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c. V1 � IR1 � (2 A)(4 �) � 8 V
V2 � IR2 � (2 A)(6 �) � 12 V
d. P4� � � � � 16 W
P6� � I 2 R2 � (2 A) 2 (8 V) 64
�
4
(6 �) � (4)(6) � 24 W
e. PE � EI � (20 V)(2 A) � 40 W
PE � P4� � P6� 40 W � 16 W � 24 W
40 W � 40 W (checks)
2
V
�
4
2 1
�
R1
f. � V ��E � V1 � V2 � 0
E � V1 � V2 20 V � 8 V � 12 V
20 V � 20 V (checks)
EXAMPLE 5.8 For the circuit of Fig. 5.18:
a. Determine V2 using Kirchhoff’s voltage law.
b. Determine I.
c. Find R1 and R3. Solutions:
a. Kirchhoff’s voltage law (clockwise direction):
�E � V3 � V2 � V1 � 0
or E � V1 � V2 � V3 and V2 � E � V1 � V3 � 54 V � 18 V � 15 V � 21 V
V2 21 V
b. I � � ��� 3 A
R2 7 �
V1 18 V
c. R1 � � ��� 6 �
I 3 A
V 3
15 V
R3 � � ��� 5 �
I 3 A
5.5 INTERCHANGING SERIES ELEMENTS
The elements of a series circuit can be interchanged without affecting
the total resistance, current, or power to each element. For instance, the
network of Fig. 5.19 can be redrawn as shown in Fig. 5.20 without
affecting I or V 2. The total resistance R T is 35 � in both cases, and I �
70 V/35 � �2 A. The voltage V 2 � IR 2 � (2 A)(5 �) � 10 V for both
configurations.
EXAMPLE 5.9 Determine I and the voltage across the 7-� resistor for
the network of Fig. 5.21.
Solution: The network is redrawn in Fig. 5.22.
RT � (2)(4 �) � 7 ��15 �
E 37.5 V
I �����2.5 A
RT 15 �
V7� � IR � (2.5 A)(7 �) � 17.5 V
INTERCHANGING SERIES ELEMENTS ⏐⏐⏐ 137
–
E
+
+
E
–
+
E –
54 V
I
75 V
75 V
10 �
R 1
V3 = 15 V
– +
R 3
R 1
+ –
V1 = 18 V
FIG. 5.18
Example 5.8.
+ V2 –
5 �
R 2
I
R 2
R 3
–
7 � V 2
+
20 �
FIG. 5.19
Series dc circuit with elements to be interchanged.
+
50 V
–
I
10 � 20 �
R 1
R 3
+
R 2 5 � V 2
–
FIG. 5.20
Circuit of Fig. 5.19 with R 2 and R 3
interchanged.
I
4 �
+ –
12.5 V
4 �
FIG. 5.21
Example 5.9.
+ V
7 �
–