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a c
v � 15 sin qt ⇒ phasor notation V � 10.605 V �0°
V V �v 10.605 V �0°
I ����� ��� �5.303 A �90°
ZC XC ��90° 2 � ��90°
and i � �2�(5.303) sin(qt � 90°) � 7.5 sin(qt � 90°)
EXAMPLE 15.6 Using complex algebra, find the voltage v for the circuit
of Fig. 15.16. Sketch the v and i curves.
Solution: Note Fig. 15.17:
i � 6 sin(qt � 60°) ⇒ phasor notation I � 4.242 A ��60°
V� IZC � (I �v)(XC ��90°) � (4.242 A ��60°)(0.5 � ��90°)
� 2.121 V ��150°
and v � �2�(2.121) sin(qt � 150°) � 3.0 sin(qt � 150°)
The phasor diagrams for the two circuits of the two preceding examples
are shown in Fig. 15.18. Both indicate quite clearly that the current
i leads the voltage v by 90°.
j
j
5.303 A
I
6 A
3 V
0
60°
Leading
10.605 V
� � 3� 2
2
90°
(a) (b)
Impedance Diagram
V
i
+
v
2�
FIG. 15.17
Waveforms for Example 15.6.
5 �
2
2.121 V
V
Leading
FIG. 15.18
Phasor diagrams for Examples 15.5 and 15.6.
3�
60°
�t
4.242 A
Now that an angle is associated with resistance, inductive reactance,
and capacitive reactance, each can be placed on a complex plane dia-
I
IMPEDANCE AND THE PHASOR DIAGRAM ⏐⏐⏐ 635
+
i = 6 sin(qt – 60°)
X C = 0.5 �
FIG. 15.16
Example 15.6.
+
v
–