Th´evenin’s and Norton’s theorems 595
Figure 33.71 Figure 33.72
Problem 15. (a) Convert the circuit to the left of terminals AB
in Figure 33.72 to an equivalent Th
´
evenin circuit by initially
converting to a Norton equivalent circuit. (b) Determine the
magnitude of the current flowing in the 1.8 C j4 impedance
connected between terminals A and B of Figure 33.72.
(a) For the branch containing the 12 V source, conversion to a Norton
equivalent network gives I
SC
1
D 12/3 D 4A and z
1
D 3 . For
the branch containing the 24 V source, conversion to a Norton
equivalent circuit gives I
SC
2
D 24/2 D 12Aandz
2
D 2 .
Thus Figure 33.73 shows a network equivalent to Figure 33.72.
From Figure 33.73, the total short-circuit current is 4 C 12 D 16 A,
and the total impedance is given by 3 ð 2/3 C2 D 1.2 . Thus
Figure 33.73 simplifies to Figure 33.74.
Figure 33.73 Figure 33.74
The open-circuit voltage across AB of Figure 33.74, E D
161.2 D 19.2 V, and the impedance ‘looking in’ at AB, z D
1.2 . Hence the Th
´
evenin equivalent circuit is as shown in
Figure 33.75.
(b) When the 1.8 C j4 impedance is connected to terminals AB of
Figure 33.75, the current I flowing is given by
I D
19.2
1.2 C 1.8 Cj4
D 3.84
6
53.13
°
A
Hence the current flowing in the .1.8Yj4/Z impedance is 3.84 A.
Figure 33.75