13.2 Impedance, Phase Angle & Power Factor

13.2 Impedance, Phase Angle & Power Factor

Quick Answer: Impedance Z is the AC opposition combining R and net reactance: for series, Z = √(R² + (X_L − X_C)²). I = V / Z. Phase angle θ from tan θ = X/R or cos θ = R/Z. Power factor = cos θ = R/Z. At series resonance X_L = X_C and Z ≈ R.

Section 13.1 fixed phase and reactance for pure elements. Module 3 topic 3.14 next demands quantitative impedance, the angle between supply voltage and current, and power factor—the bridge into true/apparent/reactive power in §13.3.

Impedance Defined

Impedance Z is the generalisation of resistance for sine AC circuits that may include L and C. Its SI unit is the ohm. Magnitude tells how large the current is for a given voltage; angle tells how far current sits from voltage in phase.

In rectangular (Module 3 working) form for a series combination:

Z = R + jX

where net reactance X = X_L − X_C (positive X → inductive; negative X → capacitive).

Magnitude:

|Z| = √(R² + X²) = √(R² + (X_L − X_C)²)

Often written simply Z for the magnitude in technician calculations.

Ohm’s law for AC (RMS):

I = V / Z

V = I Z

Worked example 1 — series RL at 400 Hz

R = 30 Ω, L = 20 mH, f = 400 Hz, V = 115 V RMS.

X_L = 2π × 400 × 0.020 = 2π × 8 ≈ 50.27 Ω.

Z = √(30² + 50.27²) = √(900 + 2527) ≈ √3427 ≈ 58.5 Ω.

I = 115 / 58.5 ≈ 1.97 A RMS.

Worked example 2 — series RC at 400 Hz

R = 40 Ω, C = 20 µF, f = 400 Hz, V = 115 V RMS.

X_C = 1 / (2π × 400 × 20 × 10⁻⁶) = 1 / (0.05027) ≈ 19.9 Ω.

Z = √(40² + 19.9²) ≈ √(1600 + 396) ≈ √1996 ≈ 44.7 Ω.

I = 115 / 44.7 ≈ 2.57 A RMS.

Worked example 3 — series RLC

R = 25 Ω, X_L = 60 Ω, X_C = 20 Ω (already at operating frequency), V = 100 V RMS.

X = 60 − 20 = 40 Ω (inductive).

Z = √(25² + 40²) = √(625 + 1600) = √2225 ≈ 47.2 Ω.

I = 100 / 47.2 ≈ 2.12 A.

Trap: Using Z = R + X_L + X_C = 105 Ω as a scalar sum is wrong — reactances cancel partially and combine under the square root with R.

Phase Angle

The angle θ between supply voltage and current (series circuit, current as reference often):

tan θ = X / R = (X_L − X_C) / R

cos θ = R / Z

sin θ = X / Z

Worked example 4 — angle from example 3. R = 25 Ω, X = 40 Ω, Z ≈ 47.2 Ω.

cos θ = 25 / 47.2 ≈ 0.529 → θ ≈ 58° lagging (inductive).

tan θ = 40/25 = 1.6 → θ = tan⁻¹(1.6) ≈ 58° (same).

Worked example 5 — capacitive angle. R = 40 Ω, X = −19.9 Ω (from example 2), Z ≈ 44.7 Ω.

cos θ = 40 / 44.7 ≈ 0.895 → θ ≈ 26.5° leading.

Power Factor

Power factor (pf) measures how effectively the apparent volt-amperes deliver true power:

pf = cos θ = P / S

For the series model used in Module 3:

pf = R / Z

State lagging or leading with the number when the stem asks for complete pf.

Worked example 6 — pf from Z. From example 1: R = 30 Ω, Z ≈ 58.5 Ω.

pf = 30 / 58.5 ≈ 0.513 lagging.

Worked example 7 — find R from pf. A series circuit has Z = 50 Ω and pf = 0.8 lagging.

R = pf × Z = 0.8 × 50 = 40 Ω.

Then |X| = √(Z² − R²) = √(2500 − 1600) = √900 = 30 Ω (inductive because lagging).

Current Calculations — Drill Pattern

Standard Module 3 path for a series branch:

1. Compute X_L = 2πfL and/or X_C = 1/(2πfC) at the given f (use 400 Hz on aircraft).
2. Net X = X_L − X_C.
3. Z = √(R² + X²).
4. I = V / Z.
5. θ from cos⁻¹(R/Z) or tan⁻¹(X/R); name lead/lag.
6. pf = cos θ = R/Z.

Worked example 8 — full 400 Hz chain. Series R = 15 Ω, L = 8 mH, C = 25 µF; V = 115 V RMS; f = 400 Hz.

X_L = 2π × 400 × 0.008 ≈ 20.11 Ω.

X_C = 1 / (2π × 400 × 25 × 10⁻⁶) ≈ 15.92 Ω.

X = 20.11 − 15.92 ≈ 4.19 Ω (slightly inductive).

Z = √(15² + 4.19²) ≈ √(225 + 17.6) ≈ √242.6 ≈ 15.6 Ω.

I = 115 / 15.6 ≈ 7.38 A.

pf = 15 / 15.6 ≈ 0.962 lagging; θ ≈ 16°.

Notice how close X_L and X_C are: Z is only slightly above R — a preview of resonance.

Resonance (Brief — Series)

Series resonance occurs when X_L = X_C:

2πfL = 1/(2πfC) → f₀ = 1 / (2π√(LC))

At that frequency:

• Net X = 0
• Z = R (minimum impedance for a series RLC)
• I = V / R (maximum current for fixed V)
• θ = 0°, pf = 1 (unity)

Worked example 9 — Z at resonance. Series R = 12 Ω, L and C such that X_L = X_C = 80 Ω at the operating frequency.

Z = √(12² + 0²) = 12 Ω ≈ R. Large equal reactances cancel; only R limits current.

Module 3 may only ask you to recognise “at resonance Z equals R” and pf is unity. Deep Q-factor bandwidth belongs with filters later; here the exam hook is Z ≈ R when X_L = X_C.

Parallel Impedance (Recognition Level)

For a simple parallel RL or RC, it is often easier to work in currents (shared V) or admittances. Equivalent impedance magnitude is not √(R² + X²) of the branch values added the series way. If a stem gives parallel data, find branch currents, phasor-sum them, then Z_eq = V / I_total. Do not force the series formula onto a parallel schematic.

Common Exam Traps

1. Forgetting to subtract X_C from X_L (or vice versa) before squaring.
2. Computing X at 50 Hz when the aircraft bus is 400 Hz.
3. Reporting pf as R/X instead of R/Z.
4. Omitting “lagging” / “leading.”
5. Claiming Z = R + X_L as a scalar.

Section Synthesis

With Z, θ, and pf fluent, §13.3 splits power into true P, apparent S, and reactive Q on the power triangle.