1.2 AC Fundamentals, Reactance, and Impedance

AEA Certified Aircraft Electronics Technician CAET

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1 Chapter 1: Basic Electrical Theory & Systems

1.1 Ohm's Law, Power, and DC Circuit Analysis 1.2 AC Fundamentals, Reactance, and Impedance 1.3 Semiconductor Devices and Power Regulation 1.4 Digital Logic Gates and Component Identification 1.5 Aircraft Electrical Power Systems: Batteries, Generation, Buses, and Circuit Protection

2 Chapter 2: CNS — Communication, Navigation & Surveillance Systems

2.1 Aviation VHF Communications and Audio Integration 2.2 Terrestrial Radio Navigation: VOR, Localizer, and Glideslope 2.3 Pulse Navigation and Surveillance: DME, Transponders, and Weather Radar 2.4 Satellite Navigation, ADS-B, and Traffic Surveillance

3 Chapter 3: Aircraft Wiring, Hardware & Installation

3.1 Aircraft Wire Selection, Sizing, and Stripping Standards 3.2 Terminations, Crimping, and Environmental Splicing 3.3 Harness Fabrication, Routing, and Lacing Standards 3.4 Shielding, Grounding, Bonding, and Antenna Installation

4 Chapter 4: Maintenance Procedures & Regulatory Requirements

4.1 14 CFR Part 43 Maintenance Records and Return to Service 4.2 Airworthiness Directives, Certifications, and Parts Traceability 4.3 Instructions for Continued Airworthiness and Systematic Troubleshooting

5 Chapter 5: Flight Instrument & Pitot-Static Systems

5.1 Pitot-Static Principles, Plumbing, and Analog Instruments 5.2 14 CFR 91.411 and 91.413 Mandatory 24-Month Testing 5.3 Gyroscopic Principles, AHRS, and Electronic Flight Displays

6 Chapter 6: Digital Databus Systems

6.1 ARINC 429 Specification, Word Layout, and Physical Signaling 6.2 Controller Area Network (CAN) and Garmin HSDB Architectures 6.3 ARINC 664/AFDX and MIL-STD-1553 Avionics Architectures

7 Chapter 7: Tools, Test Equipment & Measurement Techniques

7.1 Multimeters, Insulation Resistance Testers, and Hand Tools 7.2 Oscilloscopes, Signal Integrity, and Time-Domain Reflectometry 7.3 Avionics Ramp Test Sets and Flight Line Validation

8 Chapter 8: Safety, Housekeeping & Best Practices

8.1 Electrical Safety, Lockout/Tagout, and RF Radiation Hazards 8.2 Electrostatic Discharge (ESD) Protection and FOD Prevention 8.3 Aviation Human Factors and the Maintenance Dirty Dozen

1.2 AC Fundamentals, Reactance, and Impedance

Key Takeaways Explain This Topic AI

AC sinusoidal waveforms are quantified by instantaneous, peak (V_p), peak-to-peak (V_p-p = 2V_p), RMS (V_RMS = 0.707 × V_p), and average (V_avg = 0.637 × V_p) voltage values; AC multimeters calibrate displays strictly to RMS.

Transport and military aircraft use 400 Hz three-phase AC because the transformer core area needed for a given voltage scales inversely with frequency (A_c ∝ 1/f), so transformers, motors, and generators are much smaller and lighter than 60 Hz equivalents.

Inductive reactance (X_L = 2πfL) causes current to lag voltage by 90°, while capacitive reactance (X_C = 1 / (2πfC)) causes current to lead voltage by 90°.

Total impedance in a series RLC circuit is calculated as Z = √(R² + (X_L - X_C)²), reaching a minimum pure resistance (Z = R) at the resonant frequency f_0 = 1 / (2π√(LC)).

Aircraft instrument transformers operate via mutual inductance, stepping down 115V AC 400 Hz bus voltage to 26V AC 400 Hz to excite flight instrumentation, synchros, and resolvers.

Last updated: October 2026

1.2 AC Fundamentals, Reactance, and Impedance

Quick Answer: Alternating current (AC) voltage varies continuously in magnitude and reverses polarity periodically. For a pure sine wave, the effective root-mean-square voltage is V RMS = 0.707 × V peak V_{\text{RMS}} = 0.707 \times V_{\text{peak}} V RMS ​ = 0.707 × V peak ​ , which produces the identical heating effect as an equivalent DC voltage. Commercial transport and military aircraft standardize on 115V / 200V, 3-phase, 400 Hz AC power because magnetic core cross-sectional area in transformers, motors, and alternators scales inversely with frequency ( A c ∝ 1 / f A_c \propto 1/f A c ​ ∝ 1/ f ), making transformers, motors, and generators much smaller and lighter. In AC circuits, inductors oppose current changes with inductive reactance ( X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L , current lagging voltage by 90 ∘ 90^\circ 9 0 ∘ ), while capacitors oppose voltage changes with capacitive reactance ( X C = 1 2 π f C X_C = \frac{1}{2\pi f C} X C ​ = 2 π f C 1 ​ , current leading voltage by 90 ∘ 90^\circ 9 0 ∘ ). Total opposition to current flow is impedance ( Z = R 2 + ( X L − X C ) 2 Z = \sqrt{R^2 + (X_L - X_C)^2} Z = R 2 + ( X L ​ − X C ​ ) 2 ​ ), which reaches a minimum pure resistance ( Z = R Z = R Z = R ) at the resonant frequency f 0 = 1 2 π L C f_0 = \frac{1}{2\pi\sqrt{LC}} f 0 ​ = 2 π L C ​ 1 ​ .

AC Sine Wave Characteristics

Unlike direct current (DC), which flows unidirectionally with constant magnitude, an alternating current (AC) waveform continuously alternates in potential and direction. The standard mathematical waveform generated by aircraft alternators and inverters is the sinusoidal wave (sine wave).

An avionics technician must distinguish between four distinct voltage measurements on an AC sine wave:

Peak Voltage ( V peak V_{\text{peak}} V peak ​ or V p V_p V p ​ ): The maximum instantaneous voltage amplitude attained from the zero-volt reference line to either the positive crest or negative trough: V p = 2 × V RMS ≈ 1.414 × V RMS V_p = \sqrt{2} \times V_{\text{RMS}} \approx 1.414 \times V_{\text{RMS}} V p ​ = 2 ​ × V RMS ​ ≈ 1.414 × V RMS ​

Peak-to-Peak Voltage ( V p-p V_{\text{p-p}} V p-p ​ ): The total electrical potential swing measured from the extreme positive crest to the extreme negative trough on an oscilloscope display: V p-p = 2 × V peak = 2.828 × V RMS V_{\text{p-p}} = 2 \times V_{\text{peak}} = 2.828 \times V_{\text{RMS}} V p-p ​ = 2 × V peak ​ = 2.828 × V RMS ​

Root-Mean-Square Voltage ( V RMS V_{\text{RMS}} V RMS ​ ): Also termed effective voltage , V RMS V_{\text{RMS}} V RMS ​ represents the DC voltage equivalent that delivers identical heating power to a purely resistive load. All standard aircraft digital multimeters (DMMs), bus voltmeter gauges, and electrical specifications report AC potential in RMS unless explicitly annotated otherwise: V RMS = V peak 2 ≈ 0.707 × V peak V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} \approx 0.707 \times V_{\text{peak}} V RMS ​ = 2 ​ V peak ​ ​ ≈ 0.707 × V peak ​

Average Voltage ( V avg V_{\text{avg}} V avg ​ ): The mathematical average of all instantaneous values over a single half-cycle ( 180 ∘ 180^\circ 18 0 ∘ electrical). Over a complete 360 ∘ 360^\circ 36 0 ∘ cycle, the true algebraic average of a symmetrical sine wave is zero; therefore, average voltage is defined across one half-cycle: V avg = 2 π × V peak ≈ 0.637 × V peak = 0.900 × V RMS V_{\text{avg}} = \frac{2}{\pi} \times V_{\text{peak}} \approx 0.637 \times V_{\text{peak}} = 0.900 \times V_{\text{RMS}} V avg ​ = π 2 ​ × V peak ​ ≈ 0.637 × V peak ​ = 0.900 × V RMS ​

AC Waveform Parameter Formula Relative to V peak V_{\text{peak}} V peak ​ Formula Relative to V RMS V_{\text{RMS}} V RMS ​ 115V RMS Aircraft Bus Value

RMS (Effective) 0.707 × V peak 0.707 \times V_{\text{peak}} 0.707 × V peak ​ 1.000 × V RMS 1.000 \times V_{\text{RMS}} 1.000 × V RMS ​ 115.0 V AC

Peak ( V p V_p V p ​ ) 1.000 × V peak 1.000 \times V_{\text{peak}} 1.000 × V peak ​ 1.414 × V RMS 1.414 \times V_{\text{RMS}} 1.414 × V RMS ​ 162.6 V AC

Peak-to-Peak ( V p-p V_{\text{p-p}} V p-p ​ ) 2.000 × V peak 2.000 \times V_{\text{peak}} 2.000 × V peak ​ 2.828 × V RMS 2.828 \times V_{\text{RMS}} 2.828 × V RMS ​ 325.3 V AC

Average (Half-Cycle) 0.637 × V peak 0.637 \times V_{\text{peak}} 0.637 × V peak ​ 0.900 × V RMS 0.900 \times V_{\text{RMS}} 0.900 × V RMS ​ 103.5 V AC

The Engineering Rationale for Aircraft 400 Hz Power

Terrestrial power utilities operate at 50 Hz or 60 Hz, yet commercial transport aircraft (e.g., Boeing 737/787, Airbus A320/A350) and military aircraft standardize on 400 Hz alternating current.

Core Size and Weight Reduction

The fundamental transformer electromotive force equation dictates the magnetic core requirements:

V RMS = 4.44 × f × N × B max × A c V_{\text{RMS}} = 4.44 \times f \times N \times B_{\text{max}} \times A_c V RMS ​ = 4.44 × f × N × B max ​ × A c ​

Where:

f f f is operating frequency (Hz).

N N N is the number of winding turns.

B max B_{\text{max}} B max ​ is magnetic flux density saturation limit (Tesla).

A c A_c A c ​ is the cross-sectional area of the magnetic iron core ( m 2 m^2 m 2 ).

Rearranging for core area reveals an inverse relationship with frequency:

A c ∝ 1 f A_c \propto \frac{1}{f} A c ​ ∝ f 1 ​

By increasing operating frequency from 60 Hz 60\text{ Hz} 60 Hz to 400 Hz 400\text{ Hz} 400 Hz (a factor of 6.67 × 6.67\times 6.67 × ), the theoretical core area needed for the same voltage, turns, and flux density falls to about 15% of the 60 Hz value ( 60 / 400 = 0.15 60 / 400 = 0.15 60/400 = 0.15 ). Real components save less than that, because windings, insulation, and cooling do not shrink in proportion, but 400 Hz transformers, motors, and generators are still far smaller and lighter. That weight saving is why aircraft use 400 Hz.

High-Frequency Trade-Offs

While 400 Hz drastically reduces transformer core weight, it introduces distinct avionics installation challenges:

Skin Effect: Alternating current crowds toward the outer perimeter of a conductor. At 400 Hz, the effective conductive depth decreases, increasing AC resistance slightly compared to DC.

Inductive Reactance: Long airframe wire runs exhibit higher inductive reactance ( X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L ), causing greater line voltage drops across inductive wiring.

Capacitive Leakage: High frequency increases capacitive coupling between adjacent conductors in wire bundles, requiring shielded twisted pairs (STP) for audio and data signals.

Inductive and Capacitive Reactance

Pure resistance ( R R R ) opposes DC and AC identically. However, all reactive components (inductors and capacitors) store and release energy, creating a frequency-dependent opposition to alternating current called reactance ( X X X ) , measured in Ohms ( Ω \Omega Ω ) .

Inductive Reactance ( X L X_L X L ​ )

An inductor stores energy in an electromagnetic field. By Faraday's Law and Lenz's Law, when alternating current changes, the expanding and collapsing magnetic field induces a counter-electromotive force (back-EMF) that directly opposes the change in current. Inductive reactance is directly proportional to frequency and inductance:

X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L

Where:

f f f is frequency in Hertz (Hz).

L L L is inductance in Henries (H) .

In a purely inductive circuit, current lags applied voltage by precisely 90 ∘ 90^\circ 9 0 ∘ in phase (remembered by the mnemonic "ELI" : Voltage E E E leads current I I I in an inductor L L L ).

Capacitive Reactance ( X C X_C X C ​ )

A capacitor stores energy in an electrostatic field between conductive plates separated by a dielectric. As alternating voltage reverses, the capacitor charges and discharges, opposing any change in voltage. Capacitive reactance is inversely proportional to frequency and capacitance:

X C = 1 2 π f C X_C = \frac{1}{2\pi f C} X C ​ = 2 π f C 1 ​

Where:

f f f is frequency in Hertz (Hz).

C C C is capacitance in Farads (F) .

In a purely capacitive circuit, current leads applied voltage by precisely 90 ∘ 90^\circ 9 0 ∘ in phase (remembered by the mnemonic "ICE" : Current I I I leads voltage E E E in a capacitor C C C ). Combining both yields the classic electronics mnemonic "ELI the ICE man" .

Series RLC Impedance

When an AC circuit combines pure resistance ( R R R ), inductance ( L L L ), and capacitance ( C C C ), the total opposition to alternating current is called Impedance ( Z Z Z ) , measured in Ohms ( Ω \Omega Ω ) .

Because inductive reactance ( + 90 ∘ +90^\circ + 9 0 ∘ ) and capacitive reactance ( − 90 ∘ -90^\circ − 9 0 ∘ ) act 180 ∘ 180^\circ 18 0 ∘ out of phase with each other along the imaginary vertical axis, they directly subtract. Net reactance is:

X net = X L − X C X_{\text{net}} = X_L - X_C X net ​ = X L ​ − X C ​

Resistance and net reactance act at right angles ( 90 ∘ 90^\circ 9 0 ∘ apart). Total series impedance is calculated using the Pythagorean theorem:

Z = R 2 + ( X L − X C ) 2 Z = \sqrt{R^2 + (X_L - X_C)^2} Z = R 2 + ( X L ​ − X C ​ ) 2 ​

Ohm's Law for AC circuits is expressed as:

• I = V Z and V = I × Z I = \frac{V}{Z} \quad \text{and} \quad V = I \times Z I = Z V ​ and V = I × Z
• Worked Numerical Example

An avionics cooling blower motor circuit connected to a 115V AC, 400 Hz bus contains a winding resistance R = 30 Ω R = 30\ \Omega R = 30 Ω , an inductive reactance X L = 80 Ω X_L = 80\ \Omega X L ​ = 80 Ω , and an inline tuning capacitor with X C = 40 Ω X_C = 40\ \Omega X C ​ = 40 Ω :

Calculate net reactance: X net = 80 − 40 = 40 Ω X_{\text{net}} = 80 - 40 = 40\ \Omega X net ​ = 80 − 40 = 40 Ω (inductive).

Calculate total impedance: Z = 30 2 + 40 2 = 900 + 1600 = 2500 = 50 Ω Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega Z = 3 0 2 + 4 0 2 ​ = 900 + 1600 ​ = 2500 ​ = 50 Ω

Calculate total RMS line current: I RMS = V RMS Z = 115 V 50 Ω = 2.30 A I_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{115\text{ V}}{50\ \Omega} = 2.30\text{ A} I RMS ​ = Z V RMS ​ ​ = 50 Ω 115 V ​ = 2.30 A

Resonant Frequency ( f 0 f_0 f 0 ​ )

In any circuit containing both inductance and capacitance, there exists a unique frequency at which inductive reactance equals capacitive reactance:

X L = X C ⟹ 2 π f 0 L = 1 2 π f 0 C X_L = X_C \quad \implies \quad 2\pi f_0 L = \frac{1}{2\pi f_0 C} X L ​ = X C ​ ⟹ 2 π f 0 ​ L = 2 π f 0 ​ C 1 ​

Solving for f 0 f_0 f 0 ​ yields the resonant frequency formula :

• f 0 = 1 2 π L × C f_0 = \frac{1}{2\pi \sqrt{L \times C}} f 0 ​ = 2 π L × C ​ 1 ​
• Series vs. Parallel Resonance Behavior

Resonance Type Impedance ( Z Z Z ) at f 0 f_0 f 0 ​ Current ( I I I ) at f 0 f_0 f 0 ​ Phase Angle ( θ \theta θ ) Avionics Application

Series Resonant ( R − L − C R-L-C R − L − C ) Minimum ( Z = R Z = R Z = R ) Maximum ( I = V / R I = V / R I = V / R ) 0 ∘ 0^\circ 0 ∘ (Unity PF) Bandpass filters, IF receiver stages, antenna tuning units

Parallel Resonant (Tank) Maximum ( Z → ∞ Z \to \infty Z → ∞ ) Minimum (Line current drops) 0 ∘ 0^\circ 0 ∘ (Unity PF) RF oscillators, notch/trap filters, transmitter tank circuits

At series resonance, X L X_L X L ​ and X C X_C X C ​ completely cancel each other out ( X L − X C = 0 X_L - X_C = 0 X L ​ − X C ​ = 0 ). The circuit behaves as a pure resistance ( Z = R Z = R Z = R ), power factor reaches unity ( 1.0 1.0 1.0 ), and maximum AC current flows through the circuit.

Phase Angle and Power Factor

The phase difference between applied voltage and total circuit current is the phase angle ( θ \theta θ ) :

θ = arctan ⁡ ( X L − X C R ) \theta = \arctan\left(\frac{X_L - X_C}{R}\right) θ = arctan ( R X L ​ − X C ​ ​ )

Power Factor (PF) represents the ratio of true useful power dissipated by resistance to the total apparent power supplied by the aircraft generator:

PF = cos ⁡ θ = R Z = P true P apparent \text{PF} = \cos\theta = \frac{R}{Z} = \frac{P_{\text{true}}}{P_{\text{apparent}}} PF = cos θ = Z R ​ = P apparent ​ P true ​ ​

True Power ( P true P_{\text{true}} P true ​ ): Actual work performed and heat dissipated by resistive elements, measured in Watts (W) : P = V RMS × I RMS × cos ⁡ θ P = V_{\text{RMS}} \times I_{\text{RMS}} \times \cos\theta P = V RMS ​ × I RMS ​ × cos θ .

Apparent Power ( P apparent P_{\text{apparent}} P apparent ​ ): Total power delivered by the generator, measured in Volt-Amperes (VA) : P = V RMS × I RMS P = V_{\text{RMS}} \times I_{\text{RMS}} P = V RMS ​ × I RMS ​ .

Reactive Power ( Q Q Q ): Non-working power shuttled back and forth in reactive electromagnetic/electrostatic fields, measured in Volt-Amperes Reactive (VAR) .

In our cooling blower example ( R = 30 Ω R = 30\ \Omega R = 30 Ω , Z = 50 Ω Z = 50\ \Omega Z = 50 Ω ):

PF = R Z = 30 50 = 0.60 (60% lagging) \text{PF} = \frac{R}{Z} = \frac{30}{50} = 0.60\ \text{(60\% lagging)} PF = Z R ​ = 50 30 ​ = 0.60 (60% lagging)

A low power factor causes high circulating line currents, requiring heavier aircraft wiring without performing productive work. Some aircraft equipment, such as modern switching power supplies, includes power-factor correction to bring power factor closer to 1.0 1.0 1.0 .

Aircraft Power and Instrument Transformers

Transformers transfer alternating electrical energy between circuits through mutual electromagnetic induction across a laminated core. They do not operate on direct current (DC).

Turns Ratio and Governing Equations

The relationship between primary and secondary winding turns ( N p , N s N_p, N_s N p ​ , N s ​ ), voltages ( V p , V s V_p, V_s V p ​ , V s ​ ), and currents ( I p , I s I_p, I_s I p ​ , I s ​ ) is defined by the turns ratio :

N p N s = V p V s = I s I p \frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p} N s ​ N p ​ ​ = V s ​ V p ​ ​ = I p ​ I s ​ ​

Notice that voltage transformation is directly proportional to turns ratio, whereas current transformation is inversely proportional (to conserve power: V p I p ≈ V s I s V_p I_p \approx V_s I_s V p ​ I p ​ ≈ V s ​ I s ​ in an ideal transformer):

Step-Down Transformer: N s < N p N_s < N_p N s ​ < N p ​ , resulting in V s < V p V_s < V_p V s ​ < V p ​ and I s > I p I_s > I_p I s ​ > I p ​ .

Step-Up Transformer: N s > N p N_s > N_p N s ​ > N p ​ , resulting in V s > V p V_s > V_p V s ​ > V p ​ and I s < I p I_s < I_p I s ​ < I p ​ .

Isolation Transformer: N p = N s N_p = N_s N p ​ = N s ​ ( 1 : 1 1:1 1 : 1 ratio), providing V s = V p V_s = V_p V s ​ = V p ​ while breaking galvanic ground loops and filtering common-mode electrical noise between avionics racks.

26V AC Synchro/Resolver Excitation

A ubiquitous application of aircraft transformers is stepping down the primary 115V AC 400 Hz bus to 26V AC 400 Hz to supply reference excitation to flight director indicators, Horizontal Situation Indicators (HSI), synchros, and resolvers. An instrument transformer with 460 primary turns and 104 secondary turns yields:

V s = V p × ( N s N p ) = 115 V × ( 104 460 ) = 26.0 V AC V_s = V_p \times \left(\frac{N_s}{N_p}\right) = 115\text{ V} \times \left(\frac{104}{460}\right) = 26.0\text{ V AC} V s ​ = V p ​ × ( N p ​ N s ​ ​ ) = 115 V × ( 460 104 ​ ) = 26.0 V AC

Autotransformers utilize a single continuous tapped winding sharing turns between primary and secondary, offering superior weight savings where electrical isolation is not mandated.