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652 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS
work, I leads E by 90°. As the frequency increases, therefore, the
capacitive reactance is decreasing, and eventually R >> X C with v C �
�90°, and the angle between I and E will approach 0°. Keep in mind
as we proceed through the other frequencies that v C is the phase angle
between V C and E and that the magnitude of the angle by which I leads
E is determined by
(15.15)
f � 5 kHz Applying Eq. (15.13):
XC � � � 3.18 k�
Note the dramatic drop in XC from 1 kHz to 5 kHz. In fact, XC is now
less than the resistance R of the network, and the phase angle determined
by tan �1 (XC/R) must be less than 45°. Here,
(3.18 k�)(10 V)
VC � ���� �5.37 V
�(5� k���) 2 � �� (�3�.1�8� k���) 2 1
����
(2p)(5 � 10
XCE ��
2 2
�R� ��� X� C�
�
3 Hz)(0.01 � 10 �6 1
�
2pfC
F)
with v C ��tan �1
R
� XC
⎥ v I⎥ � 90° � ⎥ v C⎥
��tan �1
��tan �1 1.56 � �57.38°
5 k�
� 3.2 k�
f � 10 kHz
XC � 1.59 k� VC � 3.03 V vC � �72.34°
f � 15 kHz
XC � 1.06 k� VC � 2.07 V vC � �78.02°
f � 20 kHz
XC � 795.78 � VC � 1.57 V vC � �80.96°
The phasor diagrams for f � 5 kHz and f � 20 kHz appear in Fig.
15.50 to show the continuing rotation of the V C vector.
V R
I
θC
= –57.38°
V C
E
f = 5 kHz
V R
θC
= –80.96°
VC f = 20 kHz
FIG. 15.50
The phasor diagram for the circuit of Fig. 15.45 for f � 5 kHz and 20 kHz.
Note also from Figs. 15.49 and 15.50 that the vector V R and the current
I have grown in magnitude with the reduction in the capacitive
reactance. Eventually, at very high frequencies X C will approach zero
E
I
a c