254 Practical MATLAB® Applications for Engineers
and
Zj
jj
jj
TH j
14
8484
8484
64
()()
(by replacing the voltage sources by shorts).
The Thevenin’s equivalent circuit is shown in Figure 3.27.
Then,
I
V
ZZ
j
L TH j
LTH
10 5
10
105 .A
R.3.62 Note that source transformation concept for the DC case discussed in Chapter 2
can be extended to include the AC. Note that source transformation was already
employed in the example presented in R.3.61. Observe that a current source of
I = 6 + j3 with a parallel impedance of Z 2 = 8 − j4 was converted into a voltage
source of 60 V (Z 2 * I) in series with Z 2 (Figure 3.25).
R.3.63 The circuit shown in Figure 3.28 is used to illustrate Norton’s theorem and source
transformation in evaluating the current IL through the load ZL.
VTH = −10 + j 5
ZTH = 6 + j 4
IL ZL = 4 − j 4
FIGURE 3.27
Thevenin’s equivalent circuit of Figure 3.25.
Z 1 = 4 − j 2
Z 2 = 6 + j 2 Z^4 = 11 −^ j^2
Z 3 = 6 + j 2
E = 100 V ZL = 46
a
a′
IL
9
8
− j 9
FIGURE 3.28
Network of R.3.63.