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Stumped by the Code? September 2011

Sep 20, 2011 5:30 PM, Mike Holt, NEC Consultant

Your most pressing National Electrical Code (NEC) questions answered

All questions and answers are based on the 2011 NEC.

Q. I’m having trouble understanding pull box calculation requirements. Can you give me an example of how to properly perform these steps?

A. For raceways containing conductors 4 AWG and larger, the minimum dimensions of boxes and conduit bodies must comply with the following [314.28(A)]:

1. Straight pull calculations. The minimum distance from where the conductors enter the box or conduit body to the opposite wall must not be less than eight times the trade size of the largest raceway.
2. Angle pull, U pull, or splice calculations:

Angle pulls. The distance from the raceway entry of the box or conduit body to the opposite wall must not be less than six times the trade size of the largest raceway, plus the sum of the trade sizes of the remaining raceways on the same wall and row.

U pulls. When a conductor enters and leaves from the same wall of the box, the distance from where the raceways enter to the opposite wall must not be less than six times the trade size of the largest raceway, plus the sum of the trade sizes of the remaining raceways on the same wall and row.

Splices. When conductors are spliced, the distance from where the raceways enter to the opposite wall must not be less than six times the trade size of the largest raceway, plus the sum of the trade sizes of the remaining raceways on the same wall and row.

Rows. If there are multiple rows of raceway entries, each row is calculated individually and the row with the largest distance must be used, as shown in the Figure.

Distance between raceways. The distance between raceways enclosing the same conductor must not be less than six times the trade size of the largest raceway, measured from the raceways’ nearest edge to nearest edge.

Q. How many general use receptacles are allowed per circuit for dwelling units?

A. For dwelling units, 210.11(A) requires that the minimum number of general lighting and general-use receptacle branch circuits must be determined by dividing the total calculated load in amperes by the ampere rating of the circuits used.

Here’s an example problem designed to help you see this requirement more clearly.

How many 15A, 120V circuits are required for the general lighting and general-use receptacles for a dwelling having floor area of 1,500 sq ft, exclusive of an unfinished cellar not adaptable for future use [Example D1(a) in Annex D]?

Step 1: First, determine the total load in volt-amperes:

VA = 1,500 sq ft × 3VA per sq ft [Table 220.12] = 4,500VA

Step 2: Next, determine the current in amperes:

I = VA ÷ E = 4,500VA ÷ 120V = 38A

Step 3: Now we can determine the number of circuits as follows:

Number of circuits = 38A ÷ 15A = three 15A, or two 20A, 120V circuits

Note: There’s no limit to the number of receptacles on a circuit in a dwelling unit.

If the load is calculated on the volt-amperes per square foot method, the wiring system must be provided to serve the calculated load, with the loads evenly proportioned among multi-outlet branch circuits within the panelboard [210.11(A)].               

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