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Electronics Workbench
We will now examine the response of a network versus frequency rather
than time using the network of Fig. 15.79 which now appears on the
schematic of Fig. 15.112. The ac current source appears as AC – CUR-
RENT – SOURCE in the Sources tool bin next to the ac voltage source.
Note that the current source was given an amplitude of 1 A to establish
a magnitude match between the response of the voltage across the network
and the impedance of the network. That is,
V s
� ZT � � � � � � ��� V � � s �
Is 1 A
Before applying computer methods, we should develop a rough idea
of what to expect so that we have something to which to compare the
computer solution. At very high frequencies such as 1 MHz, the impedance
of the inductive element will be about 25 k� which when placed
in parallel with the 220 � will look like an open circuit. The result is
that as the frequency gets very high, we should expect the impedance of
the network to approach the 220-� level of the resistor. In addition,
since the network will take on resistive characteristics at very high frequencies,
the angle associated with the input impedance should also
approach 0 �. At very low frequencies the reactance of the inductive
element will be much less than the 220 � of the resistor, and the network
will take on inductive characteristics. In fact, at, say, 10 Hz, the
reactance of the inductor is only about 0.25 � which is very close to a
short-circuit equivalent compared to the parallel 220-� resistor. The
result is that the impedance of the network is very close to 0 � at very
low frequencies. Again, since the inductive effects are so strong at low
V s
FIG. 15.112
Obtaining an impedance plot for a parallel R-L network using Electronics Workbench.
COMPUTER ANALYSIS ⏐⏐⏐ 693