Electronics Guide

Why Three Phases

A polyphase system uses two or more sources that are out of phase with one another. Three phases combine three advantages that a single-phase system cannot offer: constant total power, less conductor material for a given power, and a rotating magnetic field that starts and runs motors without auxiliary windings.

Constant Instantaneous Power

In a single-phase circuit, instantaneous power pulsates at twice the supply frequency. With v(t) = √2 V cos ωt and i(t) = √2 I cos(ωt − θ), where θ is the angle by which the current lags the voltage, the product is

p(t) = V I cos θ + V I cos(2ωt − θ)

The first term is the average power. The second term swings the delivered power above and below that average twice every cycle, so a single-phase induction motor develops a torque component that pulsates at twice line frequency.

Now add two more phases, each delayed by a further 120°. Phases b and c contribute the same average term. Their double-frequency terms become cos(2ωt − θ − 240°) and cos(2ωt − θ − 480°), and the second equals cos(2ωt − θ − 120°). Three equal cosines spaced 120° apart sum to zero at every instant, so the total power is

p_total(t) = 3 V_ph I_ph cos θ

The total instantaneous power of a balanced three-phase load is constant, even though the power in each phase still pulsates. Reactive power still flows in each phase, but its oscillations cancel in the total. A three-phase motor therefore produces smooth torque, and a three-phase generator sees a steady mechanical load. The result requires balance: equal voltage magnitudes, exact 120° spacing, and identical impedances in all three phases.

Conductor Economy

Compare two lines of equal length that deliver the same power P at unity power factor. Both have the same voltage V between conductors and the same total I²R loss. The single-phase line has two conductors, each carrying I₁ = P / V, so its loss is 2 I₁² R₁ = 2 P² R₁ / V². The three-phase line has three conductors, each carrying I₃ = P / (√3 V), so its loss is 3 I₃² R₃ = P² R₃ / V².

Setting the two losses equal gives R₃ = 2 R₁, so each three-phase conductor needs only half the cross-sectional area of a single-phase conductor. Three half-size conductors contain 1.5 units of metal against 2 units for the single-phase pair. The three-phase line therefore needs 75 percent of the conductor material. The comparison holds the voltage between conductors constant, because that voltage sets insulation and clearance requirements. A four-wire system adds a neutral conductor, which carries only unbalance and certain harmonic currents, as later sections show.

Rotating Magnetic Fields

Place three identical windings around a motor stator with their magnetic axes 120 electrical degrees apart, and feed them from a balanced three-phase supply. Each winding produces a field that pulsates along its own axis. Adding the three pulsating fields as space vectors gives a field of constant magnitude, 3/2 times the peak field of one winding, that rotates at constant angular velocity. For a machine with P poles on a supply of frequency f, the field turns at the synchronous speed

n_s = 120 f / P

where n_s is in revolutions per minute (rpm). A four-pole machine has a synchronous speed of 1,800 rpm on a 60 Hz supply and 1,500 rpm on a 50 Hz supply. The rotating field lets an induction motor start without help.

Two-phase systems, with sources 90° apart, also deliver constant power and a rotating field, but they need four conductors, or three with a shared return that carries √2 times the phase current. A three-phase line achieves both with only three conductors of equal size.

Balanced Sources and Phase Sequence

A three-phase generator has three identical windings displaced 120 electrical degrees around its stator. As the rotor field sweeps past them, each winding generates the same sinusoidal voltage, delayed one-third of a cycle behind the winding before it. A balanced set of voltages has equal magnitudes and exactly 120° spacing. Taking phase a as the reference, a balanced set of phase voltages is

V_an = V_ph ∠ 0°, V_bn = V_ph ∠ −120°, V_cn = V_ph ∠ +120°

In rectangular form, the three unit phasors are 1, −0.5 − j0.866, and −0.5 + j0.866, which sum to zero. The voltages of a balanced set therefore add to zero at every instant, and so do the three currents of a balanced load. That identity explains why a balanced wye load needs no neutral conductor.

Positive and Negative Sequence

The phase sequence, or phase rotation, is the order in which the three voltages reach their positive peaks. In the set above, phase a peaks first, then b, then c. This order is the abc, or positive, sequence. Reversing the order gives the acb, or negative, sequence, in which V_bn leads V_an by 120°. Phase labels vary with region and application; a-b-c, A-B-C, L1-L2-L3, R-Y-B, and U-V-W (for motor terminals) are all common.

Sequence matters in practice. It sets the direction in which a three-phase motor turns, so interchanging any two of the three supply conductors reverses the motor. A generator or inverter must match the grid's sequence before it can be paralleled with it, one of the conditions covered in grid synchronization and control. Electricians therefore confirm rotation with a phase-sequence meter before connecting a motor that must not run backward.

Double-Subscript Notation

Three-phase work relies on a double-subscript convention. V_ab is the voltage of node a measured with respect to node b, so V_ab = V_an − V_bn and V_ba = −V_ab. Likewise, I_ab is the current flowing from a to b through the element that joins them.

Wye and Delta Connections

Three sources, or three load impedances, can be joined in two basic ways, named for their shapes on a circuit diagram.

Wye (Star) Connection

In a wye connection, called a star connection in IEC practice, one end of each element joins a common point called the neutral. The other ends connect to the three lines. The voltage across each element is a line-to-neutral, or phase, voltage, and the current in each element is the same current that flows in its line. A wye source can supply a four-wire system, in which a neutral conductor lets single-phase loads connect between any line and neutral. It can also supply a three-wire system with the neutral left unconnected. In many systems, the neutral point of a wye transformer winding is also bonded to earth.

Delta (Mesh) Connection

In a delta connection, the three elements form a closed loop, and each line connects to one corner of the loop. Each element sits directly between two lines, so its voltage is a line-to-line voltage. Each line current, however, is the difference of the currents in the two elements that meet at its corner. A delta has no neutral point, so a plain delta supply is a three-wire system.

A delta source works only because the three voltages around its loop sum to zero. If they do not, because the windings are mismatched or because they contain third-harmonic voltages that add instead of canceling, a current circulates around the loop even with no load connected.

Source and Load Combinations

Any source connection can feed any load connection, so four combinations exist: wye–wye, wye–delta, delta–wye, and delta–delta. Transformers add flexibility, because one transformer can have a delta winding on one side and a wye winding on the other. A delta–wye distribution transformer is a common way to derive a four-wire 208Y/120 V or 480Y/277 V system from the three phase conductors of a primary feeder.

Line and Phase Quantities

Two kinds of quantity describe every three-phase circuit. Line quantities are the voltages between lines and the currents in the lines, which instruments on the supply conductors measure. Phase quantities are the voltages across and currents through the individual source windings or load impedances. The type of connection determines how the two relate. Nameplates and specifications normally state line-to-line voltage, so "480 V, three-phase" means 480 V between lines.

Wye: Line Voltage Is √3 Times Phase Voltage

For a balanced positive-sequence wye with V_an = V_ph ∠ 0°, the double-subscript rule gives

V_ab = V_an − V_bn = V_ph [1 − (−0.5 − j0.866)] = V_ph (1.5 + j0.866) = √3 V_ph ∠ 30°

Each line-to-line voltage is √3 times the phase voltage. In positive sequence, each line voltage also leads its phase voltage by 30°: V_ab leads V_an, V_bc leads V_bn, and V_ca leads V_cn. On a phasor diagram, the three line voltages form the sides of an equilateral triangle with the neutral at its center. The line currents equal the phase currents, because each line is in series with one phase.

Delta: Line Current Is √3 Times Phase Current

In a balanced delta load, the voltage across each impedance is a line voltage, and the phase currents I_ab, I_bc, and I_ca form a balanced set. In positive sequence, I_ca = I_ab ∠ +120°, so Kirchhoff's current law at corner a gives

I_a = I_ab − I_ca = I_ab (1.5 − j0.866) = √3 I_ab ∠ −30°

Each line current is √3 times the phase current and, in positive sequence, lags its phase current by 30°. In negative sequence, both shifts change sign: line voltage lags phase voltage in a wye, and line current leads phase current in a delta.

The √3 ratios and 30° shifts hold only for balanced sets. In an unbalanced system, each line and phase quantity must be found individually with Kirchhoff's laws.

Per-Phase Analysis of Balanced Systems

A balanced three-phase circuit contains three identical copies of one single-phase problem, rotated by 120°. Per-phase analysis solves one copy and writes down the other two by symmetry.

The method rests on one fact: in a balanced wye–wye system, the load neutral and the source neutral are at the same potential, whether or not a conductor joins them. Where line impedance separates source from load, lowercase subscripts mark source terminals and capital letters mark load terminals. With source phase voltages V_an, V_bn, and V_cn measured from the source neutral n, Millman's theorem gives the voltage of the load neutral N, relative to n, as

V_Nn = (V_an Y_a + V_bn Y_b + V_cn Y_c) / (Y_a + Y_b + Y_c + Y_n)

Here Y_a, Y_b, and Y_c are the admittances of the three phase paths, including their line impedances, and Y_n is the admittance of the neutral conductor, which is zero when no neutral exists. With equal phase admittances, the numerator is one admittance times the sum of a balanced set of voltages, which is zero. No current flows between the neutral points, so an ideal wire can join them, and each phase becomes an independent single-phase circuit.

Procedure

1. Convert every delta-connected load and source to its wye equivalent. A delta load of Z_Δ per phase becomes Z_Y = Z_Δ / 3. A positive-sequence delta source with line voltage V_ab becomes a wye source with V_an = (V_ab / √3) ∠ −30°.
2. Draw the phase a circuit: the phase a source voltage, the line impedance, and the wye-equivalent load, returned through an ideal neutral. Loads in parallel combine as ordinary impedances.
3. Solve that single-phase circuit for the line current, the load voltage, and the power in phase a.
4. Obtain the phase b and phase c results by shifting every phasor by −120° and +120°, respectively, for positive sequence.
5. Convert back to delta quantities where needed with the √3 and 30° relationships, and multiply the per-phase power by three.

Worked Example: Delta Load on a 480 V Feeder

A balanced positive-sequence 480 V source feeds a delta-connected load of Z_Δ = 19.2 + j14.4 Ω per phase, which is 24 Ω ∠ 36.87° with a power factor of 0.80 lagging. The feeder has an impedance of Z_line = 0.1 + j0.2 Ω per conductor.

1. Convert the load to wye: Z_Y = Z_Δ / 3 = 6.4 + j4.8 Ω, a magnitude of 8.0 Ω.
2. Take the source phase voltage as reference: V_an = 480 / √3 = 277.1 V ∠ 0°.
3. Add the impedances in the phase path: Z_line + Z_Y = 6.5 + j5.0 Ω = 8.201 Ω ∠ 37.57°.
4. Find the line current: I_a = (277.1 V ∠ 0°) / (8.201 Ω ∠ 37.57°) = 33.79 A ∠ −37.57°.
5. Find the load's wye-equivalent phase voltage: V_AN = I_a Z_Y = 270.3 V ∠ −0.70°. The load's line voltage is √3 × 270.3 = 468.3 V, a drop of 2.45 percent from the source.
6. Find the current inside the delta by dividing the line current by √3 and advancing it 30°: I_AB = 19.51 A ∠ −7.57°.
7. Find the power. The load absorbs 3 × 33.79² × 6.4 = 21.93 kW and 3 × 33.79² × 4.8 = 16.44 kvar. The feeder dissipates 3 × 33.79² × 0.1 = 343 W. The source supplies 22.27 kW at a power factor of 0.793 lagging.

As a check, the power computed inside the delta, 3 × 19.51² × 19.2, also equals 21.93 kW. The phase b and phase c currents are 33.79 A ∠ −157.57° and 33.79 A ∠ 82.43°.

Transformers in Per-Phase Analysis

A delta–wye or wye–delta transformer shifts its secondary line voltages by an odd multiple of 30°, usually ±30°, in a direction set by the winding connections and terminal labels; the shift reverses in negative sequence. For North American units with standard terminal markings, IEEE C57.12.00 makes the low-voltage side lag the high-voltage side by 30° in positive sequence. Per-phase analysis models the unit as an equivalent wye–wye transformer, then applies the shift to the secondary phasors. In IEC notation, the vector group states the shift as a clock number: Dyn11, for example, denotes a delta high-voltage winding, a wye low-voltage winding with its neutral brought out, and a low-voltage side displaced by 11 × 30° = 330° of lag, which equals a 30° lead.

Wye–Delta Load Conversion

Any three impedances connected in delta between terminals a, b, and c can be replaced by three wye-connected impedances that behave identically at those terminals, and the reverse is also true. For a delta with impedances Z_ab, Z_bc, and Z_ca, the equivalent wye impedances from each terminal to the center point are

Z_a = Z_ab Z_ca / (Z_ab + Z_bc + Z_ca)

Z_b = Z_ab Z_bc / (Z_ab + Z_bc + Z_ca)

Z_c = Z_bc Z_ca / (Z_ab + Z_bc + Z_ca)

Each wye impedance equals the product of the two delta impedances that meet at its terminal, divided by the sum of all three. The inverse conversion is

Z_ab = (Z_a Z_b + Z_b Z_c + Z_c Z_a) / Z_c

and Z_bc and Z_ca follow the same pattern, each divided by the wye impedance at the opposite terminal. These are the resistive delta–wye formulas of DC circuit analysis with complex impedances in place of resistances. For a balanced load, all three delta impedances equal Z_Δ, and the formulas reduce to Z_Y = Z_Δ / 3, or equivalently Z_Δ = 3 Z_Y.

The equivalence holds only at the terminals. Line currents, line voltages, and total complex power are identical, but the voltage across each element and the current through it are not. In a balanced load, each delta element sees √3 times the voltage of its wye counterpart and carries 1/√3 of the current, which matters when choosing component voltage ratings.

Wye–Delta Motor Starting

A wye–delta starter exploits the same ratio. It starts a motor designed to run in delta by connecting its windings in wye, so each winding receives 1/√3 of its rated voltage and draws 1/√3 of the current it would draw in delta. In delta, the line current would have been √3 times the winding current, so the starting line current falls to one-third of its direct-on-line value. Induction motor torque at a given slip varies with the square of the applied voltage, so starting torque also falls to one-third. As the motor nears full speed, the starter reconnects the windings in delta. Soft starters reduce starting current with smoother, adjustable control.

Power in Three-Phase Circuits

Total power in a three-phase circuit is the sum of the power in its three phases.

Real, Reactive, and Apparent Power

For a balanced load with phase voltage V_ph, phase current I_ph, and impedance angle θ, the total real power P in watts, reactive power Q in volt-amperes reactive, and apparent power |S| in volt-amperes are

P = 3 V_ph I_ph cos θ = √3 V_L I_L cos θ

Q = 3 V_ph I_ph sin θ = √3 V_L I_L sin θ

|S| = 3 V_ph I_ph = √3 V_L I_L

The line-quantity forms are identical for wye and delta loads. Complex power combines the three as S = P + jQ = 3 V_ph I_ph^*, where the asterisk denotes the complex conjugate of the phase current phasor. The power factor of a balanced load is P / |S| = cos θ, lagging for inductive loads and leading for capacitive ones.

One detail causes frequent errors. The angle θ is the impedance angle of one phase, which is the angle between a phase voltage and its own phase current. It is not the angle between a line voltage and a line current, which differs by 30°. For a purely resistive load in positive sequence, V_ab leads I_a by 30°, yet the power factor is 1.0.

Consider the delta load of the worked example fed directly from the 480 V source, without the feeder. Each phase current is 480 V / 24 Ω = 20.0 A, and each line current is √3 × 20.0 = 34.64 A. The load draws

P = √3 × 480 × 34.64 × 0.80 = 23.04 kW

Q = √3 × 480 × 34.64 × 0.60 = 17.28 kvar

|S| = √3 × 480 × 34.64 = 28.80 kVA

Line Current from a kVA Rating

Transformers, generators, and uninterruptible power supplies carry kilovolt-ampere ratings, and rated line current follows from I_L = |S| / (√3 V_L). A 75 kVA transformer with a 208 V secondary delivers 208 A per line, while the same rating at 480 V delivers 90.2 A.

Power Factor Correction

Suppose the example load is to be corrected from 0.80 to 0.95 lagging. Real power stays at 23.04 kW, and the reactive power permitted at the new power factor is 23.04 × tan(cos⁻¹ 0.95) = 7.57 kvar. The capacitors must supply the difference, 17.28 − 7.57 = 9.71 kvar, or 3.24 kvar per phase. At 60 Hz, ω = 377 rad/s, so a delta-connected bank needs

C = 3,236 / (377 × 480²) = 37.3 µF per capacitor

A wye-connected bank would need 111.8 µF per capacitor, three times as much, because each capacitor sees 277 V instead of 480 V. Low-voltage three-phase capacitor units are commonly connected in delta. After correction, the line current falls from 34.64 A to 29.17 A. Because I²R loss varies with the square of current, feeder losses upstream of the capacitors fall by about 29 percent. The harmonics and power factor article covers correction methods and the distinction between displacement and distortion power factor.

Power in Unbalanced Systems

With unbalanced loads, the three phases carry different powers, but total real and reactive power remain simple sums: P = P_a + P_b + P_c and Q = Q_a + Q_b + Q_c. Apparent power, and with it power factor, is no longer uniquely defined. Arithmetic apparent power adds the magnitudes of the three phase apparent powers, |S_a| + |S_b| + |S_c|. Vector apparent power takes the magnitude of their sum, |S_a + S_b + S_c|. The two give different power factors for the same unbalanced load, so an analyzer's three-phase power factor reading depends on the definition it uses. IEEE Std 1459-2010, which defined power quantities under sinusoidal, nonsinusoidal, balanced, and unbalanced conditions, also defined an effective apparent power intended to reflect the added losses that unbalance and neutral current cause. IEEE Std 1459-2025 has since superseded that edition.

Unbalanced Loads and Neutral Current

Real three-phase systems are rarely perfectly balanced. Lighting, receptacles, and single-phase motors connect between line and neutral or between two lines, and they switch on and off independently. Once the load is unbalanced, per-phase shortcuts no longer apply, and the full three-phase circuit must be solved.

Four-Wire Wye with a Solid Neutral

When a low-impedance neutral joins the load neutral to the source neutral, the source holds each phase voltage fixed, and each phase behaves as an independent single-phase circuit. The neutral carries the phasor sum of the line currents:

I_n = I_a + I_b + I_c

Consider a 208Y/120 V system with resistive loads of 3.6 kW on phase a, 2.4 kW on phase b, and 1.2 kW on phase c. The line currents are 30 A ∠ 0°, 20 A ∠ −120°, and 10 A ∠ +120°. In rectangular form, they are 30, −10 − j17.32, and −5 + j8.66 A, which sum to I_n = 15.0 − j8.66 A, or 17.3 A ∠ −30°.

With loads of equal power factor, the neutral current can never exceed the largest line current. Loads with different power factors can push it higher, and so can the harmonic currents discussed in the next section.

Three-Wire Wye and the Floating Neutral

If the neutral conductor is absent or broken, the load neutral is free to move, and Millman's theorem with Y_n = 0 gives its displacement. The loads of the previous example have resistances of 4 Ω, 6 Ω, and 12 Ω, or conductances of 0.25, 0.1667, and 0.0833 S, which total 0.5 S:

V_Nn = (120 ∠ 0° × 0.25 + 120 ∠ −120° × 0.1667 + 120 ∠ +120° × 0.0833) / 0.5 = 34.6 V ∠ −30°

The numerator equals the solid-neutral current of 17.3 A ∠ −30°, so the neutral shift is that current divided by the total load conductance. Subtracting V_Nn from each source phase voltage gives load voltages of 91.7 V on phase a, 124.9 V on phase b, and 151.0 V on phase c. The heavily loaded phase loses nearly a quarter of its voltage, while the lightly loaded phase rises 26 percent above nominal. Real loads are not constant resistances, but the pattern holds: an open neutral overvolts the lightly loaded phases of a four-wire system and can destroy equipment connected to them. High voltage on some circuits and low voltage on others, changing as loads switch, is the classic symptom of a loose or broken neutral.

Unbalanced Delta Loads

A delta load has no neutral to shift. Its phase voltages are the line voltages, which a stiff source holds fixed, so each phase current follows directly: I_ab = V_ab / Z_ab, and likewise for the other two phases. Kirchhoff's current law at each corner then gives the line currents, such as I_a = I_ab − I_ca. The line currents of an unbalanced delta are generally not √3 times any phase current, and they are no longer 120° apart. Because each load element sees its own line voltage, however, an unbalanced delta does not disturb the voltages across the other phases the way a floating wye does.

Voltage Unbalance and Motors

Unbalanced load currents flowing through the supply impedance make the line voltages themselves unequal. NEMA defines percent voltage unbalance as 100 times the maximum deviation of a line voltage from the average of the three line voltages, divided by that average. The U.S. Department of Energy's Motor Systems Tip Sheet #7, "Eliminate Voltage Unbalance" (2012), gives an example: line voltages of 462, 463, and 455 V average 460 V, and the largest deviation, 5 V, is an unbalance of 1.1 percent.

Small voltage unbalance does disproportionate harm. The tip sheet reports that current unbalance may be 6 to 10 times the voltage unbalance. It recommends keeping unbalance at the motor terminals at or below 1 percent and notes that NEMA MG 1-2011 requires derating above that level. It estimates the added winding temperature rise as

total temperature rise = balanced temperature rise × [1 + 2 × (percent unbalance)² / 100]

so a motor with an 80°C rise runs another 6.4°C hotter at 2 percent unbalance.

The general tool for unbalanced systems is the method of symmetrical components, which Charles L. Fortescue introduced in 1918. It resolves any set of three unbalanced phasors into balanced positive-sequence, negative-sequence, and zero-sequence sets. The negative-sequence voltage explains why unbalance heats motors: it drives a field that rotates backward relative to the rotor and induces rotor currents at nearly twice line frequency. The zero-sequence current is the part that returns through the neutral, and it leads directly to the problem of triplen harmonics.

Triplen Harmonics in the Neutral

Many single-phase loads draw nonsinusoidal current. A switch-mode power supply with a capacitor-input rectifier, for example, draws current in short pulses near the peaks of the voltage waveform. Fourier analysis expresses such a current as a fundamental plus harmonics at integer multiples of the line frequency. When identical nonlinear loads sit on all three phases, each harmonic order forms its own three-phase set.

If the fundamental currents of phases b and c lag phase a by 120° and 240°, their h-th harmonics lag by h × 120° and h × 240°. For the third harmonic, those angles are 360° and 720°, so the third-harmonic currents of all three phases are exactly in phase. The same holds for every harmonic whose order is a multiple of three. These are the triplen harmonics.

Positive- and negative-sequence harmonics cancel at the neutral, just as the fundamental does. Zero-sequence harmonics add arithmetically, so with balanced loads the neutral carries three times the triplen current of each phase:

I_n = 3 √(I₃² + I₉² + I₁₅²)

Here I₃, I₉, and I₁₅ are the RMS triplen currents in each phase, and higher triplens add further terms. A limiting case shows how large the result can be. Suppose the current pulses are narrow enough that no two phases ever conduct at the same instant. The neutral current then equals whichever phase current is flowing, so i_n² = i_a² + i_b² + i_c² at every instant. Averaging over a cycle gives I_n = √3 I_ph, or 173 percent of the phase current. A neutral sized like the phase conductors can overheat while no phase overcurrent device trips, which is why designers may specify double-size neutrals in buildings dominated by electronic loads.

Delta windings handle triplens differently. Zero-sequence currents can circulate around a delta loop, but they cancel in the line currents, because I_a = I_ab − I_ca subtracts two equal, in-phase components. A delta–wye transformer feeding nonlinear loads therefore traps balanced triplen currents in its delta winding. They do not reach the primary lines, but they still heat the transformer. By the same subtraction, triplen voltages in a wye source cancel in its line-to-line voltages. Common responses to triplen currents include oversized neutrals, transformers rated for nonsinusoidal loads, zigzag transformers that give zero-sequence current a local path, and active power filters with a neutral connection.

Measuring Three-Phase Power

A wattmeter measures the average product of the voltage across its voltage circuit and the current through its current circuit. Blondel's theorem states how many such elements a polyphase system needs. In a system of N conductors, N − 1 wattmeters measure total power when each current element sits in a different conductor and all voltage elements use the remaining conductor as their reference. A three-wire system therefore needs two wattmeters, and a four-wire system needs three.

Two-Wattmeter Method

Place current elements in lines a and c, and reference both voltage elements to line b. With no neutral, i_b = −i_a − i_c, so the total instantaneous power, with phase voltages measured from any common point n, becomes

p = v_an i_a + v_bn i_b + v_cn i_c = v_ab i_a + v_cb i_c

The two wattmeters average those two terms, so their sum is the total real power for any waveform and any degree of unbalance. For a balanced positive-sequence load with impedance angle θ, the readings are

W_a = V_L I_L cos(30° + θ) and W_c = V_L I_L cos(30° − θ)

Their sum is √3 V_L I_L cos θ, and their difference gives the reactive power and the power factor angle:

Q = √3 (W_c − W_a) and tan θ = √3 (W_c − W_a) / (W_c + W_a)

For the 23.04 kW example load at 0.80 power factor, the meters read 6.53 kW and 16.51 kW. Their sum is 23.04 kW, and √3 times their difference is 17.28 kvar. Below a power factor of 0.5, θ exceeds 60° and W_a becomes negative, but the total is still the algebraic sum of the two readings. The reactive-power formulas hold only for balanced loads, and in negative sequence the two readings trade places.

Three-Wattmeter Method

A four-wire wye system uses three elements, each measuring one line current against that phase's line-to-neutral voltage. The sum gives total power, and the individual readings give the power in each phase. Power analyzers typically offer both this and the two-wattmeter connection, as power meters and analyzers describes.

Common System Voltages

A three-phase system is named by its nominal voltages. North American practice writes the line-to-line voltage first and marks a wye with a Y, so 208Y/120 V is a wye system with 208 V between lines and 120 V from line to neutral. IEC 60038 writes the line-to-neutral value first, as in 230/400 V or 120/208 V. A single value, such as 480 V, denotes a three-wire system and gives the voltage between lines. The two values of each pair differ by a factor of √3, within rounding: 120 × √3 = 207.8 V, and 480 / √3 = 277.1 V.

IEC 60038 notes that 230/400 V is the result of the evolution of 220/380 V and 240/415 V systems, which still exist in some countries, and it also lists 347/600 V and 600 V for 60 Hz systems. Distribution primaries use the same notation; Pacific Gas and Electric (PG&E) lists 4,160Y/2,400 V, 12,000Y/6,930 V, and 20,780Y/12,000 V among its standard three-phase primary service voltages.

The High-Leg Delta

The 240/120 V four-wire delta needs special care. One of its three 240 V transformer windings is center-tapped, and the tap is grounded as the neutral. Two lines measure 120 V to neutral, but the third line, the high leg, sits at the far corner of the triangle, 240 × √3 / 2 = 208 V from the tap. A 120 V load connected to the high leg would receive 208 V. The U.S. National Electrical Code (NFPA 70) therefore requires the high-leg conductor to be durably and permanently marked orange, or identified by other effective means, at each point where a connection is made if the neutral is also present. It also generally assigns the high leg to the B-phase position in switchboards and panelboards.

Nominal, Service, and Utilization Voltages

A nominal voltage names a system; the voltage that equipment actually receives varies around it. IEC 60038 states that under normal operating conditions the supply voltage should not differ from nominal by more than ±10 percent. Utility rules can be tighter. PG&E's Electric Rule 2 holds 480 V services between 456 and 504 V and cites the lower utilization voltages of ANSI C84.1, which allow for voltage drop in customer wiring. The U.S. Department of Energy's Premium Efficiency Motor Selection and Application Guide (2014) notes that NEMA members customarily rate motors at about 95.8 percent of nominal system voltage, so motors for 480 V systems are rated 460 V, and motors for 240 V systems are rated 230 V. A 208 V system calls for 200 V motors, and PG&E's rule warns that 230 V motors will not perform satisfactorily on it.

Three-Phase Equipment

Motors

The three-phase induction motor is the most direct application of the rotating field. Its rotor turns slightly slower than synchronous speed, and the fractional difference, the slip s = (n_s − n) / n_s, induces the rotor currents that produce torque. As a balanced load, the motor has a per-phase equivalent circuit, and its nameplate states its winding connection. A motor rated 400 V in delta and 690 V in wye has windings designed for 400 V each. It runs in delta on a 400 V supply or in wye on a 690 V supply, and on a 400 V supply it can use a wye–delta starter. Synchronous machines use the same rotating field with a rotor that turns in step with it. AC induction motor drives extend these ideas to variable-speed operation.

Transformers

Three-phase power can be transformed by a bank of three single-phase transformers or by one three-phase unit on a common core. A delta–wye unit provides a neutral for four-wire loads, keeps zero-sequence currents on its wye side out of its primary lines, and shifts the phase by 30°. An open-delta bank omits one of three single-phase transformers and still supplies three-phase voltages. Each remaining transformer must then carry the full line current rather than 1/√3 of it, so the bank delivers only 1/√3, or 57.7 percent, of the full bank's capacity, which is 86.6 percent of the two remaining units' combined rating. The transformers article covers the underlying principles.

Rectifiers

A six-pulse diode bridge connects its positive output, at each instant, to the most positive line and its negative output to the most negative line. The output follows the tops of the six line-to-line voltages, so its ripple frequency is six times the line frequency, and it never falls below cos 30° = 86.6 percent of the line-to-line peak. Averaging over one 60° interval gives

V_dc = (3√2 / π) V_L ≈ 1.35 V_L

for an ideal bridge with no source inductance and no smoothing capacitor, or 648 V from a 480 V supply. A capacitor-filtered DC bus, as in most drives, rises toward the line-to-line peak of 1.41 V_L at light load. The input current contains harmonics of order 6k ± 1, where k is a positive integer (the 5th, 7th, 11th, 13th, and so on), and ideally no triplens. Two bridges fed from a delta winding and a wye winding, whose voltages are 30° apart, cancel the 5th and 7th harmonics at their common input and form a twelve-pulse rectifier with harmonics of order 12k ± 1. AC-DC conversion (rectification) treats these circuits in detail.

Inverters and Drives

A three-phase inverter reverses the process with six switches in three legs, one leg per output phase. With sinusoidal pulse-width modulation, each leg produces a fundamental with a peak of at most V_dc / 2 relative to the DC midpoint, so the largest undistorted line-to-line output is √3 V_dc / (2√2) ≈ 0.612 V_dc RMS. Triplen theory offers a way past that limit. Adding the same third-harmonic component to all three phase references flattens their peaks, and because that component is identical in every phase, it cancels in the line-to-line voltages that the motor sees. Third-harmonic injection and space-vector modulation raise the linear output limit by a factor of 2/√3, about 15.5 percent, to V_dc / √2 ≈ 0.707 V_dc. From the 648 V bus of the rectifier example, the limit rises from about 397 V to 458 V. Variable frequency drives combine a rectifier, a DC bus, and an inverter to run standard motors at adjustable speed, and DC-AC conversion (inversion) covers inverter topologies and modulation.

Summary

Three-phase systems deliver constant total power to balanced loads, save conductor material, and create the rotating field that drives induction and synchronous motors. In a balanced wye, line voltage is √3 times phase voltage and leads it by 30° in positive sequence. In a balanced delta, line current is √3 times phase current and lags it by 30°. Converting deltas to wyes with Z_Y = Z_Δ / 3 lets a balanced network be solved one phase at a time, and total real power is √3 V_L I_L cos θ for either connection, where θ is the impedance angle of one phase.

Unbalance removes those shortcuts. A four-wire neutral carries the phasor sum of the line currents, an open neutral overvolts lightly loaded phases, and triplen harmonics add in the neutral instead of canceling. Two wattmeters measure total power in any three-wire circuit, and three measure it in a four-wire circuit.

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