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220 ⏐⏐⏐ SERIES-PARALLEL NETWORKS
+
E
–
R 2
I s
12 V
+
V1 –
R 1
R 2
I 1
5 �
a
+ Vab –
3 �
V3 R3 R 4
FIG. 7.17
Network of Fig. 7.16 redrawn.
5 �
7 �
R 3
E 2
R1 a
R 4
+ 18 V
+
6 �
I
6 �
FIG. 7.18
Example 7.7.
V1 –
+
V2 –
+
–
b
I 3
6 �
2 �
E 1 = – 6 V
E 1
–
6 V
+
6 �
+ V – 3
a. Note the similarities with Fig. 7.12, permitting the use of the voltage
divider rule to determine V1 and V3: R1E (5 �)(12 V) 60 V
V1 ��� ��� ���7.5 V
R1 � R2 5 ��3 � 8
R3E (6 �)(12 V) 72 V
V3 ��� ��� ���9 V
R3 � R4 6 ��2 � 8
The open-circuit voltage Vab is determined by applying Kirchhoff’s
voltage law around the indicated loop of Fig. 7.17 in the clockwise
direction starting at terminal a.
�V1 � V3 � Vab � 0
and Vab � V3 � V1 � 9 V � 7.5 V � 1.5 V
b. By Ohm’s law,
V1 7.5 V
I1 �����1.5 A
R1 5 �
V3 9 V
I3 �����1.5 A
R3 6 �
Applying Kirchhoff’s current law,
Is � I1 � I3 � 1.5 A � 1.5 A � 3 A
R 3
R 1
5 �
+ V – 1
b
a
S
S P P
EXAMPLE 7.7 For the network of Fig. 7.18, determine the voltages V1 and V2 and the current I.
Solution: It would indeed be difficult to analyze the network in the
form of Fig. 7.18 with the symbolic notation for the sources and the reference
or ground connection in the upper left-hand corner of the diagram.
However, when the network is redrawn as shown in Fig. 7.19, the
unknowns and the relationship between branches become significantly
clearer. Note the common connection of the grounds and the replacing
of the terminal notation by actual supplies.
It is now obvious that
V2 ��E1 � �6 V
The minus sign simply indicates that the chosen polarity for V2 in Fig.
7.18 is opposite to that of the actual voltage. Applying Kirchhoff’s voltage
law to the loop indicated, we obtain
�E1 � V1 � E2 � 0
–
V ab
+
FIG. 7.16
Example 7.6.
R 4
2 �
R 2
3 �
18 V
I s
–
+
E 2