When we’re dealing with three-phase power, there is another factor that must be taken into consideration and that adds another dimension to the power equation. Since each of the three phases are 120° apart from each other, they interact differently than we might expect.
When they share a common conductor, as they often do, the currents do not sum in a straightforward way; they add vectorially.
For example, if phase A and phase B have a common node, then any conductor connected to that node will carry the current from phase A and phase B. But since they are 120° apart from each other, we have to add them vectorially to find the resulting current through that conductor.
If we look at the illustration to the right, we can see that the current going through the common conductor is the vectorial sum of I1 and I2.Itotal = I1 + 12
For the sake of simplicity we’ll take the case where I1 and I2 are equal in magnitude. We can then solve for the total current by drawing a perpendicular line from the end of I1 to Itotal and breaking the triangle into two right triangles, as shown in Figure 6.6.
Since I1 and I2 are equal in magnitude and the angle between them is 120°, we know that the angle θ is 60° because the perpendicular line bisects the angle between I1 and I2.
We also know that the side opposite θ is half of Itotal for the same reason. Therefore, we can use the formula for the sine of an angle.
sinθ = Opposite side ÷Hypotenuse
sin60° = (Itotal/2 ) ÷ I1
I1 × 0.866 = Itotal 2
Itotal = 2 × I1 × 0.866Itotal = 1.732 × I1
Notice that if we had simply added I1 and I2 we would get 2 × I1 or
2 × I2, provided it’s a balanced three-phase system and I1 = I2. But since
I1 and I2 are 120° out of phase with each other, the total current in a conductor carrying both of these currents is only 1.732 × I1.
The purpose of this exercise is to illustrate how two sinewaves that are out of phase with each other, in this case the two currents, do not add in a straightforward manner. On the opposite end of the spectrum, if they are completely out of phase with each other, then they will cancel.
But if they are somewhat out of phase with each other, then the sum of the magnitude will be somewhere between 0 and twice the magnitude of one of them (assuming they are equal in magnitude).
In a three-phase system, there are two commonly used ways of wiring the three phases, as we will see later on. Both wiring methods use a common node between phases. Therefore, each of the three conductors carries the vectorial sum of two currents.
As a result, the final formula for calculating the power in a balanced three-phase system is:
Three-phase balanced AC (power watts) = Voltage (volts) ×Current amps x Power factor ×1.73,
where the power factor is the cosine of the phase angle.