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In power systems three-phase transformers are used, as they present the possibility of working with high power and high efficiency. There are several types of three-phase transformers, for example, transformers voltage elevators, voltage lowerers, distribution, etc ... In distribution transformers, as a rule, the primary is connected to the high voltage circuit using the Delta or Triangle connection. Thus, in the secondary the low voltage circuit is connected using the type Y or Star connection to four wires. This means that in the secondary, in addition to the three wires intended for phases A, B and C, we have a unique wire for the connection of neutral. Nothing prevents using a delta or triangle connection on the secondary (low voltage).

One possibility is to connect a balanced load on the secondary, which is when we have the same impedance in all three phases, thus having the same power factor. If the load is balanced, then the electric current flowing through the neutral must be equal to zero. Another possibility is to connect an unbalanced load where the impedances connected to the secondary have different values in at least one of them. In this case, the electric current flowing through the neutral is nonzero. The unbalanced circuit will be studied in another chapter. For now, let's focus on the study of the balanced circuit.

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Let's look at the four possible ways we can connect the transformer secondary together with the load, assuming the sequence is positive or straight, also known as the sequence ABC. For the inverse sequence we must exchange B for C and vice versa. As a consequence, all changes to the negative sequence should be opposite to those made to the positive sequence.

    2.1   Secundary in "Y" and Load in "Y"

This configuration is widely used in the distribution of electricity in homes. In the Figure 93-01 we present the circuit where we use a Y connection on the transformer secondary. We use the same type of connection for the load. Since the circuit is balanced, so Z_A = Z_B = Z_C.

Notice that I_linha = I_fase. As stated earlier, the neutral current must be zero. Then we can treat the circuit as if it were a single phase circuit. See the Figure 93-02 for the simplification of the circuit that will allow the current and power to be calculated in the load.

Note that we use phase A as a reference. Thus, the phase current A will have the phase angle equal to the difference between the angle of   V_AN ∠ θ_A   and   Z_A∠ θ_Z,   that is:

    2.2   Secundary in "Y" and Load in "Delta"

In the Figure 93-03 we present the circuit where we use a Y connection on the transformer secondary and the load is connected as a delta circuit. How the circuit is balanced, then Z_AB = Z_BC = Z_CA.

Note that in this case, the voltage applied to the load is the line voltage and not the phase voltage as in the previous item. So, about the load Z_AB we have the tension V_AB. If the phase voltage, rather than the line voltage, has been provided in the problem, then we must do the proper transformation. Let's recall the relationship between line voltage and phase voltage.

Note that we must add 30° to the phase voltage angle to get the correct line voltage angle and multiply its magnitude by √ 3.

Knowing the line voltage one can calculate the phase current in the load by dividing the voltage by the impedance. On the other hand, you can calculate the line current, if necessary, using the equation below.

In this case, we must multiply the phase current modulus by √ 3, beyond subtract 30° from the angle θ_F of the phase current to get the correct angle of the line current.

    2.3   Secundary in "Delta" and Load in "Y"

Let's study this configuration from a didactic point of view, as it is not widely used in practice due to the fact that we have no neutral reference in the secondary. Then, using the transformer secondary in the delta configuration, the transformer provides line voltage. The circuit is shown in the Figure 93-04. Since the load is on Y we must find the phase voltage by dividing the line voltage by √3. So to find the line current (I_L) flowing through the load (in this particular case, the phase current equals line current), we must divide the phase voltage by the phase impedance. In a balanced circuit the three currents are equal in modulus, varying only the angle.

Here we must pay attention to the angle that the phase voltage will assume. When we move from the star circuit to delta, we advance the line voltage angle by 30° to the phase voltage angle. Now as we move from the delta circuit to the star circuit we must delay the phase voltage angle (V_F) 30° relative to the line voltage angle (V_L). Therefore, we must use the equation below.

    2.4   Secundary in "Delta" and Load in "Delta"

This setting also has no reference to neutral . However, it is a configuration widely used in transmission lines that carry large amounts of energy between two distant points. As a rule, the starting point of the line has a transformer, called the voltage elevator transformer . The secondary of this transformer operates on tens, or even hundreds of thousands of volts. On the other side of the transmission line, we have another transformer, called the voltage lowering transformer, with the purpose of reducing the grid voltage to values used in the electricity distribution system. So the load on the secondary of the first transformer is the primary of the second transformer. See the Figure 93-05 for the layout of this configuration.

Note that in this configuration there is no need to change the voltage phase. We should only correct the angle of the phase and line currents. To do this, after calculating the phase current we can find the line current by multiplying the magnitude of the current phase by √3 and subtract 30° from its angle, that is, we must use eq. 93-02, repeated here.

In this equation, θ_F represents the angle of the phase current and by subtracting 30° we find the line current angle.