Code Quiz

Aug. 24, 2004
A 200A rated panelboard (load center) has wiring space on each side that measures 4 inches wide by 4 inches deep. The total length of the wiring space on each side is 36 inches. Three individual splices that each consists of two 4/0 AWG THHN/THWN ...

A 200A rated panelboard (load center) has wiring space on each side that measures 4 inches wide by 4 inches deep. The total length of the wiring space on each side is 36 inches. Three individual splices that each consists of two 4/0 AWG THHN/THWN insulated conductors are connected by split-bolt connectors are located on one side of the panelboard in the wiring space. Each split bolt connector is about 1.5625 square inches (2 cubic inches). Does the NEC permit these splices in the wiring space area of a panelboard? If they are, do they comply with the NEC requirements for splices in enclosures?

A) Yes, No
B) Yes, Yes
C) No, No
D) No, Yes

Answer: B

Explanation: In 312.8 the NEC prohibits the use of enclosures for switches or overcurrent devices as junction boxes, auxiliary gutters, or raceways for conductors that feed through or tap off to other switches or overcurrent devices, unless adequate space for this purpose is provided.

The conductors shall not fill the wiring space at any cross section to more than 40% of the cross-sectional area of the space, and the conductors, splices and taps shall not fill the wiring space at any cross section to more than 75% of the cross-sectional area of that space.

Calculation:
Step 1. Wiring space on both side is 36 inches long by 4 inches wide by 4 inches deep. Disregarding the 36 inch length for this calculation, it's 16 square inches at any cross-sectional area (4 in. x 4 in. = 16 sq. in.). Seventy-five percent of 16 is 12 square inches of useable space. Forty percent of 16 is 6.4 square inches.

Step 2. Calculate the square inch area of #4/0 AWG THHN/THWN insulated conductors, which, according to Chapter 9, Table 5, is 0.3237 square inches. The total area of the conductors for each splice would be 0.6474 square inches (2 x 0.3237 sq in. = 0.6474 sq in.). Add to that the area of a split bolt connector, which for two 4/0 AWG conductors is estimated to be 1.5625 square inches, for a total of 2.2099 square inches per splice. Therefore, the total for all three splices would be 6.6297 square inches.

Even if all three splices were made in a very tight area, according to this calculation, there's adequate space, because 6.6297 is approximately 41% full. Up to 75% full is permitted. Therefore, this example is Code-compliant.

Owen is the owner and president of National Code Seminars and the holder of master electrician certifications in 46 states.

About the Author

Steven Owen

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