This chapter consists of the items below. If you want to go directly to an item, click it.

Industrial and residential consumers use equipment that can be single phase and / or three phase. In general, the distribution networks offer the two types of voltage, and the single phase is obtained from the three-phase using only the neutral and one of the phases. As the load on the three phases changes constantly, a four-wire (neutral + 3 phase) system is used to maintain stable voltage and provide a return path for the neutral current due to load unbalance. This system is used by energy distribution companies electrical, when in low voltage, for homes, commercial rooms and small industries.

For a three-phase system to be considered unbalanced there are two possible situations: the first is that the voltages of the source are not equal in modulus and / or differ in phase (different angles); the second, in which case the load impedances are unequal (at least one of them). On this site, we will give priority to the second case which is the most likely situation to happen.

In addition, we can have four-wire systems (as mentioned earlier) or a three-wire system. Let's look at the two cases.

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For the system to have four wires, both the source and the load must be connected in the star (or "Y") configuration. By using neutral, we ensure that the source provides the load with a stable voltage. So, let's look at the circuit shown in the Figure 84-01 where, for simplicity, only the load appears.

To calculate the currents I_A, I_B and I_C, just apply the Ohm's law for each phase. Remembering that since the circuit is unbalanced, Z_A, Z_B e Z_C they're not the same. In the equations below, we must take into account the phase of both, the voltages and the impedances. Thus, we obtain:

And to calculate the current I_N (current in the neutral), it can be calculated by applying the Kirchhoff's law for currents to the node N, obtaining:

Note that in this case it does not matter if the voltage source is balanced or not, since the neutral guarantees the stability of the phase voltages on loads. Do not forget that the equation expresses a sum phasor.

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When the system has no neutral and the circuit is unbalanced, to solve the circuit we must use nodal analysis and / or mesh analysis. As we have four types of configurations, that is, star-star, star-triangle, triangle-star and triangle - triangle, let's look at each one separately.

    3.1   Star - Star Configuration

When the load in this configuration does not have the neutral, then a potential difference between the points N and N' appears. That is, there is a displacement of the neutral. This leads to different voltage-per-phase values for each load.

In the Figure 84-02 we show the circuit without the neutral wire. For the solution of this case we must find out the value of the potential difference between the points N and N', that is, V_NN'. Knowing the value of V_NN', just do the mesh equation for each phase and we find I_A, I_B and I_C. To calculate V_NN' there is a practical method called neutral displacement method. Let's study it.

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    3.1.1   Neutral Displacement Method

Let us develop an equation that allows us to calculate the voltage V_NN'. For this, we will express the line currents in terms of the admittances and the voltages in the loads. Like this:

These relationships are in accordance with the circuit shown in the Figure 84-03. We must remember that the admittances are the inverse of the impedances. Now let's apply the law of Kirchhoff to currents to the node N'.

Let's replace the line currents in this last equation with the values found previously. Soon:

On the other hand, based on the circuit of the figure shown in item 3.1, we can write the following relations:

Substituting these values into the previous equation and working algebraically the expression, we arrive at the equation sought.

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    3.2   Star - Delta Configuration

In this case we have to determine the line voltage for each phase to which the load is connected. So let's determine the phase current by dividing the line voltage by the corresponding phase impedance. To find the line currents we need to make a phasor sum of the phase currents involved in the respective line current.

Note that in the model circuit shown in the Figure 84-04, we represent the impedances with different phase angles because we are facing an unbalanced circuit. The same can occur with line voltages. Let's assume that we have   V_AB∠ θ₁,   V_BC∠ θ₂   e   V_CA ∠ θ₃. Therefore, dividing these voltages by the respective impedances we will obtain the phase currents. By doing the phasorial sum of these currents we obtain the line currents. See below the required steps.

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    3.3   Delta - Star Configuration

In this case, to solve the problem, we need to use the mesh equations for the circuit and establish a system of three equations with three unknowns. Using this method we will be able to find the solution of the problem.

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    3.4   Delta - Delta Configuration

This case falls on item 3.2 because the line voltages are supplied with their respective phases. Thus, just follow all the steps presented in said item to solve the problem, including using the equations presented to arrive at the results.

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For unbalanced circuits there is no validity of the equations studied in balanced circuits. For this, we must calculate the power in each phase, separately, and then sum them all together. See the correct procedure in the example below.

Be a circuit with load in the configuration delta, according to the circuit shown in the figure below, and with the following values of voltage and impedances:

V_AB = 100∠30°,    V_BC = 200∠-60°,     V_CA = 150∠150°,

Z_AB =10∠30°,     Z_BC =10∠45°    and     Z_CA =15∠-70°.

a) Calculate the phase and line currents.

b) Calculate the apparent, real and reactive power per phase and total.

First let's calculate the phase currents. Like this:

With these data and using the equations of item 3.2, we can calculate the line currents. So:

As we are working with an unbalanced circuit we must calculate the power for each phase. So let's calculate the apparent power for each phase in the complex form. We remind you that the apparent in complex form is calculated by the product between the voltage and the conjugate complex of the current. So:

From the complex power we can calculate the real and the reactive power, simply finding the Cartesian form of the complex apparent power.