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632 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS
i
i
X L = 3 �
+
XL = qL v = Vm sin qt
–
FIG. 15.7
Inductive ac circuit.
+
FIG. 15.8
Example 15.3.
v = 24 sin qt
–
Inductive Reactance
It was learned in Chapter 13 that for the pure inductor of Fig. 15.7, the
voltage leads the current by 90° and that the reactance of the coil XL is
determined by qL.
v � Vm sin qt ⇒ phasor form V � V �0°
By Ohm’s law,
V �0° V
I ��� � /0° � vL XL �vL XL
Since v leads i by 90°, i must have an angle of �90° associated with it.
To satisfy this condition, v L must equal �90°. Substituting v L � 90°, we
obtain
V �0° V
V
I ��� � /0° � 90° � � ��90°
XL �90° XL
XL
so that in the time domain,
V
� XL
i � �2� � � sin(qt � 90°)
The fact that v L � 90° will now be employed in the following polar
format for inductive reactance to ensure the proper phase relationship
between the voltage and current of an inductor.
Z L � X L �90° (15.2)
The boldface roman quantity Z L, having both magnitude and an
associated angle, is referred to as the impedance of an inductive element.
It is measured in ohms and is a measure of how much the inductive
element will “control or impede” the level of current through the
network (always keep in mind that inductive elements are storage
devices and do not dissipate like resistors). The above format, like that
defined for the resistive element, will prove to be a useful “tool” in the
analysis of ac networks. Again, be aware that Z L is not a phasor quantity,
for the same reasons indicated for a resistive element.
EXAMPLE 15.3 Using complex algebra, find the current i for the circuit
of Fig. 15.8. Sketch the v and i curves.
Solution: Note Fig. 15.9:
24 V
8 A
0
90°
v
� �
2
i
3� 2
2�
FIG. 15.9
Waveforms for Example 15.3.
5� �t
2
a c