6.1 Resistivity & Factors Affecting Resistance
Key Takeaways
Resistance of a uniform conductor is R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area
For a given material and temperature, doubling length doubles R; doubling area halves R
Resistivity ρ is a material property (ohm·metre); copper has low ρ, nichrome and carbon have much higher ρ
Most metallic conductors have a positive temperature coefficient of resistance — R rises as temperature rises
Aircraft wiring, bonding straps, and high-resistance joints are practical applications of the same R = ρL/A and temperature rules tested on CAAS Module 3
6.1 Resistivity & Factors Affecting Resistance
Quick Answer: For a uniform conductor, R = ρL/A. Resistance rises with length and resistivity, falls as cross-sectional area increases, and usually rises with temperature in metals. Resistivity ρ is the material property (ohm·metre); Module 3 expects you to rearrange the formula, compare materials, and apply temperature-coefficient ideas to conductors and resistors.
CAAS SAR-66 Module 3 topic 3.7 Resistance / resistor starts with why a wire or resistor has the ohm value it has. Topic 3.3 already defined resistance as opposition to current; here you quantify that opposition from geometry and material, then link temperature behaviour to maintenance symptoms such as hot joints and rising feeder drop on a 28 V DC bus.
The Resistance Formula: R = ρL/A
For a conductor of uniform cross-section:
R = ρL / A
| Symbol | Meaning | Typical SI unit |
|---|---|---|
| R | Resistance | ohm (Ω) |
| ρ (rho) | Resistivity of the material | ohm·metre (Ω·m) |
| L | Length of the conductor | metre (m) |
| A | Cross-sectional area | square metre (m²) |
Read the formula as three independent levers:
- Material sets ρ.
- Length sets how much material the current must travel through.
- Area sets how wide the conduction path is.
Direct and inverse relationships (exam gold)
- R ∝ L — double the length → double the resistance (same material, same A, same temperature).
- R ∝ 1/A — double the cross-sectional area → half the resistance.
- R ∝ ρ — higher resistivity material → higher resistance for identical size.
Worked example 1 — rearrange for resistivity. A copper bar is 2.0 m long, has A = 1.0 × 10⁻⁶ m², and measures R = 0.034 Ω at the stated temperature. Find ρ.
ρ = RA / L = (0.034 × 1.0 × 10⁻⁶) / 2.0 = 1.7 × 10⁻⁸ Ω·m (order of magnitude of copper).
Worked example 2 — length change. A bonding strap has R = 0.020 Ω. An alternative strap of the same material and cross-section is 1.5 times as long. New resistance = 0.020 × 1.5 = 0.030 Ω.
Worked example 3 — area change. A feeder cable has R = 0.080 Ω. Replacing it with a cable of twice the cross-sectional area (same length and material) gives R = 0.080 / 2 = 0.040 Ω. Lower R means less I²R heating and less voltage drop for the same load current.
Worked example 4 — combined change. Original wire: R₁ = ρL / A. New wire: twice as long and half the area → R₂ = ρ(2L) / (A/2) = 4 × (ρL/A) = 4R₁. Combined geometry mistakes are common on Module 3 — multiply the factors carefully.
Resistivity: The Material Fingerprint
Resistivity (ρ) is resistance of a sample of unit length and unit cross-sectional area. It lets you compare materials without arguing about wire size.
| Material (typical order) | Relative resistivity | Aviation / training use |
|---|---|---|
| Silver | Very low | Best conductor; costly — rarely used as bulk wiring |
| Copper | Very low | Primary aircraft power and signal wiring |
| Aluminium | Low (higher than Cu) | Weight-sensitive power runs; watch joints and oxide |
| Tungsten | Higher | Lamp filaments (high melting point) |
| Nichrome (Ni-Cr alloy) | High | Heating elements, some high-stability resistors |
| Carbon / graphite | High (and NTC behaviour often) | Resistor compositions, brushes |
| Insulators (glass, plastics) | Extremely high | Insulation — not used as intentional conductors |
You do not need to memorise every textbook ρ table to the last digit for CAAS Module 3, but you must know the ranking logic: copper and aluminium are low-ρ conductors; nichrome and carbon are much higher; good insulators have enormous ρ. If a stem says “same dimensions, which has highest resistance?” pick the highest-ρ material listed.
Conductivity is the reciprocal idea: materials with low resistivity are high-conductivity conductors. Mentally: low ρ ↔ easy current ↔ low R for a given L and A.
Cross-Section and Length in Aircraft Context
Aircraft electrical practice is full of R = ρL/A decisions even when nobody writes the Greek letter:
- Undersized cables (small A) raise R → voltage drop and heat under load.
- Long feeder runs (large L) raise R → remote equipment may see less than bus voltage.
- Bonding straps must be short and adequate in section so airframe equipotential bonding stays low-resistance.
- Corroded or loose joints add contact resistance that is not in the ideal ρL/A wire calculation — but the symptom (heat, drop) looks the same.
Voltage-drop check: Drop in a feeder ≈ I × R_feeder. At 40 A through 0.05 Ω, drop = 2.0 V. On a nominal 28 V DC bus, the load may see about 26 V. Raising A (thicker cable) or shortening L cuts R and the drop.
Temperature and Resistance
Most pure metallic conductors have a positive temperature coefficient of resistance (PTC behaviour in the broad sense): as temperature rises, lattice vibration increases, electron drift is impeded more, and R increases.
A linear approximation often taught for Module 3 is:
R_t = R₀ [1 + α(t − t₀)]
where:
- R_t = resistance at temperature t
- R₀ = resistance at reference temperature t₀
- α (alpha) = temperature coefficient of resistance (°C⁻¹)
Worked example 5 — temperature rise. A copper winding has R₀ = 10.0 Ω at 20 °C. If α ≈ 0.004 /°C and the winding reaches 70 °C:
R₇₀ = 10.0 [1 + 0.004(70 − 20)] = 10.0 [1 + 0.20] = 12.0 Ω.
A 20% rise in resistance for a 50 °C rise is the sort of magnitude Module 3 expects you to recognise — metals warm up, resistance climbs.
Positive vs negative temperature coefficient (preview of §6.4)
| Behaviour | Resistance vs temperature | Typical materials / devices |
|---|---|---|
| Positive TCR | R rises as T rises | Most metals, PTC thermistors |
| Negative TCR | R falls as T rises | Carbon (often), NTC thermistors |
Carbon composition resistors and NTC thermistors are the classic “resistance falls when hot” counter-examples. Do not assume every resistor behaves like copper wire.
Factors Summary Table
| Factor | If it increases… | Effect on R (others fixed) |
|---|---|---|
| Length L | Longer path | R increases |
| Area A | Wider path | R decreases |
| Resistivity ρ | “Worse” conductor material | R increases |
| Temperature (most metals) | Hotter conductor | R increases |
Exam Scenarios to Rehearse
Scenario A — Same wire, cut in half. Half the length → half the resistance of that piece (and half the voltage drop for the same current through that segment).
Scenario B — Parallel paths vs area. Two identical conductors in parallel behave like roughly twice the conducting area → about half the resistance of one alone (link forward to series/parallel in §6.3).
Scenario C — Hot connector. Measured millivolt drop across a joint rises under load as contact resistance heats — temperature and effective R climb together until cleaned or retorqued.
Scenario D — Material swap. Replacing copper with nichrome of identical L and A multiplies R by ρ_nichrome / ρ_copper (large factor) — useful for heaters, disastrous for power feeders.
Master R = ρL/A, the four factors, and the metallic temperature trend. Those tools unlock colour-code wattage limits, series/parallel networks, and thermistor bridges later in this chapter.
A uniform conductor obeys R = ρL/A. If length is doubled and cross-sectional area is halved, what happens to resistance?
Resistance is unchanged
Resistance is halved
Resistance is doubled
Resistance becomes four times the original value
Which change, taken alone, decreases the resistance of a metallic conductor of fixed material?
Increasing length while keeping cross-sectional area constant
Increasing cross-sectional area while keeping length constant
Replacing copper with a higher-resistivity alloy of the same size
Raising temperature for a conductor with a positive temperature coefficient
A copper coil measures 5.0 Ω at 20 °C. Using R_t = R₀[1 + α(t − t₀)] with α = 0.004 /°C, what is the resistance at 45 °C?
4.5 Ω
5.0 Ω
5.5 Ω
6.0 Ω
Two samples have identical length and cross-sectional area. Sample X has much higher resistivity than sample Y. Which statement is correct?
Sample X has higher resistance than sample Y
Sample X has lower resistance than sample Y
Both samples must have identical resistance because geometry matches
Resistance depends only on temperature, not on resistivity
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