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Engineering electricity: How are aircraft electrical and avionics systems engineered to power flight controls, lighting, communications and navigation?

Describe the architecture of an aircraft electrical system, identify the role of generators, batteries and bus bars, calculate electrical loads and voltage drops, and outline the role of fly-by-wire avionics

A focused answer to the HSC Engineering Studies Aeronautical Engineering dot point on aircraft electrical and avionics systems. Generators and bus bars, the 787 More Electric Aircraft architecture, fly-by-wire flight controls, voltage drop and load calculations, and worked HSC-style past exam questions.

Generated by Claude Opus 4.89 min answerUpdated 2026-07-03

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Aircraft electrical systems run on 115 V AC three-phase at 400 Hz (or 235 V on the 787), with 28 V DC for avionics. Each engine drives a generator; an APU and battery provide backup, with a RAT as the last-resort emergency source. Essential loads sit on a bus fed from multiple independent sources; non-essential loads are shed first. The More Electric Aircraft architecture of the Boeing 787 replaces bleed air and hydraulics with electrical equivalents, saving fuel and mass at the cost of much greater generation capacity (1450 kVA versus around 240 kVA on an older twin). Fly-by-wire flight controls use computers to translate sidestick input into actuator commands with envelope protection, using $V = IR$ for DC voltage drop and $S = \sqrt{3} V_L I_L$ for three-phase AC load calculations.

In plain English

Think of an airliner as a small flying city with its own power station. Each engine spins a generator the way a hydro dam spins a turbine, an auxiliary power unit is the backup generator in the basement, and the battery is like an uninterruptible power supply that keeps the lights on for a few minutes if everything else trips out. The "essential bus bar" is the hospital-grade circuit that never gets switched off, no matter which generator fails, while the "non-essential bus" (cabin lights, entertainment, galley ovens) is the first thing load-shed in a power squeeze, just like a suburb's non-critical loads get shed before a hospital does in a blackout. Fly-by-wire is simply replacing the mechanical steering cables running the length of the fuselage with electrical wires and a computer in between, the same shift a modern car made when it swapped a physical throttle cable for a drive-by-wire pedal sensor, except the aircraft's computer also refuses to let the pilot bend the wings past their design limit.

What this dot point is asking

NESA wants you to describe how an aircraft electrical and avionics system is organised, identify the role of generators, batteries and bus bars, perform basic load and voltage-drop calculations, and outline how fly-by-wire flight control systems work. This builds on the propulsion and structures content in jet engine fundamentals and composite materials in aircraft.

The answer

Architecture of an aircraft electrical system

A typical airliner electrical system has:

Engine-driven generators. One main generator per engine (typically 90 to 120 kVA on a 737, 250 kVA on a 787). Driven by the accessory gearbox at constant speed by an integrated drive generator (IDG) or, on the 787, at variable speed with frequency conversion in the bus controller (VFG).
Auxiliary power unit (APU). A small gas turbine in the tail with its own generator, used on the ground and as a backup in flight.
Ram air turbine (RAT). A small wind-driven generator that deploys from the fuselage in emergency, powering essential flight instruments and controls.
Batteries. Sealed lead-acid or lithium-ion. Provide power during engine start and as final backup. The 787 main battery is a 32 V, 65 Ah lithium-ion unit.
Bus bars. Distribution rails for AC and DC power. Essential and non-essential loads are split so non-essential loads can be shed if a generator fails.
Transformer-rectifier units (TRUs). Convert 115 V AC three-phase (or 235 V on the 787) to 28 V DC for avionics.

Standard voltages

Most large airliners use 115 V AC three-phase at 400 Hz for main distribution. The higher frequency (versus 50 Hz mains) allows smaller transformers and motors, saving mass. The Boeing 787 raised the standard to 235 V AC three-phase, allowing the same power at lower current and reducing wire mass.

DC distribution is 28 V for most avionics and lighting.

Fly-by-wire flight controls

Traditional aircraft used mechanical cables and pushrods from the control column to the hydraulic actuators at the control surfaces. Modern airliners use fly-by-wire (FBW):

Sidestick or yoke position is read by transducers.
A flight control computer translates pilot input into desired aircraft response.
The computer sends electrical signals to the actuators at the flight controls.
The actuators (hydraulic on most aircraft, electric on the 787) move the surfaces.

Advantages: lower mass (no cables); envelope protection (the computer prevents pilots from over-stressing the airframe); auto-trim and ride-quality enhancement.

The flagship Airbus FBW programmes are the A320 family and the A380; Boeing implemented FBW on the 777, 787 and 747-8. Triplicated or quadruplicated computers and sensors provide fault tolerance.

Load and voltage drop calculations

For a DC system, Ohm's law gives the voltage drop along a wire:

V_{\text{drop}} = I R

where $R = \rho l / A$ depends on wire length and cross-section. Aircraft wiring uses copper or, in the 787, aluminium for high-current runs to save mass.

For a three-phase AC system, the apparent power is:

S = \sqrt{3} V_L I_L

Active power is $P = S \cos\phi$ where $\phi$ is the power-factor angle.

Generation capacity: older twins versus the 787

Bus bar redundancy

Essential systems (flight instruments, hydraulics, fly-by-wire, communication) are powered from an essential bus that can be fed from any generator, the APU, the battery or the RAT. Non-essential systems (galleys, in-flight entertainment, cabin lighting) are on separate buses and are shed first during a generator failure.

The 787 architecture is unusual for using such a large electrical generation capacity (1450 kVA, four generators). This is the More Electric Aircraft (MEA) concept: replace traditional bleed-air, hydraulic and pneumatic systems with electrical equivalents.

Australian context

The Boeing 787-9 Dreamliners operated by Qantas use MEA architecture; the Airbus A380 fleet (still in service for Qantas international routes) uses traditional bleed-air pressurisation and hydraulic primary flight controls but FBW. Royal Australian Air Force F-35A Lightning II combat aircraft use a fully FBW flight control system with quadruply redundant computers.

Worked example

Example 1. A 1.5 m cable carries 50 A DC at 28 V. The copper wire has cross-section 6 mm^2 and resistivity $1.7 \times 10^{-8}$ ohm m.

Resistance: $R = \rho l / A = 1.7 \times 10^{-8} \times 1.5 / (6 \times 10^{-6}) = 4.25 \times 10^{-3}$ ohm.

Voltage drop: $V = IR = 50 \times 4.25 \times 10^{-3} = 0.21$ V. About 0.8 percent of bus voltage, well within the typical 2 percent limit.

Marker's note: state the percentage explicitly and compare it with the 2 percent limit; a numeric answer alone without the comparison is an incomplete response to a "calculate and comment" style question.

Example 2. A generator is rated 90 kVA at 115 V AC three-phase. Find the rated line current.

I_L = \frac{S}{\sqrt{3} V_L} = \frac{90 \times 10^3}{\sqrt{3} \times 115} = \frac{90{,}000}{199.2} = 452 \text{ A}

Marker's note: convert kVA to VA before substituting, and keep $\sqrt{3}$ (about 1.732) to at least three significant figures, otherwise the final current can be out by several amps.

Common traps

Mixing 28 V DC and 115 V AC: DC powers avionics and lighting; AC powers motors, pumps and heating. They run on different buses.
Treating 400 Hz as a mains-style frequency: 400 Hz is chosen for mass reasons (smaller transformers and motors), but it limits cable lengths because of skin-effect and inductive losses.
Forgetting the redundancy: Aircraft electrical systems have at least four independent generation sources for the essential bus.
Treating fly-by-wire as the only difference between modern and old aircraft: FBW is the headline change, but MEA architecture, lithium batteries and higher voltages are equally significant on the 787.
Forgetting to convert kVA to VA (or vice versa): A missed factor of 1000 in $S = \sqrt{3} V_L I_L$ produces a current out by a factor of 1000, an easy mark to lose in a multi-step calculation.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC style5 marksA Boeing 787 main electrical generator produces 250 kVA at 235 V AC three-phase. Calculate the rated current per phase. Outline two engineering advantages of the 'More Electric Aircraft' architecture used on the 787 compared with earlier airliners.

Show worked answer →

Rated current. For a balanced three-phase system, the apparent power is

S = \sqrt{3} V_L I_L

where $V_L$ is line-to-line voltage and $I_L$ is the line current.

I_L = \frac{S}{\sqrt{3} V_L} = \frac{250 \times 10^3}{\sqrt{3} \times 235} = \frac{250{,}000}{407} = 614 \text{ A}

Each generator carries about 614 A per phase at full load.

Two engineering advantages of More Electric Aircraft.

Engine bleed-air system eliminated. The Boeing 787 takes air for cabin pressurisation from electric compressors driven from the main electrical buses, not by bleeding hot compressed air from the engine compressor. This raises engine efficiency by about 1 to 2 percent (the engines no longer give up high-pressure air mid-compression) and reduces airframe maintenance because there is no high-temperature bleed-air ducting through the wing root.
Electrically actuated brakes and many flight controls. The 787 uses electric brakes (resistive heating element actuators) instead of hydraulic brakes, removing the need for a hydraulic system in the main landing gear and saving about 270 kg per aircraft. Many secondary flight controls are also electric, simplifying the hydraulic distribution. The trade-off is that the aircraft needs more generator capacity (1450 kVA total, four generators) than the equivalent older airliner.

Markers reward (1) correct application of the three-phase apparent power formula, (2) consistent units (kVA and V), (3) two distinct engineering advantages of MEA architecture, and (4) at least one quantitative benefit (mass saving, efficiency improvement).

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA Boeing 737 engine-driven generator supplies 100 kVA at 115 V AC three-phase. Calculate the rated line current.

Show worked solution →

Apply the three-phase apparent power formula.

I_L = \frac{S}{\sqrt{3} V_L} = \frac{100 \times 10^3}{\sqrt{3} \times 115} = \frac{100{,}000}{199.2} = 502 \text{ A}

Marking criteria: 1 mark for correctly rearranging $S = \sqrt{3} V_L I_L$ , 1 mark for correct substitution with consistent units (W and V, not kVA), 1 mark for the correct answer (502 A) with unit.

foundation3 marksA 2.0 m aluminium cable carries 40 A DC at 28 V. The cable has cross-sectional area 10 mm^2 and resistivity

2.8 \times 10^{-8}

ohm m. Calculate the voltage drop and state whether it is within the typical 2 percent aerospace limit.

Show worked solution →

Step 1: resistance.

R = \frac{\rho l}{A} = \frac{2.8 \times 10^{-8} \times 2.0}{10 \times 10^{-6}} = 5.6 \times 10^{-3} \text{ ohm}

Step 2: voltage drop.

V = IR = 40 \times 5.6 \times 10^{-3} = 0.224 \text{ V}

Step 3: check against limit. $0.224 / 28 = 0.8\%$ , which is below the 2 percent limit, so the cable is acceptable.

Marking criteria: 1 mark for correct resistance calculation, 1 mark for correct voltage drop, 1 mark for correctly comparing the percentage against the 2 percent limit and stating a conclusion.

Show worked solution →

(a) Rated line current.

I_L = \frac{S}{\sqrt{3} V_L} = \frac{250 \times 10^3}{\sqrt{3} \times 235} = 614 \text{ A}

(b) Why 787 capacity is so much larger. The 787 replaces systems that were traditionally pneumatic (bleed-air cabin pressurisation and wing anti-ice) and hydraulic (brakes, some flight controls) with electrically powered equivalents. Electric cabin-air compressors, electric brakes and electric flight-control actuators all draw continuous electrical load that older airliners supplied from engine bleed air or hydraulic pumps instead. Because more of the aircraft's total power budget is now electrical rather than pneumatic or hydraulic, the electrical generation capacity has to scale up to match, which is why the 787's 1450 kVA dwarfs the 737's 240 kVA despite carrying roughly twice the passengers, not ten times.

Marking criteria: 1 mark for correct substitution, 1 mark for correct current (614 A), 1 mark for identifying at least one system moved from pneumatic/hydraulic to electric, 1 mark for explaining the causal link between MEA design and higher generation capacity, 1 mark for quantitative comparison using the table data.

core4 marksExplain, with reference to bus bar architecture, what happens electrically if one of an aircraft's two main engine-driven generators fails in flight.

Show worked solution →

Aircraft electrical systems split loads onto essential and non-essential bus bars. The essential bus (flight instruments, hydraulics, fly-by-wire computers, communication and navigation) can be fed from any generator, the APU, the battery, or the RAT, giving it multiple independent power paths. The non-essential bus (galleys, in-flight entertainment, some cabin lighting) is fed only from the main generators.

If one main generator fails, a bus tie automatically (or the crew manually) reconfigures the essential bus to draw from the remaining healthy generator, so flight-critical systems are unaffected. Non-essential loads on the affected side may be shed (load-shedding) to keep the remaining generator within its rated capacity, since one generator alone typically cannot supply both essential and non-essential demand simultaneously.

Marking criteria: 1 mark for identifying the essential/non-essential bus split, 1 mark for stating that the essential bus can be fed from multiple independent sources, 1 mark for describing automatic reconfiguration (bus tie) to the remaining generator, 1 mark for explaining load-shedding of non-essential loads to stay within the remaining generator's rated capacity.

core4 marksCompare mechanical flight controls with fly-by-wire flight controls, referring to mass, redundancy, and envelope protection.

Show worked solution →

Mass: Mechanical controls use cables, pulleys and pushrods running the length of the fuselage and wings, adding significant mass. Fly-by-wire replaces these with lightweight electrical wiring, reducing structural mass.
Redundancy: A mechanical system's redundancy comes from duplicated cable runs, which is heavy and still vulnerable to a single jam or severed run affecting a whole control path. Fly-by-wire achieves redundancy electronically, with triplicated or quadruplicated flight control computers and sensors that vote on the correct output, tolerating a computer or sensor failure without losing control.
Envelope protection: Mechanical controls give the pilot direct, unfiltered authority over the control surfaces, so an aggressive input can overstress the airframe. Fly-by-wire computers interpret the pilot's input and limit the commanded response so the aircraft cannot exceed its structural or aerodynamic limits (envelope protection), a capability a purely mechanical system cannot provide.

Marking criteria: 1 mark for correct mass comparison, 1 mark for correct redundancy comparison (cable duplication versus computer/sensor voting), 1 mark for correctly explaining envelope protection, 1 mark for a coherent comparison rather than two disconnected descriptions.

exam6 marksAssess the engineering trade-offs of the Boeing 787's More Electric Aircraft architecture compared with the traditional bleed-air and hydraulic architecture of earlier airliners.

Show worked solution →

This is a 6-mark ASSESS: markers reward a supported judgement, not just a list of features of each architecture.

Plan. Thesis: MEA trades increased electrical mass/complexity for net gains in fuel efficiency and maintainability, a worthwhile trade-off for a long-range widebody. Traditional bleed-air pressurisation wastes already-compressed engine air and needs heavy high-temperature ducting. MEA's electric cabin-air compressors and electric brakes remove that ducting and one hydraulic circuit (saving about 270 kg), lifting engine efficiency 1 to 2 percent. The cost is four times the generation capacity (1450 kVA versus roughly 240 kVA), adding generator/wiring mass and making the electrical system a more critical single point of failure, mitigated by the RAT and independent buses.

Model paragraph (excerpt). Removing bleed air from the 787's pressurisation and anti-ice systems avoids diverting already-compressed core air, lifting engine efficiency by roughly 1 to 2 percent and eliminating heavy bleed ducting. Replacing hydraulic brakes with electric brakes removes an entire hydraulic circuit, saving about 270 kg per aircraft. These savings require four times the generation capacity of an equivalent older twin (1450 kVA versus roughly 240 kVA), making the electrical system itself more critical, which the 787 mitigates with independent essential buses, a battery and a RAT. For an aircraft flying long sectors for two to three decades, compounding fuel savings outweigh the added electrical mass, though the case is less clear-cut for a short-haul narrowbody with fewer flight hours to amortise the cost.

Marker's note: top-band answers (1) weigh a cost against a benefit rather than listing features, (2) use a quantitative figure (mass, efficiency, kVA), (3) acknowledge MEA's genuine engineering cost, and (4) end with an explicit judgement tied to aircraft type.

exam5 marksA maintenance engineer measures the voltage at the start and end of an essential-bus feeder cable on a light twin-engine aircraft under full load: 28.0 V at the generator end, 27.3 V at the avionics end. The cable is rated to carry up to 60 A. (a) Calculate the percentage voltage drop. (b) Evaluate whether this cable meets the typical 2 percent aerospace design limit, and (c) recommend one engineering change if it does not.

Show worked solution →

(a) Percentage voltage drop.

V_{drop} = 28.0 - 27.3 = 0.7 \text{ V}

\% \text{ drop} = \frac{0.7}{28.0} \times 100 = 2.5\%

(b) Evaluation against the limit. The typical aerospace design limit is 2 percent. A measured drop of 2.5 percent EXCEEDS this limit, meaning avionics at the far end of the cable could be receiving marginal voltage under full load, risking incorrect operation of sensitive instruments.

(c) Recommended engineering change. Since $V = IR$ and $R = \rho l / A$ , the drop can be reduced by increasing the cable's cross-sectional area (a heavier-gauge conductor lowers $R$ for the same length and current), or by shortening/re-routing the cable run if practical. Increasing cross-sectional area is the standard fix, at the cost of added cable mass, and is preferred over reducing the load because the avionics load is fixed by the aircraft's equipment.

Marking criteria: 1 mark for correct voltage drop (0.7 V), 1 mark for correct percentage (2.5%), 1 mark for correctly comparing against the 2 percent limit and concluding it fails, 1 mark for a valid engineering recommendation (larger cross-section or shorter run), 1 mark for justifying the recommendation using $R = \rho l/A$ or noting the mass trade-off.