PW-B1 · Power and power factor

The power triangle: active, reactive, apparent

When a load draws current out of step with the voltage, power splits into a part that does work and a part that only goes back and forth. Reshape for yourself how those two and their total (apparent power) tie into a right triangle, and learn that the power factor is the ratio of the base to the hypotenuse of this triangle.

Drag φ to reshape the triangle

The hypotenuse (apparent S) has a fixed length. Dragging the vertex changes the phase angle φ, and the base (active P = S cosφ) and the height (reactive Q = S sinφ) change together: at φ of 0 everything is active, and as φ grows the reactive part grows.

Drag the vertex to change φ. The first view is power factor 0.8, so P : Q : S = 0.8 : 0.6 : 1, a 4 : 3 : 5 right triangle.

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

If the current is in phase with the voltage, φ = 0. The current is large when the voltage is high and 0 when the voltage is 0, so the power v·i always flows in one direction. All of the power is then active power P = VI, net work such as light, heat or rotation, and the triangle becomes a flat line of height 0.

The out-of-step part only sloshes

If the current lags by φ, split the current into two components. The component in line with the voltage makes active power, and the one at right angles to the voltage makes reactive power Q = VI sinφ. The power made by the reactive component changes direction every quarter cycle, going back and forth between load and source, so its net work is 0. This is the round trip in which coils and capacitors store energy for a moment and hand it back.

The three bind into a right triangle

Active P and reactive Q meet at a right angle, and the hypotenuse is the apparent power S = VI. So S² = P² + Q², and the power factor is cosφ = P/S = base/hypotenuse. Wires and transformers must carry the full hypotenuse S, so when the reactive part is large and the power factor is low, the equipment bears a large S even though the active P that does the work is small. That is why the power factor affects electricity tariffs and equipment capacity.

ObserveS² = P² + Q²

Apparent² = active² + reactive².

ChooseP = S · ?

Active power is the base: the hypotenuse times cosφ.

Fill inQ = S · ?

Reactive power is the height: the hypotenuse times sinφ.

On your owncosφ = P / ?

The power factor is active power divided by apparent power.

Back to the first screen

The hypotenuse S had a fixed length, yet as φ grew the base, active P, shrank and the height, reactive Q, grew. Even with the same apparent power, a low power factor means the active power that does the work is small. The φ in the P = √3 V_L I_L cosφ seen in A3 is exactly the angle between the hypotenuse and the base of this triangle, and power is not a single size but a bundle of three with direction.

Power triangle: when a load draws current φ out of step with the voltage, power splits into a right triangle. Active power P = S cosφ (net work), reactive power Q = S sinφ (only exchanged · net 0), apparent power S = VI (hypotenuse · the size equipment must carry); S² = P² + Q² and the power factor is cosφ = P/S.

The next step

Reactive power Q does no net work yet enlarges the apparent S and takes up equipment capacity. Then it is a matter of reducing Q: the next unit (PW-B2) improves the power factor with capacitors. The lagging reactive power drawn by inductive loads is cancelled by the leading reactive power a capacitor supplies, lowering the height of the triangle and pulling the hypotenuse down toward the base.