The Three Phase Generator

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

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