What Is the Neutral Conductor?

The neutral conductor carries the return current in a single-phase or multi-wire circuit and helps balance current in 3-phase systems. It may or may not carry full load current — and that affects how it’s sized.

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1. Is the Neutral a Current-Carrying Conductor?

Ask yourself:

  • Is the neutral expected to carry current under normal operation?
    • Yes? --- It must be full-sized (same as the phase conductors)
    • No (e.g., balanced 3-phase loads)? --- It may be reduced

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2. Use the Same Ampacity Table as for the Hot Conductors

If the neutral is current-carrying (which it often is in residential or single-phase loads), size it based on the calculated load using the appropriate ampacity table (e.g., CEC Table 2 for copper in raceways).

Example:

  • If your calculated neutral load is 100A, and you're using 75°C rated conductors, you might select #3 AWG copper (from Table 2).

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3. Consider Unbalanced Loads and Harmonics

  • Residential: The neutral often carries almost as much current as the phase — so same size as phase conductors.
  • 3-phase with nonlinear loads (like computers or VFDs): The neutral may carry more than expected due to harmonics — often it must be full size or oversized.

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4. Reduced Neutral? Only When Allowed

Rule 4-018: How to Size the Neutral Conductor‍

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Example 1: 240/120V with Two 500A Loads

Determine the maximum unbalanced load

  • If loads are different, the maximum unbalanced load is the higher of the two loads
  • In this case, it is 500A

Rule 4-018 Demand Factors

  • ⚠️ You do not reduce the neutral or apply demand factors if:
    • There is discharge lighting, and
    • Anything with harmonics (fed by a 3-phase 4-wire system)
  • ⚠️ You cannot touch 200A of the load

Calculation:

  • 500A – 200A = 300A
  • Apply 70% de-rating to anything over the 200A
    • 300A * 0.7 = 210A
  • Add back what we weren’t allowed to touch earlier
    • 210A + 200A = 410A
  • Size the neutral based on 410A
    • (Rule 4-006 applies)
    • Table 2 (75°C) → 600kcmil

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Example 2: 240/120V with Two 500A Loads and 100A of Discharge Lighting

Calculation:

  • Start with 500A
  • ⚠️ Can’t touch 200A and can’t touch discharge lighting at 100A
  • 500A – 200A – 100A = 200A
  • Apply 70% de-rating
    • 200A * 70% = 140A
  • Add 200A + 100A (untouchable loads) + 140A = 440A
  • Table 2 (75°C) → 700kcmil