Real, reactive and apparent power, power factor

Why it matters

Wattmeters, energy meters and power analysers are core instrumentation, and you cannot design or read them without separating real, reactive and apparent power. Utilities penalise low power factor. Cables, transformers and switchgear are rated in VA, not watts. Power-factor correction is one of the most common calculations in industry.

Key ideas

Instantaneous power. With v = √2·V·cos ωt and i = √2·I·cos(ωt − θ) (V and I are RMS values, θ is the angle by which the current lags), the instantaneous power is: p(t) = V·I·cos θ + V·I·cos(2ωt − θ). It has a constant part plus a part that oscillates at twice the supply frequency.

Real (active, average) power P. The average of p(t): P = V·I·cos θ, in watts (W). It is consumed only by resistance, P = I²R, and it is converted into heat, light or mechanical work. Pure L and C absorb zero average power.

Reactive power Q. Q = V·I·sin θ, in volt-ampere reactive (var). It measures energy that flows back and forth between the source and the fields of L and C each half-cycle without being consumed: Q = I²X.

• Inductive loads absorb Q (Q > 0, lagging current).
• Capacitors supply Q (Q < 0 when absorbed, leading current).

Apparent power S. |S| = V·I, in volt-amperes (VA). This sets the current, and therefore the rating of conductors and equipment.

Complex power. S = V·I* = P + jQ, where I* is the complex conjugate of the current phasor (RMS phasors). The power triangle has P on the real axis, Q on the imaginary axis and |S| as the hypotenuse: |S|² = P² + Q². Complex power is conserved: the total S delivered by the sources equals the sum of S absorbed by all elements, so P and Q each balance separately. Apparent powers |S| do not add.

Power factor. pf = P/|S| = cos θ, where θ is the impedance angle (the angle between V and I). Always state lagging (inductive) or leading (capacitive), because cos θ alone does not tell you which. pf lies between 0 and 1. Note that pf is not efficiency.

Why low pf is bad. For a fixed P at a fixed V, the line current is I = P/(V·pf). A low pf means more current, higher I²R losses in the lines, larger voltage drop and larger equipment. That is why tariffs penalise it.

Power-factor correction. Put a capacitor (or a synchronous condenser) in parallel with an inductive load. It supplies part of the load's Q locally. P is unchanged, the source Q drops, the line current drops, and the load voltage and current through the load itself are unaffected. A series capacitor would change the load voltage and is not used for correction.

Measurement link. An electrodynamometer or digital wattmeter measures P = V·I·cos θ directly. A varmeter measures Q. In three-phase systems, the two-wattmeter method gives both P and Q (see the three-phase topic).

Formulas

• P = V·I·cos θ = I²R = V_R²/R (W).
• Q = V·I·sin θ = I²X (var).
• |S| = V·I = I²|Z| (VA).
• S = V·I* = P + jQ.
• |S|² = P² + Q².
• pf = cos θ = P/|S| = R/|Z|.
• Correction capacitor: Qc = P·(tan θ1 − tan θ2) and C = Qc/(ω·V²) (F), with V the RMS voltage across the capacitor.

All voltages and currents here are RMS. If peak phasors are used, S = ½·V·I*.

Worked examples

Example 1 (standard): powers of an RL load. Given: 230 V RMS, 50 Hz, supplying Z = 12 + j16 Ω.

1. |Z| = √(12² + 16²) = 20 Ω, so I = 230/20 = 11.5 A.
2. |S| = 230 × 11.5 = 2645 VA.
3. P = I²R = 11.5² × 12 = 1587 W.
4. Q = I²X = 11.5² × 16 = 2116 var (absorbed, inductive).
5. pf = 12/20 = 0.6 lagging.
6. Check: √(1587² + 2116²) = 2645 VA ✓.

Answer: P = 1587 W, Q = 2116 var, |S| = 2645 VA, pf = 0.6 lagging.

Example 2 (GATE level): power-factor correction. Given: a single-phase 230 V, 50 Hz load takes 5 kW at 0.7 pf lagging. Find the parallel capacitor that raises the pf to 0.95 lagging, and the line current before and after.

1. Initial reactive power: θ1 = cos⁻¹(0.7) = 45.57°, so Q1 = 5000 × tan 45.57° = 5101 var.
2. Target reactive power: θ2 = cos⁻¹(0.95) = 18.19°, so Q2 = 5000 × tan 18.19° = 1643 var.
3. Capacitor must supply: Qc = 5101 − 1643 = 3458 var.
4. Capacitance: C = Qc/(ω·V²) = 3458/(2π × 50 × 230²) = 3458/(314.16 × 52900) = 208 µF.
5. Line current before: 5000/(230 × 0.7) = 31.1 A. After: 5000/(230 × 0.95) = 22.9 A.

Answer: C ≈ 208 µF; the line current falls from 31.1 A to 22.9 A, with P unchanged.

Common mistakes

• Using peak values in V·I·cos θ without the ½ factor.
• Forgetting to conjugate the current: S = V·I*, not V·I.
• Adding apparent powers of loads with different power factors. Add P and Q separately, then combine.
• Taking θ as the angle of the voltage alone instead of the angle between voltage and current.
• Stating a pf without "lagging" or "leading".
• Correcting to unity pf when the question asks for a specific value, or using the full load Q instead of the difference Q1 − Q2.
• Equating power factor with efficiency.

For GATE IN

Typical questions: P, Q, S and pf for a given impedance or for given v(t) and i(t) expressions; the capacitance for pf correction; the effect of correction on line current; the total power of several parallel loads by complex-power addition; wattmeter readings. Practise converting time-domain expressions to RMS phasors, and summing loads in P/Q form.

Quick check

1. v = 100·cos(ωt) V and i = 10·cos(ωt − 60°) A. What is the average power?
2. A load takes 800 W and 600 var. What are |S| and the pf?
3. Does adding a parallel capacitor change the real power drawn by an inductive load?
4. What is the unit of reactive power?

Answers: 1. ½ × 100 × 10 × cos 60° = 250 W. 2. 1000 VA, 0.8 lagging. 3. No; only Q and the line current change. 4. var (volt-ampere reactive).