PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 253

R.3.61 The example shown in Figure 3.25 uses the Thevenin’s theorem to calculate the cur-

rent IL, assuming that the load is ZL = 4 – j4 Ω.

ANALYTICAL Solution

Source transformation as presented in Chapter 2 can be extended to AC circuits as

indicated as follows:

Note that the current source I, in parallel with the impedance Z 2 , can be trans-

formed into a voltage source in series with Z 2 as shown as follows (see R.3.62):

Therefore,

E 3 = I * Z 2 = (6 + j3)(8 − j4) = 60 V

The circuit of Figure 3.25 is transformed into the circuit shown in Figure 3.26, where

the load ZL is removed, following Thevenin’s theorem. Then,

I 1 40 60

16

100

16

25

4

  

 
AA 625.

Therefore, VTH = (8 − j4)6.25 + j30 − 60 = − 10 + j5 V

E 1 = 40 V

I = 6 + j3 A

Z 1 = 8 + j 4

Z 2 = 8 − j 4

Z 3 = 1 + j 4

E 2 = 30 90 °V

ZL = 4 − j 4

IL

FIGURE 3.25

Network of R.3.61.

E 1 = 40 V

E 2 = 30

Z 1 = 8 + j 4

Z 2 = 8 − j 4

Z 3 = 1 + j 4

90 °V

VTH

I 1

−

+

E 3 = 60 V

FIGURE 3.26

Thevenin’s model of the circuit diagram of Figure 3.25.