What this calculator does
Give it any two of voltage, current, resistance and power, and it returns all four. That is the classic Ohm’s law solve, and it is the reason this page exists. What it does beyond that is the reason to keep the page open: it also splits the result into the three powers that matter on AC — apparent (VA), real (W) and reactive (var) — reports the phase angle, scales the result across a bank of identical units, and gives you the energy in kWh over a run time you enter.
All four number fields are always on the form. The “which two do you know” select decides which pair the calculator reads; the other two are ignored and solved for. That is deliberate. A form that hides fields makes you re-enter numbers every time you flip the problem around, and it hides the structure of the thing you are learning. Set the select, fill the two fields it names, and read the rest off the datasheet. The field hints tell you which are live.
The formula
Ohm’s law itself relates three quantities:
V = I × Z I = V / Z Z = V / I
V is RMS voltage in volts, I is RMS current in amps, and Z is the impedance magnitude in ohms. On DC, or on AC into a purely resistive load, Z is just the resistance R and the distinction never arises. On AC into anything with a coil or a capacitor in it, Z = V / I is still exactly true, but it is no longer the resistance — more on that below.
Power is where AC parts company with DC. Multiply RMS volts by RMS amps and you get the apparent power:
S = V × I volt-amperes, VA
Only the part of the current that is in phase with the voltage does any work. The cosine of the angle between them is the power factor, and it gives the real power:
P = S × cos φ = V × I × PF watts, W
The quadrature part does no net work — it pushes energy into a magnetic or electric field for a quarter cycle and takes it back the next. That is the reactive power:
Q = S × sin φ = S × √(1 − PF²) volt-amperes reactive, var
Because the two components are at right angles, the three powers form a right triangle:
S² = P² + Q² φ = arccos(PF) = arctan(Q / P)
Set PF = 1 and every relation above collapses to the DC set you already know: Q = 0, S = P = V × I = I²R = V²/R.
Energy is then simply power multiplied by time, for however many identical units are on the circuit:
E = n × P × t / 1000 kWh, with P in W and t in hours
Reading the result
The power triangle, and why it is a triangle
Take the voltage as the reference phasor, pointing along the positive real axis. On an inductive load the current phasor lags it by an angle φ. Resolve that current into two components: I cos φ, in line with the voltage, and I sin φ, at ninety degrees to it.
Multiply each component by V and you have the triangle. V × I cos φ is real power — the in-phase product, the part with a non-zero average over the cycle, the part that becomes heat, light and shaft torque. V × I sin φ is reactive power — the quadrature product, whose instantaneous value swings symmetrically positive and negative so that its average over a full cycle is exactly zero. V × I is the hypotenuse, the apparent power, and it is the only one of the three the conductor can tell apart, because the conductor only ever sees the total current.
That last sentence is the whole practical point. Reactive power is real current in real copper. It does no work, it earns you nothing, and it heats the cable exactly as much as working current would.
The same decomposition applies to the impedance itself: R = Z cos φ and X = Z sin φ, so the impedance triangle and the power triangle are the same triangle scaled by I². This is why the tool labels its resistance output “resistance / |Z|” — at PF 1 it is the resistance, and below PF 1 it is the impedance magnitude V / I, of which the actual resistance is only the cos φ part.
One thing the magnitudes cannot tell you is the direction of the reactive power. An inductive load (motors, transformers, chokes) draws lagging current; a capacitive one (long cable runs, capacitor banks, some electronic front ends) draws leading current. Both give the same power factor magnitude and the same Q here. Reactive compensation depends on the sign, so keep track of it yourself.
Why power factor matters for cable sizing
Rearrange the real-power relation for current:
I = P / (V × PF)
The current a load draws is inversely proportional to its power factor. A 6.4 kW load at PF 1.0 on 400 V draws 16 A. The same 6.4 kW at PF 0.8 draws 20 A — a quarter more copper, a quarter more volt-drop, and 1.25² ≈ 1.56× the I²R losses in the cable. At PF 0.5 it is double the current and four times the losses.
So: size cables, protective devices, switchgear and transformers on apparent power, and size the tariff conversation on real power. Every kVA rating on a nameplate exists because the equipment is limited by current and by heat, neither of which knows or cares about phase angle.
Why it matters for billing
A domestic kWh meter measures real energy only, so a poor power factor costs a small consumer nothing directly. Commercial and industrial tariffs are different: most utilities either bill a kVA maximum demand charge, meter kvarh separately, or apply a penalty below a threshold power factor — commonly 0.90 or 0.95. In all three cases the payback on capacitor banks or active correction is calculated from exactly the Q figure this tool reports.
Typical values
- Power factor: incandescent and resistive heating 1.00; modern LED drivers and PC power supplies with active correction 0.90–0.99; induction motors 0.80–0.90 at full load, but as low as 0.20–0.40 at no load; welding sets 0.50–0.70; unfiltered rectifier loads 0.60–0.75. A whole-installation figure of 0.85–0.95 is normal; below 0.85 expect a utility to notice.
- Voltage: low voltage means up to 1000 V AC in most regimes; above that, high-voltage design, clearance and competency rules apply. Common nominal LV values are 100–127 V and 200–240 V single phase, 380–480 V three phase.
- Resistance: a domestic heating element is tens of ohms; a motor winding is fractions of an ohm; a fault loop is milliohms. If this tool hands you an impedance below about 0.01 Ω for a load, check for a unit slip.
Worked example
A three-phase-fed machine bay has three identical single-phase loads, each drawing 20 A at 400 V with a measured power factor of 0.80 lagging. They run 10 hours a day.
Per unit:
S = 400 × 20 = 8000 VA
P = 8000 × 0.80 = 6400 W
sin φ = √(1 − 0.80²) = √0.36 = 0.60
Q = 8000 × 0.60 = 4800 var
Check the triangle closes: 6400² + 4800² = 40 960 000 + 23 040 000 = 64 000 000, and √64 000 000 = 8000 VA. It is a 3-4-5 triangle scaled by 1600. The phase angle is arccos(0.80) = 36.87°.
The impedance magnitude is Z = 400 / 20 = 20 Ω, of which the resistive part is R = 20 × 0.80 = 16 Ω and the reactive part X = 20 × 0.60 = 12 Ω. Confirm against the current: I²R = 400 × 16 = 6400 W and I²X = 400 × 12 = 4800 var. Same answers, different route.
Across three units the real power is 6400 × 3 = 19 200 W, and over ten hours:
E = 19 200 × 10 / 1000 = 192 kWh
Worth noting what the cable carries: 20 A per unit, not the 16 A that 6400 W at unity power factor would have needed. Correcting these loads to PF 0.95 would drop the current to 6400 / (400 × 0.95) = 16.8 A and cut the cable losses by about 30 %.
FAQ
Is the resistance output the DC resistance of my load?
Only at PF 1. Otherwise it is the impedance magnitude, |Z| = V / I. The resistive component is |Z| × PF and the reactive component is |Z| × √(1 − PF²). Both are reported implicitly by the power triangle: P = I²R and Q = I²X.
What power factor should I enter for DC?
- There is no phase angle on DC, so apparent, real and reactive collapse to
P = V × I,Q = 0. The same is true for any purely resistive AC load — a heater, an incandescent lamp, a kettle.
Can I use this for a three-phase circuit?
The relations here are single-phase, or per-phase. For a balanced three-phase load use S = √3 × V_LL × I_L, so enter one phase’s figures (V_LN and the line current) and multiply the total by three, or set the unit count to 3 as the worked example does. Do not enter line-to-line voltage against line current here — that is the commonest error in this whole subject.
Why does my meter reading not match the kWh figure? Three usual reasons: the load does not run at full output for the whole period; the power factor is not constant with load, especially on motors; and the nameplate wattage is often input power at rated conditions rather than what the machine actually draws. Use a measured current for anything you intend to bill against.
Can power factor be greater than 1?
No. It is a cosine. A displayed value above 1 means a measurement error or a distorted, non-sinusoidal waveform — in which case what you want is the true power factor, which includes a distortion component in addition to the displacement cos φ used here. On a heavily harmonic load the two differ substantially.
This tool provides indicative figures for design and checking. Circuit design, protection and installation must comply with the standards applicable in your jurisdiction — for example IEC 60364, NEC or AS/NZS 3000.