On the other hand, the phase and line current are equal in modulus. Since the angle of the impedances are equal, the phase of the current will depend only on the angle of the phase voltage considered, that is:

Where θ_AN represents the phase voltage angle V_AN and, φ_ZA is the angle of impedance Z_A. In an analogous way we do to the other line currents.

These equations also serve for unbalanced circuits. For balanced circuits we should not forget that   φ_ZA = φ_ZB = φ_ZC.

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    2.1.1   Power Calculation

From the calculations of the phase voltage and phase current we can calculate the power in the load. Leaving of the basic principle that power is the product of the voltage by the electric current, for the phase A, the apparent power modulus is given by:

Using the standard nomenclature where V_F = V_AN and I_F = I_A, the apparent power modulus in the three phases will be the sum of the power in each phase. Therefore, the modulus of the apparent power total in the three-phase load will be:

Pay attention to the fact that we should use the RMS values to V_F and to I_F. For the Y configuration, we know that the line current equals the phase current, that is, it can be written as V_F = V_L /√3 . Therefore, we can write the modulus of the apparent power total in the three-phase load as a function of line current and line voltage, such that:

We should use the RMS values for V_L and I_L. Note that for the computation of the modulus of the apparent power it does not matter the phases (angles) of the voltages and currents.

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    2.2   Balanced Loads in "Delta"

For balanced circuits connected in the Delta configuration, see the Figure 82-02, we calculate the phase currents by dividing the line voltage by the corresponding impedance. Then the modulus of the apparent power of a single phase will be given by:

To express the modulus of the apparent power as a function of V_L and I_L we must transform the phase current into line current. For this, we use the equation below:

We must be aware of subtracting 30° from the angle θ_F of the phase current to get the correct angle of the line current, in addition to multiplying by √3 its magnitude.

On the other hand, to express the modulus of the total apparent power of the circuit, we simply use the absolute values (or modulus) of V_L and I_L. So:

"Note that for both the "Y" and the "Delta" connections, we arrive at the same equation to calculate the total apparent power modulus in the load, simply knowing the line voltage and the line current of one phase, regardless of which type of configuration the load is connected."

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Now let's look at how we can compute the complex power. For the computation of the complex power we must always find the difference between the angle of the voltage and the angle of the current. However, for balanced circuits, this difference represents the φ angle of the impedance of the circuit.

Then the apparent total power in the load, where the line current modulus and line voltage modulus expressed in effective value (RMS), will be given by:

In doing so, it should be clear that when we find the complex power automatically, we are also calculating the real power and the reactive power, that is:

and:

Then, with this information we can write the complex power in its cartesian form, or:

Thus, the real part of the complex power is the real power (or average, or active, or RMS), while the imaginary part is the reactive power. For a circuit with inductive predominance the reactive power is positive (+jQ) and for a circuit with capacitive predominance the reactive power is negative ( - jQ).

We can also express the apparent power in the polar form, or:

Where the apparent power modulus can be written, in an alternative way, as:

And the angle φ is given by:

In the Figure 82-03 we show the power triangle, where we can interpret all the equations studied in this item. Notice that on the axis of the Real we find the active power P and on the axis of the Imaginary we have the reactive power Q. For a positive value of φ we have a positive value of Q (inductive circuit). And for a negative value of φ we find a negative value for Q (capacitive circuit) since we know that sin (- φ) = - sin φ. Perfectly according to the chart in the figure below. On the other hand, the powers S , P and Q form a triangle rectangle where we can apply the theorem of Pythagoras and trigonometry, tools enough to prove all the equations shown in this item. Check !!!!.

We should not forget that cos φ is the power factor of the circuit. Case φ > 0 the power factor is said to be inductive and if φ < 0 then the power factor is said capacitive.