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Quizzes on the Code

Aug 1, 2001 12:00 PM, By James Stallcup, Jr., NEC and OSHA Consultant

Choose the best answer:

What size octagonal box must you use to support a lighting fixture supplied with two No. 14-2 (with ground) nonmetallic-sheathed cables (NMCs) with a fixture stud, hickey, and two No. 14 fixture wires? Neutrals are spliced with a pigtail and NMC connectors.

(a) 4 in. x 1 1/4 in.
(b) 4 in. x 1 1/2 in.
(c) 4 in. x 2 1/8 in.
(d) 4 in. x 2 1/4 in.
What size octagonal box must you use to support a lighting fixture supplied with two No. 12-2 (with ground) NMCs with a fixture stud, two cable clamps, two pigtails, and two No. 18 fixture wires (no fixture canopy)?

(a) 4 in. x 1 1/4 in.
(b) 4 in. x 1 1/2 in.
(c) 4 in. x 2 1/8 in.
(d) 4 in. x 2 1/4 in.
What size square box must you use where four No. 14-2 and four No. 12-2 (with ground) NMCs are spliced and routed to loads?

(a) 4 in. x 1 1/4 in.
(b) 4 in. x 1 1/2 in.
(c) 4 11/16 in. x 1 1/2 in.
(d) 4 11/16 in. x 2 1/8 in.
What size device box with cable clamps must you use for two No. 14-2 (with ground) NMCs and one duplex receptacle?

(a) 3 in. x 2 in. x 1 1/2 in.
(b) 3 in. x 2 in. x 3 1/2 in.
(c) 4 in. x 2 1/8 in. x 1 1/2 in.
(d) 4 in. x 2 1/8 in. x 2 1/8 in.

answers and discussion

1. (c), 4 in. x 2 1/8 in. Determine the total fill space of boxes containing same-size conductors by adding the number of conductors to the number of additional conductors for each fitting or device:

Step 1: 2 No.14 hots=2 conductors
2 No.14 neutrals = 2 conductors
2 No.14 grounds = 1 conductor
1 fixture stud = 1 conductor
1 hickey = 1 conductor
1 pigtail = 0 conductors
2 connectors = 0 conductors
2 No.14 fixture wires = 2 conductors
Total = 9 conductors

Step 2: Select the proper size box per Table 370-16(a).
Nine No.14 conductors require a 4 in. x 2 1/8 -in. octagonal box.

2. (c), 4 in. x 2 1/8 in. Size boxes containing conductors that are not the same size based on the volume (in cu. in.) of each size conductor listed in Table 370-16(b):

Step 1:
2 No.12 hots = 2.25 cu. in. x 2 = 4.5 cu. in.
2 No.12 neutrals = 2.25 cu. in. x 2 = 4.5 cu. in.
2 No.12 EGCs = 2.25 cu. in. x 1 = 2.25 cu. in.
1 fixture stud = 2.25 cu. in. x 1 = 2.25 cu. in.
2 cable clamps = 2.25 cu. in. x 1 = 2.25 cu. in.
2 pigtails = 0
2 No.18 fixture wires = 1.5 cu. in. x 2 = 3 cu. in.
Total=18.75 cu. in.

Step 2: Select the proper size box per Table 370-16(a).
An 18.75 cu. in. total requires a 4 in. x 2 1/8-in. octagonal box.

3. (d), 4 11/16 in. x 2 1/8 in. Sec. 370-16(b) refers to Table 370-16(b), which requires you to use the volume (in cu. in.) of each conductor in the box to determine the fill space in cu. in.:

Step 1:
4 No.14 hots = 2 cu. in. x 4 = 8 cu. in.
4 No.14 neutrals = 2 cu. in. x 4 = 8 cu. in.
4 No.12 hots = 2.25 cu. in. x 4 = 9 cu. in.
4 No.12 neutrals = 2.25 cu. in. x 4 = 9 cu. in.
4 No.12 EGCs = 2.25 cu. in. x 1 = 2.25 cu. in.
Total = 36.25 cu. in.

Step 2: Select the proper size box per Table 370-16(a).
A 36.25 cu. in. total requires a 4 11/16 in. x 2 1/8-in. square box.

4. (b), 3 in. x 2 in. x 3 1/2 in. Determine the total fill space of a box containing the same size conductors by adding the number of conductors and additional conductors for each fitting or device:

Step 1:
2 No.14 hots = 2 conductors
2 No.14 neutrals = 2 conductors
2 No.14 EGCs = 1 conductor
1 receptacle = 2 conductors
2 cable clamps = 1 conductor
Total = 8 conductors

Step 2: Select the proper size box per Table 370-16(a).
Eight No.14 NMC conductors require a 3 in. x 2 in.3 1/2-in. box.