PW-B1 · Power and power factor

The power triangle: active, reactive, apparent

When a load pulls its current out of step with the voltage, power splits into a part that does work and a part that only sloshes back and forth. Reshape the right triangle that binds those two with their resultant (apparent power), and learn which side of it the power factor is.

Drag φ and reshape the triangle

The hypotenuse (apparent S) is fixed in length. Drag the vertex to change the phase angle φ. The base (active P = S cosφ) and height (reactive Q = S sinφ) change together. At φ = 0 it is all active; as φ grows the reactive part rises.

Drag the vertex to set φ.

Power factor cosφ and the three powers (S = 1)

cosφ ≈ 0.80

P ≈ 0.80   Q ≈ 0.60   S = 1   φ ≈ 37°

In step, all of it is active

When the current is in phase with the voltage, φ = 0. The current is large when the voltage is high and zero when the voltage is zero, so the power v·i flows in one direction only. All of it is active power P = VI — real work as light, heat or rotation. The triangle is a flat horizontal line of zero height.

The out-of-step part only sloshes

When the current lags by φ, split it into two components. The component aligned with the voltage makes active power; the component at right angles makes reactive power Q = VI sinφ. The power of the reactive component flows to the load for half a cycle and back to the source for the other half, so its net work is zero. This is the back-and-forth of coils and capacitors storing energy briefly and returning it.

The three bind into a right triangle

Active P and reactive Q meet at a right angle, and their hypotenuse is the apparent power S = VI. So S² = P² + Q² and the power factor is cosφ = P/S = base over hypotenuse. Wires and transformers must withstand the hypotenuse S — if the reactive part is large and the power factor low, the equipment carries a big S even though the working active P is small. This is why power factor drives billing and equipment rating.

ObserveS² = P² + Q²

Apparent² = active² + reactive².

ChooseP = S · ?

Active is the base, hypotenuse times cosφ.

Fill inQ = S · ?

Reactive is the height, hypotenuse times sinφ.

On your owncosφ = P / ?

The power factor is active over apparent.

Back to the first screen

The hypotenuse S held fixed, yet the more φ grew the smaller the base, active P, and the larger the height, reactive Q. It means that for the same apparent power, a low power factor leaves little actual working power. The cosφ in the P = √3 V_L I_L cosφ of A3 is exactly this angle between the hypotenuse and the base. Power is not one magnitude but a bundle of three with direction.

The power triangle — when a load pulls its current out of step with the voltage by φ, power splits into a right triangle. Active power P = S cosφ (real work), reactive power Q = S sinφ (only sloshing, net zero), apparent power S = VI (the hypotenuse, what the equipment must withstand). S² = P² + Q², power factor cosφ = P/S.

The next step

Reactive power Q does no real work yet swells the apparent S and ties up equipment. So reduce that Q. The next unit (PW-B2) corrects the power factor with a capacitor. The lagging reactive drawn by a coil load is cancelled by the leading reactive a capacitor supplies, lowering the height of the triangle and pulling the hypotenuse down toward the base.