PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

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