13.1 Phase Relationships in R, L, C Circuits

13.1 Phase Relationships in R, L, C Circuits

Quick Answer: Pure R: V and I in phase. Pure L: V leads I by 90° (ELI); X_L = 2πfL. Pure C: I leads V by 90° (ICE); X_C = 1/(2πfC). Series circuits share current; parallel circuits share voltage. Combine phasors, not arithmetic sums of peaks, when R, L, and C sit together.

CAAS SAR-66 Module 3 topic 3.14 R, C and L Circuits follows AC theory (3.13). You already know sinusoids, RMS, and phase language. This chapter asks what happens when those waves drive resistance, inductance, and capacitance—alone and in combination. Phase shift and reactance are the foundation for impedance, power factor, and the power triangle in §§13.2–13.3.

Pure Resistance — Voltage and Current in Phase

Ohm’s law still holds with RMS (or peak) values for a linear resistor on sine AC:

I = V / R

V = I R

Because v = iR at every instant, the voltage and current waveforms reach zero and peak together. Phase angle between V and I is θ = 0°. Instantaneous power is always non-negative for a pure resistor (or zero at the zeros); average power is P = V_RMS I_RMS = I_RMS² R.

Worked example 1 — resistive branch at 400 Hz. A heater of R = 23 Ω is fed from 115 V RMS aircraft AC (frequency does not change R).

I = 115 / 23 = 5.0 A RMS. Voltage and current are in phase. P = 115 × 5.0 = 575 W.

Pure Inductance — Voltage Leads Current by 90°

An ideal inductor stores energy in its magnetic field. For sinusoidal current, the self-induced voltage is proportional to di/dt. Differentiating a sine advances it by 90°, so:

Voltage leads current by 90° (equivalently, current lags voltage by 90°).

Memory aid ELI: voltage E across L, then I — E before I.

The opposition of an ideal inductor to sine AC is inductive reactance:

X_L = 2πfL = ωL

Ohm’s-law form for magnitude: V_L = I X_L (RMS with RMS).

Worked example 2 — X_L at 400 Hz. L = 10 mH = 0.010 H; f = 400 Hz.

X_L = 2π × 400 × 0.010 = 2π × 4 = 25.13 Ω ≈ 25.1 Ω.

Worked example 3 — same L at 50 Hz (trap comparison). X_L = 2π × 50 × 0.010 ≈ 3.14 Ω. At aircraft frequency the same choke looks about eight times “stiffer” (400/50 = 8).

Worked example 4 — current in pure L. 115 V RMS across that 10 mH inductor at 400 Hz.

I = V / X_L = 115 / 25.13 ≈ 4.58 A RMS. Current lags voltage by 90°. Ideal average true power is zero (energy oscillates to and from the field each cycle).

Pure Capacitance — Current Leads Voltage by 90°

An ideal capacitor’s current is proportional to dv/dt. Differentiating voltage advances current by 90°, so:

Current leads voltage by 90° (equivalently, voltage lags current by 90°).

Memory aid ICE: current I through C, then E — I before E.

Capacitive reactance:

X_C = 1 / (2πfC) = 1 / (ωC)

V_C = I X_C (RMS form).

Worked example 5 — X_C at 400 Hz. C = 10 µF = 10 × 10⁻⁶ F; f = 400 Hz.

X_C = 1 / (2π × 400 × 10 × 10⁻⁶) = 1 / (2π × 4 × 10⁻³) = 1 / (0.02513) ≈ 39.8 Ω.

Worked example 6 — same C at 50 Hz. X_C = 1 / (2π × 50 × 10 × 10⁻⁶) ≈ 318 Ω. Higher frequency lowers X_C — the capacitor passes AC more easily at 400 Hz than at utility frequency.

Worked example 7 — current in pure C. 115 V RMS across 10 µF at 400 Hz.

I = 115 / 39.8 ≈ 2.89 A RMS. Current leads voltage by 90°. Ideal average true power is again zero.

Reactance Summary Table

Exam discipline: Never add X_L and X_C as if they were the same kind of opposition without a sign/phasor rule — they oppose each other in series (next section). Always use f = 400 Hz on aircraft stems unless told otherwise.

Series Combination Overview

In a series R, L, C path there is one current through every element. Element voltages differ in phase:

• V_R in phase with I
• V_L leads I by 90°
• V_C lags I by 90° (or V_C is 180° from V_L along the reactive axis)

Supply voltage is the phasor sum of the element voltages — not V_R + V_L + V_C as scalars. Net reactive voltage is often written V_L − V_C (magnitudes) along the ±j axis, then combined with V_R:

V = √(V_R² + (V_L − V_C)²)

Worked example 8 — series intuition at 400 Hz. Series R = 30 Ω, X_L = 40 Ω, X_C = 10 Ω; I = 2 A RMS.

V_R = 2 × 30 = 60 V; V_L = 80 V; V_C = 20 V.

Net reactive drop = 80 − 20 = 60 V.

V_supply = √(60² + 60²) = 60√2 ≈ 84.9 V. Current lags the supply voltage (inductive net) — full Z and θ follow in §13.2.

Parallel Combination Overview

In a parallel R, L, C network the same voltage appears across each branch. Branch currents differ in phase:

• I_R in phase with V
• I_L lags V by 90°
• I_C leads V by 90°

Total current is the phasor sum of branch currents:

I = √(I_R² + (I_C − I_L)²) (common Module 3 form when comparing capacitive and inductive branch currents)

Worked example 9 — parallel sketch at 400 Hz. 115 V RMS across parallel R = 57.5 Ω and L with X_L = 57.5 Ω.

I_R = 115 / 57.5 = 2.0 A (in phase with V).

I_L = 115 / 57.5 = 2.0 A (lags V by 90°).

I_total = √(2² + 2²) = 2√2 ≈ 2.83 A. Supply current lags voltage — inductive parallel mix.

Series-Parallel Overview

Real aircraft wiring often mixes topologies: a series RL feed into a parallel capacitive filter branch, or a resistor in series with a parallel LC tank. Method for Module 3:

1. Reduce each pure series or parallel group to an equivalent impedance (or admittance) using phasors.
2. Combine the equivalents step by step toward the source.
3. Keep one reference phasor clear — usually series current or parallel voltage.

You do not need every exotic network on the first pass; you do need to refuse scalar addition of out-of-phase voltages or currents.

Aircraft 400 Hz Context

Section Synthesis

Lock ELI / ICE and the two reactance formulae with 400 Hz drills. Section 13.2 packs them into impedance Z, phase angle, and power factor.