Three-Phase Power

In one sentence: Three-phase power tells you how much useful power a machine takes from three AC wires when the three wires share the load equally.

$P = \sqrt{3}\, V_L I_L \cos\varphi$

Read it out loud: “P equals the square root of 3, times V sub L, times I sub L, times cosine phi. Real power P equals the square root of three times the line voltage V sub L times the line current I sub L times the cosine of the phase angle phi.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Three-phase AC: A three-phase supply carries three AC voltages on three wires. Each one is a wave like the single-phase AC in a house, but the three waves are timed so that their peaks follow one another, each one third of a cycle after the last. A third of a cycle is 120 degrees of phase. Factories and large motors draw from all three wires at once, and each one of the three waves is called a phase.
Line voltage $V_L$ and line current $I_L$: The small letter $L$ stands for line, the wires that run from the supply to the machine. $V_L$ is the voltage between two of the three line wires. $I_L$ is the current in one line wire. On a 400 V European supply, the voltage between any two lines is 400 V, while the voltage from one line to the neutral wire is about 231 V, because $400 / \sqrt{3} \approx 231$ .
Square roots, like $\sqrt{3}$: The square root of a number is the value that gives the number when it is multiplied by itself. The square root of 3, written $\sqrt{3}$ , is about 1.732, because $1.732 \times 1.732 \approx 3$ . The sign $\sqrt{\ }$ is called the radical sign. Your calculator has a square root key that gives this value directly.
Sine, $\sin$ , and the 120 degree spacing: The sine, written $\sin$ , turns an angle into a number from $-1$ to $1$ . For example, $\sin 60^\circ = \frac{\sqrt{3}}{2}$ , about 0.866. Two equal voltages that are 120 degrees apart give a difference whose size is $2 \sin 60^\circ = \sqrt{3}$ times one of them. That is where the $\sqrt{3}$ in the formula comes from.
A balanced load: A balanced load draws the same current in each of the three wires, and each of its three parts has the same power factor. Motors built for a three-phase supply are designed to be balanced. If one wire carries much more current than the others, the load is unbalanced, and the formula in this page is no longer exact.
Kilo and powers of ten, like $10^{3}$: A power of ten is ten multiplied by itself a set number of times. $10^{3}$ is $10 \times 10 \times 10 = 1000$ . The prefix kilo (k) means 1,000, so a kilowatt (kW) is 1,000 watts. Dividing by $10^{3}$ moves the decimal point three places to the left, so 5542.6 W is $5.5426$ kW. Multiplying by $10^{3}$ does the opposite.

What each symbol means

Symbol	Meaning
$P$	Real power of the whole three-phase load, the useful power that all three wires deliver together. It is measured in watts (W), and one watt is one joule of energy each second. A motor that uses 5.5 kW uses energy at the rate of 5,500 joules every second.
$\sqrt{3}$	The square root of 3, about 1.732. It is a pure number, so it has no unit. It appears here because the three phases are spaced 120 degrees apart, and the voltage between two line wires is $\sqrt{3}$ times the voltage from one line to neutral.
$V_L$	Line voltage, the voltage between two of the three line wires, measured in volts (V). One volt is one joule of energy for each coulomb of charge. On a 400 V supply, the voltage between any two of the line wires is 400 V.
$I_L$	Line current, the current in one of the three line wires, measured in amperes (A). One ampere is one coulomb of charge passing a point each second. In a balanced motor each wire carries the same current, so one number describes all three.
$\cos\varphi$	Power factor of each phase, the cosine of the phase angle between voltage and current in that phase. It has no unit, because it is a ratio, and it is always a number from 0 to 1. For a balanced load the same value applies to all three phases, so one number is enough.

The formula piece by piece

$V_L$: Start with the line voltage, the voltage between two of the line wires, in volts. It is the push the supply applies between the wires, and it is the number you read with a meter placed across two of the wires.
$I_L$: Next, the line current, the current in one wire, in amperes. It tells you how much charge each line wire carries every second, which sets how thick the wires and the fuses must be.
$\sqrt{3}\, V_L I_L$: Multiply the line voltage by the line current, then by $\sqrt{3}$ . For a balanced load this product is the total apparent power $S$ , in volt-amperes, that the three wires carry together. It is the three-phase version of volts times amperes for a single wire.
$\cos\varphi$: The power factor, which is the share of that total apparent power that does useful work. It is a number from 0 to 1, and for a balanced load it is the same in each of the three phases.
$P = \sqrt{3}\, V_L I_L \cos\varphi$: The whole formula in one sentence: the real power of a balanced three-phase load equals the square root of 3, times the line voltage, times the line current, times the power factor.
$P = S \cos\varphi$: The same law with the three-phase apparent power $S = \sqrt{3}\, V_L I_L$ as the starting point. Multiply the total apparent power by the power factor, and you get the real power in watts.

What the formula is doing, and why

A three-phase supply uses three alternating voltages, each one third of a cycle apart, carried on three wires. Factories and large motors use it because a balanced three-phase load draws steady power. With a single-phase supply and a resistive load, the power drops to zero twice in every cycle. With three phases the three pulses fill each other’s gaps, so for a balanced load the total power stays constant.

Why $\sqrt{3}$ ? The formula uses line values: the voltage between two line wires and the current in one line wire. The voltage between two lines is not the voltage from one line to neutral. Each line-to-line voltage is the difference between two phase voltages that are 120 degrees apart. Subtracting two equal waves 120 degrees apart gives a wave $\sqrt{3}$ times as big, and that is where the factor comes from.

Why multiply by the power factor? The same reasoning as for one phase applies to each phase. Only the in-step part of each phase’s current delivers net energy, and the cosine of the phase angle is that in-step share. For a balanced load all three phases contribute the same amount, so the total real power is three times one phase’s share, which works out to $\sqrt{3}\, V_L I_L \cos\varphi$ .

Star and delta. The formula works for both star (wye) and delta connections, because it uses the line values you can measure on the wires. Inside the load the voltages and currents are different, but you do not need those internal values. In a star load, the line current equals the current in each winding. In a delta load it does not, but the line values in the formula still give the right total.

When it does not apply. The formula assumes a balanced load: equal currents and equal power factors in all three phases. An unbalanced load, such as one big single-phase heater connected to only one wire, breaks that assumption. In that case the power must be found for each phase separately and then added up.

An everyday picture

Picture three rowers in a boat, each pulling with the same strength and taking turns in an even rhythm. Because each one pulls the same amount at its own moment, the boat moves forward smoothly instead of lurching. A balanced three-phase load works the same way: three equal contributions, spaced evenly in time, add up to a steady total. If one rower pulls much harder than the others, the boat lurches, and an unbalanced load behaves the same way.

Worked example: A 400 V three-phase motor drawing 10 A at a power factor of 0.8

A balanced three-phase motor is supplied at a line voltage of $V_L = 400$ V. Each of the three line wires carries $I_L = 10$ A, and the power factor is $\cos\varphi = 0.8$ . We want the real power $P$ . On the way we will also find the apparent power $S$ , first in volt-amperes and then in kilovolt-amperes.

Find the factor for three phases. The formula starts with $\sqrt{3}$ , the number that gives 3 when multiplied by itself. That number is about 1.732, so $\sqrt{3} \approx 1.732$ . This factor turns the line voltage and the line current into the total for all three wires.
Find the apparent power in volt-amperes. Multiply the factor by the line voltage and the line current: $S = \sqrt{3} \times 400\ \text{V} \times 10\ \text{A} \approx 6928\ \text{VA}$ . The numbers 400 and 10 multiply to 4000, and 1.732 times 4000 is 6928. The unit is volt-amperes because volts times amperes gives volt-amperes.
Convert volt-amperes to kilovolt-amperes. The prefix kilo means 1,000, so we divide by 1,000 to change VA into kVA. Dividing by $10^{3}$ moves the decimal point three places to the left, so $6928\ \text{VA} = 6.928\ \text{kVA}$ . The number is the same size, only written with a bigger unit.
Multiply by the power factor. Real power is the share of $S$ that does useful work, so we multiply by 0.8, which is 80 percent: $P = S \times 0.8 = 6928.2\ \text{VA} \times 0.8 \approx 5542.6\ \text{W}$ . We keep the unrounded 6928.2 here, so the rounding error does not grow.
Convert watts to kilowatts. Dividing by 1,000 again moves the decimal point three places to the left, so $5542.6\ \text{W} = 5.5426\ \text{kW} \approx 5.54\ \text{kW}$ . The kilowatt is the easier number to read for a machine of this size.
Check the units. Volts times amperes gives volt-amperes, and both $\sqrt{3}$ and the power factor have no units. So $S$ is in volt-amperes, and after multiplying by the power factor, $P$ is in watts. One watt is one joule of energy each second, so the answer is a rate of energy use.
Say the answer in everyday terms. 5.54 kW is the rate of energy use of a little over five and a half 1 kW electric heaters running at once. The three line wires each carry 10 A, and the supply must be able to deliver 6.93 kVA to meet that demand.

Answer: The motor uses about 5.54 kW of real power. Its apparent power is about 6.93 kVA, and with a power factor of 0.8 the real power is 80 percent of that.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using the voltage from one line to neutral as $V_L$ . The formula needs the voltage between two line wires. In a star-connected supply that voltage is $\sqrt{3}$ times larger, so using the smaller value gives a power that is about 42 percent too small.
Using 3 instead of $\sqrt{3}$ . That overstates the power by about 73 percent, because 3 is about 1.73 times $\sqrt{3}$ .
Using the formula for an unbalanced load. It assumes the three phases carry equal currents and have equal power factors. If they do not, add up the power of each phase separately.
Mixing up line current and phase current in a delta load. The formula uses the current in the line wire, which differs from the current inside each winding. Using the winding current gives the wrong answer for a delta load.

Where you will meet it

Industrial motors and pumps, which are rated in kilowatts and often run on a three-phase supply.
Power distribution in factories and large buildings, where three-phase supply delivers heavy loads.
Sizing three-phase cables, breakers and transformers, which must carry the line current and the full apparent power.
Large electric vehicle chargers and heat pumps, which are often wired for a three-phase supply.

Questions people ask

What is line voltage?

Line voltage is the voltage between two line wires. On a common European three-phase supply of 400 V, the voltage from each line to the neutral wire is about 231 V, while the voltage between any two lines is 400 V.

Why is three-phase power steady?

The three voltages are spaced a third of a cycle apart. For a balanced load, the power from the three phases adds up to a steady total. Single-phase power, by contrast, rises and falls with every half cycle.

Is the formula different for a delta connection?

The same formula works for both star and delta loads, because it uses the line values you measure on the wires. The voltages and currents inside the load are different, but you do not need them.

Why do the line current and power factor both appear?

The line current sets how much current flows in the wires, and the power factor sets how much of the apparent power is real power. Together they give the real power the load uses.

AC Power Factor: Power factor is a number from 0 to 1 that tells you what share of the power an AC device draws is turned into useful work.
RMS Voltage: RMS voltage is the steady voltage that heats a resistor as much as an alternating voltage does, and for a sine wave you get it by dividing the peak voltage by the square root of 2.
Electric Power: Electric power tells you how fast a device uses energy. You get it by multiplying the voltage, the push, by the current, the flow, and the answer is measured in watts.