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266 ⏐⏐⏐ METHODS OF ANALYSIS AND SELECTED TOPICS (dc)
�6 2�
�4 6� 36 � 8 28
I2 � ––––––– � –––––– � –– � 2 A
14 14 14
I3 � I1 � I2 ��1 � 2 � 1 A
It is now important that the impact of the results obtained be understood.
The currents I1, I2, and I3 are the actual currents in the branches
in which they were defined. A negative sign in the solution simply
reveals that the actual current has the opposite direction than initially
defined—the magnitude is correct. Once the actual current directions
and their magnitudes are inserted in the original network, the various
voltages and power levels can be determined. For this example, the
actual current directions and their magnitudes have been entered on the
original network in Fig. 8.24. Note that the current through the series
elements R1 and E1 is 1 A; the current through R3, 1 A; and the current
through the series elements R2 and E2, 2 A. Due to the minus sign in the
solution, the direction of I1 is opposite to that shown in Fig. 8.21. The
voltage across any resistor can now be found using Ohm’s law, and the
power delivered by either source or to any one of the three resistors can
be found using the appropriate power equation.
R 1
E 1
+
–
+
–
I1 = 1 A
2 �
2 V
I3 = 1 A
R3 4 �
–
+
6 V
–
Applying Kirchhoff’s voltage law around the loop indicated in Fig.
8.24,
� V ��(4 �)I3 � (1 �)I2 � 6 V � 0
or (4 �)I3 � (1 �)I2 � 6 V
and (4 �)(1 A) � (1 �)(2 A) � 6 V
4 V � 2 V � 6 V
6 V � 6 V (checks)
EXAMPLE 8.10 Apply branch-current analysis to the network of Fig.
8.25.
Solution: Again, the current directions were chosen to match the
“pressure” of each battery. The polarities are then added and Kirchhoff’s
voltage law is applied around each closed loop in the clockwise
direction. The result is as follows:
loop 1: �15 V � (4 �)I1 � (10 �)I3 � 20 V � 0
loop 2: �20 V � (10 �)I3 � (5 �)I2 � 40 V � 0
+
I2 = 2 A
R2 FIG. 8.24
Reviewing the results of the analysis of the network of Fig. 8.21.
E 2
–
+ 1
�
N A