AC Power Explained: kW, kvar, kVA & Power Factor

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Imagine a motor drawing electricity from the grid. Some of that electricity becomes useful work, while some circulates between the supply and magnetic fields. The cables and transformer must carry the total current — not just the part doing useful work.

That is why AC systems use three connected quantities: active power (P), reactive power (Q) and apparent power (S).

Try the calculator →Read first, or test the formulas as you go.

01 · Overview

The three kinds of AC power — in plain English

For a simple sinusoidal AC circuit, it helps to picture three different questions:

Active power

How much real work or heat is produced?

Reactive power

kvar

How much power moves back and forth with magnetic or electric fields?

Apparent power

kVA

What total volt-ampere loading must the supply carry?

The three quantities are not added like ordinary numbers. In sinusoidal conditions, P and Q form perpendicular components of a triangle, and S is its hypotenuse.

02 · Active power

P — Power that performs useful work

Active power (P) is the average rate at which electrical energy is converted into work, light, heat or other energy. It is measured in watts (W) or kilowatts (kW).

A resistive heater uses active power to make heat. A motor draws active electrical power and converts part of it into shaft power, with the rest lost as heat, friction and other losses.

P=U\times I\times\cos\varphi\qquad\text{(single phase)}

In everyday language: P is the power that can do the job. But P alone does not tell you how much current flows through an AC cable.

03 · Reactive power

Q — Power associated with energy storage

Reactive power (Q) is associated with energy temporarily stored and returned by magnetic and electric fields. Its unit is the volt-ampere reactive (var), usually expressed as kvar.

Motors, transformers and reactors need magnetic fields, so their currents typically lag the voltage (inductive behavior). Capacitors do the opposite: their current typically leads the voltage.

Q=U\times I\times\sin\varphi\qquad\text{(single phase)}

Inductive loadQ positive (typical convention)Capacitive loadQ negative (typical convention)

Reactive power is not simply “wasted electricity.” It is part of how AC fields operate; however, it can increase current and loading without increasing useful kW. For a given active power, reducing unnecessary reactive power can reduce current.

Calculator note: KiloVolt currently reports the magnitude |Q|. It does not ask whether the load is inductive or capacitive, so the result does not indicate the sign of Q.

04 · Apparent power

S — What the supply must be able to carry

Apparent power (S) combines the voltage and the total RMS current. It is measured in VA or kVA. Transformers and many generators have kVA ratings because their electrical loading depends strongly on voltage and current.

S=U\times I\qquad\text{(single phase)}

S=\sqrt{3}\times U_{LL}\times I_L\qquad\text{(balanced three phase)}

For example, a load may do 8 kW of real work while requiring 10 kVA from the supply. The difference comes from its power factor, not from an extra 2 kW of energy consumption.

05 · Geometry of power

How P, Q and S fit together

Imagine P drawn horizontally and positive inductive Q drawn vertically. S completes the triangle. The angle φ (phi) is the angle between the P axis and S in this diagram.

The power triangleInductive (lagging) example

P and Q are perpendicular components of complex power. S is the magnitude of their combination; the triangle applies to sinusoidal AC.

S^{2}=P^{2}+Q^{2}

S=\sqrt{P^{2}+Q^{2}}

\cos\varphi=\frac{P}{S}

This triangle is especially useful when you know two powers and need to calculate the third. For capacitive loads, signed Q is below the horizontal axis; the same magnitude relationship still holds.

06 · Understanding AC

Why voltage and current are not always in phase

Alternating voltage and current rise and fall continuously. With a pure resistor, they rise and fall together. With inductors and capacitors, one waveform shifts in time relative to the other. That shift is called the phase angle φ.

A phasor diagram represents sinusoidal voltage and current as arrows. Their angular separation shows φ; a waveform diagram shows the same relationship across time.

Explore the phase angle

Choose a type of load to see how current moves relative to the voltage reference.

Phasor diagramTypical inductive load

Phasors show *relative phase*, not instantaneous voltage or current. The vector lengths are illustrative, not to scale.

Waveforms over timeSame angle, time domain

Both traces are normalized so the timing difference is easy to see. Their real amplitudes may be different.

Inductive · current lags: Current reaches the same point in its cycle after the voltage. Motors and transformers commonly behave this way. For this illustration, |φ| = 35° and displacement PF ≈ 0.82.

A lagging current reaches its positive peak later than voltage. A leading current reaches it earlier. The diagrams use voltage as the 0° reference and assume sinusoidal steady-state conditions.

07 · Power factor

What does cos φ actually tell us?

For sinusoidal voltage and current, power factor describes how much of the apparent power becomes active power. A power factor of 1 means P and S are equal in magnitude; the current is in phase with voltage.

\mathrm{PF}=\cos\varphi=\frac{P}{S}

1.00In phase

0.90Small angle

0.70More current for same kW

Suppose the electrical active power stays at 10 kW. At PF = 1, the apparent power is 10 kVA. At PF = 0.8, it becomes 12.5 kVA — so, at a fixed voltage, the supply current is 25% higher.

S=\frac{10}{0.8}=12.5\ \mathrm{kVA}

Important distinction: With distorted waveforms (for example, some power electronic loads), true power factor is not necessarily just cos φ. Harmonics also affect the RMS current. The present calculator assumes sinusoidal voltage and current.

08 · Connection type

Single-phase vs. three-phase calculations

In a typical European 230/400 V system, one phase to neutral is nominally 230 V (L-N). Between two different phases it is nominally 400 V (L-L). These are different voltages, so the calculator changes both the label and its default value when you change the system.

Single-phase AC

Use the voltage between line and neutral for a standard line-to-neutral load:

P=U_{LN}\times I\times\cos\varphi

I=\frac{P\times1000}{U_{LN}\times\cos\varphi}

Balanced three-phase AC

Use line-to-line voltage and line current for the total three-phase power:

P=\sqrt{3}\times U_{LL}\times I_L\times\cos\varphi

I_L=\frac{P\times1000}{\sqrt{3}\times U_{LL}\times\cos\varphi}

The √3 factor comes from the 120° separation between phases in a balanced three-phase supply. It is not a correction factor for cable losses or efficiency.

09 · Motors and machines

Power factor is not the same as efficiency

This is one of the most common misunderstandings. Efficiency (η) compares useful mechanical output power with electrical active input power. Power factor compares electrical active input power with apparent power.

\eta=\frac{P_{\mathrm{mechanical\ output}}}{P_{\mathrm{electrical\ input}}}

P_{\mathrm{electrical\ input}}=\frac{P_{\mathrm{mechanical\ output}}}{\eta}

For a motor marked 11 kW on its shaft-output rating, you first account for efficiency to find the electrical active input. Then use power factor to find kVA and current.

What does HP on a motor nameplate mean?

HP (horsepower) on many motor nameplates means rated mechanical output power at the shaft. It is not the electrical power drawn from the supply. KiloVolt uses mechanical (imperial) horsepower, with the conversion shown below.

1\ \mathrm{HP}=0.7457\ \mathrm{kW}

P_{\mathrm{electrical}}=\frac{\mathrm{HP}\times0.7457}{\eta}

For instance, a 10 HP motor operating at 90% efficiency delivers approximately 7.46 kW at the shaft and draws approximately 8.29 kW of electrical active power at its rated condition. Power factor is then used to calculate apparent power and current.

HP vs PS/CV: Mechanical HP ≈ 745.7 W, while metric horsepower (PS or CV) ≈ 735.5 W. Check the nameplate unit before entering the value.

How to use KiloVolt: If the entered kW is already the electrical input, leave efficiency at 100%. If it is a mechanical output rating, enter the motor efficiency. HP is interpreted as motor mechanical output and efficiency is required. Do not apply efficiency twice.

10 · Worked examples

Let's calculate two real-world examples

EXAMPLE A · THREE-PHASE MOTOR

11 kW motor, 400 V, PF 0.85, efficiency 90%

The 11 kW is mechanical shaft output, so electrical input power is higher.

1. Electrical input P

P_{\mathrm{in}}=\frac{11}{0.90}=12.22\ \mathrm{kW}

2. Apparent power S

S=\frac{12.22}{0.85}\approx14.38\ \mathrm{kVA}

3. Line current I

I_L=\frac{12.22\times1000}{\sqrt{3}\times400\times0.85}\approx20.75\ \mathrm{A}

4. Reactive power magnitude

|Q|=\frac{11}{0.90}\times\frac{\sqrt{1-0.85^2}}{0.85}\approx7.57\ \mathrm{kvar}

Displayed intermediate values are rounded. The final values are based on unrounded inputs.

EXAMPLE B · SINGLE-PHASE LOAD

2 kW electrical load, 230 V, PF 0.90

Here the 2 kW is already the electrical input, so no efficiency adjustment is needed.

1. Apparent power S

S=\frac{2}{0.90}\approx2.22\ \mathrm{kVA}

2. Current I

I=\frac{2000}{230\times0.90}\approx9.66\ \mathrm{A}

3. Reactive power magnitude

|Q|=2\times\tan(\cos^{-1}(0.90))\approx0.97\ \mathrm{kvar}

11 · Practical advice

Common mistakes to avoid

Confusing kW and kVA

They are only numerically equal when power factor equals 1 under the stated assumptions.

Using 230 V in a 400 V three-phase formula

Use the correct line-to-line or line-to-neutral voltage and the corresponding equation.

Applying motor efficiency to electrical-input kW

Efficiency only converts between useful output and active electrical input.

Assuming Q is always positive

Inductive and capacitive systems can have opposite reactive-power signs. This calculator shows magnitude.

Using power formulas as a complete cable design

Actual cable and protection selection also requires installation conditions, temperature, voltage drop, fault levels, disconnection checks and applicable standards.

PUT IT INTO PRACTICE

Now try it with your own numbers

Switch between single-phase and three-phase, adjust the power factor and view every calculation step with real values.

Open KiloVolt power calculator →

Scope: Introductory explanation for balanced, sinusoidal steady-state AC calculations. Actual installations, harmonic-rich loads and protection studies require additional methods, verified equipment data and checks against applicable requirements.