PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

122 Practical MATLAB® Applications for Engineers

Then

60 V = 20 I 1 − 5(−12)

60 = 20 I 1 + 60

60 60

20

−



I 1 0

Then

Va a’ = 5 Ω ∗ 12 A = 60 V

b. The node equation solution is referred to as the circuit diagram of Figure 2.16. The

node equation for node a shown in Figure 2.16 is given below while the reference

node is a’ (grounded).

12 A
 60

1

5

1

15

1

15









Va

12 


4

15

Va 4

12 4

4

15


Va

16

15

4

Va

Va = 60 V or Va a’ = 60 V

c. The voltage source transformation into a current source is shown by the circuit

diagram of Figures 2.17 and 2.18, where fi rst the transformation is illustrated

(Figure 2.17), followed by the currents’ source combinations (Figure 2.18).

Then

Va a’ = ( 1 5 || 5 ) 16

Vaa′




(15 (^5) 16)
60 V
(^155)
FIGURE 2.16
Loop model of network of R.2.83.
RL = 5 Ω
RL = 5 Ω
R = 15 Ω
Is = 12 A
Vs = 60 V
a
a′
I 1
I 2