The Three Phase Generator
The three-phase generator of Fig. 1(a) has three induction coils
placed $120^\circ$ apart on the stator, as shown symbolically by Fig. 1(b). Since the three coils have an equal number of turns, and each coil rotates with the same angular velocity, the voltage induced across each coil will have the same peak value, shape, and frequency.

As the shaft
of the generator is turned by some external means, the induced voltages
$e_{AN}$, $e_{BN}$, and $e_{CN}$ will be generated simultaneously, as shown in Fig. 2. Note the $120^\circ$ phase shift between waveforms and the similarities in appearance of the three sinusoidal functions.
In particular, note that
This is shown at $ \omega t=0 $ in Fig. 2, where it is also evident that when one induced voltage is zero, the other two are $ 86.6 \% $ of their positive or negative maximums. In addition, when any two are equal in magnitude and sign $ \left(\right. $ at $ \left.0.5 E_{m}\right) $, the remaining induced voltage has the opposite polarity and a peak value.
The sinusoidal expression for each of the induced voltages of Fig. 2 is
The phasor diagram of the induced voltages is shown in Fig. 3, where the effective value of each is determined by
$$\begin{array}{l}E_{A N}=0.707 E_{m(A N)} \\
E_{B N}=0.707 E_{m(B N)} \\
E_{C N}=0.707 E_{m(C N)}\end{array}$$
and
$$\begin{array}{l}\mathbf{E}_{A N}=E_{A N} \angle 0^{\circ} \\
\mathbf{E}_{B N}=E_{B N} \angle-120^{\circ} \\
\mathbf{E}_{C N}=E_{C N} \angle+120^{\circ}\end{array}$$
By rearranging the phasors as shown in Fig. 4 and applying a law
of vectors which states that the vector sum of any number of vectors
drawn such that the “head” of one is connected to the “tail” of the
next, and that the head of the last vector is connected to the tail of the
first is zero, we can conclude that the phasor sum of the phase voltages
in a three-phase system is zero. That is,

Fig. 1: (a) Three-phase generator; (b) induced voltages of a three-phase generator
Fig. 2: Phase voltages of a three-phase generator.
$$ \bbox[10px,border:1px solid grey]{\begin{split}
e_{A N}&=E_{m(A N)} \sin \omega t \\
e_{B N}&=E_{m(B N)} \sin \left(\omega t-120^{\circ}\right)\\ e_{C N}&=E_{m(C N)} \sin \left(\omega t-240^{\circ}\right)=E_{m(C N)} \sin \left(\omega t+120^{\circ}\right)
\end{split}} \tag{1}$$
Fig. 3: Phasor diagram for the phase voltages of a
three-phase generator.
Fig. 4: Demonstrating that the vector sum of the
phase voltages of a three-phase generator
is zero.
$$ \bbox[10px,border:1px solid grey]{E_{AN} + E_{BN} + E_{CN} = 0} \tag{2}$$
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