Real, Reactive & Apparent Power and the Power Factor

The water that does no work

Imagine pushing a child on a swing. Most of your effort goes into the useful job — keeping the swing arcing higher. But part of every push just compresses your own arm and springs back; that energy borrows from you on the way in and pays you back on the way out, netting nothing yet still tiring your muscles. An AC grid feeding a motor does exactly this. Every cycle, energy floods into the motor's magnetic field, then drains back out into the wires. Over a full cycle the net transfer is zero, but the back-and-forth still demands real current in the cables, and that current still heats them.

This splits the power flowing into any AC load into two characters. Real power (symbol P, unit the watt, W) is the part that actually leaves for good and turns into heat, torque, or light — the part you can never get back. Reactive power (symbol Q, unit the volt-ampere reactive, VAR) is the swinging-arm part: it surges out of the source and back every cycle, doing no net work, parked temporarily in magnetic fields (inductors) or electric fields (capacitors). Crucially, reactive power is not 'fake' — it is genuine current in genuine wires. It is just work-free.

Instantaneous power  p(t) = v(t)·i(t)  when V and I are 60 deg apart

 v(t) ~~~~/\~~~~~~~~~/\~~~~~~~~~/\~~~~  voltage
 i(t) ~~~~~~/\~~~~~~~~~/\~~~~~~~~~/\~~  current (lags by 60 deg)

 p(t)      __        __        __
        __/  \__   _/  \__   _/  \__      p dips BELOW zero here:
  0 ---/--------\-/--------\-/--------\-- energy flowing back to source!
      /          V          V

  Average of p(t)  = P   (the part above the axis wins) -> real power
  Swing about that = Q   (the dip-below-zero) -> reactive power

Instantaneous power when current lags voltage. Whenever the curve dips below zero, energy is flowing back out of the load — that returning energy is what reactive power accounts for.

Three sides of one triangle

Here is the elegant part: real power P and reactive power Q are at right angles to each other. P is the in-phase component of current times voltage; Q is the quarter-cycle-shifted (90°) component. Two perpendicular quantities add like the legs of a right triangle, and their hypotenuse is the apparent power S — the simple product of the RMS voltage and RMS current you would read off a voltmeter and an ammeter, with no regard for phase. Its unit is the volt-ampere (VA) to keep it visibly distinct from the watt.

                 ____
                 |S| = apparent power  (VA)  = V_rms * I_rms
               /  |
         |S| /    |
           /      | Q = reactive power (VAR) = V_rms*I_rms*sin(theta)
         /        |
       / theta    |
      +-----------+
        P = real power (W) = V_rms*I_rms*cos(theta)

  Pythagoras:   |S|^2 = P^2 + Q^2
  Phase angle:  theta = angle between V and I  =  atan2(Q, P)

  Q > 0  inductive load  (current LAGS voltage)  -> triangle tips up
  Q < 0  capacitive load (current LEADS voltage) -> triangle tips down
  Q = 0  pure resistance (in phase)              -> triangle is flat: S = P

The power triangle. The single angle theta — the phase angle between voltage and current — sets all three sides at once.

The angle θ at the base is the very same phase angle between voltage and current you met building AC circuits, and it is also the angle of the load's impedance. That is why one number ties everything together. An inductive load (motors, transformers, fluorescent ballasts) makes current lag, so θ is positive and Q is positive — the triangle tips upward. A capacitive load makes current lead, so θ and Q go negative and the triangle tips down. This sign convention is what lets a capacitor's reactive power cancel an inductor's, a fact we will cash in shortly.

Power factor: how much of your current actually works

Divide the useful side by the hypotenuse and you get the single most quoted number in power engineering: the [[ee-power-factor|power factor]], PF = P / |S| = cos θ. It is a fraction between 0 and 1 that answers one blunt question — of all the current you are drawing, what fraction is doing real work? A PF of 1.0 is perfect: every amp earns its keep. A PF of 0.7 means almost a third of your current is sloshing uselessly. A PF of 0 (a perfect inductor or capacitor) means all of it is, and you draw current while consuming no power at all.

Engineers tack on a direction word so the bare number is not ambiguous. Lagging power factor means current lags voltage — an inductive load, the overwhelmingly common case in industry. Leading power factor means current leads — a capacitive load. 'A 0.8 lagging PF' is shorthand for 'inductive, with current 36.9° behind the voltage'. Because cosine is even, the PF number alone can't tell lagging from leading, so the word matters.

Worked example: a 230 V single-phase induction motor delivering 4 kW
               of mechanical work, running at PF = 0.70 lagging.

  Real power needed      P    = 4000 W
  Apparent power         |S|  = P / PF = 4000 / 0.70  = 5714 VA
  Line current           I    = |S| / V = 5714 / 230  = 24.8 A

Now compare a hypothetical PF = 1.0 motor doing the SAME 4 kW work:

  Apparent power         |S|  = 4000 / 1.0 = 4000 VA
  Line current           I    = 4000 / 230 = 17.4 A

  Extra current forced by the poor PF:  24.8 / 17.4 = 1.43  -> +43%

  Wire heating loss scales as I^2 (P_loss = I^2 * R_wire):
     (24.8 / 17.4)^2 = 2.03  -> the cables waste ~2x the heat
     for the very same 4 kW of useful work delivered.

Low power factor is expensive twice over: 43% more current to carry, and because losses go as I-squared, roughly double the heat wasted in the supply cables.

Why utilities care — and how they make you pay

Step back to the three-phase transmission system feeding a city. Every generator, every line, every transformer is sized in MVA — apparent power — because that is what determines the current it must survive without melting. Reactive power flowing through those assets is pure overhead: it occupies thermal headroom, it drops voltage along long lines (sagging voltage at the far end is one of reactive power's nastiest side effects), and it earns the utility nothing on the energy meter. A grid full of lagging loads is a grid forced to build infrastructure it cannot bill for.

So utilities push the cost back onto the offenders. Large industrial customers are metered not just in kWh but in kVA demand or in kVARh, and a tariff clause penalizes any month where the average power factor drops below a threshold — commonly 0.90 or 0.95. The penalty can be a flat surcharge, a multiplier on the demand charge, or a direct charge per kVARh of reactive energy consumed. For a factory running hundreds of motors, a chronically bad power factor turns into a five- or six-figure annual line item.

Fixing it: the capacitor bank

The cure is beautifully direct. An inductive load draws lagging, positive reactive power Q. A capacitor draws *leading*, negative reactive power. Park a capacitor in parallel with the load and its negative Q subtracts from the load's positive Q, shrinking the resultant reactive power toward zero. On the triangle, Q collapses, the hypotenuse |S| swings down toward the base P, and the power factor climbs back toward 1 — all while the real power P, and therefore the actual work done, is untouched. The reactive energy that used to swing all the way back to the generator now just swings between the capacitor and the motor next door, a few metres of cable instead of a few kilometres of transmission line.

Start from the load's existing real and reactive power. Our 4 kW motor at PF 0.70 lagging: P = 4 kW, θ = arccos(0.70) = 45.6°, so Q_load = P·tan θ = 4·tan(45.6°) = 4.08 kVAR (inductive, +).
Pick a target power factor — say 0.95 lagging. The new angle is θ' = arccos(0.95) = 18.2°, and the reactive power we are willing to keep is Q_target = P·tan θ' = 4·tan(18.2°) = 1.31 kVAR.
The capacitor must supply the difference: Q_cap = Q_load − Q_target = 4.08 − 1.31 = 2.77 kVAR. Size the capacitance from Q_cap = V²·ωC, so C = Q_cap / (V²·ω) = 2770 / (230²·2π·50) ≈ 167 µF.
Re-check the result. New apparent power |S'| = P / 0.95 = 4.21 kVA, new line current I' = 4210 / 230 = 18.3 A — down from 24.8 A. We just cut the current by 26% and dodged the utility penalty without changing the work done one watt.

Before correction          After adding 2.77 kVAR capacitor
-----------------          --------------------------------
        |S|=5.71 kVA               |S'|=4.21 kVA
       /|                         /|
      / | Q=4.08                 / | Q'=1.31 kVAR
     /  | kVAR                  /  |
    /th |          ===>        /th'|
   +----+                     +----+
    P=4 kW                     P=4 kW   (unchanged!)
   th=45.6 deg  PF=0.70       th'=18.2 deg  PF=0.95
   I = 24.8 A                  I = 18.3 A   (-26%)

The capacitor injects -2.77 kVAR; the motor's +4.08 kVAR
is mostly cancelled locally, so the supply only carries +1.31 kVAR.

Power-factor correction in pictures: the capacitor's negative Q eats most of the motor's positive Q, collapsing the triangle and the line current while real power holds firm.

The bookkeeping of the grid

Zoom out one last time. Everything in this guide — P, Q, |S|, PF — is just energy accounting for alternating current, and the accounts must balance everywhere at once. The reactive power you compute for one motor, multiplied across a continent of motors, transformers, long lines (which are themselves slightly inductive), and capacitor banks, is what control-room operators watch minute by minute. Too much net reactive demand and far-end voltages sag; too little and they rise. Keeping the grid's reactive ledger balanced is, quite literally, a full-time job for both engineers and the markets that price reactive power as a separate service.