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TDLR ampacity and derating practice

A conductor's ampacity starts at a table value, then gets knocked down for heat and for crowding it with other current-carrying wires. This is a TDLR Calculations-part question type, now scored on its own. Here is the pattern, one worked example, and an original question to try.

Last reviewed June 2026

One worked example

Eight current-carrying 12 AWG THHN share a conduit. THHN 12 AWG starts at 30 amps in the 90 C column. What's the adjusted ampacity?

  1. 1
    Name the problem

    More than three current-carrying conductors, so an adjustment factor applies.

  2. 2
    Pick the rule

    Start at the 90 C ampacity, then multiply by the count adjustment from Table 310.15(C)(1). 7 to 9 conductors is 70%.

    Adjusted = base × adjustment

  3. 3
    Pull the numbers

    Base 30 A. Eight conductors means 70%.

  4. 4
    Run the arithmetic

    Take 70% of the base.

    Adjusted = 30 × 0.70 = 21 A

  5. 5
    Check it

    21 amps. And the terminations cap you at the 75 C column, so check that before you size the breaker.

Try an original question

Sample question · original

Six current-carrying 10 AWG THHN share a conduit. 10 AWG THHN is 40 amps in the 90 C column. With six conductors the adjustment factor is 80%.

What is the adjusted ampacity?

  • A40 amps
  • B35 amps
  • 32 amps
  • D28 amps

Answer C. Start at the 90 C ampacity of 40 amps, then multiply by the 80% adjustment for 4 to 6 conductors. 40 x 0.80 = 32 amps (Table 310.15(C)(1)). Check the termination column before sizing the breaker.

  • NEC 2023 Table 310.16
  • NEC 2023 Table 310.15(C)(1)

Why the other answers tempt you

  • A: 40 amps applies no adjustment. More than three current-carrying conductors triggers the count adjustment.
  • B: 35 amps is the 75 C value with no adjustment. You start the math from the 90 C column, then adjust.
  • D: 28 amps uses the 7-to-9 conductor factor (70%) instead of the 4-to-6 factor (80%).

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