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This final relationship between current levels is incorrect, however,
as can be demonstrated by taking the total current determined by the
superposition theorem and squaring it as follows:
I 2 T � (I1 � I2) 2 � I 2 1 � I 2 2 � 2I1I2 which is certainly different from the expression obtained from the addition
of power levels.
In general, therefore,
the total power delivered to a resistive element must be determined
using the total current through or the total voltage across the element
and cannot be determined by a simple sum of the power levels
established by each source.
EXAMPLE 9.1 Determine I 1 for the network of Fig. 9.4.
Solution: Setting E � 0 V for the network of Fig. 9.4 results in the
network of Fig. 9.5(a), where a short-circuit equivalent has replaced the
30-V source.
I
I
I
3 A
R 1
I′ 1
6 �
+
E 30 V
–
(a) (b)
FIG. 9.5
(a) The contribution of I to I 1; (b) the contribution of E to I 1.
As shown in Fig. 9.5(a), the source current will choose the shortcircuit
path, and I′ 1 � 0 A. If we applied the current divider rule,
RscI (0 �)I
I′ 1 ����� �0 A
Rsc � R1 0 ��6 �
Setting I to zero amperes will result in the network of Fig. 9.5(b), with
the current source replaced by an open circuit. Applying Ohm’s law,
E 30 V
I″ 1 �����5 A
R1 6 �
Since I′ 1 and I″ 1 have the same defined direction in Fig. 9.5(a) and (b),
the current I1 is the sum of the two, and
I1 � I′ 1 � I″ 1 � 0 A � 5 A � 5 A
Note in this case that the current source has no effect on the current
through the 6-� resistor. The voltage across the resistor must be fixed
at 30 V because they are parallel elements.
R 1
I″ 1
6 �
SUPERPOSITION THEOREM ⏐⏐⏐ 323
E
+
30 V I 3 A
–
FIG. 9.4
Example 9.1.
R 1
I 1
6 �