University Physics II · Inductance and AC Circuits · 11.8
AC Power & Power Factor
A load can draw heavy current and convert almost none of it. Multiply voltage by current instant by instant, average over a cycle, and cos φ appears — the number that separates what the wires carry from what the load uses.
Build the model
Connect the measurement to the mechanism.
Multiply the instantaneous voltage and current in an AC circuit and the product is not steady: p(t) = ½V₀I₀[cos φ + cos(2ωt − φ)], a constant offset plus a ripple at twice the drive frequency. The ripple averages to zero over a cycle, so P = V(rms)I(rms) cos φ, and the factor cos φ = R/Z carries the whole story. Resistance converts energy irreversibly; inductance and capacitance only borrow it, storing energy in ½Li² and ½Cv² for a quarter cycle and handing every joule back, which is why P = I(rms)²R with no reactive term.
So a load can pull a large current, stress every conductor and transformer between it and the generator, and still convert little of what it draws. The current is real even when the power is not, and because line heating goes as I(rms)²R(line), a poor power factor is paid for in the wires.
- Simple definition
- The average power in an AC circuit is P = V(rms)I(rms) cos φ, where φ is the phase of the current relative to the voltage. The power factor cos φ = R/Z is the fraction of the apparent power that is actually dissipated.
- Example
- A motor on a 230 V rms supply draws 5.0 A rms at cos φ = 0.60. The wiring carries 1150 V·A, but only 690 W is converted; the rest is reactive exchange, borrowed and handed back every quarter cycle.
A constant offset plus a ripple at twice the drive frequency, dipping negative unless φ = 0.
V₀ and I₀ are peak values in V and A; p in W
The 2ω ripple averages to zero over a cycle, leaving the resistor's dissipation.
P in W, with V(rms) = V₀/√2 and I(rms) = I₀/√2
The impedance triangle read as a ratio; φ is the phase of the current relative to the voltage.
dimensionless, 0 ≤ cos φ ≤ 1; φ > 0 means the current lags
Conductors and transformers are rated for S, but only P is converted by the load.
S in V·A, Q in var, P in W — the units keep the three apart
Halving the power factor quadruples the line heating for the same delivered watts.
R(line) is the round-trip conductor resistance in Ω
Sizes the capacitor that cancels the reactive power an inductive load would otherwise draw.
C in F, ω = 2πf in rad s⁻¹; the capacitor sits in parallel with the load
Power is a product, and the product ripples
Phasors give you the steady state: with v(t) = V₀ cos ωt across a load, the current is i(t) = I₀ cos(ωt − φ), where φ comes from the impedance. Power is not a phasor — it is the ordinary product p(t) = v(t)i(t), taken instant by instant. Use cos A cos B = ½[cos(A − B) + cos(A + B)] and it separates cleanly: p(t) = ½V₀I₀[cos φ + cos(2ωt − φ)]. A constant term, and a ripple at twice the drive frequency. The ripple is observable — a filament on a 50 Hz supply brightens and dims at 100 Hz, and a single-phase motor delivers a torque that pulses at 100 Hz too. Note the size of the ripple, ½V₀I₀, against the constant offset, ½V₀I₀ cos φ. Whenever φ ≠ 0 the ripple is the larger of the two, so p(t) goes negative over part of every cycle. During those intervals the load is not taking energy from the source; it is pushing energy back.
The cycle average, and where cos φ comes from
Average p(t) over one full period. The 2ω term completes two whole cycles in that time, so it averages to exactly zero and only the constant survives: P = ½V₀I₀ cos φ. Substitute V(rms) = V₀/√2 and I(rms) = I₀/√2 — the substitution that made rms worth defining in the first place — and the halves are absorbed: P = V(rms)I(rms) cos φ. The factor cos φ is the power factor. It is not an efficiency and not a fudge; it is the cosine of the phase angle between current and voltage, and for a series circuit the impedance triangle reads it off directly as cos φ = R/Z. Put numbers on it: R = 30 Ω, X(L) = 60 Ω and X(C) = 20 Ω give Z = √(30² + 40²) = 50 Ω, so cos φ = 0.60. On a 120 V rms supply, I(rms) = 2.40 A and P = 120 × 2.40 × 0.60 = 173 W. Forget the cos φ and you would claim 288 W — 67% too high.
Reactance borrows and returns
Set φ = 90°, a pure inductor. Then cos φ = 0 and P = 0 — yet the current is not zero, and the instantaneous power is not zero either. It is p(t) = ½V₀I₀ sin 2ωt, a pure oscillation at 2ω: for a quarter cycle the source pours energy into the magnetic field, raising the stored ½Li² to its peak of ½LI₀², and over the next quarter it pulls every joule back out. A capacitor does the same with ½Cv², a quarter cycle out of step, which is why an inductor and a capacitor can trade with each other. Nothing is dissipated in either case. That is why the average power in any series RLC circuit can be written P = I(rms)²R, with no reactive term anywhere: dissipation lives in the resistance alone. Check it on the same circuit — (2.40 A)² × 30 Ω = 173 W, identical to V(rms)I(rms) cos φ, because V(rms) cos φ is just I(rms)R.
Three powers, three units
The three quantities are kept apart by their units rather than by their symbols. Apparent power S = V(rms)I(rms) is quoted in volt-amperes: it is what conductors, breakers and transformers must be sized for, because they respond to current and are indifferent to phase. Real power P = S cos φ is in watts — what the load converts and what the meter bills. Reactive power Q = S sin φ carries the unit var precisely to flag that it is not watts; it measures the amplitude of the borrow-and-return exchange. The three form a right triangle, S² = P² + Q², which is the impedance triangle multiplied through by I(rms)². For the circuit above: S = 288 V·A, P = 173 W, and Q = 288 × 0.80 = 230 var, confirmed by I(rms)²(X(L) − X(C)) = 5.76 × 40 = 230 var. Q is signed — positive for a lagging inductive load, negative for a leading capacitive one — and that sign is what makes cancellation possible.
Correcting the power factor
An inductive load's Q is positive and a capacitor's is negative, so a capacitor wired across the load lets the two swap energy with each other instead of with the generator. In parallel, not in series, because the load's voltage must not change. Size it from the reactive powers. A factory draws P = 20.0 kW at 400 V rms and 50 Hz with cos φ₁ = 0.70 lagging, so it pulls I = 20000/(400 × 0.70) = 71.4 A and needs Q₁ = P tan φ₁ = 20.4 kvar. At the target cos φ₂ = 0.95 the supply need only provide Q₂ = P tan φ₂ = 6.57 kvar, so the capacitor must furnish ΔQ = 13.8 kvar. A capacitor's reactive power is V(rms)²ωC, so C = ΔQ/(ωV(rms)²) = 13830/(314 × 400²) = 275 μF. The current falls to 20000/(400 × 0.95) = 52.6 A. The real power is untouched — the machines do the same work — but the supply carries 26% less current.
What the grid pays, and where the model stops
Follow that current back out to the line. Delivering P at V(rms) demands I(rms) = P/(V(rms) cos φ), so the conductor heating is P(loss) = I(rms)²R(line) = R(line)P²/(V(rms) cos φ)², inverse in cos²φ. Put the factory on a feeder of 0.10 Ω round trip: 71.4 A wastes 510 W before correction and 52.6 A wastes 277 W after — 233 W saved by a component that consumes nothing itself. That 1/cos²φ, and the 1/V(rms)² sitting beside it, is why utilities meter reactive power for industrial customers and why transmission runs at high voltage. Two limits. Everything here assumes sinusoids: rectifiers and switching supplies draw distorted current whose harmonics carry no average power against a sinusoidal voltage, so the true power factor P/S drops below the displacement factor cos φ. And over-correcting swings the load capacitive, which raises the current again and can resonate with the supply inductance.
Change one variable at a time
Make the relationship visible.
Hold the current at 5.0 A and drag φ from 0° to 90°: the wires keep carrying 1150 V·A the whole way, but the power curve sinks as far below zero as it rises above and the average line drops to nothing.
POWER FACTOR cos φ0.60
REAL POWER P692 W
REACTIVE POWER Q918 var
APPARENT POWER S1150 V·A
Live interpretationPOWER FACTOR cos φ: 0.60. REAL POWER P: 692 W. REACTIVE POWER Q: 918 var. APPARENT POWER S: 1150 V·A
Catch the common trap
Explain before calculating.
A series RLC circuit with R = 30 Ω, X(L) = 60 Ω and X(C) = 20 Ω is driven from a 120 V rms supply. What average power does the source deliver to it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn induction motor on a 230 V rms supply draws 8.0 A rms at a power factor of 0.75 lagging. Find the apparent power the wiring must carry, the real power the motor converts, and the reactive power.
- Apparent power needs no phase at all: S = V(rms)I(rms) = 230 × 8.0 = 1840 V·A. That is the figure the cable, the breaker and any transformer must be rated for.
- Real power is that product cut by the power factor: P = S cos φ = 1840 × 0.75 = 1380 W.
- Reactive power needs sin φ: sin φ = √(1 − 0.75²) = √0.4375 = 0.661, so Q = S sin φ = 1840 × 0.661 = 1220 var.
- Check the triangle S² = P² + Q²: √(1380² + 1220²) = √(3.393 × 10⁶) = 1842 V·A, back to S within the rounding.
AnswerS = 1.84 kV·A, P = 1.38 kW, Q = 1.22 kvar. The wiring carries 8.0 A whatever the phase does; only three-quarters of what it carries is converted.
MediumA series circuit of R = 40.0 Ω, L = 0.150 H and C = 60.0 µF is driven from a 120 V rms, 60.0 Hz supply. Find the power factor and the average power delivered.
- Reactances at ω = 2π(60.0) = 377 rad s⁻¹: X(L) = ωL = 377 × 0.150 = 56.5 Ω, and X(C) = 1/(ωC) = 1/(377 × 60.0 × 10⁻⁶) = 44.2 Ω.
- Net reactance X(L) − X(C) = 56.5 − 44.2 = 12.3 Ω, so Z = √(40.0² + 12.3²) = √1752 = 41.9 Ω.
- Power factor straight off the impedance triangle: cos φ = R/Z = 40.0/41.9 = 0.956, lagging because X(L) exceeds X(C).
- Current: I(rms) = V(rms)/Z = 120/41.9 = 2.87 A.
- P = V(rms)I(rms) cos φ = 120 × 2.87 × 0.956 = 329 W, and I(rms)²R = 2.87² × 40.0 = 329 W agrees — dissipation lives in R alone.
Answercos φ = 0.956 lagging; P = 329 W out of S = 344 V·A, with Q = 101 var borrowed and returned each quarter cycle.
HardA workshop draws 12.0 kW at 240 V rms, 50.0 Hz with a power factor of 0.650 lagging, fed down a feeder of 0.200 Ω round trip. Find the line current, the parallel capacitance that raises the power factor to 0.920, and the feeder loss before and after.
- Line current before: I = P/(V(rms) cos φ₁) = 12000/(240 × 0.650) = 76.9 A.
- Reactive power now and wanted: tan φ₁ = 1.169 for cos φ₁ = 0.650 and tan φ₂ = 0.4260 for cos φ₂ = 0.920, so Q₁ = 12.0 × 1.169 = 14.03 kvar and Q₂ = 12.0 × 0.4260 = 5.11 kvar. The capacitor must supply the difference, ΔQ = 8.92 kvar.
- A capacitor's reactive power is V(rms)²ωC, so C = ΔQ/(ωV(rms)²) = 8920/(314.2 × 240²) = 8920/(1.810 × 10⁷) = 4.93 × 10⁻⁴ F = 493 µF, wired across the load so its voltage is untouched.
- Line current after: I = 12000/(240 × 0.920) = 54.3 A.
- Feeder loss goes as I²: before, 76.9² × 0.200 = 1.18 kW; after, 54.3² × 0.200 = 590 W. The capacitor dissipates nothing itself and saves 593 W, about 5% of the works' entire load.
AnswerI falls from 76.9 A to 54.3 A; C = 493 µF in parallel; the feeder loss falls from 1.18 kW to 0.59 kW. The 12.0 kW the machines actually convert never changes.