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BLD-ELEC05-02 · Lesson 10 · 24 clock hours · lab

RLC Circuits, Resonance, and Power Factor

Objective. Students will calculate impedance, power factor, true power, reactive power, and apparent power for a series RLC circuit at a given frequency, then verify results by measurement and identify the resonant frequency.

RLC Circuits, Resonance, and Power Factor

What Is an RLC Circuit?

A series RLC circuit contains a resistor (R), an inductor (L), and a capacitor (C) connected end-to-end in a single loop. When AC voltage is applied, all three components oppose current flow, but they do so in fundamentally different ways. The resistor dissipates energy as heat. The inductor stores energy in a magnetic field and causes current to lag voltage by up to 90°. The capacitor stores energy in an electric field and causes current to lead voltage by up to 90°. Because inductive and capacitive effects are opposite in phase, they partially cancel each other.

Impedance in a Series RLC Circuit

Impedance (Z) is the total opposition to AC current, measured in ohms. It combines resistance and reactance using the Pythagorean relationship:

Z = √(R² + (X_L − X_C)²)

Where:

The difference (X_L − X_C) is called the net reactance (X). When X is positive the circuit is net inductive; when negative it is net capacitive.

Resonance

Resonance occurs at the frequency where X_L exactly equals X_C, so net reactance equals zero. At that point impedance equals R alone — its minimum possible value — and current reaches its maximum. The resonant frequency is:

f_r = 1 / (2π√(LC))

Below resonance the circuit behaves capacitively (current leads). Above resonance it behaves inductively (current lags). At resonance voltage across L and voltage across C are equal in magnitude and opposite in phase, so they cancel; only the resistor drop appears across the source.

The Power Triangle: P, Q, and S

AC circuits with reactive components require three power quantities:

These three form a right triangle with S as the hypotenuse, P as the base, and Q as the vertical leg. This is called the power triangle.

Power Factor

Power factor (PF) is the ratio of true power to apparent power:

PF = P / S = cos θ

where θ is the phase angle between voltage and current. PF ranges from 0 to 1. A PF of 1.0 (unity) means all power is doing useful work. Low power factor means the source must supply excess current to deliver the same real power, increasing losses in conductors and transformers. Utilities often penalize commercial customers for low power factor.

Power-Factor Correction

Adding capacitance in parallel with an inductive load reduces the net reactive power drawn from the source. The capacitor supplies reactive power locally, so the line current drops even though the load's true power stays the same. This is power-factor correction — a practice you will encounter whenever you work on commercial motor installations or large HVAC equipment.

Lab Safety Reminders

Before energizing any circuit on the bench, observe the following:

Connecting Theory to the Trade

Every motor, transformer, and fluorescent ballast on a jobsite is an inductive load with a lagging power factor. Electricians size conductors and overcurrent protection based on apparent power (VA), not just watts. An undersized conductor in a low-PF installation overheats even if the kilowatt load seems modest. Understanding the power triangle helps you explain to a building owner why corrective capacitors save money on their utility bill and prevent nuisance tripping.

Key terms9

Impedance (Z)The total opposition to AC current in a circuit, combining resistance and reactance, measured in ohms. Z = √(R² + (X_L − X_C)²).
Inductive reactance (X_L)The opposition to AC current caused by an inductor, equal to 2πfL ohms. It increases with frequency.
Capacitive reactance (X_C)The opposition to AC current caused by a capacitor, equal to 1/(2πfC) ohms. It decreases with frequency.
Resonant frequency (f_r)The frequency at which X_L equals X_C, net reactance is zero, impedance is at its minimum (equal to R), and circuit current is at its maximum.
True power (P)The power actually consumed and converted to useful work or heat, measured in watts (W). P = V·I·cos θ.
Reactive power (Q)Power continuously exchanged between the source and reactive components without doing useful work, measured in volt-amperes reactive (VAR).
Apparent power (S)The total power the source must supply, the product of RMS voltage and RMS current, measured in volt-amperes (VA). S = V·I = √(P² + Q²).
Power factor (PF)The ratio of true power to apparent power: PF = P/S = cos θ. Ranges from 0 to 1; unity (1.0) means all supplied power does useful work.
Power-factor correctionAdding capacitance in parallel with an inductive load to reduce reactive power drawn from the source, lowering line current and improving efficiency.

Worked examples4

1. A series RLC circuit has R = 30 Ω, L = 80 mH, and C = 50 µF. The source frequency is 60 Hz. Find X_L, X_C, net reactance X, and impedance Z.
  1. Calculate X_L: X_L = 2π × 60 × 0.080 = 2π × 4.8 = 30.16 Ω ≈ 30.2 Ω
  2. Calculate X_C: X_C = 1 / (2π × 60 × 0.000050) = 1 / (0.018850) = 53.05 Ω ≈ 53.1 Ω
  3. Net reactance: X = X_L − X_C = 30.2 − 53.1 = −22.9 Ω (net capacitive)
  4. Impedance: Z = √(30² + (−22.9)²) = √(900 + 524.4) = √1424.4 = 37.7 Ω
Answer: X_L ≈ 30.2 Ω, X_C ≈ 53.1 Ω, X ≈ −22.9 Ω (capacitive), Z ≈ 37.7 Ω
2. Using the circuit from Example 1 (Z = 37.7 Ω, R = 30 Ω) with a source voltage of 120 V RMS at 60 Hz, find the current I, true power P, reactive power Q, apparent power S, and power factor PF.
  1. Current: I = V / Z = 120 / 37.7 = 3.18 A
  2. True power: P = I² × R = (3.18)² × 30 = 10.11 × 30 = 303.4 W
  3. Reactive power: Q = I² × |X| = (3.18)² × 22.9 = 10.11 × 22.9 = 231.5 VAR (leading, because net capacitive)
  4. Apparent power: S = V × I = 120 × 3.18 = 381.6 VA (verify: √(303.4² + 231.5²) = √(92,052 + 53,592) = √145,644 ≈ 381.6 VA ✓)
  5. Power factor: PF = P / S = 303.4 / 381.6 = 0.795 leading
Answer: I ≈ 3.18 A, P ≈ 303 W, Q ≈ 232 VAR (leading), S ≈ 382 VA, PF ≈ 0.795 leading
3. Find the resonant frequency for a circuit with L = 80 mH and C = 50 µF, then confirm that at resonance the impedance equals R = 30 Ω.
  1. f_r = 1 / (2π√(LC)) = 1 / (2π × √(0.080 × 0.000050))
  2. LC = 0.080 × 0.000050 = 0.000004 H·F; √(LC) = √(0.000004) = 0.002
  3. f_r = 1 / (2π × 0.002) = 1 / 0.012566 = 79.6 Hz
  4. At f_r: X_L = 2π × 79.6 × 0.080 = 40.0 Ω; X_C = 1/(2π × 79.6 × 0.000050) = 40.0 Ω; X_L − X_C = 0
  5. Z = √(R² + 0²) = R = 30 Ω ✓
Answer: Resonant frequency f_r ≈ 79.6 Hz; at resonance Z = 30 Ω (equal to R alone)
4. A 240 V, 60 Hz motor draws 8 A and has a power factor of 0.72 lagging. Find true power P, apparent power S, and reactive power Q. A capacitor bank is added that supplies 900 VAR. What is the new power factor?
  1. Apparent power: S = V × I = 240 × 8 = 1,920 VA
  2. True power: P = S × PF = 1,920 × 0.72 = 1,382 W
  3. Reactive power (original): Q = √(S² − P²) = √(1,920² − 1,382²) = √(3,686,400 − 1,909,924) = √1,776,476 ≈ 1,333 VAR lagging
  4. After correction: Q_new = 1,333 − 900 = 433 VAR lagging
  5. New apparent power: S_new = √(P² + Q_new²) = √(1,382² + 433²) = √(1,909,924 + 187,489) = √2,097,413 ≈ 1,448 VA
  6. New power factor: PF_new = P / S_new = 1,382 / 1,448 = 0.954 lagging
Answer: P ≈ 1,382 W, S ≈ 1,920 VA, Q ≈ 1,333 VAR lagging; after correction PF ≈ 0.954 lagging

The math in this lesson2

Learners compute RLC reactances and impedance, true, reactive and apparent power, resonant frequency, and power factor.
A-SSE.1G-SRT.8
Brush up in the math halls at newSchoolWorks.com:
In class: Instructor-led derivation of the power triangle (P, Q, S) and power factor formula; students solve four increasingly complex RLC problems using a scientific calculator. · Lab build: students assemble a series RLC circuit, measure voltage and current at three frequencies (below, at, and above resonance), calculate impedance and power factor at each, and record in the lab log; they identify minimum impedance (resonance) from the data table. · Power-factor correction discussion and demo: instructor adds a capacitor in parallel with an inductive load and students re-measure current to observe reduction, connecting the concept to real utility billing and motor loads seen on jobsites.
Materials: Resistors, inductors, and capacitors; AC bench power supply and function generator; true-RMS DMM; clamp-on ammeter; breadboard or trainer; scientific calculator; lab log; NCCER Level 1 and Level 2 textbooks; power-triangle reference card
Check: Calculated vs. measured impedance within ±10%; power factor correct in sign and magnitude; resonant frequency identified and within ±15% of formula value; lab log complete and signed by instructor.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

NCCER Electrical — AC Theory
1. In a series RLC circuit at 60 Hz, X_L = 45 Ω, X_C = 45 Ω, and R = 20 Ω. What is the impedance?
NCCER Electrical — AC Theory
2. Which unit is used to express reactive power (Q)?
NCCER Electrical — AC Theory
3. A series RLC circuit has R = 10 Ω, X_L = 30 Ω, and X_C = 16 Ω. What is the impedance?
NCCER Electrical — AC Theory
4. A load draws 800 W of true power and 600 VAR of reactive power. What is the apparent power?
NCCER Electrical — AC Theory
5. Inductive reactance in a circuit increases when frequency:
NCCER Electrical — AC Theory / State Journeyman knowledge
6. A motor operating at a low lagging power factor causes which of the following problems on the distribution system?
OSHA 10 — Electrical Safety
7. Before handling a capacitor that was recently energized on the bench, a student should:
NCCER Electrical — AC Theory
8. A 120 V, 60 Hz circuit has Z = 24 Ω and R = 19.2 Ω. What is the power factor?
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