7.2 Power Dissipation & Calculations

7.2 Power Dissipation & Calculations

Quick Answer: In a pure resistor, electrical power becomes heat at rate P = VI = I²R = V²/R. Energy (heat) over time is W = P t. Conductors and poor joints heat by the same I²R rule — high current makes heat rise fast even when R looks “small.”

Topic 3.8 expects more than naming the watt. You must calculate dissipation, choose resistor ratings, and explain why cables and connections get hot under load. Everything below assumes steady DC (or instantaneous values that Module 3 treats like DC).

The Three Power Forms for Resistance

Start from P = V I and Ohm’s law V = I R (or I = V / R):

Exam habit: Write the knowns, pick one formula, substitute carefully. Do not invent a fourth hybrid like “V²I” — that is wrong.

Worked example 1 — all three forms agree

A resistor has R = 8.0 Ω, I = 3.0 A. Then V = IR = 24 V.

• P = VI = 24 × 3.0 = 72 W
• P = I²R = 9.0 × 8.0 = 72 W
• P = V²/R = 576 / 8.0 = 72 W

Same physics, three algebraic doors.

Worked example 2 — P = I²R only

Current through a 12 Ω resistor is 2.0 A. Voltage is not given.

P = I²R = 4.0 × 12 = 48 W.

Worked example 3 — P = V²/R only

A 220 Ω bleed resistor sees 110 V across it.

P = V²/R = 12 100 / 220 = 55 W.

Worked example 4 — rearrange for R

A resistor dissipates 20 W at 40 V.

From P = V²/R → R = V²/P = 1 600 / 20 = 80 Ω.

Worked example 5 — rearrange for I

A 5.0 Ω resistor dissipates 45 W.

From P = I²R → I² = P/R = 45/5.0 = 9.0 → I = 3.0 A (magnitude).

Power Dissipation in Resistors

A resistor converts electrical energy to thermal energy (heat) at rate P. The part must reject that heat to ambient without exceeding its temperature rating.

Wattage rating is the manufacturer’s continuous dissipation limit (often at a stated ambient). Colour bands give ohms and tolerance — not watts. Physical size is only a rough cue.

Selection rule for Module 3: calculated continuous P must be less than the rating, preferably with margin (training rule of thumb: do not run a resistor at 100% of its rating continuously if a larger part is available).

Worked example 6 — rating check. A 100 Ω resistor drops 20 V continuously.

P = V²/R = 400 / 100 = 4.0 W.

A 0.25 W or 0.5 W film resistor is unsuitable. Choose at least a 5 W (or higher) wirewound/ceramic class part, then apply any derating your procedure requires.

Worked example 7 — series string dissipation. Two series resistors on 24 V: R1 = 30 Ω, R2 = 50 Ω. Total R = 80 Ω. Current I = 24/80 = 0.30 A.

• P1 = I²R1 = 0.09 × 30 = 2.7 W
• P2 = I²R2 = 0.09 × 50 = 4.5 W
• Total = 7.2 W (also P = VI = 24 × 0.30)

Each resistor needs its own adequate wattage rating.

Worked example 8 — parallel branch. Two 40 Ω resistors in parallel across 20 V. Each sees 20 V.

P_each = V²/R = 400 / 40 = 10 W per resistor. Combined load power = 20 W. Parallel does not split voltage; it splits current — each still needs a ≥10 W capable part if continuous.

Heating of Conductors (I²R Losses)

Aircraft wiring and bonding paths have small but non-zero resistance. Heat generated in a conductor segment is still:

P_loss = I² R_conductor

Even if R_conductor is only 0.05 Ω, a large current produces serious heat:

Worked example 9 — feeder heating. Feeder R = 0.050 Ω, I = 40 A.

P = I²R = 1 600 × 0.050 = 80 W of heat along that feeder resistance — comparable to a small heater, concentrated in cable and terminations.

Voltage drop on the same feeder: V_drop = IR = 40 × 0.050 = 2.0 V. On a 28 V bus the load may see ~26 V, while the missing 2 V × 40 A = 80 W appears as heat — consistent with P = V_drop × I.

Worked example 10 — joint heating. A corroded connector adds 0.20 Ω at 25 A.

P = I²R = 625 × 0.20 = 125 W at the joint — a classic cause of melted insulation and intermittent faults. Cleaning/retorquing reduces R and therefore I²R heat.

Design intuition Module 3 wants:

• Heat scales with I² — doubling current quadruples heating for the same R.
• Undersized wire (higher R) increases both drop and heat.
• Long runs increase R (from R = ρL/A in topic 3.7) and therefore I²R loss.

Work and Energy Calculations from Power

With roughly constant power:

Energy W = P t (joules if P in watts and t in seconds)

Worked example 11 — resistor heat energy. A 10 Ω resistor carries 2.0 A for 5.0 minutes.

P = I²R = 4 × 10 = 40 W.

t = 5 × 60 = 300 s.

Energy = 40 × 300 = 12 000 J = 12 kJ converted to heat.

Worked example 12 — kWh style (optional conversion). A 500 W load runs for 2.0 hours.

Energy = 0.500 kW × 2.0 h = 1.0 kWh = 1.0 × 3.6 × 10⁶ = 3.6 MJ.

Use kWh only when the stem invites it; otherwise stay in J and s.

Worked example 13 — find time. How long for a 25 W heater to deliver 50 kJ?

t = W/P = 50 000 / 25 = 2 000 s ≈ 33.3 minutes.

Worked example 14 — battery-side energy estimate. A 24 V load draws 10 A for 15 minutes at constant rates.

P = 24 × 10 = 240 W.

Energy = 240 × (15 × 60) = 240 × 900 = 216 000 J = 216 kJ drawn from the supply (idealised, ignoring internal battery losses).

Choosing Formulas Under Exam Pressure

Worked example 15 — √ form. Need voltage across a 16 Ω resistor dissipating 64 W.

V = √(P R) = √(64 × 16) = √1024 = 32 V.

Practical Aircraft Scenarios

Scenario A — Dim lights + warm cable. High feeder I²R drop and heat: check gauge, length, terminations, load current.

Scenario B — Resistor smoke on a PCB. Computed P exceeds marking; replace with correct wattage and correct ohms.

Scenario C — Parallel lamps. Each lamp’s dissipation uses its own V and R (or rated V and I); total supply power is the sum.

Scenario D — Short-circuit fault. Current surges; I²R in fault path skyrockets — protection must open before conductors reach damaging temperatures.

Common Calculation Traps

• Using P = V/R or P = IR (those are not power).
• Forgetting to square I or V in I²R and V²/R.
• Applying V²/R with the supply voltage when the resistor only sees a fraction of supply (series networks).
• Comparing wattage ratings without computing actual P.
• Leaving time in minutes inside W = P t.

Drill the three resistive power formulas, energy W = Pt, and conductor I²R heating until substitutions are automatic. That closes CAAS Module 3 topic 3.8 Power and supports later motor, generator, and AC power ideas.