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TDLR motor calculation practice

Motor wiring sizes off the table full-load current, not the nameplate, and the conductor carries 125% of it. 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

A 10 HP, 230-volt, three-phase motor. Table 430.250 gives a full-load current of 28 amps. Size the branch-circuit conductor.

  1. 1
    Name the problem

    Single motor conductor. Use the table FLC and the 125% rule.

  2. 2
    Pick the rule

    Branch-circuit conductors for a single motor carry 125% of the table full-load current (430.22).

    Conductor = 1.25 × FLC

  3. 3
    Pull the numbers

    Table FLC is 28 amps.

  4. 4
    Run the arithmetic

    Add the 25%.

    Conductor = 1.25 × 28 = 35 A

  5. 5
    Check it

    35 amps. Size the wire for 35, even though the nameplate might read lower.

Try an original question

Sample question · original

A 10 HP, 240-volt, single-phase motor. Table 430.248 gives a full-load current of 50 amps. Size the branch-circuit conductor.

What is the minimum conductor ampacity?

  • A50 amps
  • 62.5 amps
  • C40 amps
  • D70 amps

Answer B. 430.22 requires branch-circuit conductors for a single motor to be 125% of the table full-load current. 1.25 x 50 = 62.5 amps. Use the table FLC, never the nameplate.

  • NEC 2023 430.22
  • NEC 2023 Table 430.248

Why the other answers tempt you

  • A: 50 amps is the table value with no 125%. A single-motor conductor carries 125% of the full-load current.
  • C: 40 amps applies an 80% derate instead of the 125% increase. Motor conductors go up, not down.
  • D: 70 amps is the next standard breaker size, not the required conductor ampacity.

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