Reactive Power in Motion

How much useful power can a fixed kVA rating carry? This demo holds apparent power at 100 VA while you change the phase angle between voltage and current.

Source

Load

Blue shows useful delivery. Orange (lagging) and purple (leading) show energy exchange. The particles are a visual analogy, not charges or separate kinds of electricity.

Drag to rotate · scroll to zoom

Same rating, different useful power

Try this: choose Resistive, then Inductive. The rating stays at 100 VA, but useful power falls from 100 W to about 71 W.

Power factor Leading current (capacitive)

Power factor
Useful power P
Reactive power Q
Apparent power S

Power factor is P/S, not the fraction of energy lost. For correction at a fixed useful load, try Power Factor Correction.

Power triangle

P and Q use the same scale. Drag to pan; scroll to zoom.

Power components

The blue component averages to P. The orange or purple exchange component averages to zero. Their sum is instantaneous power.

Equations and assumptions

The power triangle

This is a sinusoidal, single-phase model. Voltage and current magnitudes stay fixed, so S = V_rmsI_rms stays at 100 VA. Changing their relative phase changes P and Q.

P = S cos φ · Q = S sin φ · S² = P² + Q² · PF = P/S

Positive Q means current lags voltage; negative Q means it leads. The voltage and current chart uses normalized amplitudes to make the phase difference easy to see.

What the power chart shows

With v(t) = V_max sin(ωt), the total instantaneous power is:

p(t) = P[1 − cos(2ωt)] − Q sin(2ωt)

The chart separates the nonnegative real-power component from the oscillating exchange component. Only the latter averages to zero. Q itself is a signed reactive-power quantity measured in var.

Why reactive power matters

Inductors and capacitors store energy and return it during each cycle. This exchange does no net work, but the associated current uses capacity in cables and transformers. At a fixed useful load and voltage, more current also means higher resistive losses.

The model leaves out harmonics, line losses, and changes in voltage. MIT’s AC circuit notes give the derivation.