PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

124 Practical MATLAB® Applications for Engineers

The two single source networks of Figure 2.20 are solved for the voltage drop

across R = 5 Ω labeled Va 1 and Va 2 and the voltage Va a’ is then the algebraic sum

(Va 1 + Va 2 ). The steps involved are indicated as follows:

a. Circuit of Figure 2.15 with Is = 0

b. Circuit of Figure 2.15 with Vs = 0

The voltage Va 1 for the circuit shown in Figure 2.20a is given by

Va 1

20

 15

(5*60)

V (using the voltage divider rule)

The voltage Va 2 for the circuit shown in Figure 2.20b is given by

Va 2

15 5

 45



(15 (^5) 12)
V
Since Va 1 and Va 2 have the same polarity
Va = Va 1 + Va 2 = 15 V + 45 V= 60 V

6. The solution using Thevenin’s theorem is indicated by the circuit diagram of
   Figure 2.21. (Note that the load is RL = 5 Ω.)

FIGURE 2.20
Superposition models of the network of R.2.83. (a) Circuit of Figure 2.15 with Is = 0, (b) Circuit of Figure 2.15
with Vs = 0.

Vs = 60 V

15

5

12 A

5

15

Va 1 Va 2

(a) (b)

FIGURE 2.21
Thevenin’s models of the network of R.2.83.

a

a′

Vs = 60 V

Is = 12 A

R = 15 Ω