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E BC
The length x is
C
B
+ –
E CN
E BN
N EAN –
–
FIG. 22.6
Line and phase voltages of the Y-connected
three-phase generator.
�3�
x � EAN cos 30° � �
2
EAN �3�
and EAB � 2x � (2) �EAN � �3�EAN 2
Noting from the phasor diagram that v of EAB � b � 30°, the result is
EAB � EAB �30° � �3�EAN �30°
and ECA � �3�ECN �150°
EBC � �3�EBN �270°
In words, the magnitude of the line voltage of a Y-connected generator
is �3� times the phase voltage:
E L � �3�E f
(phase voltage)
+
EAB (line voltage)
(22.4)
with the phase angle between any line voltage and the nearest phase
voltage at 30°.
In sinusoidal notation,
eAB � �2�EAB sin(qt � 30°)
eCA � �2�ECA sin(qt � 150°)
and eBC � �2�EBC sin(qt � 270°)
The phasor diagram of the line and phase voltages is shown in Fig. 22.8.
If the phasors representing the line voltages in Fig. 22.8(a) are rearranged
slightly, they will form a closed loop [Fig. 22.8(b)]. Therefore, we can
conclude that the sum of the line voltages is also zero; that is,
E CA
E AB � E CA � E BC � 0
A
+
(22.5)
THE Y-CONNECTED GENERATOR ⏐⏐⏐ 981
120°
E CN
E BN
α = 60°
α α = 60°
120°
E NB
x β = 30°
E AN
EAB x
FIG. 22.7
Determining a line voltage for a
three-phase generator.