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Ip = 10 A ∠ 0° 1 � 2 �
Solutions:
a. Re � Rp � a 2 Rs � 1 ��(2) 2 (1 �) � 5 �
Xe � Xp � a 2 Xs � 2 ��(2) 2 (2 �) � 10 �
b. The transformed equivalent circuit appears in Fig. 21.21.
aVL � (Ip)(a 2 RL) � 2400 V
Thus,
VL � � � 1200 V
and
Vg � Ip(Re � a 2 2400 V 2400 V
� �
a 2
RL � j Xe) � 10 A(5 ��240 ��j 10 �) � 10 A(245 ��j 10 �)
Vg � 2450 V � j 100 V � 2452.04 V �2.34° � 2452.04 V �2.34°
I p = 10 A ∠ 0° R e
V g
+
–
+
V g
–
5 �
X e
10 �
aV L
R p
X p
2 : 1
Ideal transformer
FIG. 21.20
Example 21.7.
a 2 R L = (4)(60 �) = 240 �
FIG. 21.21
Transformed equivalent circuit of Fig. 21.20.
c. For R e and X e � 0, V g � aV L � (2)(1200 V) � 2400 V.
Therefore, it is necessary to increase the generator voltage by
52.04 V (due to R e and X e) to obtain the same load voltage.
21.7 FREQUENCY CONSIDERATIONS
For certain frequency ranges, the effect of some parameters in the
equivalent circuit of the iron-core transformer of Fig. 21.15 should not
be ignored. Since it is convenient to consider a low-, mid-, and highfrequency
region, the equivalent circuits for each will now be introduced
and briefly examined.
For the low-frequency region, the series reactance (2pfL) of the primary
and secondary leakage reactances can be ignored since they are
+
–
X s
2 �
FREQUENCY CONSIDERATIONS ⏐⏐⏐ 951
R s
1 �
R L
60 �
+
V L
–