boylistad

Th
The total current through the 4-� resistor (Fig. 9.9) is
I3 � I″ 3 � I′ 3 � 4 A � 1.5 A � 2.5 A (direction of I″ 3)
EXAMPLE 9.3
a. Using superposition, find the current through the 6-� resistor of the
network of Fig. 9.10.
+
E 36 V
–
R 1
12 �
R 2
I 2
6 � I
FIG. 9.10
Example 9.3.
b. Demonstrate that superposition is not applicable to power levels.
Solutions:
a. Considering the effect of the 36-V source (Fig. 9.11):
E E 36 V
I′ 2 ������� �2 A
RT R1 � R2 12 ��6 �
Considering the effect of the 9-A source (Fig. 9.12):
Applying the current divider rule,
R1I (12 �)(9 A) 108 A
I″ 2 � � ��� �� � 6 A
R1 � R2 12 ��6 � 18
The total current through the 6-� resistor (Fig. 9.13) is
I2 � I′ 2 � I″ 2 � 2 A � 6 A � 8 A
R2 6 � I′ 2 = 2 A I″ 2 = 6 A R2 6 � I2 = 8 A
Same direction
FIG. 9.13
The resultant current for I 2.
b. The power to the 6-� resistor is
Power � I 2 R � (8 A) 2 (6 �) � 384 W
The calculated power to the 6-� resistor due to each source, misusing
the principle of superposition, is
P1 � (I′ 2) 2 R � (2 A) 2 (6 �) � 24 W
P2 � (I″ 2) 2 R � (6 A) 2 (6 �) � 216 W
P1 � P2 � 240 W � 384 W
9 A
SUPERPOSITION THEOREM ⏐⏐⏐ 325
E
I' 3 = 1.5 A
4 �
I" 3 = 4 A
FIG. 9.9
The resultant current for I 3.
+
36 V
–
R 1
12 �
I� 2
Current source replaced
by open circuit
R 2
6 �
FIG. 9.11
The contribution of E to I 2.
R 1
12 �
I″ 2
R 2
6 �
FIG. 9.12
The contribution of I to I 2.
I = 9 A