PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 255

ANALYTICAL Solution

Replacing ZL 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






100

7

27()A

and

I

I

N

j

2

7A

Then ZTH 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

E = 100 V Z^2 = 6 + j^2

a

a′

IN

FIGURE 3.29

Norton’s model of the circuit diagram of Figure 3.28.

E = 100 V

Z 1 = 4 − j 2

I Z 5 = 4 − j 4

I

I

I

FIGURE 3.30

Simplifi ed version of Figure 3.29.