For a 4 in. × 11/2 in. square box, how many 6 AWG TW conductors are allowed according to Article 314?

To determine how many 6 AWG TW conductors can fit in a 4-inch by 1.5-inch square box according to Article 314 of the National Electrical Code (NEC), one must consider the box fill calculations.

First, the cross-sectional area of a 6 AWG conductor is 0.3951 square inches. Article 314 specifies that each conductor requires a certain amount of cubic volume based on its size. For conductors, an important value to note is that the minimum box volume for conductors is determined by their area, and a certain amount of space is also allotted for each conductor to avoid overcrowding.

A typical 4-inch by 1.5-inch box provides a volume of 12 cubic inches. To find out how many 6 AWG conductors can be accommodated without exceeding the box fill limit, we calculate as follows:

• Each 6 AWG TW conductor occupies approximately 2.5 cubic inches in terms of box fill.

• Therefore, in a box with 12 cubic inches of volume, dividing the total volume by the volume taken by each conductor yields:

( \frac{12 \text{ cubic inches}}{2.5 \text{ cubic inches per conductor}}