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P q
s
Thus,
PT 600 W
Fp �����0.6 leading (C)
ST 1000 VA
The power triangle is shown in Fig. 19.18.
Since ST � VI � 1000 VA, I � 1000 VA/100 V � 10 A; and since v
of cos v � Fp is the angle between the input voltage and current:
I � 10 A ��53.13°
The plus sign is associated with the phase angle since the circuit is predominantly
capacitive.
EXAMPLE 19.2
a. Find the total number of watts, volt-amperes reactive, and voltamperes,
and the power factor F p for the network of Fig. 19.19.
+
E = 100 V ∠0°
–
R
X L
I 6 �
7 �
FIG. 19.19
Example 19.2.
X C
15 �
b. Sketch the power triangle.
c. Find the energy dissipated by the resistor over one full cycle of the
input voltage if the frequency of the input quantities is 60 Hz.
d. Find the energy stored in, or returned by, the capacitor or inductor
over one half-cycle of the power curve for each if the frequency of
the input quantities is 60 Hz.
Solutions:
E 100 V �0°
100 V �0°
a. I ������ ��� ZT 6 ��j 7 ��j 15 � 10 � ��53.13°
� 10 A �53.13°
VR � (10 A �53.13°)(6 � �0°) � 60 V �53.13°
VL � (10 A �53.13°)(7 � �90°) � 70 V �143.13°
VC � (10 A �53.13°)(15 � ��90°) � 150 V ��36.87°
PT � EI cos v � (100 V)(10 A) cos 53.13° � 600 W
� I 2 R � (10 A) 2 (6 �) � 600 W
� � �600 W
ST � EI � (100 V)(10 A) � 1000 VA
� I 2 ZT � (10 A) 2 (10 �) � 1000 VA
� � �1000 VA
QT � EI sin v � (100 V)(10 A) sin 53.13° � 800 VAR
� QC � QL � I 2 (XC � XL) � (10 A) 2 (100 V)
(15 ��7 �) � 800 VAR
2
E
�
10 �
2
(60 V)
�
ZT
2
V
�
6
2 R
�
R
THE TOTAL P, Q, AND S ⏐⏐⏐ 861
P T = 600 W
S T = 1000 VA
53.13° = cos –1 0.6
Q T = 800 VAR (C)
FIG. 19.18
Power triangle for Example 19.1.