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630 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS
i
+
5 � v = 100 sin qt
–
FIG. 15.2
Example 15.1.
In phasor form,
v � V m sin qt ⇒ V � V �0°
a c
where V � 0.707Vm. Applying Ohm’s law and using phasor algebra, we have
V �0° V
I ��� � /0° � vR R �vR R
Since i and v are in phase, the angle associated with i also must be 0°.
To satisfy this condition, vR must equal 0°. Substituting vR � 0°, we
find
V �0° V
V
I ��� � /0° � 0° � � �0°
R �0° R
R
so that in the time domain,
V
i � �2� �� � sin qt
R
The fact that vR � 0° will now be employed in the following polar
format to ensure the proper phase relationship between the voltage and
current of a resistor:
Z R � R �0°
(15.1)
The boldface roman quantity Z R, having both magnitude and an
associated angle, is referred to as the impedance of a resistive element.
It is measured in ohms and is a measure of how much the element will
“impede” the flow of charge through the network. The above format
will prove to be a useful “tool” when the networks become more complex
and phase relationships become less obvious. It is important to
realize, however, that Z R is not a phasor, even though the format R�0°
is very similar to the phasor notation for sinusoidal currents and voltages.
The term phasor is reserved for quantities that vary with time, and
R and its associated angle of 0° are fixed, nonvarying quantities.
EXAMPLE 15.1 Using complex algebra, find the current i for the circuit
of Fig. 15.2. Sketch the waveforms of v and i.
Solution: Note Fig. 15.3:
100 V
20 A
0
i
�
2
v
3� � 2
FIG. 15.3
Waveforms for Example 15.1.
2�
�t �