256 Practical MATLAB® Applications for Engineers
Then,ZTH = [(Z 1 ||Z 2 ) + Z 3 ] ||Z 4 ΩZ||Z Z
jj
12 3 jj
()()
..
6242
106 2 88 16 Ωand
Z
jj
j
TH j
(. .)*()
..88 16 11 2
19 8 3 644
98
9ΩFinally, the Norton’s equivalent circuit is shown in Figure 3.32, and the Thevenin’s
equivalent circuit is shown in Figure 3.33. The current IL can then easily be evaluated
from either circuit, as illustrated as follows:I
jj
L jj
(( ) ( ))*()
()()()()44 9 8 9 7
449 89 469 89A (from FFigure 3.32)I
j
L j
(/)(/)300 9 100 9
1030
910
9A (from Figure 3.33)ZjTH^44
98
9Ωa′Z 1 = 4 – j 2 Z^2 = 6 + j^2 Z 4 = 11 + j 2 ZTHaZ 3 = 6 + j 2FIGURE 3.31
ZTH model for the circuit diagram of Figure 3.28.
IN = 7 + j (^) ZL=
9
8
9
46
−j
ZTH =
9
8
9
44
+j
IL
FIGURE 3.32
Norton’s equivalent circuit of Figure 3.28.