In a single-gang box containing two ungrounded conductors, one neutral, and one equipment grounding conductor, how do you determine box fill?

• A. Count all conductors equally as 1, regardless of size, then multiply by a fixed factor.
• B. Count each ungrounded conductor as 1, the neutral as 1, all equipment grounding conductors together as 1, and any internal clamp or device yoke as applicable; multiply by the volume per conductor.
• C. Ignore internal clamps and device yokes when calculating box fill.
• D. Only count conductors that are connected to the device; ignore others.

Answer: (B)

Box-fill calculation follows NEC rules: you count conductors that originate outside the box and are terminated or spliced inside, then multiply that total by the volume per conductor, which depends on the largest conductor present.

In this scenario, each of the two ungrounded conductors counts as one, and the neutral also counts as one, giving three so far. All equipment grounding conductors in the box count collectively as a single conductor, adding one more. If there are no internal clamps or device yokes, the total count is four. Multiply that by the volume per conductor for the largest conductor in the box to get the minimum box volume required. If a device yoke or internal clamp were present, they would add additional counts accordingly.

The other options don’t fit because they either oversimplify the counting, ignore clamps or yokes, or restrict counting only to conductors attached to a device.