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(4,0)�((9,�7)�(8,6)) �1 * (11.4��37.87)(10�36.87) ENTER<br />

(10.689E0,276.413E�3)<br />

Ans � Pol ENTER<br />

(10.692E0�1.481E0)<br />

FIG. 16.18<br />

Using Mathcad to determine the total impedance for the network of Fig.16.16.<br />

Function Name-arg. Finally the variable is written again and the<br />

equal sign selected to obtain an angle of 1.478°. The computer solution<br />

of 10.693 � �1.478° is an excellent verification of the theoretical<br />

solution of 10.684 � �1.5°.<br />

Calculator Another opportunity to demonstrate the versatility of the<br />

calculator! For the above operation, however, one must be aware of the<br />

priority of the mathematical operations, as demonstrated in the calculator<br />

display below. In most cases, the operations are performed in the<br />

same order they would be performed longhand.<br />

E 100 V �0°<br />

b. I ����� � 9.36 A ��1.5°<br />

ZT 10.684 � �1.5°<br />

R 10.68 �<br />

c. Fp � cos vT � � ��� � 1<br />

ZT 10.684 �<br />

(essentially resistive, which is interesting, considering the complexity<br />

of the network)<br />

ILLUSTRATIVE EXAMPLES ⏐⏐⏐ 719

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