Conductor sizing and ampacity

Why it matters

A conductor that is too small for its load overheats. Insulation ages, cracks and eventually fails, and that is how many electrical fires start. Sizing the conductor correctly is one of the most frequent calculations an electrician does, and one of the most tested on the journeyman exam. The steps are always the same: find the load, apply the continuous-load rule, look up the ampacity, then correct and adjust it for the real conditions.

Awareness level only. This lesson is paper and calculator practice. Selecting and installing conductors on a real job is done under a licensed electrician's supervision, with the code edition your jurisdiction has adopted.

Step 1: the load

Start from the load the circuit will carry. If it is continuous (maximum current for three hours or more), the general rule is to size the conductor at 125% of the continuous load plus 100% of any non-continuous load. A 32 A continuous EV charger load needs a conductor good for at least 32 × 1.25 = 40 A before any corrections.

Step 2: the base ampacity

Ampacity tables in Article 310 give a base ampacity for each conductor size (AWG or kcmil), material (copper or aluminium) and insulation temperature rating (60 °C, 75 °C, 90 °C), under standard conditions: a set ambient temperature and no more than three current-carrying conductors together.

Two limits apply before you even correct anything:

• Terminal ratings (110.14(C)). The final ampacity cannot exceed what the equipment terminals are rated for. Many terminals are rated 75 °C, and smaller equipment may be rated only 60 °C. You may use the 90 °C column as the starting point for derating, but the result must not exceed the ampacity at the terminal rating.
• Small-conductor rule (240.4(D)). For the smallest common sizes (14, 12 and 10 AWG copper), the overcurrent device is capped at specific values regardless of the table. Look these up in your code book.

Step 3: correct for temperature

Hotter surroundings mean less room for the conductor's own heat. Article 310 gives temperature correction factors: in ambient temperatures above standard, multiply the base ampacity by a factor less than 1. Conductors on rooftops or in hot attics are the classic case.

Step 4: adjust for bundling

When more than three current-carrying conductors share a raceway or cable for a significant length, they cannot shed heat as well. Article 310 gives adjustment factors that reduce the ampacity as the count rises. Neutrals that carry only unbalanced current and equipment grounding conductors usually do not count, but there are exceptions, so read the section.

The formula

All of it combines into one line:

adjusted ampacity = base ampacity × temperature correction factor × adjustment factor

The conductor is acceptable when the adjusted ampacity is at least the required ampacity from Step 1 (and the terminal and overcurrent limits are met).

Worked example (illustrative values)

The numbers below are practice values made up for this course, chosen to look realistic. They are not the values in the NEC tables, which are copyrighted and change between editions. On the job and in the exam, use the tables in your code book.

A 30 A continuous load runs in a raceway with six current-carrying conductors through a space that runs hot.

• Required ampacity: 30 × 1.25 = 37.5 A.
• Illustrative temperature factor: 0.91. Illustrative adjustment factor for 4–6 conductors: 0.80. Combined: 0.91 × 0.80 = 0.728.
• Try a conductor with an illustrative base ampacity of 38 A: 38 × 0.728 = 27.7 A. Too small.
• Try the next size up, illustrative base 52 A: 52 × 0.728 = 37.9 A. That is at least 37.5 A, so it passes this check.

Then confirm the terminal limit (the conductor's ampacity at the terminal temperature rating must also carry 37.5 A) and choose an overcurrent device that protects the conductor.

Common mistakes

• Forgetting the 125% for continuous loads, or applying it twice.
• Applying correction and adjustment to a column the terminals cannot use, then not checking the terminal limit.
• Counting equipment grounding conductors as current-carrying.
• Using a factor table from another edition.
• Treating the derated value as the breaker size without checking the overcurrent rules in Article 240.

Quick recap

Required ampacity comes from the load (125% if continuous). Adjusted ampacity = base × temperature factor × bundling factor. Pick the smallest conductor whose adjusted ampacity meets the requirement, then check terminal ratings and overcurrent limits. Always take the real values from your edition's tables (NFPA 70; checked on 2026-10-23).

Try it

Open the Do tab. The conductor picker is written for you except for one line: the adjusted-ampacity formula.