13.3 True, Apparent & Reactive Power

13.3 True, Apparent & Reactive Power

Quick Answer: True power P = VI cos θ (watts) — average energy converted per second, usually to heat in R. Apparent power S = VI (VA). Reactive power Q = VI sin θ (VAR). They form a power triangle with pf = P/S = cos θ. Ideal L and C exchange energy with the source but average P = 0; only R dissipates true power in the ideal model.

Impedance and power factor (§13.2) tell you the angle between V and I. Module 3 topic 3.14 finishes by naming the three power quantities that angle creates—and by stating clearly what dissipates energy in R, L, and C circuits.

Why Three “Powers”?

On DC, P = VI is unambiguous. On AC with phase shift, the product of RMS voltage and RMS current is not always the average heating power. Part of the VI product may represent energy that flows into magnetic or electric fields on one quarter-cycle and returns on the next. Module 3 therefore splits:

Also: P = S cos θ, Q = S sin θ, and pf = P / S.

True Power (P)

True power is the average power over a cycle—the rate at which electrical energy is converted irreversibly (heat in resistors, mechanical work in motors after accounting for losses, and so on).

For any linear sine circuit:

P = V_RMS × I_RMS × cos θ

In the series RLC idealisation, only the resistor dissipates average power:

P = I_RMS² × R

Worked example 1 — from V, I, pf. V = 115 V, I = 4.0 A, pf = 0.8 lagging.

P = 115 × 4.0 × 0.8 = 368 W.

Worked example 2 — from I and R. Series circuit, I = 2.5 A, R = 40 Ω.

P = (2.5)² × 40 = 6.25 × 40 = 250 W.

Frequency does not appear explicitly once I and R are known—but frequency shaped Z and therefore I.

Apparent Power (S)

Apparent power is the product of RMS voltage and RMS current as if you ignored phase:

S = V_RMS × I_RMS

Generators, transformers, and cable ratings are often stated in VA or kVA because heating in windings and wiring depends on current magnitude, and insulation stress depends on voltage, regardless of pf. A load drawing 10 A from 115 V needs 1150 VA of apparent capacity even if pf is only 0.5 and true power is 575 W.

Worked example 3 — S then P. V = 115 V, I = 2.5 A, θ such that pf = 0.6.

S = 115 × 2.5 = 287.5 VA.

P = 287.5 × 0.6 = 172.5 W.

Reactive Power (Q)

Reactive power quantifies the amplitude of the energy oscillation associated with inductance and capacitance:

Q = V_RMS × I_RMS × sin θ

Worked example 4 — Q from S and θ. S = 287.5 VA, pf = 0.6 → cos θ = 0.6 → sin θ = 0.8 (because sin² + cos² = 1).

Q = 287.5 × 0.8 = 230 VAR (lagging if the load is inductive).

Worked example 5 — 400 Hz inductive branch. Pure inductor, V = 115 V, X_L = 46 Ω at 400 Hz.

I = 115 / 46 = 2.5 A; θ = 90°; cos θ = 0; sin θ = 1.

P = 0 W; S = 115 × 2.5 = 287.5 VA; Q = 287.5 VAR.

The Power Triangle

Draw P along the horizontal (in-phase) axis and Q along the vertical (quadrature) axis. The hypotenuse is S:

S² = P² + Q²

tan θ = Q / P

cos θ = P / S

This triangle is similar to the impedance triangle (R, X, Z): multiply the impedance triangle by I² (or by I and use V drops) and you recover the power triangle.

Worked example 6 — close the triangle. P = 300 W, Q = 400 VAR.

S = √(300² + 400²) = √(90000 + 160000) = √250000 = 500 VA.

pf = 300 / 500 = 0.6.

Worked example 7 — aircraft numeric. Series R = 30 Ω, X = 40 Ω, V = 115 V at 400 Hz (Z = 50 Ω from §13.2 pattern).

I = 115 / 50 = 2.3 A.

P = I²R = (2.3)² × 30 ≈ 5.29 × 30 ≈ 159 W.

Q = I²X ≈ 5.29 × 40 ≈ 212 VAR.

S = VI = 115 × 2.3 = 264.5 VA.

Check: √(159² + 212²) ≈ √(25281 + 44944) ≈ √70225 ≈ 265 VA ✓.

pf = 30/50 = 0.6 lagging.

Dissipation in R, L, and C Circuits

Module 3 expects a clear energy story—not only formulae.

Pure resistance

Current through R produces Joule heating I²R every instant (averaged over the cycle for AC). Energy leaves the electrical system as heat. P > 0, Q = 0, S = P, pf = 1.

Pure inductance (ideal)

Energy is stored in the magnetic field as current rises and returned to the source as current falls. Over a full cycle the average energy transfer is zero: P = 0, Q = S = VI, pf = 0 lagging. A real coil has winding resistance, so a small true power appears as copper (and core) loss—modelled by adding R in series with X_L.

Pure capacitance (ideal)

Energy is stored in the electric field as voltage rises and returned as voltage falls. Average P = 0, Q = S (leading), pf = 0 leading. Real capacitors have small dielectric/ESR losses—again a small P.

Combined RLC

If X_L and X_C both exist, their reactive powers oppose; net Q = I²(X_L − X_C) in the series model. At resonance net Q → 0 and S → P.

Worked example 8 — where the watts go. Series RLC at 400 Hz: I = 3 A, R = 20 Ω, X_L = 50 Ω, X_C = 30 Ω.

P = 9 × 20 = 180 W (all in R).

Q_net = 9 × (50 − 30) = 180 VAR inductive.

S = √(180² + 180²) = 180√2 ≈ 255 VA.

Ideal L and C still contribute zero to the 180 W; they only shape the 180 VAR net.

Aircraft Maintenance Angle

Low power factor means large current (and large S) for a given true power—more I²R loss in feeders, larger generator loading in VA, hotter cables. Capacitive correction and load management appear in later systems topics; Module 3’s job is to compute P, S, Q and state that watts live in resistance, while L and C shuttle VARs.

Section Synthesis

Master P = VI cos θ, S = VI, Q = VI sin θ, the triangle check S² = P² + Q², and the dissipation rule (ideal L and C: P = 0). That closes syllabus 3.14 and prepares transformer loading language in topic 3.15.