Box Fill Calculations Made Simple (NEC 314.16)

Box fill questions look intimidating, but they're pure addition once you know the rules. NEC 314.16(B) is one of the most reliable sources of points on the journeyman exam — here's the whole method.

Step 1: Know the per-conductor volumes

From Table 314.16(B), each conductor is assigned a free volume based on its size:

• 14 AWG → 2.00 in³
• 12 AWG → 2.25 in³
• 10 AWG → 2.50 in³
• 8 AWG → 3.00 in³
• 6 AWG → 5.00 in³

Step 2: Count everything correctly

This is where people slip. The rules:

• Each conductor that enters and doesn't leave counts as one.
• Each device (yoke/strap) — a switch or receptacle — counts as two conductors of the largest size connected to it (314.16(B)(4)).
• One or more internal cable clamps count as one conductor of the largest size in the box (314.16(B)(2)).
• All equipment grounding conductors together count as one (314.16(B)(5)).

Step 3: Add it up

Say a box has six 12 AWG conductors, one receptacle, and internal clamps:

• 6 conductors × 2.25 = 13.50 in³
• 1 device = 2 × 2.25 = 4.50 in³
• clamps = 1 × 2.25 = 2.25 in³
• Total = 20.25 in³

Then compare that to your box's cubic-inch capacity. If the box is too small, size up. Done.

Make it muscle memory

Every box fill question follows the same rhythm: assign volumes → count conductors, devices, and clamps → add → compare to the box. Drill a dozen and it becomes automatic. Our Calc-Mastery engine generates unlimited box-fill problems with the worked solution shown — so on exam day you're not thinking, you're just adding.