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Th<br />

Z1 � R1 � j XL1 � 6 ��j 8 �<br />

Z2 � R2 � j XC � 3 ��j 4 �<br />

Z3 ��j XL2 � j 5 �<br />

Step 3 (Fig. 18.30):<br />

Z1Z2 (10 � �53.13°)(5 � ��53.13°)<br />

ZTh � Z3 ���j 5 ������<br />

Z1 � Z2 (6 ��j 8 �) � (3 ��j 4 �)<br />

50 �0°<br />

50 �0°<br />

� j 5 ���j 5 ���<br />

9 � j 4 9.85 �23.96°<br />

� j 5 � 5.08 ��23.96° � j 5 � 4.64 � j 2.06<br />

ZTh � 4.64 ��j 2.94 � � 5.49 � �32.36°<br />

Z 1<br />

FIG. 18.30<br />

Determining the Thévenin impedance for the network of Fig. 18.28.<br />

Step 4 (Fig. 18.31): Since a-a′ is an open circuit, IZ3 � 0. Then ETh is<br />

the voltage drop across Z2: Z<br />

ETh � �<br />

2E<br />

(voltage divider rule)<br />

Z2 � Z1 +<br />

E 1<br />

–<br />

�<br />

Z 2<br />

Z 3<br />

(5 � ��53.13°)(10 V �0°)<br />

���<br />

9.85 � �23.96°<br />

a<br />

Z Th<br />

50 V ��53.13°<br />

ETh ��� �5.08 V ��77.09°<br />

9.85 �23.96°<br />

Z 1<br />

Z 2<br />

Z 3<br />

I Z3 = 0<br />

a�<br />

+<br />

E Th<br />

a<br />

–<br />

a�<br />

FIG. 18.31<br />

Determining the open-circuit Thévenin voltage for the network of Fig. 18.28.<br />

THÉVENIN’S THEOREM ⏐⏐⏐ 801

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