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640 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS
X C = 8 Ω
R = 6 �
+
j
j
v R –
R = 6 �
θ T = 53.13°
Z T = 10 Ω
30 V VR
I 50 V
53.13°
36.87°
40 V
X C = 8 �
FIG. 15.32
Impedance diagram for the series R-C circuit
of Fig. 15.30.
V C
E
+
+
FIG. 15.33
Phasor diagram for the series R-C circuit
of Fig. 15.30.
+
v C –
i = 7.07 sin(qt + 53.13°)
FIG. 15.30
Series R-C ac circuit.
only to form the ratio of the total resistance to the magnitude of the
input impedance. For the case at hand,
R 3 �
Fp � cos v �����0.6 lagging
ZT 5 �
as found above.
R-C
Refer to Fig. 15.30.
Phasor Notation
i � 7.07 sin(qt � 53.13°) ⇒ I � 5 A �53.13°
Note Fig. 15.31.
R = 6 �
V
I = 5 �53.13°
+ R –
+
I E
–
Z T
X C = 8 Ω
+ VC –
FIG. 15.31
Applying phasor notation to the circuit of Fig. 15.30.
ZT ZT � Z1 � Z2 � 6 � �0° � 8 � ��90° � 6 ��j8 �
and ZT � 10 � ��53.13°
Impedance diagram: As shown in Fig. 15.32.
E
V R and V C
E � IZ T � (5 A �53.13°)(10 � ��53.13°) � 50 V �0°
VR � IZR � (I �v)(R �0°) � (5 A �53.13°)(6 � �0°)
� 30 V �53.13°
VC � IZC � (I �v)(XC ��90°) � (5 A �53.13°)(8 � ��90°)
� 40 V ��36.87°
Kirchhoff’s voltage law:
� V � E � VR � VC � 0
or E � V R � V C
which can be verified by vector algebra as demonstrated for the R-L
circuit.
Phasor diagram: Note on the phasor diagram of Fig. 15.33 that the
current I is in phase with the voltage across the resistor and leads the
voltage across the capacitor by 90°.
a c