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Current carrying capacity of cable

Short answer

Current carrying capacity is the max continuous current a cable can carry within its temperature limit. Factors: conductor material/size, insulation, ambient temp, grouping, installation method, laying depth, soil resistivity, harmonics. To size a 3.5 core cable, calculate full-load current, derate using product of correction factors, select tabulated rating ≥ derated current, then check voltage drop and short-circuit withstand.

Define current carrying capacity of cable. Write down the factors that affect the current carrying capacity of cable. Describe the method of calculation while determining the size of 3.5 core cable.

Current carrying capacity

The current carrying capacity (ampacity) of a cable is the maximum continuous current it can carry under given installation conditions without the conductor exceeding its permissible temperature (70 °C for PVC, 90 °C for XLPE insulation).

Factors affecting current carrying capacity

  1. Conductor material and size: copper carries about 1.3 times the current of aluminium of the same size; larger area gives lower resistance.
  2. Type of insulation: XLPE (90 °C) allows more current than PVC (70 °C).
  3. Ambient temperature (air or ground): a higher ambient leaves less temperature margin, so the rating is reduced.
  4. Grouping: cables laid close together heat each other; derating factors apply for the number of cables and their spacing.
  5. Method of installation: in air, on trays, in conduit, in ducts or buried directly; each method dissipates heat differently.
  6. Depth of laying: deeper cables dissipate heat less well.
  7. Thermal resistivity of soil: dry or sandy soil conducts heat poorly, lowering the rating.
  8. Number of loaded cores and harmonics: harmonic currents (e.g. from drives) load the neutral and increase heating.
  9. Load cycle: cables with cyclic loads can be rated higher than continuous ones.

Method of sizing a 3.5 core cable

  1. Find the full-load current:
I = P / (√3 × V × cos φ × η)
  1. Find the total derating factor (product of temperature, grouping, depth and soil factors from the standard/maker's tables).
  2. Required tabulated rating = I / derating factor. Select the smallest size whose rating is equal to or greater than this.
  3. Check voltage drop:
VD = √3 × I × L × (R cos φ + X sin φ)

It should normally be within about 5% (often 3% for motor feeders). 5. Check short-circuit withstand: A ≥ I_sc × √t / k (k = 115 for Cu PVC, 76 for Al PVC, 94 for Al XLPE). 6. Choose the size that meets all three conditions; the neutral is half size (3.5 core).

Example

A 75 kW, 400 V motor, cos φ = 0.85, η = 0.92, fed by a 100 m Al PVC armoured cable buried in ground. Derating: temperature 0.89, grouping 0.8. Fault level 10 kA for 0.2 s.

I = 75000 / (1.732×400×0.85×0.92) = 138.4 A
Total derating = 0.89 × 0.8      = 0.712
Required rating = 138.4 / 0.712  = 194.4 A

From the manufacturer's table, a 3.5 × 120 mm² Al armoured cable (rating about 215 A in ground, R ≈ 0.32 Ω/km, X ≈ 0.07 Ω/km) is selected. (Exact ratings vary by maker.)

VD = 1.732 × 138.4 × 0.1 × (0.32×0.85 + 0.07×0.527)
   = 7.41 V  = 1.85 % of 400 V   (OK, < 5%)
Short circuit: A ≥ 10000 × √0.2 / 76 = 58.8 mm²  (OK)

Answer: 3.5 × 120 mm² Al armoured cable satisfies current, voltage drop and short-circuit conditions.

✓ Answer checked by Claude Opus 5.5 (high effort)

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