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?
Which change, taken alone, decreases the resistance of a metallic conductor of fixed material?
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?
Two samples have identical length and cross-sectional area. Sample X has much higher resistivity than sample Y. Which statement is correct?