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S
S P P
E
72 V 72 V
I5 ��� ��� ���3 mA
R(1,2,3)�4 � R5 12 k��12 k� 24 k�
with
R7�(8,9)E (4.5 k�)(72 V) 324 V
V7 ��� ��� ���19.6 V
R7�(8,9) � R6 4.5 k��12 k� 16.5
V7 19.6 V
I6 �����4.35 mA
R7�(8,9) 4.5 k�
and Is � I5 � I6 � 3 mA � 4.35 mA � 7.35 mA
Since the potential difference between points a and b of Fig. 7.22 is
fixed at E volts, the circuit to the right or left is unaffected if the network
is reconstructed as shown in Fig. 7.24.
R 2
8 k�
R 1
4 k�
R 3
12 k�
R 4
24 k�
R 5
12 k�
E
I 5
I 5
72 V
E
I 6
R 6
12 k�
FIG. 7.24
An alternative approach to Example 7.9.
We can find each quantity required, except I s, by analyzing each circuit
independently. To find I s, we must find the source current for each
circuit and add it as in the above solution; that is, I s � I 5 � I 6.
EXAMPLE 7.10 This example demonstrates the power of Kirchhoff’s
voltage law by determining the voltages V1, V2, and V3 for the network
of Fig. 7.25. For path 1 of Fig. 7.26,
E1 � V1 � E3 � 0
and V1 � E1 � E3 � 20 V � 8 V � 12 V
For path 2,
E2 � V1 � V2 � 0
and V2 � E2 � V1 � 5 V � 12 V � �7 V
indicating that V2 has a magnitude of 7 V but a polarity opposite to that
appearing in Fig. 7.25. For path 3,
V3 � V2 � E3 � 0
and V3 � E3 � V2 � 8 V � (�7 V) � 8 V � 7 V � 15 V
Note that the polarity of V2 was maintained as originally assumed,
requiring that �7 V be substituted for V2. 7.3 LADDER NETWORKS
A three-section ladder network appears in Fig. 7.27. The reason for
the terminology is quite obvious for the repetitive structure. Basically
two approaches are used to solve networks of this type.
9 k� +
72 V
R V
7
7 3 k�
–
I6 R9 +
E 1
–
+
E 1
–
LADDER NETWORKS ⏐⏐⏐ 223
6 k�
20 V
20 V
R 8
E 2 +
–
+
V3 5 V
V2 –
R1 +
R2 R3 E3 +
–
FIG. 7.25
Example 7.10.
1
E 2 +
–
+
V3 –
+
+
V 1
5 V V2 –
R1 +
R3 R2 E3 +
–
2
3
–
V 1
–
8 V
–
8 V
FIG. 7.26
Defining the paths for Kirchhoff’s voltage law.