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a c
v � 24 sin qt ⇒ phasor form V � 16.968 V �0°
V V �v
16.968 V �0°
I �������� �5.656 A ��90°
ZL XL �90°
3 � �90°
and i � �2�(5.656) sin(qt � 90°) � 8.0 sin(qt � 90°)
EXAMPLE 15.4 Using complex algebra, find the voltage v for the circuit
of Fig. 15.10. Sketch the v and i curves.
Solution: Note Fig. 15.11:
i � 5 sin(qt � 30°) ⇒ phasor form I � 3.535 A �30°
V � IZL � (I �v)(XL �90°) � (3.535 A �30°)(4 � ��90°)
� 14.140 V �120°
and v ��2�(14.140) sin(qt � 120°) � 20 sin(qt � 120°)
5.656 A
j
I
– �
2
90°
30°
5 A
16.968 A
Leading
20 V
0
υ
�
2
V
i
+
�
3� 2
V
14.140 V
2�
FIG. 15.11
Waveforms for Example 15.4.
The phasor diagrams for the two circuits of the two preceding examples
are shown in Fig. 15.12. Both indicate quite clearly that the voltage
leads the current by 90°.
j
3.535 A
FIG. 15.12
Phasor diagrams for Examples 15.3 and 15.4.
�t
Leading
30°
I
IMPEDANCE AND THE PHASOR DIAGRAM ⏐⏐⏐ 633
+
i = 5 sin(qt + 30°)
X L = 4 �
FIG. 15.10
Example 15.4.
+
v
–