1.2 Ohm's Law and Electrical Power
Key Takeaways
- Ohm's Law expresses the direct relationship between voltage (E), current (I), and resistance (R): I = E/R, E = I * R, and R = E/I.
- Electrical power (P), measured in Watts, represents the rate of electrical energy consumption or conversion: P = E * I = I^2 * R = E^2 / R.
- Electrical energy (W), measured in Joules or Watt-hours (Wh), is the total work performed over time: W = P * t.
- Avionics system power calculations must account for nominal 28 VDC aircraft bus voltage variations (typically 24.0 VDC battery to 28.5 VDC generator output) to ensure correct wire sizing and heat dissipation management.
1.2 Ohm's Law and Electrical Power
Quick Takeaway: Ohm's Law establishes the fundamental proportionalities of DC circuits: $I = E/R$, $E = I \cdot R$, and $R = E/I$. Electrical power ($P$), measured in Watts, defines the rate of energy dissipation: $P = E \cdot I = I^2 R = E^2 / R$. In aircraft 28 VDC systems, calculating total bus current and power dissipation ensures correct circuit breaker sizing, wire gauge selection, and thermal management for flight-critical electronics.
Ohm's Law Fundamentals and Mathematical Relationships
Formulated by German physicist Georg Simon Ohm in 1827, Ohm's Law is the foundational relationship governing direct current electrical circuits. Ohm discovered that the current ($I$) flowing through a conductor between two points is directly proportional to the potential difference or voltage ($E$) across the two points and inversely proportional to the resistance ($R$) between them.
Mathematically, Ohm's Law is expressed in three equivalent forms:
-
Solving for Current ($I$): Interpretation: Increasing system voltage increases current flow proportionally, whereas increasing circuit resistance reduces current flow proportionally.
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Solving for Voltage / Potential Drop ($E$): Interpretation: The voltage drop across any resistive component equals the current passing through it multiplied by its resistance.
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Solving for Resistance ($R$): Interpretation: The effective resistance of a circuit or load can be determined by measuring the applied voltage and resulting current draw.
Ohm's Law Triangle:
/ E \
/-------\
/ I | R \
-----------
E = I * R | I = E / R | R = E / I
In aircraft applications, standard nominal DC bus voltages are 14.0 VDC (light general aviation aircraft powered by 12 V lead-acid batteries) and 28.0 VDC (turboprop, business jet, and transport category aircraft powered by 24 V lead-acid or NiCd batteries, with generator output set to 28.5 VDC).
Electrical Power and Energy Equations
While Ohm's Law defines static voltage, current, and resistance, Watt's Law defines the work performed by electricity per unit of time. Electrical Power ($P$) is measured in Watts (W), where one Watt represents an energy transfer rate of one Joule per second ($1\text{ W} = 1\text{ J/s}$).
Power Wheel Formulas
By combining Ohm's Law ($E = IR$) with the basic power equation ($P = EI$), we derive three primary formulas for electrical power:
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Standard Power Equation: Used when system voltage and load current are known.
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Joule Heating / Resistance-Current Equation: Used when load current and conductor resistance are known. This equation illustrates why high current leads to severe heating in wiring harnesses ($I^2 R$ losses).
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Voltage-Resistance Equation: Used when system voltage and component internal resistance are known.
Mechanical Power Conversion
In aircraft electrical motor applications (e.g., flap actuators, fuel boost pumps, landing gear hydraulic power packs), electrical power is converted into mechanical horsepower (HP):
Electrical Energy ($W$)
Electrical Energy ($W$) represents total work performed over a given time interval ($t$):
SI units measure energy in Joules (J) ($1\text{ J} = 1\text{ Watt} \cdot \text{second}$). In aviation battery capacity and power management, energy is commonly expressed in Watt-hours (Wh) or Kilowatt-hours (kWh):
Summary Matrix of DC Formulas
The following table summarizes all twelve mathematical combinations for solving $E$, $I$, $R$, and $P$ in DC circuits:
| Unknown Quantity | Formula 1 (Basic) | Formula 2 (Ohm Substitution) | Formula 3 (Power Substitution) | Standard Unit |
|---|---|---|---|---|
| Voltage ($E$) | $E = I \cdot R$ | $E = \frac{P}{I}$ | $E = \sqrt{P \cdot R}$ | Volts ($V$) |
| Current ($I$) | $I = \frac{E}{R}$ | $I = \frac{P}{E}$ | $I = \sqrt{\frac{P}{R}}$ | Amperes ($A$) |
| Resistance ($R$) | $R = \frac{E}{I}$ | $R = \frac{E^2}{P}$ | $R = \frac{P}{I^2}$ | Ohms ($\Omega$) |
| Power ($P$) | $P = E \cdot I$ | $P = I^2 \cdot R$ | $P = \frac{E^2}{R}$ | Watts ($W$) |
Worked Avionics Calculation: Step-by-Step 28 VDC Essential Bus Load Analysis
Problem Scenario
An avionics technician is auditing the electrical load on an aircraft's 28.0 VDC Essential Bus during emergency battery-only operation. The bus supplies three parallel loads:
- Primary Flight Display (PFD): Consumes $112\text{ Watts}$ of power.
- VHF Com/Nav Transceiver: Consumes $84\text{ Watts}$ during transmit.
- Avionics Cooling Fan: Possesses an internal operating resistance of $14.0\ \Omega$.
Calculate:
- (a) Current drawn by each load.
- (b) Total current drawn from the 28.0 VDC essential bus.
- (c) Total power consumed by the essential bus loads.
- (d) Total thermal energy dissipated over a 45-minute battery endurance flight (in Watt-hours and Joules).
+28.0 VDC Essential Bus -----------------------------------------+
| | |
[PFD] [VHF Com] [Cooling Fan]
112 W 84 W R = 14.0 Ohm
| | |
Airframe Ground Return -------+--------------+--------------+
Step-by-Step Solution
Step 1: Calculate individual load currents.
- For the PFD ($P_1 = 112\text{ W}$, $E = 28.0\text{ V}$):
- For the VHF Transceiver ($P_2 = 84\text{ W}$, $E = 28.0\text{ V}$):
- For the Avionics Cooling Fan ($R_3 = 14.0\ \Omega$, $E = 28.0\text{ V}$):
Step 2: Calculate total essential bus current ($I_{\text{total}}$).
- Since the loads are connected in parallel across the bus:
Step 3: Calculate total power consumed ($P_{\text{total}}$).
- Using $P = E \cdot I_{\text{total}}$:
- Alternatively, summing individual powers ($P_{\text{Fan}} = E \cdot I_{\text{Fan}} = 28 \times 2 = 56\text{ W}$):
Step 4: Calculate total energy consumed ($W$) during 45 minutes ($0.75\text{ hours}$).
- Energy in Watt-hours:
- Energy in Joules ($1\text{ Wh} = 3,600\text{ Joules}$):
Avionics Trap: The $I^2 R$ Quadratic Heating Trap and 28V vs 14V Aircraft Architecture
A critical design mistake made by entry-level technicians is failing to appreciate the quadratic nature of $I^2 R$ power loss in aircraft wire harnesses. If current through a wire doubles, the power dissipated as heat increases by a factor of $2^2 = 4$. If current triples, heating increases by $3^2 = 9$.
This mathematical reality explains why transport-category aircraft use 28 VDC (or higher AC voltages like 115 VAC) rather than 14 VDC. To deliver $2,800\text{ Watts}$ of power to a hydraulic pump motor:
- At 14 VDC, current draw is $I = 2800 / 14 = 200\text{ Amperes}$.
- At 28 VDC, current draw is $I = 2800 / 28 = 100\text{ Amperes}$.
Halving the current reduces heat dissipation in feeder cables by $75%$ ($1/4\text{ of original } I^2 R\text{ loss}$), allowing aircraft designers to use significantly lighter gauge wire (e.g., AWG 2 instead of AWG 0000), saving hundreds of pounds of airframe weight while dramatically improving thermal safety margins.
An aircraft landing light operating on a 28 VDC bus draws a current of 7.5 Amperes. What is the electrical resistance of the light filament?
An avionics cooling fan draws 2.5 Amperes from a 28 VDC supply. How much power is consumed by the fan, and how much energy does it consume during 3 hours of continuous operation?
If the current passing through an aircraft wire harness increases from 5 Amperes to 10 Amperes while maintaining constant resistance, by what factor does the heat power dissipated (I^2 R loss) in the wire increase?