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real and imaginary axes and finding the length of the resultant vector ZT and angle vT. Or, by using vector algebra, we obtain
ZT � Z1 � Z2 � R �0° � XL �90°
� R � jXL � 4 ��j8 �
ZT � 8.944 � �63.43°
EXAMPLE 15.8 Determine the input impedance to the series network
of Fig. 15.23. Draw the impedance diagram.
Solution:
ZT � Z1 � Z2 � Z3 � R �0° � XL �90° � XC ��90°
� R � jXL � jXC � R � j(XL � XC) � 6 ��j(10 ��12 �) � 6 ��j2 �
ZT � 6.325 � ��18.43°
The impedance diagram appears in Fig. 15.24. Note that in this
example, series inductive and capacitive reactances are in direct opposition.
For the circuit of Fig. 15.23, if the inductive reactance were
equal to the capacitive reactance, the input impedance would be purely
resistive. We will have more to say about this particular condition in a
later chapter.
For the representative series ac configuration of Fig. 15.25 having
two impedances, the current is the same through each element (as it
was for the series dc circuits) and is determined by Ohm’s law:
ZT � Z1 � Z2 E
and I � �� (15.5)
Z
The voltage across each element can then be found by another application
of Ohm’s law:
(15.6a)
(15.6b)
Kirchhoff’s voltage law can then be applied in the same manner as it
is employed for dc circuits. However, keep in mind that we are now
dealing with the algebraic manipulation of quantities that have both
magnitude and direction.
E � V1 � V2 � 0
or E � V1 � V2 (15.7)
The power to the circuit can be determined by
where v T is the phase angle between E and I.
T
V 1 � IZ 1
V 2 � IZ 2
P � EI cos v T
(15.8)
Z T
+
E
–
SERIES CONFIGURATION ⏐⏐⏐ 637
X L = 8 �
j
Z T
v T
R = 4 �
+
FIG. 15.22
Impedance diagram for Example 15.7.
Z 1
R = 6 �
X C – X L = 2 Ω
Z 2
X L = 10 �
FIG. 15.23
Example 15.8
j
X L = 10 Ω
R = 6 Ω
Z T
X C = 12 Ω
Z 3
X C = 12 �
θT
FIG. 15.24
Impedance diagram for Example 15.8.
I
Z T
+
V 1 –
Z 1
FIG. 15.25
Series ac circuit.
Z 2
+
+
V 2
–