# Y connected Generator connected to a Y connected load

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Loads connected to three-phase supplies are of two types: the Y and the $\Delta$. If a Y-connected load is connected to a Y-connected generator, the system is symbolically represented by $Y-Y$. The physical setup of such a system is shown in Fig. 1.
Fig. 1: Y-connected generator with a Y-connected load.
If the load is balanced, the neutral connection can be removed without affecting the circuit in any manner; that is, if
$$Z_1 = Z_2 = Z_3$$
then $I_N$ will be zero. (This will be demonstrated in Example 1.) Note that in order to have a balanced load, the phase angle must also be the same for each impedance—a condition that was unnecessary in dc circuits when we considered balanced systems.
In practice, if a factory, for example, had only balanced, three-phase loads, the absence of the neutral would have no effect since, ideally, the system would always be balanced. The cost would therefore be less since the number of required conductors would be reduced. However, lighting and most other electrical equipment will use only one of the phase voltages, and even if the loading is designed to be balanced (as it should be), there will never be perfect continuous balancing since lights and other electrical equipment will be turned on and off, upsetting the balanced condition. The neutral is therefore necessary to carry the resulting current away from the load and back to the Y-connected generator. This will be demonstrated when we consider unbalanced Y-connected systems. We shall now examine the four-wire Y-Y-connected system. The current passing through each phase of the generator is the same as its corresponding line current, which in turn for a Y-connected load is equal to the current in the phase of the load to which it is attached:
$$\mathbf{I}_{\phi \mathrm{g}}=\mathbf{I}_{L}=\mathbf{I}_{\phi L}$$
For a balanced or an unbalanced load, since the generator and load have a common neutral point, then
$$\mathbf{V}_{\phi}=\mathbf{E}_{\phi}$$
In addition, since $\mathbf{I}_{\phi L}=\mathbf{V}_{\phi} / \mathbf{Z}_{\phi}$, the magnitude of the current in each phase will be equal for a balanced load and unequal for an unbalanced load. You will recall that for the Y-connected generator, the magnitude of the line voltage is equal to $\sqrt{3}$ times the phase voltage. This same relationship can be applied to a balanced or an unbalanced four-wire Y-connected load:
$$E_{L}=\sqrt{3} V_{\phi}$$
For a voltage drop across a load element, the first subscript refers to that terminal through which the current enters the load element, and the second subscript refers to the terminal from which the current leaves. In other words, the first subscript is, by definition, positive with respect to the second for a voltage drop. Note Fig. 2, in which the standard double subscripts for a source of voltage and a voltage drop are indicated.
Example 1: The phase sequence of the Y-connected generator in Fig. 2 is $ABC$.
a. Find the phase angles v2 and v3.
b. Find the magnitude of the line voltages.
c. Find the line currents.
d. Verify that, since the load is balanced, $I_N = 0$.
Fig. 2: Example 1.
Solution:
a. For an $A B C$ phase sequence, $$\theta_{2}=-120^{\circ} \text { and } \theta_{3}=+120^{\circ}$$ b. $E_{L}=\sqrt{3} E_{\phi}=(1.73)(120 \mathrm{~V})=208 \mathrm{~V}$. Therefore, $$E_{A B}=E_{B C}=E_{C A}=\mathbf{2 0 8} \mathrm{V}$$ c. $\mathbf{V}_{\phi}=\mathbf{E}_{\phi}$. Therefore, \begin{aligned}\mathbf{V}_{a n} &=\mathbf{E}_{A N} \quad \mathbf{V}_{b n}=\mathbf{E}_{B N} \quad \mathbf{V}_{c n}=\mathbf{E}_{C N} \\\mathbf{I}_{\phi L}=\mathbf{I}_{a n}=\frac{\mathbf{V}_{a n}}{\mathbf{Z}_{a n}} &=\frac{120 \mathrm{~V} \angle 0^{\circ}}{3 \Omega+j 4 \Omega}=\frac{120 \mathrm{~V} \angle 0^{\circ}}{5 \Omega \angle 53.13^{\circ}} \\&=24 \mathrm{~A} \angle-53.13^{\circ} \\\mathbf{I}_{b n}=\frac{\mathbf{V}_{b n}}{\mathbf{Z}_{b n}} &=\frac{120 \mathrm{~V} \angle-120^{\circ}}{5 \Omega \angle 53.13^{\circ}}=24 \mathrm{~A} \angle-173.13^{\circ} \\\mathbf{I}_{c n}=\frac{\mathbf{V}_{c n}}{\mathbf{Z}_{c n}} &=\frac{120 \mathrm{~V} \angle+120^{\circ}}{5 \Omega \angle 53.13^{\circ}}=24 \mathrm{~A} \angle 66.87^{\circ}\end{aligned} and, since $\mathbf{I}_{L}=\mathbf{I}_{\phi L}$ $$I_{A a}=I_{a n}=24 \mathrm{~A} \angle-53.13^{\circ}$$ $$I_{B b}=I_{b n}=24 \mathrm{~A} \angle-173.13^{\circ}$$ $$I_{C c}=I_{c n}=24 \mathrm{~A} \angle 66.87^{\circ}$$ d. Applying Kirchhoff's current law, we have $$\mathbf{I}_{N}=\mathbf{I}_{A a}+\mathbf{I}_{B b}+\mathbf{I}_{C c}$$ In rectangular form, $$\mathbf{I}_{A a}=24 \mathrm{~A} \angle-53.13^{\circ}=14.40 \mathrm{~A}-j 19.20 \mathrm{~A}$$ $$\mathbf{I}_{B b}=24 \mathrm{~A} \angle-173.13^{\circ}=-22.83 \mathrm{~A}-j 2.87 \mathrm{~A}$$ $$\mathbf{I}_{C c}=24 \mathrm{~A} \angle-66.87^{\circ}=9.43 \mathrm{~A}-j 22.07 \mathrm{~A}$$ $$\sum (I_{Aa} + I_{Bb} + I_{Cc} ) = 0 + j0$$ and $I_N$ is in fact equal to zero, as required for a balanced load.

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