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Power Factor Calculator

A 15 kW load pulling 22 amps at 480 volts runs at a power factor of 0.82. Apparent power is 1.732 x 480 x 22 = 18,290 volt-amperes, and 15,000 / 18,290 = 0.82. Correcting it to 0.95 takes about 5.54 kVAR of capacitors.

Pick what your meter gives you. The target power factor is only used for the capacitor size.

Enter what the meter or the nameplate gives you and get the power factor, the apparent power, the reactive power and the capacitor size for a target power factor.

By the Lelifi editors ·

Key takeaways

Where a 0.82 power factor falls on the usual scalePoor0Fair0.7Good0.850.9510.82
The bands are a field guide, not a rule; the penalty threshold is set by each utility tariff.
How much capacitor a 15 kW load needskW = 15present PF = 0.82, so tan of the angle = 0.698target PF = 0.95, so tan of the angle = 0.3290.698 - 0.329 = 0.36915 x 0.369 = 5.54 kVAR of capacitors= 5.54 kVAR brings 15 kW to a 0.95 power factorThe tangents come from the phase angle, not from the PF itself.
The same tangent difference works for any load size, with kW as the multiplier.
On this page
  1. What power factor measures
  2. The formulas, by circuit type
  3. Worked example: a 1,500 watt load on 120 volts
  4. Worked example: a 15 kW motor on 480 volts, three phase
  5. Sizing capacitors to hit a target
  6. What a low power factor costs
  7. Reading a motor nameplate and running at part load
  8. Mistakes, harmonics and overcorrection

What power factor measures

Power factor is the ratio of the power that does work to the power the utility has to deliver. A resistive load such as a space heater or an incandescent lamp uses almost all of what it draws, so its power factor sits near 1.0. A motor, a transformer or a bank of fluorescent ballasts needs a magnetic field to run, and the current that builds that field sloshes back and forth every cycle without doing work. The meter still has to carry it.

The result is two different numbers for the same load. Real power, measured in kilowatts, is what the equipment turns into motion, heat or light. Apparent power, measured in kilovolt-amperes, is what the wires and the transformer carry. Divide the first by the second and you have the power factor, a number between 0 and 1 that is also the cosine of the angle between voltage and current.

The gap between the two is reactive power, measured in kVAR. It is not lost energy in the billing sense, since it circulates rather than being consumed, but it forces every wire, breaker and transformer upstream of the load to be bigger than the real work would require. That is the practical reason to measure it.

The formulas, by circuit type

Four relationships cover almost every case, and the only trap is the voltage. On a three phase circuit the voltage in the formula is line to line, the one written on the nameplate as 480 or 208, and the 1.732 is the square root of three. Using line to neutral voltage there, or forgetting the 1.732, is the most common way to get an answer that is off by 73 percent.

What you knowFormulaExample
kW and kVAPF = kW / kVA15 / 18.29 = 0.82
Watts, volts and amps, single phasePF = W / (V x A)1,500 / 1,680 = 0.89
Watts, volts and amps, three phasePF = W / (1.732 x V x A)15,000 / 18,290 = 0.82
kW and PFkVA = kW / PF15 / 0.82 = 18.29
kVA and kWkVAR = square root of (kVA squared minus kW squared)10.47 at 18.29 and 15

Watts and kilowatts are the same quantity with a factor of 1,000 between them, so pick one and stay with it. A 15 kW load is 15,000 watts. Mixing the two inside one formula is the second most common error, and it usually shows up as a power factor of 0.0008 or 820.

For a whole building the utility supplies kVA at the meter, so the monthly demand reading often gives the apparent power directly. Real power comes from the kWh meter and the hours of operation, or from a power quality logger that records both at once.

Worked example: a 1,500 watt load on 120 volts

A portable heater is rated at 1,500 watts and a clamp meter shows 14 amps on a 120 volt circuit. Apparent power is 120 x 14 = 1,680 volt-amperes, which is 1.68 kVA. Real power divided by apparent power is 1,500 / 1,680 = 0.893, so this heater runs at a power factor of about 0.89.

A resistive heater should be closer to 1.0, and the gap here comes from the small fan motor, the tolerance of the clamp meter and the fact that the element resistance changes as it heats. Readings on a plug in meter drift the same way, so treat a single spot reading as an estimate.

The gap costs 180 volt-amperes of capacity on a circuit that may be limited to 15 or 20 amps. On a 15 amp breaker at 120 volts the circuit can deliver 1,800 volt-amperes, so this heater sits close to the limit even though its real power is 1,500 watts.

Worked example: a 15 kW motor on 480 volts, three phase

A motor control center shows 22 amps per phase on a 480 volt three phase feeder, and the load is known to draw 15 kW. Start with apparent power: 1.732 x 480 x 22 = 18,290 volt-amperes, which is 18.29 kVA. Then 15 / 18.29 = 0.820.

Reactive power is the third side of the triangle. 18.29 x 18.29 = 334.5 and 15 x 15 = 225, so the difference is 109.5, and 109.5 ^ 0.5 = 10.47 kVAR. That is the magnetizing current the motor needs and the part a capacitor bank can supply locally.

Check the arithmetic backwards: 15 kW of real power with 10.47 kVAR of reactive power gives 18.29 kVA, and 15 / 18.29 returns 0.82. If a nameplate says 0.82 at full load, this motor is running at its rated point, which is what you want to see. A reading well below nameplate on a loaded motor points to a problem, and a reading above it usually means the motor is oversized for the job.

Sizing capacitors to hit a target

A capacitor bank does not reduce the real power a load uses. It supplies the reactive power locally, so the utility no longer has to send it down the line. The size needed is the difference between the reactive power at the present power factor and the reactive power at the target, which is the same as kW times the difference of two tangents.

For the 15 kW motor at 0.82, the phase angle is 34.9 degrees and its tangent is 0.698. At a target of 0.95 the angle is 18.2 degrees and the tangent is 0.329. The difference is 0.369, and 15 x 0.369 = 5.54 kVAR. A 5 kVAR fixed bank or a 7.5 kVAR step gets close, and most plants round up to a standard size.

Present power factorkVAR per 100 kW to reach 0.95Typical load
0.60100.46Idle or lightly loaded induction motors
0.7069.15Older motor heavy shop
0.7555.32Mixed motors and drives
0.8042.13Common uncorrected industrial reading
0.8529.11Partly corrected plant
0.9015.56Already close to the usual target

Those multipliers come from the same tangent difference, so a 100 kW load at 0.70 needs 100 x 0.6915 = 69.15 kVAR to reach 0.95. The US Department of Energy guide on reducing power factor cost publishes the same kind of table for field work, and utility engineers carry a version of it in their trucks.

What a low power factor costs

Every kilowatt of real power needs 1 / PF kilovolt-amperes of capacity. At 0.80 that is 1.25 kVA per kW, and at 0.60 it is 1.67 kVA per kW, which means the same work needs 67 percent more transformer and wire capacity. Put differently, 75 kW of load at 0.80 is 93.75 kVA and 112.8 amps at 480 volts; at 0.95 it is 78.95 kVA and 95.0 amps.

The current drop matters more than the headline number suggests. Line losses rise with the square of current, so 112.8 / 95.0 = 1.19 and 1.19 x 1.19 = 1.41, which means the losses in the conductors fall by about 29 percent after correction. Heat in cables, breaker temperatures and transformer loading all follow.

Power factorkVA per kWAmps for 100 kW at 480 V three phase
0.601.667200.5
0.701.429171.8
0.751.333160.4
0.801.250150.4
0.851.176141.5
0.901.111133.7
0.951.053126.6
1.001.000120.3

Many US utilities also charge for a low reading directly, through a power factor clause, a kVA demand charge or a ratchet that holds the peak for a year. The threshold and the penalty live in the utility's own rate schedule, and the numbers differ enough between utilities that no single figure applies everywhere. Read the tariff before promising a client savings.

Reading a motor nameplate and running at part load

A motor nameplate gives the power factor at rated load, not at every load. A 10 hp motor at 230 volts, 24 amps and 0.87 power factor is producing 10 x 0.746 = 7.46 kW of shaft power, and at 90 percent efficiency it draws 7.46 / 0.90 = 8.29 kW from the line. The apparent power at that point is 1.732 x 230 x 24 = 9,561 volt-amperes, and 8.29 / 9.56 = 0.867, close to the nameplate figure.

Run the same motor at half load and the power factor falls, often to 0.6 or lower, because the magnetizing current stays roughly the same while the working current drops. That is why an oversized motor is doubly wasteful: it runs at poor efficiency and poor power factor at the same time, and the utility sees the second one on the demand meter.

Efficiency and power factor are separate numbers and should not be confused. Efficiency compares shaft power to electrical input. Power factor compares real power to apparent power. A motor can have good efficiency and a poor power factor, which is exactly what happens at light load.

Mistakes, harmonics and overcorrection

Overcorrection is the first risk. A capacitor sized for full load stays connected when the motor is idle, and then the current leads the voltage instead of lagging it. Leading power factor can raise voltage at the motor terminals, and at the wrong combination of capacitance and inductance it can set up a resonance that amplifies harmonic currents. Switched banks that follow the motor, or a controller that steps capacitors in as load rises, avoid the problem.

Harmonics are the second. A true RMS meter reads a lower power factor than a meter that only looks at the fundamental, because distorted current carries apparent power that does no work. Variable frequency drives and switch mode power supplies are the usual sources. The fix for distortion is filtering or a reactor, not more capacitors, and a capacitor bank on the output of a drive is a mistake that damages both.

The Pacific Gas and Electric power factor guide walks through the same definitions and a worked correction, and it is a useful second opinion when a measured value looks odd. Two independent readings, taken at the same load, settle most arguments about a single number.

Frequently asked questions

What is a good power factor?

Above 0.95 is treated as good in most utility tariffs, and 0.90 is the common cutoff for a penalty. A reading below 0.85 usually means heavily loaded motors or a lightly loaded oversized motor.

How do I calculate power factor from kW and kVA?

Divide kW by kVA. A 15 kW load drawing 18.29 kVA runs at 15 / 18.29 = 0.82.

How do I find power factor from volts and amps?

Single phase: watts divided by volts times amps. Three phase: watts divided by 1.732 times line to line volts times amps. A 1,500 watt load at 120 volts and 14 amps is 1,500 / 1,680 = 0.89.

Why does three phase use 1.732?

1.732 is the square root of three, which comes from combining three phases 120 degrees apart. For 480 volts and 22 amps, 1.732 x 480 x 22 = 18,290 volt-amperes.

How many kVAR do I need to correct a 15 kW load from 0.82 to 0.95?

About 5.54 kVAR. The tangent at 0.82 is 0.698 and at 0.95 it is 0.329, so 15 x 0.369 = 5.54.

What does a 0.82 power factor cost in current?

Current rises by 1 / 0.82, which is 22 percent above the ideal. A 75 kW load draws 112.8 amps at 0.82 against 95.0 amps at 0.95, and line losses are 41 percent higher.

Can power factor be above 1?

No. Real power cannot exceed apparent power, so the ratio stays at or below 1.0. A meter showing more than 1 has a scaling error or is reading a leading current with the sign flipped.

What is the difference between kW, kVA and kVAR?

kW is working power, kVA is working plus reactive power as the wires see it, and kVAR is the reactive part alone. At 15 kW and 18.29 kVA the reactive part is 10.47 kVAR.

Does a capacitor bank reduce my kWh bill?

No. It reduces kVA and current, so it can cut demand charges, line losses and penalty clauses, but the kilowatt hours of real work stay the same.

Why did my power factor drop after installing variable frequency drives?

Drives draw current in short pulses, which adds harmonic distortion. A meter that measures true RMS reports a lower power factor even when the fundamental is fine, and the cure is a reactor or filter rather than capacitors.

How do I convert kVA to amps on a three phase circuit?

Amps = kVA x 1,000 / (1.732 x volts). For 100 kVA at 480 volts: 100,000 / 831.4 = 120.3 amps.

Is a 0.87 nameplate power factor the value at all loads?

No. It applies at rated load. At half load the same motor often falls to 0.6 or lower, because the magnetizing current does not shrink with the load.

What happens if I overcorrect a motor?

The current goes from lagging to leading, voltage can rise, and a resonant bank can amplify harmonics. Keep the bank close to the reactive power at full load and switch it with the motor.

Spotted a wrong number or a broken result?

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