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TDLR box fill practice

Box fill adds up the cubic inches every conductor, device, clamp, and ground takes up, then checks it against the box. 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 4 by 4 by 1.5 inch square box has four 12 AWG conductors, one device, the grounds, and an internal clamp. Does it have room?

  1. 1
    Name the problem

    This is a box fill. Every conductor, device, clamp, and ground takes up cubic inches. We add them all up and compare to what the box holds.

  2. 2
    Count the conductors

    Each insulated conductor that enters the box counts once, sized by its own AWG (the volume table in 314.16(B)). One that passes straight through counts once too; one that never leaves the box, like a pigtail, doesn't count.

    4 × 12 AWG = 9.00 cu in

    Subtotal: 9.00 cu in

  3. 3
    Add devices, grounds, and clamps

    Up to four grounds count as one, sized by the largest; each ground beyond four adds a quarter. Each device yoke counts as two, sized by the largest conductor connected to the device. All the internal clamps together count as one, sized by the largest conductor in the box.

    Yokes: 1 × 2 × 2.25 = 4.50 cu in

    Grounds: 1 × 2.25 (12 AWG) = 2.25 cu in

    Clamp: 1 × 2.25 = 2.25 cu in

  4. 4
    Add it up

    That total is the volume the box has to give you.

    Required = 9.00 + 4.50 + 2.25 + 2.25

    Required = 18.00 cu in

  5. 5
    Compare to the box

    The 4 in. square × 1-1/2 in. holds 21.00 cu in. You need 18.00. The box has room. You are clear.

    Box 21.00 cu in vs needed 18.00 cu in

Try an original question

Sample question · original

A device box holds six 12 AWG conductors, one duplex receptacle, and the equipment grounds. Each 12 AWG counts as 2.25 cubic inches.

What is the minimum box volume required?

  • A15.75 cubic inches
  • B18.00 cubic inches
  • 20.25 cubic inches
  • D22.50 cubic inches

Answer C. Count six conductors, plus 2 for the duplex receptacle yoke, plus 1 for the grounds together (four or fewer), which is 9. 9 x 2.25 = 20.25 cubic inches (the volume table in 314.16(B)).

  • NEC 314.16(B) volume table
  • NEC 314.16(B)(1) through (B)(5)

Why the other answers tempt you

  • A: 15.75 forgets the device. A duplex receptacle on a yoke counts as two conductors.
  • B: 18.00 counts the device as one instead of two, or skips the grounds.
  • D: 22.50 counts the grounds as two. Up to four equipment grounds together count as one conductor.

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