig. 18.81, for maximum power transfer to the load,
Z
N
R
2
I
N
k
2
V
i
R
1
Z
N
Z
Th
R
2
1
k
1
k
2
R
2
R
1
I
N
k
2
V
i
R
1
I
N
E
Th
Z
Th
k
2
R
2
V
i
R
1
k
1
k
2
R
2
R
1
R
2
R
1
k
1
k
2
R
2
Z
N
E
g
I
g
R
1
11 h2R
2
1 h
E
Th
= E
Th
? v
Th
s
Z
Th
Z
L
Z
Th
? v
Th
z
= Z
L
? v
L
FIG. 18.81
Defining the conditions for maximum power transfer to a load.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 812
(18.16)
or, in rectangular form,
(18.17)
The conditions just mentioned will make the total impedance of the cir-
cuit appear purely resistive, as indicated in Fig. 18.82:
Z
T
(R
jX) (R j X)
and (18.18) Z
T
2R
R
L
R
Th
and
j X
load
j X
Th
Z
L
Z
Th
andu
L
u
Th
Z
Z
Th
= R
Th
± jX
Th
Z
L
E
Th
=
E
Th
?
v
Th
s
+
–
Z
T
= R
±
jX
I
FIG. 18.82
Conditions for maximum power transfer to Z
L
.
Since the circuit is purely resistive, the power factor of the circuit un-
der maximum power conditions is 1; that is,
(maximum power transfer) (18.19)
The magnitude of the current I in Fig. 18.82 is
The maximum power to the load is
and (18.20)
EXAMPLE 18.19 Find the load impedance in Fig. 18.83 for maximum
power to the load, and find the maximum power.
Solution: Determine Z
Th
[Fig. 18.84(a)]:
13.33 36.87° 10.66 j 8
Z
Th
Z
1
Z
2
Z
1
Z
2
110 53.13°218 90°2
6 j 8 j 8
80 36.87°
6 0°
Z
2
j X
L
j 8
Z
1
R j X
C
6 j 8 10 53.13°
P
max
E
Th
2
4R
P
max
I
2
R a
E
Th
2R
b
2
R
I
E
Th
Z
T
E
Th
2R
F
p
1
E = 9 V ? 0°
R
6
+
–
X
C
8
X
L
8 Z
L
FIG. 18.83
Example 18.19.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 813
(a)
Z
2
Z
1
Z
Th
(b)
E
+
Z
2
+
–
E
Th
Z
1
–
FIG. 18.84
Determining (a) Z
Th
and (b) E
Th
for the network external to the load in Fig. 18.83.
and Z
L
13.3 36.87° 10.66 j 8
To find the maximum power, we must first find E
Th
[Fig. 18.84(b)], as
follows:
(voltage divider rule)
Then
EXAMPLE 18.20 Find the load impedance in Fig. 18.85 for maximum
power to the load, and find the maximum power.
Solution: First we must find Z
Th
(Fig. 18.86).
Z
1
j X
L
j 9 Z
2
R 8
Converting from a to a Y (Fig. 18.87), we have
The redrawn circuit (Fig. 18.88) shows
j 3
3 90°1 j 3 8 2
j 6 8
Z
Th
Z
1
Z
1
1Z
1
Z
2
2
Z
1
1Z
1
Z
2
2
Z
1
Z
1
3
j 3 Z
2
8
P
max
E
Th
2
4R
112 V2
2
4110.66 2
144
42.64
3.38 W
18 90°219 V 0°2
j 8 6 j 8
72 V 90°
6 0°
12 V 90°
E
Th
Z
2
E
Z
2
Z
1 R
8
Z
L
E =
10 V
? 0°
+
–
X
L
9
X
L
9
9
X
L
FIG. 18.85
Example 18.20.
Z
Th
Z
1
Z
2
Z
1
Z
1 1
2
3
FIG. 18.86
Defining the subscripted impedances for the network
in Fig. 18.85.
Z
Th
Z
2
1
2
3
Z
1
Z
1 Z
1
FIG. 18.87
Substituting the Y equivalent for the upper
configuration in Fig. 18.86.
Z
Th
Z
1
Z
1
Z
2
Z
1
FIG. 18.88
Determining Z
Th
for the network in Fig. 18.85.
boy30444_ch18.qxd 3/24/06 2:58 PM Page 814
and Z
L
0.72 j 5.46
For E
Th
, use the modified circuit in Fig. 18.89 with the voltage source
replaced in its original position. Since I
1
0, E
Th
is the voltage across the
series impedance of and Z
2
. Using the voltage divider rule gives us
and
If the load resistance is adjustable but the magnitude of the load reac-
tance cannot be set equal to the magnitude of the Thévenin reactance,
then the maximum power that can be delivered to the load will occur
when the load reactance is made as close to the Thévenin reactance as
possible and the load resistance is set to the following value:
(18.21)
where each reactance carries a positive sign if inductive and a negative
sign if capacitive.
The power delivered will be determined by
(18.22)
where (18.23)
The derivation of the above equations is given in Appendix G of the
text. The following example demonstrates the use of the above.
EXAMPLE 18.21 For the network in Fig. 18.90:
a. Determine the value of R
L
for maximum power to the load if the load
reactance is fixed at 4 .
b. Find the power delivered to the load under the conditions of part (a).
c. Find the maximum power to the load if the load reactance is made
adjustable to any value, and compare the result to part (b) above.
R
av
R
Th
R
L
2
P E
Th
2
>4R
av
R
L
2R
Th
2
1X
Th
X
load
2
2
25.32 W
P
max
E
Th
2
4R
18.54 V2
2
410.72 2
72.93
2.88
W
E
Th
8.54 V 16.31°
18.54 20.56°2110 V 0°2
10 36.87°
E
Th
1Z
1
Z
2
2E
Z
1
Z
2
Z
1
1j 3 8 2110 V 0°2
8 j 6
Z
1
Z
Th
0.72 j 5.46
j 3 0.72 j 2.46
j 3
25.62 110.56°
10 36.87°
j 3 2.56 73.69°
j 3
13 90°218.54 20.56°2
10 36.87°
E
Th
Z
1
Z
1
Z
2
Z
1
+
–
I
1
= 0
E
+
–
FIG. 18.89
Finding the Thévenin voltage for the network in
Fig. 18.85.
boy30444_ch18.qxd 3/24/06 2:59 PM Page 815
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