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Electrical Circuits and Sensors · Module 6 of 10

AC Power, Power Factor, and Resonance

In AC circuits not all the power does useful work. Real power heats and drives; reactive power sloshes back and forth. The power triangle keeps them straight, and resonance is where reactance disappears.

01

Readiness check

This module quantifies AC power. Tick only what you can do closed-notes.

  • Find the impedance angle of an RLC branch.
  • Recall the cosine and sine of 36.9° (0.8 and 0.6).
  • Use a right triangle to relate two legs and a hypotenuse.
  • Recall ω = 2πf and the reactances ωL and 1/ωC.
  • Distinguish energy dissipated from energy stored.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit impedance and phase in Module 5.
3 or more weak itemsReview energy storage in Module 4.
02

The core idea

Apparent power is the product of rms voltage and current; the phase angle splits it into real power that does work and reactive power that only circulates. Power factor is the cosine of that angle.

S = Vrms Irms, P = S cos θ, Q = S sin θpf = cos θ = P/Sresonance: ω0 = 1/√(LC)

In an AC circuit the instantaneous power swings as voltage and current rise and fall. Averaged over a cycle, the real power P = VrmsIrms cos θ is what actually heats a resistor or turns a motor, where θ is the phase between voltage and current. The reactive power Q = VrmsIrms sin θ flows into and out of inductors and capacitors without net work. Their vector sum is the apparent power S = VrmsIrms, the product the supply must actually deliver. These three form a right triangle, and the power factor cos θ measures how much of the apparent power is real. At resonance the inductive and capacitive reactances cancel, the circuit looks purely resistive, and the power factor is one.

The skill works when: you find the phase angle first, then read P, Q, and S off the same triangle.
The skill breaks down when: apparent power is treated as real power, or resonance is sought by setting impedance, not reactance, to zero.
The concept. The power triangle: real power along the base, reactive power up the side, apparent power as the hypotenuse. The angle θ is the same phase that separates voltage and current, and its cosine is the power factor.
03

The skills, taught in order

Five skills build the three powers, the power factor, and the resonance condition that removes reactance.

6.1 Instantaneous and average power

Instantaneous power p(t) = v(t)i(t) pulses at twice the source frequency. What matters for heating and work is its average over a cycle, the real power P = VrmsIrms cos θ, measured in watts.

6.2 Real, reactive, and apparent power

Real power P (watts) does work. Reactive power Q = VrmsIrms sin θ (volt-amps reactive) only exchanges energy with the fields. Apparent power S = VrmsIrms (volt-amps) is their magnitude, the load the supply must carry.

QuantitySymbolUnitFormula
Real powerPWVrms Irms cos θ
Reactive powerQVARVrms Irms sin θ
Apparent powerSVAVrms Irms
Power factorpfnonecos θ = P/S

The three powers and the factor that links them. They satisfy S2 = P2 + Q2.

6.3 Power factor

The power factor cos θ is the fraction of apparent power that is real. A purely resistive load has pf = 1; a reactive load has pf below 1, forcing the supply to carry extra current. Power-factor correction adds reactance of the opposite sign to push it back toward one.

6.4 Series resonance

In a series RLC circuit the reactances XL = ωL and XC = 1/(ωC) cancel at the resonant frequency ω0 = 1/√(LC). There the impedance is purely R, the current is maximum, and voltage and current are in phase, giving a power factor of one.

6.5 Quality factor and bandwidth

The quality factor Q0 = ω0L/R measures how sharp the resonance is. The half-power bandwidth is BW = ω0/Q0 = R/L: a high Q0 gives a narrow, selective peak, the basis of tuning and filtering.

Engineering connection: a resonant sensor, such as a quartz crystal or a vibrating element, reports a measured quantity as a shift in its resonant frequency.

04

Worked example 1: the power triangle of a load

A load draws 5 A rms from a 120 V rms supply, with the current lagging the voltage by 36.87°. Find the real, reactive, and apparent power, and the power factor.

Figure 1. The 36.87 degree lag gives a 0.8 power factor. The real power is the base of the triangle, the reactive power its height, and the apparent power its hypotenuse: a scaled 3-4-5.
  1. ProblemFind P, Q, S, and the power factor for the load in Figure 1.
  2. Given / findVrms = 120 V, Irms = 5 A, θ = 36.87° lagging. Find P, Q, S, pf.
  3. AssumptionsSteady sinusoidal operation; the angle is the phase of current behind voltage.
  4. ModelS = VrmsIrms; P = S cos θ; Q = S sin θ; pf = cos θ.
  5. EquationsS = Vrms IrmsP = S cos θ, Q = S sin θpf = cos θ
  6. SolveS = 120 × 5 = 600 VA. With cos 36.87° = 0.8 and sin 36.87° = 0.6: P = 600 × 0.8 = 480 W, Q = 600 × 0.6 = 360 VAR. pf = 0.8 lagging.
  7. CheckS2 = P2 + Q2: 4802 + 3602 = 230400 + 129600 = 360000 = 6002. The triangle closes.
  8. ConclusionOnly 480 of the 600 VA does real work; the rest circulates. Correcting the power factor toward 1 would cut the current the supply must carry.
Result. S = 600 VA, P = 480 W, Q = 360 VAR, pf = 0.8 lagging.
05

Worked example 2: series resonance

A series RLC circuit has R = 5 Ω, L = 25 mH, and C = 40 µF. Find the resonant angular frequency, the resonant frequency in hertz, the quality factor, and the bandwidth.

Figure 2. At resonance the reactances cancel, the impedance is purely R, and the current peaks. A quality factor of 5 makes the peak relatively sharp, with a half-power bandwidth of 200 rad/s.
  1. ProblemFind ω0, f0, Q0, and BW for the RLC circuit in Figure 2.
  2. Given / findR = 5 Ω, L = 25 mH, C = 40 µF. Find ω0, f0, Q0, BW.
  3. AssumptionsSeries RLC; resonance is where XL = XC.
  4. Modelω0 = 1/√(LC); f0 = ω0/2π; Q0 = ω0L/R; BW = ω0/Q0.
  5. Equationsω0 = 1/√(LC)Q0 = ω0L/RBW = ω0/Q0 = R/L
  6. SolveLC = 0.025 × 40×10−6 = 1×10−6, so ω0 = 1/√(10−6) = 1000 rad/s. f0 = 1000/2π = 159 Hz. Q0 = 1000 × 0.025/5 = 5. BW = 1000/5 = 200 rad/s.
  7. CheckBW = R/L = 5/0.025 = 200 rad/s agrees. At ω0, XL = 1000 × 0.025 = 25 Ω and XC = 1/(1000 × 40×10−6) = 25 Ω, so they cancel.
  8. ConclusionThe circuit resonates at 159 Hz, where it looks purely resistive. A quality factor of 5 sets how sharply it selects that frequency.
Result. ω0 = 1000 rad/s, f0 = 159 Hz, Q0 = 5, BW = 200 rad/s.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Apparent treated as realPower billed as VA used as watts"Did I multiply by cos θ?"Real power is S cos θ, not S.
Ignoring power factorSupply current underestimated"What is the phase angle?"A low pf raises the current for the same real power.
Resonance from impedanceω0 sought where Z = 0"Which reactances cancel?"Set XL = XC; the impedance is R, not zero.
Q and bandwidth invertedHigh Q paired with wide band"Does a sharp peak mean narrow or wide?"BW = ω0/Q0: high Q gives a narrow band.
07

Practice ladder

Level 1 · Direct skill

A load takes 230 V rms and 4 A rms at a power factor of 1. Find the real power.

Show answer

P = VrmsIrmscos θ = 230 × 4 × 1 = 920 W.

Level 2 · Mixed concept

A motor draws 10 A from 240 V at pf 0.7 lagging. Find S, P, and Q.

Show answer

S = 240 × 10 = 2400 VA. P = 2400 × 0.7 = 1680 W. Q = 2400 × sin(45.6°) = 2400 × 0.714 = 1714 VAR.

Level 3 · Independent problem

A series RLC circuit has L = 10 mH and C = 1 µF. Find its resonant frequency in hertz.

Show answer

ω0 = 1/√(LC) = 1/√(10−8) = 10000 rad/s. f0 = 10000/2π = 1592 Hz.

Transfer task | Real engineering

A factory pays for apparent power. Explain how adding capacitors to a motor load reduces the bill without changing the real power delivered.

What good work looks like

Capacitors supply the reactive power the inductive motor needs, raising the power factor toward 1. The real power P is unchanged, but the apparent power S = P/pf falls, cutting the current and the charge for apparent power.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my P, Q, and S satisfy the power triangle."
"Give me three loads; I will find the power factor of each."
"Compute the power for me." Building the triangle from the phase angle is the skill.
"What is the resonant frequency?" Setting the reactances equal yourself is the point.

Portfolio task

Take a real reactive load (a motor or a fluorescent fixture), find its three powers and power factor from rated values, and propose a correction capacitor.

Must include: S, P, Q, a power factor, and a correction that raises it toward 1.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define real, reactive, and apparent power.

P = S cos θ does work; Q = S sin θ circulates; S = VrmsIrms is their magnitude.

2. What is the power factor?

cos θ = P/S, the fraction of apparent power that is real.

3. State the resonance condition.

XL = XC, giving ω0 = 1/√(LC).

4. What is the impedance at resonance?

Purely resistive, R, with current maximum and in phase.

5. Relate quality factor and bandwidth.

BW = ω0/Q0; a high Q0 gives a narrow band.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the power triangle and resonance from a blank page.
+3 daysFind P, Q, S for three new loads.
+7 daysCarry the idea of gain into operational amplifiers, Module 7.
+30 daysReuse resonance when reading a resonant sensor.
10

Textbook mapping

This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.

Topic in this moduleWhere to read more
Real, reactive, and apparent powerAlexander & Sadiku, Chapter 11
Power factor and correctionAlexander & Sadiku, Chapter 11
Series resonance and quality factorAlexander & Sadiku, Chapter 14

Chapter numbers refer to the 4th edition. The power and resonance relations are standard, so any recent edition will align closely.