Three-phase circuits
Nearly all of the world’s electrical energy is generated, transported and distributed as three-phase. It is not a gratuitous complication: with three phases the power arrives steady instead of pulsating, less copper is needed for the same job, and the rotating field that drives motors appears for free.
01Why three and not one
In single-phase, the instantaneous power pulsates at 100 Hz and passes through zero a hundred times per second. With three balanced phases, the sum of the three instantaneous powers is constant: the motor doesn’t vibrate and the torque is smooth.
Transporting the same power with three three-phase conductors requires considerably less material than with two single-phase ones, because the current per conductor is lower and the neutral carries almost no current.
Three coils placed 120° apart and fed by three voltages 120° apart produce a magnetic field that rotates by itself. From this comes the induction (asynchronous) motor, the most widely used motor on the planet, which doesn’t need brushes or electronics to start.
02The three voltages
A three-phase alternator has three windings spaced 120° mechanical apart on the stator. As the rotor turns, it induces in each one a sine wave of the same amplitude and frequency, but shifted by 120° with respect to the other two.
- Phase voltage (Vph): between a phase and the neutral. In Argentina, 220 V.
- Line voltage (VL): between two phases. In Argentina, 380 V.
- The ratio between them is not 2 but √3 = 1.732, because the two voltages are 120° out of phase and add vectorially: 220 × 1.732 = 381 V.
- The system is therefore called 380/220 V, and those two numbers appear together on any distribution panel.
03Wye and delta
The three loads —or the three windings— can be connected in two ways, and that choice changes voltages, currents and power.
| Wye (Y) | Delta (Δ) | |
|---|---|---|
| Voltage across each load | Vph = VL/√3 | Vph = VL |
| Current in the line | IL = Iph | IL = √3 · Iph |
| Neutral | Exists and can be brought out | None |
| Power with the same load | P | 3 P |
| Where it is used | Residential distribution, motor starting | Motors at running speed, power transformers |
In wye:
- Each resistor receives Vph = 380/1.732 = 219.4 V
- Iph = 219.4/100 = 2.19 A, and since IL = Iph, 2.19 A also flows in the line
- P = 3 × 219.4 × 2.19 = 1444 W
In delta:
- Each resistor receives the full 380 V
- Iph = 380/100 = 3.8 A; in the line, IL = 1.732 × 3.8 = 6.58 A
- P = 3 × 380 × 3.8 = 4332 W
The same load delivers three times more power in delta, and draws three times more line current. That factor of 3 is exactly the basis of the wye-delta starter: the motor is started in wye so that it draws one third of the current, and once it is up to speed it is switched to delta.
04Unbalanced loads and the role of the neutral
With three identical loads, the sum of the three currents is zero and the neutral does not conduct. In a real installation —one house per phase, single-phase outlets spread around, motors that start and stop— that is never exact, and the neutral carries the difference.
If the neutral is interrupted in an unbalanced installation, the common point is no longer at zero and shifts: the more lightly loaded phases end up with overvoltage —easily 300 V or more— and the more heavily loaded ones with low voltage. Equipment burns out throughout the house or building, and it is one of the most destructive faults in low-voltage distribution.
Hence two rules: the neutral never gets a fuse and is never disconnected on its own, and in a panel it is tightened with the same care as the phases, because a loose neutral produces exactly the same effect intermittently.
It is a modern and counterintuitive phenomenon. Electronic loads —switching power supplies, LED lamps, computers— generate a third harmonic (150 Hz) that is in phase on all three phases, not 120° apart. So, instead of canceling, they add up in the neutral: in an office building you can measure more current in the neutral than in any phase, with the conductor sized to carry very little. That is why installations with a lot of electronic load use a neutral of the same size as the phases, or larger.
05Power in three-phase systems
7.5 kW (10 HP) motor on 380 V, with cos φ = 0.85 and efficiency η = 0.88:
- Power drawn from the mains: Pabs = 7500/0.88 = 8523 W
- I = Pabs/(√3 · V · cos φ) = 8523/(1.732 × 380 × 0.85) = 8523/559.4 = 15.2 A
- The direct-on-line starting current is 6 to 8 times that: between 91 A and 122 A, for a few seconds. That is the reason for everything studied in motor starting.
In a three-wire system, two wattmeters are enough to measure the total power: P = W₁ + W₂. And their readings tell you something more:
- Equal readings → cos φ = 1.
- One reading of zero → cos φ = 0.5.
- One negative reading → cos φ less than 0.5. The total power is the difference.
- In addition, tan φ = √3 · (W₁ − W₂)/(W₁ + W₂), which makes it possible to obtain the power factor without a power factor meter.
06How it reaches the installation
| Item | Value in Argentina |
|---|---|
| Low-voltage distribution | 380 V between phases, 220 V between phase and neutral, 50 Hz |
| Residential service | Single-phase: one phase + neutral. Three-phase: three phases + neutral (for workshops, buildings and higher power demands) |
| Grounding (earthing) system | TT: neutral grounded at the utility’s transformer and exposed metal parts connected to the user’s own ground rod |
| Standard colors (AEA) | Phases brown, black and red; neutral light blue; protective conductor green-yellow |
| Medium voltage | 13.2 kV or 33 kV in urban distribution; the neighborhood transformer steps down to 380/220 |
All the detail on conductors, protective devices and panels is developed in electrical installations, and the transformer that makes that change of voltage possible, in transformers.
07In the lab
At a three-phase outlet in the workshop, measure the three voltages between phases and the three between phase and neutral. Calculate the ratio of each pair: it should come out at 1.73. Also note how much the three phases differ from one another: in a real network a 2 or 3 % difference is normal, and that already explains part of the neutral current.
With three identical incandescent lamps connected first in wye and then in delta on the same network (with suitable protection), observe the brightness and measure the line current. In delta they light up much more and draw three times more current. It is the experiment that establishes the 1:3 power ratio.
With the three lamps in wye with a neutral, use the clamp meter to measure the current in each phase and in the neutral: it should be almost zero. Then disconnect one lamp and measure the neutral again: it now carries practically the current of one phase. Remove two and measure again. This makes it clear what the neutral does and why its cross-section matters.
With a phase sequence meter (or the classic circuit of two lamps and a capacitor), determine the sequence. Then run a small three-phase motor, swap any two conductors and verify that it turns the other way. It is the simplest and most commonly used maneuver in the electrical workshop.
The voltage between phases is far more than double that of a household outlet, and the short-circuit current at a panel is thousands of amperes. Always work de-energized, verifying the absence of voltage with an instrument, with the panel tagged and locked out, and with personal protective equipment. No lab in this topic is done with the installation energized, except for measurement with a suitable instrument and under supervision.
08Common mistakes
| Symptom | Usual cause |
|---|---|
| 440 V was expected between phases and there is 380 | It was multiplied by 2 instead of by √3. The voltages add vectorially, not arithmetically. |
| The motor turns the wrong way | Phase sequence reversed. Swap any two conductors. |
| A 380/660 motor connected in delta on 380 V burns out | You must read the nameplate: on 380 V that motor goes in wye. In delta, each winding would receive √3 times its rated voltage. |
| Odd voltages in a building, burned-out equipment | Cut or loose neutral with unbalanced loads: the neutral point shifts and overvoltages appear. |
| The neutral carries more current than the phases | Third harmonic from electronic loads, which adds up instead of canceling. |
| The calculated power is triple the real one | Phase voltage was used with line current, or the √3 was forgotten. It is best to always use line values. |
| A wattmeter reads negative and seems broken | It is normal with cos φ < 0.5 in the two-wattmeter method: that reading must be subtracted. |
| The protection trips when switching from wye to delta | The changeover was made while the motor was still slow: the delta current adds to the starting current. The timer’s time setting must be respected. |
09Self-assessment
Why is the line voltage √3 times the phase voltage and not double?
Because the two phase voltages that are subtracted are 120° out of phase: the subtraction is vectorial. The resulting magnitude is 2·Vph·cos(30°) = √3·Vph = 1.732·Vph.
Three 50 Ω resistors in wye on 380 V: what are the line current and the power?
Vph = 219.4 V; I = 219.4/50 = 4.39 A (the same in phase and in line); P = 3 × 219.4 × 4.39 = 2890 W.
The same three resistors in delta: what changes?
Each one receives 380 V: Iph = 7.6 A, IL = 1.732 × 7.6 = 13.2 A and P = 3 × 380 × 7.6 = 8664 W: exactly triple.
How much current does the neutral carry with the three phases balanced?
Zero. The three currents are 120° out of phase and their vector sum cancels at every instant.
What happens if the neutral is cut with unbalanced loads?
The neutral point of the load shifts: lightly loaded phases end up with overvoltage (it can exceed 300 V) and heavily loaded ones with low voltage. Equipment is damaged. That is why the neutral gets no fuse and is never disconnected on its own.
A 4 kW motor, 380 V, cos φ 0.82 and η 0.86: what current does it draw?
Pabs = 4000/0.86 = 4651 W. I = 4651/(1.732 × 380 × 0.82) = 4651/539.7 = 8.6 A.
Why does wye-delta starting reduce the current to one third?
Because in wye each winding receives the line voltage divided by √3, so its current falls by that factor, and in addition the line current equals the phase current. Combining both effects, the line current turns out three times smaller than in delta. The starting torque also drops to one third: that is why it is only suitable for starting unloaded or with a light load.
In the two-wattmeter method, W₁ = 3 kW and W₂ = 1 kW. What are the total power and cos φ?
P = 3 + 1 = 4 kW. tan φ = √3 × (3 − 1)/(3 + 1) = 1.732 × 0.5 = 0.866 → φ = 40.9° → cos φ = 0.756.
Why does the third harmonic add up in the neutral?
Because shifting the fundamental by 120° amounts to shifting the third harmonic by 360°, that is, nothing: the third harmonics of the three phases end up in phase with one another and add arithmetically in the neutral conductor.
How do you reverse the direction of rotation of a three-phase motor?
By swapping any two phase conductors. That reverses the sequence and with it the direction of the rotating field.