Direct Current and Transient Analysis
125
Note that the load R L is removed from the circuit of Figure 2.21 and the open
voltage Vaa’ is V TH . The evaluation of V TH is given by
Vaa’ = V TH = 60 V + 15 Ω * 12 A = 60 V+ 180 V = 240 V
Then R TH = 15 Ω is obtained by setting all the sources to zero.
The resulting Thevenin’s equivalent circuit is shown in Figure 2.22.
Finally, Vaa’ is given by
Vaa′ ϭ
ϩ
ϭ
( *
)
5 240
15 5
60 V
g. Norton’s solution is indicated by the equivalent circuit diagram model of Figure 2.23.
The short current I N is then evaluated below
I N
A
V
15
A
A
A
ϭ
ϩ
ϭ
ϩ
ϭ
12
60
12
4
16
Ω
The resulting Norton’s equivalent circuit is shown in Figure 2.24.
V
Vaa
I
R
N
L
ϭ
ϭ
′ (
)
15 5
""
V
Vaa
R L
(15 * 5 * 16)
V
ϭ
ϭ
ϩ
ϭ
′
15 5
60
FIGURE 2.22
The Thevenin’s equivalent circuit of the network of R.2.83.
a
V TH = 240 V
R L = 5 Ω
R TH = 15 Ω
a ′
FIGURE 2.23
The Norton’s equivalent model of the network of R.2.83.
V s = 60 V
15 Ω
I s = 12 A
I N
a ′
a
CRC_47760_CH002.indd 125
CRC_47760_CH002.indd 125
7/23/2008 1:38:40 PM
7/23/2008 1:38:40 PM