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812 ⏐⏐⏐ NETWORK THEOREMS (ac)
Z1 Solution:
Steps 1 and 2 (Fig. 18.62):
+
E
Z2 Z1 � R1 � j XL � 3 ��j 4 ��5 � �53.13°
Z2 ��j XC ��j 5 �
–
Step 3 (Fig. 18.63):
Norton
Z1Z2 (5 � �53.13°)(5 � ��90°) 25 � ��36.87°
ZN ������ ���
Z1 � Z2 3 ��j 4 ��j 5 �
3 � j 1
FIG. 18.62
Assigning the subscripted impedances to the
25 � ��36.87°
��� �7.91 � ��18.44° � 7.50 ��j 2.50 �
3.16 ��18.43°
network of Fig. 18.61.
Step 4 (Fig. 18.64):
E 20 V �0°
IN � I1 ����� �4 A ��53.13°
Z1 5 � �53.13°
Z 1
Z 2 ZN
FIG. 18.63
Determining the Norton impedance for the
network of Fig. 18.61.
7.50 � – j2.50 �
+
E
–
Z 1
FIG. 18.64
Determining I N for the network of Fig. 18.61.
Step 5: The Norton equivalent circuit is shown in Fig. 18.65.
IN = 4 A ∠ – 53.13°
ZN R 6 � IN = 4 A ∠ – 53.13°
RL 6 �
XC 2.50 �
FIG. 18.65
The Norton equivalent circuit for the network of Fig. 18.61.
I 1
Z 2
I N
I N
R 7.50 �
EXAMPLE 18.15 Find the Norton equivalent circuit for the network
external to the 7-� capacitive reactance in Fig. 18.66.
I = 3 A ∠ 0°
R 1
X C1
R 2
1 �
2 �
4 �
FIG. 18.66
Example 18.15.
X L
5 �
X C2 = 7 �
Th