Electrical Engineering (EE) Exam > Electrical Engineering (EE) Notes > Network Theory (Electric Circuits) > Detailed Notes: Network Theorems in A.C. circuits - 2
Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
|
68 videos|85 docs|62 tests
|
About this Document
|
Dec 22, 2024 Last updated |
Document Description: Detailed Notes: Network Theorems in A.C. circuits - 2 for Electrical Engineering (EE) 2024 is part of Network Theory (Electric Circuits) preparation.
The notes and questions for Detailed Notes: Network Theorems in A.C. circuits - 2 have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Detailed Notes: Network Theorems in A.C. circuits - 2 covers topics
like and Detailed Notes: Network Theorems in A.C. circuits - 2 Example, for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Detailed Notes: Network Theorems in A.C. circuits - 2.
Introduction of Detailed Notes: Network Theorems in A.C. circuits - 2 in English is available as part of our Network Theory (Electric Circuits)
for Electrical Engineering (EE) & Detailed Notes: Network Theorems in A.C. circuits - 2 in Hindi for Network Theory (Electric Circuits) course.
Download more important topics related with notes, lectures and mock test series for Electrical Engineering (EE)
Exam by signing up for free. Electrical Engineering (EE): Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Description
Full syllabus notes, lecture & questions for Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE) - Electrical Engineering (EE) | Plus excerises question with solution to help you revise complete syllabus for Network Theory (Electric Circuits) | Best notes, free PDF download
Information about Detailed Notes: Network Theorems in A.C. circuits - 2
In this doc you can find the meaning of Detailed Notes: Network Theorems in A.C. circuits - 2 defined & explained in the simplest way possible. Besides explaining types of
Detailed Notes: Network Theorems in A.C. circuits - 2 theory, EduRev gives you an ample number of questions to practice Detailed Notes: Network Theorems in A.C. circuits - 2 tests, examples and also practice Electrical Engineering (EE)
tests
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Exam
,Viva Questions
,video lectures
,Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
,Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
,Summary
,ppt
,Sample Paper
,mock tests for examination
,past year papers
,MCQs
,Semester Notes
,shortcuts and tricks
,study material
,Previous Year Questions with Solutions
,Detailed Notes: Network Theorems in A.C. circuits - 2 | Network Theory (Electric Circuits) - Electrical Engineering (EE)
,Free
,Extra Questions
,practice quizzes
,Important questions
,Objective type Questions
;
Additional Information about Detailed Notes: Network Theorems in A.C. circuits - 2 for Electrical Engineering (EE) Preparation
Detailed Notes: Network Theorems in A.C. circuits - 2 Free PDF Download
The Detailed Notes: Network Theorems in A.C. circuits - 2 is an invaluable resource that delves deep into the core of the Electrical Engineering (EE) exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Detailed Notes: Network Theorems in A.C. circuits - 2 now and kickstart your journey towards success in the Electrical Engineering (EE) exam.
Importance of Detailed Notes: Network Theorems in A.C. circuits - 2
The importance of Detailed Notes: Network Theorems in A.C. circuits - 2 cannot be overstated, especially for Electrical Engineering (EE) aspirants.
This document holds the key to success in the Electrical Engineering (EE) exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Detailed Notes: Network Theorems in A.C. circuits - 2
Detailed Notes: Network Theorems in A.C. circuits - 2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Detailed Notes: Network Theorems in A.C. circuits - 2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Detailed Notes: Network Theorems in A.C. circuits - 2 Notes on EduRev are your ultimate resource for success.
Detailed Notes: Network Theorems in A.C. circuits - 2 Electrical Engineering (EE) Questions
The "Detailed Notes: Network Theorems in A.C. circuits - 2 Electrical Engineering (EE) Questions" guide is a valuable resource for all aspiring students preparing for the
Electrical Engineering (EE) exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Detailed Notes: Network Theorems in A.C. circuits - 2 on the App
Students of Electrical Engineering (EE) can study Detailed Notes: Network Theorems in A.C. circuits - 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Detailed Notes: Network Theorems in A.C. circuits - 2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Detailed Notes: Network Theorems in A.C. circuits - 2 is prepared as per the latest Electrical Engineering (EE) syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup