Electrical Trade & Electronics DC Circuit Fundamentals IEEE Std 1459-2010 power quantity definitions

Electrical Power Calculator (Watts, VA and kW)

This calculator separates the three numbers people routinely confuse: watts, volt-amperes and volt-amperes reactive. Give it a voltage and either a current or a resistance, tell it whether the circuit is DC or single-phase AC, and it returns real power (the part that does work and that your meter bills), apparent power (the part your conductors, breakers and transformers must actually carry), and reactive power (the part that shuttles back and forth doing neither). On DC the three collapse into one number; on AC they do not, and the gap is what power factor measures.

Calculator

This calculator runs in your browser. Enable JavaScript for live results — the inputs, formula and worked example below remain fully readable without it.

Inputs this calculator takes, with typical values
InputWhat to enterExample
Circuit typePick DC for batteries, solar strings and bench supplies; single-phase AC for ordinary branch circuits.Single-phase AC
What you knowChoose the pair you have measured or read from the nameplate.Voltage and current
VoltageUse the RMS voltage on AC — 120 V or 240 V for a typical North American branch circuit, 230 V in most of Europe.240 V
CurrentUse the RMS current on AC, as read by a clamp meter on one conductor.20 A
Resistance or impedance magnitudeOn DC this is plain resistance; on AC enter the magnitude of the impedance, |Z|.12 Ω
Power factorRead it from the motor or equipment nameplate; use 1.00 for heaters, incandescent lamps and other purely resistive loads.0.85
Hours of operation per dayUsed only to convert the real power into daily energy; set it to 0 if you do not need that figure.4 h

It returns

  • Real power — The power that does work and that a kilowatt-hour meter registers.
  • Real power in kilowatts
  • Apparent power — What the conductors, breaker and transformer have to be sized for.
  • Reactive power
  • Phase angle between voltage and current
  • Circuit current
  • Circuit voltage
  • Energy used per day

The formula

P=VIcosφ
S2=P2+Q2
P=I2R=V2R

In plain text: S = V · I ; P = S · PF ; Q = √(S² − P²)

  • PReal (active) power — the part that does work (watts (W))
  • SApparent power — what the conductors must carry (volt-amperes (VA))
  • QReactive power — exchanged with the source, does no net work (volt-amperes reactive (VAR))
  • VRMS voltage across the load (volts (V))
  • IRMS current through the load (amperes (A))
  • PFPower factor, cos φ for a sinusoidal waveform (0 to 1)
  • φPhase angle by which current lags or leads voltage (degrees)

On DC the power factor is 1 by definition, so S = P and Q = 0. The relationship Q = √(S² − P²) assumes sinusoidal voltage and current; with harmonic distortion, apparent power also contains a distortion component.

Updated Category DC Circuit Fundamentals Verified against published test cases Reading time 11 min

Watts, volt-amperes and VARs are three different quantities

On a DC circuit, power is simply voltage times current, and there is one number. On an AC circuit there are three, because voltage and current are sinusoids that need not peak at the same instant. IEEE Std 1459-2010 is the document that fixes the definitions, and its distinctions are the ones every nameplate, breaker schedule and utility bill relies on.

Real power, P, measured in watts, is the average of the instantaneous product of voltage and current over a cycle. It is energy that leaves the circuit as heat, light, torque or sound. This is what a kilowatt-hour meter integrates and what you pay for on a residential tariff.

Apparent power, S, measured in volt-amperes, is the plain product of RMS voltage and RMS current, with no regard for phase. It is the number your conductors, terminations, breakers, transformers and generators must be sized to, because copper heats on current alone and does not care whether that current is in phase with anything.

Reactive power, Q, measured in volt-amperes reactive, is the component that flows into the load during part of the cycle and back out during the rest. Inductance — motors, transformers, ballasts, solenoids — stores energy in a magnetic field and returns it; capacitance does the same with an electric field but with the opposite sign. Reactive power does no net work, yet it occupies conductor capacity all the same.

The three are related by a right triangle: S² = P² + Q². Power factor is the cosine of the angle at the origin, PF = P/S. A power factor of 1 collapses the triangle to a line and the three quantities become numerically identical, which is exactly what happens on DC and on a purely resistive AC load.

How each formula is built, and when to use which

Start from apparent power, because it needs the fewest assumptions: S = V·I with both values RMS. Nothing about the load's nature enters into it. If you clamp a meter on a conductor and read the voltage at the panel, you have S immediately.

Real power then needs one more piece of information — how far out of phase the current is. That is the power factor: P = V·I·PF. On a sinusoidal system with no harmonic distortion, PF = cos φ, where φ is the phase angle. Nameplates on motors and HVAC equipment print this figure; resistive loads such as heaters, water heaters, incandescent lamps and resistance-element ovens have a power factor of 1.00 and need no correction.

Reactive power falls out of the triangle: Q = √(S² − P²). The square root always returns a positive magnitude, so this form does not tell you whether the load is inductive (lagging) or capacitive (leading). Almost every industrial and commercial load is lagging, because motors and transformers dominate.

When you know resistance rather than current, the substitutions P = I²·R and P = V²/R follow directly from Ohm's law — see the Ohm's law calculator for the full set of rearrangements. On an AC circuit the corresponding quantity is impedance magnitude |Z|, which gives you apparent power as S = V²/|Z|; you still need the power factor to get watts.

Worked example: a 240 V motor drawing 20 A at 0.8 power factor

An electrician clamps 20 A on the supply to a single-phase motor fed from a 240 V circuit. The nameplate gives a power factor of 0.80. Work out every quantity by hand.

  1. Apparent power. S = 240 × 20 = 4,800 VA = 4.8 kVA. This is the figure the conductors and the breaker see.
  2. Real power. P = 4,800 × 0.80 = 3,840 W = 3.84 kW. This is the mechanical output plus the motor's own losses, and the figure the meter bills.
  3. Reactive power. Q = √(4,800² − 3,840²) = √(23,040,000 − 14,745,600) = √8,294,400 = 2,880 VAR. Notice the 3-4-5 proportions: 2,880 : 3,840 : 4,800 reduces to 3 : 4 : 5.
  4. Phase angle. φ = arccos(0.80) = 36.87°. The current peaks 36.87 electrical degrees after the voltage.
  5. Continuous-load sizing. If the motor runs three hours or more at a time, the National Electrical Code requires the conductors and overcurrent device to carry 125% of the load: 20 × 1.25 = 25 A.
  6. Energy at four hours a day. 3.84 kW × 4 h = 15.36 kWh per day.

Now the instructive part. Suppose you correct the power factor to 0.95 without changing the mechanical work being done. Real power stays at 3,840 W, but apparent power falls to 3,840 ÷ 0.95 = 4,042 VA, and the current falls to 4,042 ÷ 240 = 16.84 A. The same work, 3.16 A less conductor current. That is the entire argument for power factor correction, and the power factor correction capacitor calculator sizes the capacitor bank that does it.

How to read the result: which number sizes what

Use each output for the job it is meant for, and the confusion disappears.

Size conductors, breakers, transformers and generators from apparent power or current, never from watts. A 5 kVA load at 0.7 power factor delivers 3.5 kW but still draws the full 5 kVA of current. This is why transformers and UPS units are rated in kVA and not in kW — the manufacturer does not know your power factor, and the limiting factor is winding heating, which follows current.

Bill and budget energy from real power. Multiply kilowatts by hours to get kilowatt-hours. Residential tariffs almost universally meter real energy only, which is why domestic customers are not directly penalised for poor power factor. Commercial and industrial tariffs frequently are, either through a kVA demand charge or an explicit power-factor adjustment — read your own rate schedule rather than assuming.

As a rough orientation for power factor: purely resistive loads sit at 1.00; a modern induction motor at full load typically runs in the mid-0.8s and drops substantially at light load, which is why an oversized motor is doubly wasteful. Small unfiltered switch-mode supplies can be well below 0.7 with heavy harmonic distortion. Always prefer the nameplate figure, or a measurement with a true-RMS power meter, over a rule of thumb.

Finally, watch the phase-angle output as a plausibility check. If you enter a power factor of 0.5 you are claiming a 60° phase shift, which is an extremely reactive load. Values that low usually mean the nameplate has been misread or the load was measured unloaded.

Reference: current drawn at common voltages and power factors

Current for a 1,000 W real load, computed as I = P ÷ (V × PF). Multiply by your own kilowatts to scale.
Supply voltagePF 1.00PF 0.95PF 0.85PF 0.70
120 V8.33 A8.77 A9.80 A11.90 A
208 V4.81 A5.06 A5.66 A6.87 A
230 V4.35 A4.58 A5.11 A6.21 A
240 V4.17 A4.39 A4.90 A5.95 A
277 V3.61 A3.80 A4.25 A5.16 A
480 V2.08 A2.19 A2.45 A2.98 A

Single-phase only. For a three-phase load divide by an additional √3 and use the line-to-line voltage — the three-phase power calculator does that for you.

Mistakes that make a power calculation wrong

  • Treating VA and W as interchangeable. They are equal only at unity power factor. Sizing a generator from watts when the load is a motor bank is how you buy a generator that trips on overload.
  • Using peak instead of RMS values. On a sinusoid, peak is √2 times RMS. Feeding peak volts into these formulas overstates power by 41% for one substitution and 100% if you do it to both.
  • Assuming a nameplate power factor applies at partial load. Induction motors lose power factor badly as load falls; the nameplate figure is a full-load figure.
  • Applying single-phase formulas to a three-phase circuit. Three-phase power adds a factor of √3 when you use line-to-line voltage. Use the three-phase power calculator instead.
  • Ignoring harmonic distortion. With non-sinusoidal current, PF is no longer simply cos φ, and Q = √(S² − P²) lumps distortion in with reactive power. IEEE Std 1459-2010 exists precisely to separate those terms.
  • Forgetting the continuous-load multiplier. The Code requires branch-circuit conductors and overcurrent devices serving a continuous load to be sized at 125% of that load.
  • Reading impedance as resistance. On AC the ohms field here is |Z|, the magnitude. A load with 12 Ω of impedance at 0.85 power factor has about 10.2 Ω of resistance and 6.3 Ω of reactance.

Where this fits, and when to use a different tool

This calculator covers DC and balanced single-phase AC with a sinusoidal waveform. Three other situations need their own treatment.

Three-phase circuits introduce a √3 factor between line and phase quantities: P = √3 · VLL · IL · PF. Almost all commercial motor loads are three-phase, so reach for the three-phase power calculator and, for the motor itself, the motor full load amps calculator.

Reactive components analysed individually — a capacitor's or inductor's contribution at a given frequency — belong in the RLC impedance calculator, which gives you resistance and reactance separately rather than just the magnitude.

Operating cost takes the real power figure and multiplies it by hours and by your tariff. The appliance energy cost calculator handles the tariff structure, including tiered and time-of-use rates.

One historical note that still bites: apparent power is written in volt-amperes rather than watts deliberately, so that a specification cannot be misread. When a transformer is stamped 75 kVA, that is a thermal limit on the windings. Loading it with 75 kW of real power at 0.8 power factor means 93.75 kVA of current — 25% over the rating — and the transformer will run hot regardless of how efficiently the load turns watts into work. Size from the transformer kVA sizing calculator with apparent power, not watts.

Key terms

RMS
Root mean square — the equivalent DC value that would produce the same heating in a resistor. For a sinusoid, RMS equals peak divided by √2. All AC voltage and current ratings are RMS unless stated otherwise.
Power factor
The ratio of real power to apparent power, between 0 and 1. For undistorted sinusoids it equals cos φ. Described as lagging for inductive loads and leading for capacitive ones.
Volt-ampere (VA)
The unit of apparent power. Numerically identical to a watt but reserved for the product of RMS volts and RMS amps, so that a rating cannot be confused with real power.
VAR
Volt-ampere reactive, the unit of reactive power. Utilities meter VAR-hours separately on large services.
Continuous load
A load whose maximum current is expected to persist for three hours or more. The National Electrical Code requires conductors and overcurrent devices serving one to be sized at 125% of the load current.
Distortion power
The part of apparent power that arises from harmonic content rather than phase shift. Separated out explicitly in IEEE Std 1459-2010; lumped into Q by the simple triangle used here.

Frequently asked questions

What is the difference between watts and VA?

Watts measure the power that actually does work; VA measure the product of RMS volts and RMS amps regardless of phase. They are equal only when voltage and current are in phase, which happens on DC and on purely resistive AC loads. For anything with a motor, transformer or electronic supply, VA exceeds watts, and the ratio between them is the power factor. Size conductors and transformers from VA, and bills from watts.

What power factor should I use if I do not have a nameplate?

Use 1.00 for anything purely resistive — electric heaters, resistance ovens, incandescent lamps, water heaters, toasters. For anything else, measure rather than guess: a clamp-on power meter that reads true power and true RMS current gives you the ratio directly. Assuming a value introduces an error straight into the watts figure, and the direction of that error is not predictable without knowing the load.

Does poor power factor increase my electricity bill?

On a residential meter, usually not directly — most domestic tariffs meter kilowatt-hours of real energy only. On commercial and industrial tariffs it frequently does, through a kVA demand charge or an explicit power-factor adjustment clause. There is also an indirect cost regardless of tariff: extra current means extra I²R heating in your own conductors and less spare capacity in your service. Check your published rate schedule for the exact treatment.

Why does the calculator give the same number three times on DC?

Because on DC the voltage and current do not alternate, so there is no phase angle to shift them apart and no reactive exchange with the source. Apparent power equals real power by definition and reactive power is exactly zero. Capacitors and inductors in a DC circuit matter only during transients — for that behaviour see the RC time constant calculator rather than a steady-state power figure.

How do I convert this to kilowatt-hours?

Divide the real power in watts by 1,000 to get kilowatts, then multiply by the hours of operation. A 3,840 W motor running four hours a day uses 3.84 × 4 = 15.36 kWh per day, or about 461 kWh in a 30-day month. This page reports the daily figure directly from the hours input. Note that only real power converts to kilowatt-hours; VA and VAR do not.

Can I use this for a three-phase motor?

Not directly. Three-phase power is √3 × Vline-to-line × Iline × PF, which is about 1.732 times what this page would give for the same voltage and current readings. You can get the right answer here by entering the phase voltage and phase current of one leg and multiplying the result by three, but the three-phase power calculator is the correct tool and avoids the line-versus-phase trap entirely.

Why is my measured current higher than watts divided by volts?

Because that division only works at unity power factor. Current is real power divided by voltage and by power factor, so at 0.8 the current is 25% higher than the naive calculation suggests. Harmonic distortion pushes it higher still: a current waveform rich in harmonics has a higher RMS value for the same real power, which is why non-linear loads can overheat a shared neutral.

Is reactive power wasted energy?

Not in itself — over a complete cycle, the net energy transferred by the reactive component is zero, so it is not consumed. What it does cost you is capacity and conduction losses: carrying that current still heats the conductors according to I²R, and it occupies capacity in every transformer, cable and switchgear between the source and the load. That is the real economic argument for correcting it locally.

References

  • IEEE Std 1459-2010, Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions — Institute of Electrical and Electronics Engineers
  • NFPA 70, National Electrical Code, 2023 editionNational Fire Protection Association
  • Standard Handbook for Electrical Engineers, 17th ed. — McGraw-Hill (Beaty & Fink)
  • The Art of Electronics, 3rd ed. — Cambridge University Press (Horowitz & Hill)