PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 125

Note that the load RL is removed from the circuit of Figure 2.21 and the open

voltage Va a’ is VTH. The evaluation of VTH is given by

Va a’ = VTH = 60 V + 15 Ω (^) 12 A = 60 V+ 180 V = 240 V
Then RTH = 15 Ω is obtained by setting all the sources to zero.
The resulting Thevenin’s equivalent circuit is shown in Figure 2.22.
Finally, Va a’ is given by
Vaa′




( )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 IN is then evaluated below
IN A
V
15



12 A A A
60
12 4 16
Ω
The resulting Norton’s equivalent circuit is shown in Figure 2.24.
VVaaRNL

′ ()15 5 I
VVaaRL
(15 (^5) 16)


V

′

15 5
60
FIGURE 2.22
The Thevenin’s equivalent circuit of the network of R.2.83.
a
VTH = 240 V
RL = 5 Ω
RTH = 15 Ω
a′
FIGURE 2.23
The Norton’s equivalent model of the network of R.2.83.
Vs = 60 V
15 Ω
Is = 12 A
a′ IN
a