Direct Current and Transient Analysis 123
d. The current source transformation into a voltage source is illustrated by the circuit
diagram of Figure 2.19. Then
I^6060
20
0 A
therefore
Va a’ = 60 V
e. The superposition solution is illustrated by the two circuit diagrams shown in
Figure 2.20.
FIGURE 2.17
The voltage source is transformed in the network of R.2.83.
a
R = 15 Ω
60/15 Is^ = 12 A
= 4 A
a′
RL = 5 Ω
FIGURE 2.18
Current sources are added (combine into an equivalent source) in the network of Figure 2.17.
4 +12 = 16 A R = 15 Ω RL = 5 Ω
a′
a
FIGURE 2.19
Current source transforms into a voltage source in the network of R.2.83.
a
a′
Vs = 60 V 12 * 5 = 60 V
I
R = 15 Ω RL = 5 Ω