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a c
f � 0 Hz
XC � � ⇒ very large value
Applying the open-circuit equivalent for the capacitor based on the
above calculation will result in the following:
VC � E � 10 V �0°
If we apply Eq. (15.13), we find
X 2 C >> R 2
1 1
� �
2p(0)C 0
and �R� 2 ��� X� 2 C� � �X� 2 C� � X C
and VC � � �E
with vC ��tan �1
��tan �1 XCE X
�� �
CE
2 2
�R� ��� X� C� XC
�
R
0 � 0°
XC
verifying the above conclusions.
f � 1 kHz Applying Eq. (15.13):
XC � � � 15.92 k�
�R� 2 ��� X� 2 C� � �(5� k���) 2 � �� (�1�5�.9�2� k���) 2 1
����
(2p)(1 � 10
� � 16.69 k�
XCE (15.92 k�)(10)
and VC ��� ��� �9.54 V
2 2
�R� ��� X� C� 16.69 k�
Applying Eq. (15.14):
3 Hz)(0.01 � 10 �6 1
�
2pfC
F)
v C ��tan �1
R
� XC
��tan �1
5 k�
� 15.9 k�
��tan �1 0.314 � �17.46°
and VC � 9.53 V ��17.46°
As expected, the high reactance of the capacitor at low frequencies has
resulted in the major part of the applied voltage appearing across the
capacitor.
If we plot the phasor diagrams for f � 0 Hz and f � 1 kHz, as shown
in Fig. 15.49, we find that VC is beginning a clockwise rotation with an
increase in frequency that will increase the angle vC and decrease the
phase angle between I and E. Recall that for a purely capacitive net-
I = 0 A
f = 0 Hz
E
V C
θI
= 90°
θC
= 0°
V R
I
θI
θC
f = 1 kHz
FREQUENCY RESPONSE OF THE R-C CIRCUIT ⏐⏐⏐ 651
E
–17.46°
FIG. 15.49
The phasor diagram for the circuit of Fig. 15.45 for f � 0 Hz and 1 kHz.
V C