Every aircraft technician working on electrical systems must be comfortable with the concept of electric power. Whether you are sizing a circuit breaker, evaluating generator load, or diagnosing why a component keeps failing, the ability to calculate wattage and understand how power relates to voltage, current, and resistance is a foundational skill. The FAA Aviation Maintenance Technician Handbook — General (FAA-H-8083-30) dedicates substantial coverage to basic electricity precisely because errors in electrical work carry serious consequences: overheated wiring, tripped breakers, or, in the worst case, in-flight electrical fires.
This article walks through electric power and wattage calculations from the ground up, connecting the math to real aircraft scenarios so that you can both pass the AMT General knowledge test and apply the concepts confidently on the shop floor.
What Is Electric Power?
In the physical sciences, power is defined as the rate at which work is done or energy is transferred. In an electrical circuit, power represents how quickly electrical energy is converted into another form — heat, light, mechanical motion, or radio waves. The standard unit of electrical power is the watt (W), named after Scottish inventor James Watt. One watt equals one joule of energy transferred per second.
The relationship between power, voltage, and current is expressed in the fundamental power formula:
P = E × I
Where P is power in watts, E is electromotive force (voltage) in volts, and I is current in amperes. This deceptively simple equation is the starting point for nearly every electrical load calculation you will encounter as an AMT.
How Wattage Calculations Work
Because aircraft electrical systems obey Ohm's Law — which states that E = I × R (voltage equals current times resistance) — the basic power formula can be expanded into two additional useful forms by substituting Ohm's Law relationships:
- P = I² × R — useful when you know current and resistance but not voltage directly.
- P = E² ÷ R — useful when you know voltage and resistance but not current directly.
Together, these three expressions are often displayed in a power wheel or Ohm's Law circle that AMT students use to solve for any unknown quantity. If you know any two of the four variables (P, E, I, R), you can solve for the remaining two.
Worked Example 1: Landing Light Circuit
Suppose a landing light is rated at 100 watts and the aircraft uses a 28-volt DC bus (common in many turbine and larger piston aircraft). How much current does the landing light draw?
Rearranging P = E × I for current: I = P ÷ E
I = 100 W ÷ 28 V = approximately 3.57 amperes
This tells the technician that the circuit wiring and protective device (fuse or circuit breaker) must be rated to handle at least 3.57 amps continuously, plus an appropriate safety margin per the aircraft manufacturer's specifications and advisory circular guidance.
Worked Example 2: Resistor Dissipation
A current-limiting resistor in a dimmer circuit has a resistance of 10 ohms and carries a current of 2 amperes. How many watts must it be rated to dissipate safely?
Using P = I² × R:
P = (2)² × 10 = 4 × 10 = 40 watts
A resistor selected for this application must be rated for at least 40 watts. In practice, technicians often select components rated at twice the calculated dissipation to provide a thermal safety margin, consistent with FAA-accepted maintenance practices.
Power in Aircraft Electrical Systems — Why It Matters
Aircraft electrical systems typically operate at either 14 volts DC (light single-engine aircraft) or 28 volts DC (many twins, turboprops, and military-derived designs), with some modern aircraft using 115-volt AC or even 270-volt DC architectures. The voltage level directly affects how much current is required to deliver a given amount of power — and current is the primary driver of wire heating.
This is why higher-voltage systems are favored in larger aircraft: for the same 1,000-watt load, a 28-volt system must carry about 35.7 amperes, while a 115-volt AC system carries only about 8.7 amperes. Lower current allows the use of smaller-gauge, lighter wire — a critical advantage in aerospace design where weight is always a concern.
From a maintenance standpoint, understanding power helps you:
- Verify that wire gauge is appropriate for the load (reference AC 43.13-1B for ampacity charts).
- Confirm that circuit breaker or fuse ratings match the calculated maximum circuit current.
- Diagnose overheating components by checking whether actual power dissipation exceeds component ratings.
- Evaluate generator and alternator load budgets to ensure total electrical demand does not exceed supply capacity.
Total Aircraft Electrical Load
A practical application of wattage calculations that appears on the AMT General test is computing total electrical load. When multiple loads share a bus, their individual power demands add together. If the bus voltage is known, the total current draw can be found:
I_total = P_total ÷ E_bus
For example, consider a 28-volt bus with the following loads active simultaneously: navigation lights at 40 W, instrument panel lighting at 60 W, a comm radio at 50 W, and a GPS at 30 W. Total power equals 40 + 60 + 50 + 30 = 180 watts. Total current = 180 ÷ 28 = approximately 6.43 amperes. The bus feed wire and any master protection device must comfortably handle this combined draw.
In practice, an aircraft's electrical load analysis (ELA) is a formal document that accounts for every possible load combination, including worst-case simultaneous operation scenarios. The FAA requires load analyses for certificated aircraft, and technicians modifying electrical systems must ensure the analysis remains valid after any additions.
Power Factor in AC Circuits
The wattage calculations described above apply directly to pure DC circuits and to purely resistive AC loads. However, many aircraft AC loads — electric motors, transformer-rectifier units, and inductive components — introduce a phase difference between voltage and current. This is characterized by the power factor (PF), a dimensionless number between 0 and 1.
In AC circuits with reactive loads, the formula becomes:
P = E × I × PF
The apparent power (measured in volt-amperes, or VA) is E × I, while true power (measured in watts) is E × I × PF. A power factor of 1.0 means the load is purely resistive and all apparent power becomes true power. A lower power factor means the circuit must carry more current than the wattage alone would suggest, which must be accounted for in wire and breaker sizing.
Key Numbers and Rules
- P = E × I — the fundamental power formula; memorize this first.
- P = I² × R and P = E² ÷ R — derived forms used when different variables are known.
- 14 V DC — nominal bus voltage for light single-engine aircraft.
- 28 V DC — nominal bus voltage for many twin-engine and turbine aircraft.
- 115 V AC, 400 Hz — common AC bus standard in many transport-category aircraft.
- Higher voltage = lower current for the same wattage = lighter wire gauge possible.
- Wire ampacity and breaker ratings must accommodate continuous load plus appropriate margin per AC 43.13-1B and manufacturer data.
- In AC circuits with reactive components, use P = E × I × PF for true power.
- 1 horsepower (mechanical) ≈ 746 watts — a conversion sometimes tested in relation to electric motor output.
Memory Aid
A classic study tool for the Ohm's Law and power relationships is the PIE / EIR wheel (also called the power triangle). Write a circle divided into sections with P on top, E on the lower left, and I on the lower right for the power formula — and E on top, I on the lower left, and R on the lower right for the Ohm's Law formula. To find any variable, cover it with your finger; the remaining symbols show you the operation. For power: cover P and you see E × I. Cover E and you see P ÷ I. Cover I and you see P ÷ E. The same logic applies to the EIR circle. This visual shortcut helps you quickly select the right formula variant under test-time pressure.
Common Test Traps
- Mixing up watts and volt-amperes. The FAA test may present an AC circuit problem where you must distinguish between apparent power (VA) and true power (W). Remember that watts require the power factor to be applied; VA does not.
- Forgetting to square the current in P = I² × R. A very common arithmetic error: students multiply I × R first and then multiply by I again instead of squaring I before multiplying by R. Work the exponent first.
- Confusing 14-volt and 28-volt system current values. A 100-watt load draws about 7.14 A on a 14-volt system but only about 3.57 A on a 28-volt system. The FAA may present both voltages in answer choices; confirm which bus voltage the question specifies.
- Ignoring the 746 W/hp conversion. Questions about electric motor output may require converting horsepower to watts or vice versa. Know that 1 hp ≈ 746 W, and do not confuse electrical input power with mechanical output power (motors are not 100% efficient).
- Assuming all loads add to 100% simultaneously. While the basic calculations assume all listed loads operate together, real ELA work categorizes loads as continuous, intermittent, or emergency — a nuance the FAA may test in a conceptual question about load analysis methodology.
Mastering electric power and wattage calculations gives you a durable analytical framework that applies across virtually every aircraft electrical system you will inspect, repair, or modify throughout your career. These relationships are not abstract math — they are the language the electrons speak, and reading that language accurately keeps aircraft and crews safe.