# The Y Connected Generator   Whatsapp  If the three terminals denoted N of Fig. 1 are connected together, the generator is referred to as a Y-connected three-phase generator (Fig. 2). Fig. 1: induced voltages of a three-phase generator.
As indicated in Fig. 2, the Y is inverted for ease of notation and for clarity. The point at which all the terminals are connected is called the neutral point. If a conductor is not attached from this point to the load, the system is called a Y-connected, three-phase, three-wire generator. If the neutral is connected, the system is a Y-connected, three-phase, four-wire generator. The function of the neutral will be discussed in detail when we consider the load circuit. Fig. 2: Y-connected generator
The three conductors connected from $A, B$, and $C$ to the load are called lines. For the Y-connected system, it should be obvious from Fig. 2 that the line current equals the phase current for each phase; that is,
$$\mathbf{I}_{L}=\mathbf{I}_{\phi \mathrm{g}}$$
where $\phi$ is used to denote a phase quantity and $g$ is a generator parameter.
The voltage from one line to another is called a line voltage. On the phasor diagram (Fig. 3) it is the phasor drawn from the end of one phase to another in the counterclockwise direction.
Applying Kirchhoff's voltage law around the indicated loop of Fig. 3, we obtain
$$\mathbf{E}_{A B}-\mathbf{E}_{A N}+\mathbf{E}_{B N}=0$$
or
$$\mathbf{E}_{A B}=\mathbf{E}_{A N}-\mathbf{E}_{B N}=\mathbf{E}_{A N}+\mathbf{E}_{N B}$$
The phasor diagram is redrawn to find $\mathbf{E}_{A B}$ as shown in Fig. 4. Since each phase voltage, when reversed ( $\mathbf{E}_{N B}$ ), will bisect the other two, $\alpha=$ $60^{\circ}$. The angle $\beta$ is $30^{\circ}$ since a line drawn from opposite ends of a rhombus will divide in half both the angle of origin and the opposite angle. Lines drawn between opposite corners of a rhombus will also bisect each other at right angles. Fig. 3: Line and phase voltages of the Y-connected three-phase generator. Fig. 4: Determining a line voltage for a three-phase generator.
The length $x$ is
$$x=E_{A N} \cos 30^{\circ}=\frac{\sqrt{3}}{2} E_{A N}$$
and
$$E_{A B}=2 x=(2) \frac{\sqrt{3}}{2} E_{A N}=\sqrt{3} E_{A N}$$
Noting from the phasor diagram that $\theta$ of $\mathbf{E}_{A B}=\beta=30^{\circ}$, the result is
$$\mathbf{E}_{A B}=E_{A B} \angle 30^{\circ}=\sqrt{3} E_{A N} \angle 30^{\circ}$$
and
$$\begin{array}{l}\mathbf{E}_{C A}=\sqrt{3} E_{C N} \angle 150^{\circ} \\\mathbf{E}_{B C}=\sqrt{3} E_{B N} \angle 270^{\circ}\end{array}$$
In words, the magnitude of the line voltage of a Y-connected generator is $\sqrt{3}$ times the phase voltage:
$$E_{L}=\sqrt{3} E_{\phi}$$
with the phase angle between any line voltage and the nearest phase voltage at $30^{\circ}$. In sinusoidal notation,
$$\begin{array}{l}e_{A B}=\sqrt{2} E_{A B} \sin \left(\omega t+30^{\circ}\right) \\e_{C A}=\sqrt{2} E_{C A} \sin \left(\omega t+150^{\circ}\right) \\e_{B C}=\sqrt{2} E_{B C} \sin \left(\omega t+270^{\circ}\right)\end{array}$$  Fig. 5: (a) Phasor diagram of the line and phase voltages of a three-phase generator; (b) demonstrating that the vector sum of the line voltages of a three-phase system is zero.
The phasor diagram of the line and phase voltages is shown in Fig. 5. If the phasors representing the line voltages in Fig. 5(a) are rearranged slightly, they will form a closed loop [Fig. 5(b)]. Therefore, we can conclude that the sum of the line voltages is also zero; that is,
$$\mathbf{E}_{A B}+\mathbf{E}_{C A}+\mathbf{E}_{B C}=0$$

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