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P q<br />

s<br />

Motor:<br />

h � Pi � � �4548.78 W � P3 Fp � 0.72 lagging<br />

P3 � S3 cos v S3 � � �6317.75 VA<br />

Also, v � cos �1 0.72 � 43.95°, so that<br />

Q3 � S3 sin v � (6317.75 VA)(sin 43.95°)<br />

� (6317.75 VA)(0.694) � 4384.71 VAR (L)<br />

Capacitive load:<br />

I � � � �13.87 A �53.13°<br />

P4 � I 2 R � (13.87 A) 2 • 9 ��1731.39 W<br />

Q4 � I 2 XC � (13.87 A) 2 • 12 ��2308.52 VAR (C)<br />

S4 � �P� 2 4��� Q� 2 4� � �(1�7�3�1�.3�9� W�) 2 � �� (�2�3�0�8�.5�2� V�A�R�) 2 �<br />

� 2885.65 VA<br />

Fp � � �0.6 leading<br />

b. PT � P1 � P2 � P3 � P4 � 720 W � 6400 W � 4548.78 W � 1731.39 W<br />

� 13,400.17 W<br />

QT ��Q1 � Q2 � Q3 � Q4 � 0 � 0 � 4384.71 VAR (L) � 2308.52 VAR (C)<br />

� 2076.19 VAR (L)<br />

ST � �P� 2 T��� Q� 2 T� � �(1�3�,4�0�0�.1�7� W�) 2 � �� (�2�0�7�6�.1�9� V�A�R�) 2 �<br />

� 13,560.06 VA<br />

Fp � � �0.988 lagging<br />

v � cos �1 Po Po 5(746 W)<br />

� � ��<br />

Pi h 0.82<br />

P3 4548.78 W<br />

� ��<br />

cos v 0.72<br />

E 208 V �0° 208 V �0°<br />

� �� ��<br />

Z 9 ��j 12 � 15 � ��53.13°<br />

P4 1731.39 W<br />

� ��<br />

S4 2885.65 VA<br />

PT 13.4 kW<br />

� ��<br />

ST 13,560.06 VA<br />

0.988 � 8.89°<br />

Note Fig. 19.22.<br />

S T = 13,560.06 VA<br />

8.89°<br />

PT = 13.4 kW<br />

Q T = 2076.19 VAR (L)<br />

FIG. 19.22<br />

Power triangle for Example 19.3.<br />

S T<br />

13,559.89 VA<br />

c. ST � EI I � � ��� � 65.19 A<br />

E<br />

208 V<br />

Lagging power factor: E leads I by 8.89°, and<br />

I � 65.19 A ��8.89°<br />

THE TOTAL P, Q,AND S ⏐⏐⏐ 863

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