Alternating Current Analysis
255
ANALYTICAL Solution
Replacing Z L by a short (Norton’s theorem) results in the circuit diagram shown in
Figure 3.29.
Observe that Z 2 and Z 3 are connected in parallel, then
Z 5 = Z 2 | |Z 3 = Z 2 / 2 = 3 + j
The circuit of Figure 3.29 is further simplifi ed and redrawn in Figure 3.30.
Then,
I
j
j
ϭ Ϫ
ϭ
ϩ
100
7
2 7
(
)A
and
I
I
j
N ϭ ϭ ϩ
2
7
A
Then Z TH can be evaluated from the circuit shown in Figure 3.31, where the voltage
source of Figure 3.28 is replaced by a short circuit.
Z 1 = 4 − j 2
Z 3 = 6 + j 2
Z 2 = 6 + j 2
E = 100 V
a
a′
I N
FIGURE 3.29
Norton’s model of the circuit diagram of Figure 3.28.
E = 100 V
Z 1 = 4 − j2
Z 5 = 4 − j4
I
I
I
I
FIGURE 3.30
Simplifi ed version of Figure 3.29.
CRC_47760_CH003.indd 255
CRC_47760_CH003.indd 255
7/23/2008 1:27:39 PM
7/23/2008 1:27:39 PM