Let us now consider the balanced three-phase power system.

**Contents**show

We begin by examining the instantaneous power absorbed by the load.

*It is better to read about what is three phase circuit first.*

After we learn about three phase circuit, we will learn:

- Balanced three phase voltage
- Balanced three phase power
- Unbalanced three phase power
- Three phase power measurement

## Balanced Three-Phase Power Formula

This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are

(1) |

where the factor √2 is necessary because V_{p} has been defined as the rms value of the phase voltage.

If **Z**_{Y} = Z∠θ, the phase currents lag behind their corresponding phase voltages by θ. Thus,

(2) |

where I_{p} is the rms value of the phase current.

The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is,

(3) |

Applying the trigonometric identity

(4) |

gives

(5) |

Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does.

This result is true whether the load is Y- or ∆-connected.

This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later.

Since the total instantaneous power is independent of time, the average power per phase P_{p} for either the ∆-connected load or the Y-connected load is p/3, or

(6) |

and the reactive power per phase is

(7) |

The apparent power per phase is

(8) |

The complex power per phase is

(9) |

where **V**_{p} and **I**_{p} are the phase voltage and phase current with magnitudes V_{p} and I_{p}, respectively.

The total average power is the sum of the average powers in the phases:

(10) |

For a Y-connected load, I_{L} = I_{p} but V_{L} = √3V_{p}, whereas for a ∆-connected load, I_{L} = √3I_{p} but V_{L} = V_{p}.

Thus, Equation.(10) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is

(11) |

and the total complex power is

(12) |

where **Z**_{p} = Z_{p}∠θ is the load impedance per phase. (**Z**_{p} could be **Z**_{Y} or **Z**∆)

Alternatively, we may write Equation.(12) as

(13) |

Remember that V_{p}, I_{p}, V_{L}, and I_{L} are all rms values and that θ is the angle of the load impedance or the angle between the phase voltage and the phase current.

A second major advantage of three-phase systems for power distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage V_{L} and the same absorbed power P_{L}.

We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resistivity ρ), of the same length *l*, and that the loads are resistive (i.e., unity power factor).

For the two-wire single-phase system in Figure.(1a), I_{L} = P_{L}/V_{L}, so the power loss in the two wires is

(14) |

For the three-wire three-phase system in Figure.(1b), I’_{L} = |**I**_{a}| = |**I**_{b}| = |**I**_{c}| = P_{L}/√3V_{L} from Equation.(10)

Figure 1. Comparing the power loss in (a) a single-phase system, and (b) a three-phase system. |

The power loss in the three wires is

(15) |

Equations.(14) and (15) show that for the same total power delivered P_{L} and same line voltage V_{L},

(16) |

R = ρ*l*/πr^{2} and R’ = ρ*l*/πr^{2}, where r and r’ are the radii of the wires. Thus,

(17) |

If the same power loss is tolerated in both systems, then r^{2} = 2r’^{2}. The ratio of material required is determined by the number of wires and their volumes, so

(18) |

since r^{2} = 2r’^{2}. Equation.(18) shows that the single-phase system uses 33 percent more material than the three-phase system or that the three-phase system uses only 75 percent of the material used in the equivalent single-phase system.

In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system.

Read also : balanced wye-wye connection

## Power Formula for Balanced System Examples

For better understanding let us review the examples below:

1. Refer to the circuit in Figure.(2). Determine the total average power, reactive power, and complex power at the source and at the load.

Figure 2 |

*Solution:*

It is sufficient to consider one phase, as the system is balanced. For phase *a*,

Thus, at the source, the complex power supplied is

The real or average power supplied is −2087 W and the reactive power is −834.6 VAR.

At the load, the complex power absorbed is

where **Z**_{p} = 10 + j8 = 12.81∠38.66◦ and **I**_{p} = **I**_{a} = 6.81∠−21.8◦.

Hence

The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR.

The difference between the two complex powers is absorbed by the line impedance (5 − j2) Ω.

To show that this is the case, we find the complex power absorbed by the line as

which is the difference between **S**_{s} and **S**_{L}, that is, **S**_{s} + **S**_{l} + **S**_{L} = 0, as expected.

2. A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.

*Solution:*

The apparent power is

Since the real power is

the power factor is

3. Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Figure.(3a).

Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging.

Assuming the *abc* sequence, determine:

(a) the complex, real, and reactive powers absorbed by the combined load,

(b) the line currents, and

(c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.

Figure 3 |

*Solution:*

(a) For load 1, given that P_{1} = 30 kW and cos θ_{1} = 0.6, then sin θ_{1} = 0.8.

Hence,

and Q_{1} = S_{1} sin θ_{1} = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is

(3.1) |

For load 2, if Q_{2} = 45 kVAR and cos θ_{2} = 0.8, then sin θ_{2} = 0.6. We find

and P_{2} = S_{2} cos θ_{2} = 75(0.8) = 60 kW. Therefore the complex power due to load 2 is

(3.2) |

From Equations.(3.1) and (3.2), the total complex power absorbed by the load is

(3.3) |

which has a power factor of cos 43.36◦ = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR.

(b) Since S = √3V_{L}I_{L}, the line current is

(3.4) |

We apply this to each load, keeping in mind that for both loads, V_{L} = 240 kV. For load 1,

Since the power factor is lagging, the line current lags the line voltage by θ_{1} = cos^{−1} 0.6 = 53.13 ◦. Thus,

For load 2,

and the line current lags the line voltage by θ_{2} = cos^{−1} 0.8 = 36.87 ◦.

Hence,

The total line current is

Alternatively, we could obtain the current from the total complex power using Equation.(3.4) as

and

which is the same as before. The other line currents, **I**_{b2} and **I**_{ca}, can be obtained according to the *abc* sequence (i.e., **I**_{b} = 297.82∠−163.36◦ mA and **I**_{c} = 297.82∠76.64◦ mA).

(c) We can find the reactive power needed to bring the power factor to 0.9 lagging,

where P = 90 kW, θ_{old} = 43.36◦, and θ_{new} = cos^{−1} 0.9 = 25.84◦.

Hence,

This reactive power is for the three capacitors. For each capacitor, the rating Q’_{C} = 13.8 kVAR. The required capacitance is

Since the capacitors are ∆-connected as shown in Figure.(3b), V_{rms} in the above formula is the line-to-line or line voltage, which is 240 kV.

Thus,