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EXAMPLE 25.7 The input to the circuit of Fig. 25.24 is the following:
e � 12 � 10 sin 2t
a. Find the current i and the voltages vR and vC. b. Find the rms values of i, vR, and vC. c. Find the power delivered to the circuit.
Solutions:
a. Redraw the original circuit as shown in Fig. 25.25. Then apply
superposition:
+
12 V
–
+
10 sin 2t
–
i
v R
R = 3 �
v C
X C = 1
�C
CIRCUIT RESPONSE TO A NONSINUSOIDAL INPUT ⏐⏐⏐ 1137
=
1
1 = 4 �
(2 rad/s)( 8F)
FIG. 25.25
Circuit of Fig. 25.24 with the components of the Fourier series input.
1. For the 12-V dc supply portion of the input, I � 0 since the
capacitor is an open circuit to dc when vC has reached its final
(steady-state) value. Therefore,
VR � IR � 0 V
2. For the ac supply,
and VC � 12 V
and
and
Z � 3 ��j 4 ��5 � ��53.13°
10
�� V �0°
E
�2�
2
I � � �
——
��A ��53.13°
Z
5 � ��53.13°
�2�
2
VR � (I �v)(R �0°) ���
A ��53.13°�
(3 � �0°)
�2�
6
� � V ��53.13°
�2�
2
VC � (I �v)(XC ��90°) � � � A ��53.13° � (4 � ��90°)
�2�
8
� � V ��36.87°
�2�
In the time domain,
i � 0 � 2 sin(2t � 53.13°)
Note that even though the dc term was present in the expression
for the input voltage, the dc term for the current in this circuit is
zero:
vR � 0 � 6 sin(2t � 53.13°)
and vC � 12 � 8 sin(2t � 36.87°)
+
e
–
i
v R
R = 3 �
1
C = F
8
FIG. 25.24
Example 25.7.
v C