Conduit Fill Practice Questions
Here are 30 original conduit fill questions with worked answers, from Table 1 basics to mixed-size sizing, nipples, and cables. These are unofficial exercises; use the supplied reference values and check the edition named in your exam bulletin.
Table 1 and its notes
Question 1 of 30 · CF-01 · Fill percentage for three or more conductors
Four THHN conductors will be pulled into a 40-foot run of rigid metal conduit. What is the maximum percentage of the conduit's cross-sectional area the conductors may occupy?
- A. 31%
- B. 40%
- C. 53%
- D. 60%
Show answer and explanation
Answer: B. 40%
Table 1 allows 40% when a raceway holds more than two conductors.
Why not the others:
- A: 31% is the limit for exactly two conductors.
- C: 53% is for one conductor (or one cable treated as one).
- D: 60% applies only to nipples 24 inches or shorter between enclosures. A 40-foot run isn't a nipple.
References: Table 1 and notes (2017).
Question 2 of 30 · CF-02 · Fill percentage for exactly two conductors
A complete 35-foot conduit run will hold exactly two insulated conductors of different sizes and nothing else. Which fill limit applies?
- A. 31%
- B. 40%
- C. 53%
- D. 60%
Show answer and explanation
Answer: A. 31%
Two conductors get the lowest limit in Table 1: 31%. Different sizes change the area math, not the count. Memorize the order: one = 53, two = 31, three or more = 40.
Why not the others:
- B: 40% needs more than two conductors. Mixed sizes don't raise the count.
- C: 53% is for a single conductor.
- D: Nothing in the stem makes this a short nipple.
References: Table 1 and notes (2017).
Question 3 of 30 · CF-03 · Nipple allowance
A 16-inch electrical metallic tubing (EMT) nipple (measured without its connectors) runs between two adjacent enclosures. What is the maximum fill?
- A. 40%
- B. 53%
- C. 60%
- D. 100%
Show answer and explanation
Answer: C. 60%
Note 4 lets a conduit or tubing nipple up to 24 inches long, installed between boxes, cabinets, or similar enclosures, be filled to 60% of its total area. The 2023 wording measures the nipple without its connectors. The same note says the ampacity adjustment factors for more than three current-carrying conductors don't have to be applied there.
Why not the others:
- A: 40% is the normal limit for complete runs with more than two conductors that don't qualify for the nipple allowance.
- B: 53% is the one-conductor limit.
- D: Note 4 permits 60%, not 100%, for this nipple.
References: Table 1 and notes (2017); 2023 Note 4 excerpt.
Question 4 of 30 · CF-04 · Equipment grounding conductors
How are equipment grounding conductors handled in a conduit fill calculation?
- A. They're left out because they don't carry current
- B. They're counted only if they're insulated
- C. Each one counts as half a conductor
- D. They're counted at their actual size, insulated or bare
Show answer and explanation
Answer: D. They're counted at their actual size, insulated or bare
Note 3 says equipment grounding and bonding conductors must be included, using their actual dimensions. A bare one may use the bare-conductor dimensions in Table 8 (Note 8).
Why not the others:
- A: This mixes up fill with ampacity adjustment. Adjustment counts current-carrying conductors. Fill counts everything that takes up space in the pipe.
- B: Bare grounding conductors count too, just at their smaller bare size.
- C: There's no half-conductor rule. Use the conductor's real area.
References: Table 1 and notes (2017); Table 8 (2017).
Question 5 of 30 · CF-05 · Sleeves for physical protection
A short piece of EMT is used only as a sleeve to protect an exposed cable from physical damage. How do the Table 1 fill limits apply to that sleeve?
- A. They aren't intended to apply to it
- B. The 40% limit applies
- C. The 53% limit applies
- D. The 60% nipple limit applies
Show answer and explanation
Answer: A. They aren't intended to apply to it
Note 2 says Table 1 applies to complete conduit or tubing systems. It isn't intended for sections of conduit or tubing used to protect exposed wiring from physical damage.
Why not the others:
- B: 40% is for complete raceway systems with three or more conductors.
- C: 53% is the single-conductor limit for a complete system.
- D: The 60% allowance is for nipples between enclosures, not protective sleeves.
References: Table 1 and notes (2017).
Question 6 of 30 · CF-06 · One multiconductor cable
A single round cable with an overall jacket contains five insulated conductors. It is the only thing in a complete 40-foot conduit run. Which fill percentage applies?
- A. 31%
- B. 40%, because the cable holds more than two conductors
- C. 53%
- D. 60%
Show answer and explanation
Answer: C. 53%
Note 9 treats a multiconductor cable as one conductor when you figure fill, and you use the cable's outside dimensions. One conductor means 53%. Separate wires bundled without an overall covering are not a cable; count those one by one.
Why not the others:
- A: 31% is for two conductors or cables.
- B: The conductors inside the jacket don't count separately.
- D: A 40-foot run isn't a nipple.
References: Table 1 and notes (2017).
Question 7 of 30 · CF-07 · Where bare conductor sizes are listed
Which Chapter 9 table is specifically named in Note 8 for bare-conductor dimensions where bare conductors are permitted?
- A. Table 4
- B. Table 5
- C. Table 5A
- D. Table 8
Show answer and explanation
Answer: D. Table 8
Note 8 lets you use Table 8 dimensions where the NEC permits bare conductors.
Why not the others:
- A: Table 4 lists raceway sizes and areas.
- B: Table 5 lists insulated conductors.
- C: Table 5A lists compact building wire and includes a bare-conductor diameter column, but it is not the table named in Note 8.
References: Table 1 and notes (2017); Tables 5 and 5A (2017); Table 8 (2017).
Maximum conductors of one size
For Questions 8–26 and 29–30, assume a complete raceway run longer than 24 inches containing only the listed wires or cables; calculate fill only. Insulated-wire values are for the noncompact Table 5 construction, not Table 5A. AWG means American Wire Gauge.
Question 8 of 30 · CF-08 · Same-size count
What is the maximum number of 12 AWG THHN conductors permitted in trade size 3/4 EMT?
- A. 9
- B. 16
- C. 17
- D. 26
Show table values
3/4 EMT at 40% = 0.213 in² (Table 4). 12 AWG THHN = 0.0133 in² (Table 5).
Show answer and explanation
Answer: B. 16
Work: 0.213 ÷ 0.0133 ≈ 16.02 → 16
Divide the raceway's 40% area by one conductor's area. The result is 16.02. The decimal (.02) is below 0.8, so the answer stays at 16. Use the unrounded quotient for the 0.8 test; the displayed quotient is rounded only for readability.
Why not the others:
- A: 9 is the 1/2 EMT answer (0.122 ÷ 0.0133 ≈ 9.17).
- C: 17 rounds up without meeting the 0.8 threshold.
- D: 26 is the 1 EMT answer (0.346 ÷ 0.0133 ≈ 26.02).
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 9 of 30 · CF-09 · The 0.8 rule
What is the maximum number of 14 AWG THHN conductors permitted in trade size 3/4 EMT?
- A. 12
- B. 16
- C. 21
- D. 22
Show table values
3/4 EMT at 40% = 0.213 in². 14 AWG THHN = 0.0097 in².
Show answer and explanation
Answer: D. 22
Work: 0.213 ÷ 0.0097 ≈ 21.96 → decimal ≥ 0.8 → 22
The division gives 21.96. Note 7 says that when all the conductors have the same total cross-sectional area, including insulation and the decimal is 0.8 or more, you use the next whole number. So the answer is 22.
Why not the others:
- A: 12 is the 1/2 EMT answer.
- B: 16 is the 12 AWG answer.
- C: 21 drops the decimal. "Always round down" is wrong when every conductor has the same overall area and the decimal meets Note 7's 0.8 threshold.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 10 of 30 · CF-10 · Use the right raceway row
What is the maximum number of 10 AWG THHN conductors permitted in trade size 1-1/4 rigid metal conduit (RMC)?
- A. 17
- B. 28
- C. 29
- D. 30
Show table values
1-1/4 RMC at 40% = 0.610 in². 10 AWG THHN = 0.0211 in².
Show answer and explanation
Answer: C. 29
Work: 0.610 ÷ 0.0211 ≈ 28.91 → 29
The division gives 28.91. The decimal is .91, which meets the 0.8 threshold, so the answer is 29.
Why not the others:
- A: 17 is the 1 RMC answer (0.355 ÷ 0.0211 ≈ 16.82 → 17).
- B: 28 either drops the decimal or uses the 1-1/4 EMT area (0.598 ÷ 0.0211 ≈ 28.34). RMC and EMT of the same trade size have different areas.
- D: 30 goes past the one-step round-up Note 7 allows.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 11 of 30 · CF-11 · Insulation type changes the area
What is the maximum number of 10 AWG THW conductors permitted in trade size 1 polyvinyl chloride (PVC) Schedule 40?
- A. 13
- B. 14
- C. 15
- D. 16
Show table values
1 PVC Schedule 40 at 40% = 0.333 in². 10 AWG THW = 0.0243 in².
Show answer and explanation
Answer: A. 13
Work: 0.333 ÷ 0.0243 ≈ 13.70 → 13
The division gives 13.70. The decimal (.70) is below 0.8, so the answer is 13. The listed 10 AWG THW area is larger than the 10 AWG THHN area, so fewer fit.
Why not the others:
- B: 14 rounds at 0.5. The NEC's threshold is 0.8.
- C: 15 uses the THHN area (0.333 ÷ 0.0211 ≈ 15.78).
- D: 16 uses the THHN area and then rounds up anyway.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 12 of 30 · CF-12 · XHHW count
What is the maximum number of 8 AWG XHHW conductors permitted in trade size 1 EMT?
- A. 7
- B. 8
- C. 9
- D. 10
Show table values
1 EMT at 40% = 0.346 in². 8 AWG XHHW = 0.0437 in².
Show answer and explanation
Answer: B. 8
Work: 0.346 ÷ 0.0437 ≈ 7.92 → 8
The division gives 7.92. The decimal meets the 0.8 threshold, so the answer is 8.
Why not the others:
- A: 7 drops the decimal.
- C: 9 uses the 8 AWG THHN area (0.346 ÷ 0.0366 ≈ 9.45).
- D: 10 has no basis in the math.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 13 of 30 · CF-13 · Schedule 80 PVC is smaller inside
What is the maximum number of 6 AWG THHN conductors permitted in trade size 1-1/4 PVC Schedule 80?
- A. 9
- B. 10
- C. 11
- D. 12
Show table values
1-1/4 PVC Schedule 80 at 40% = 0.495 in². 6 AWG THHN = 0.0507 in².
Show answer and explanation
Answer: A. 9
Work: 0.495 ÷ 0.0507 ≈ 9.76 → 9
The division gives 9.76. The decimal is below 0.8, so the answer is 9. Schedule 80's thicker wall leaves less room inside than Schedule 40.
Why not the others:
- B: 10 rounds at 0.5.
- C: 11 is the Schedule 40 answer (0.581 ÷ 0.0507 ≈ 11.46).
- D: 12 has no basis in the math.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 14 of 30 · CF-14 · When the total looks slightly over
Three 2 AWG THHN conductors, and nothing else, are specified for a trade size 1 EMT run. Which statement is correct?
- A. Not permitted: 3 × 0.1158 = 0.3474 in², which is more than 0.346 in²
- B. Permitted: 0.346 ÷ 0.1158 ≈ 2.99, and Note 7 lets the count round up to 3
- C. Permitted only if the run is 24 inches or shorter
- D. Not permitted: three conductors must use the 53% column
Show table values
1 EMT at 40% = 0.346 in². 2 AWG THHN = 0.1158 in².
Show answer and explanation
Answer: B. Permitted: 0.346 ÷ 0.1158 ≈ 2.99, and Note 7 lets the count round up to 3
Work: 0.346 ÷ 0.1158 ≈ 2.99 → 3
All three conductors have the same total cross-sectional area, including insulation, so Note 7 applies. The quotient is about 2.99, with a fractional part greater than 0.8, so 3 are permitted.
Why not the others:
- A: Comparing total area to the allowable area is the right method when sizes are mixed. With all-same-size conductors, Note 7's count rule applies.
- C: The answer doesn't depend on the nipple allowance.
- D: Three conductors use the 40% column.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 15 of 30 · CF-15 · Larger conductors, same rule
What is the maximum number of 4 AWG THHN conductors permitted in trade size 1-1/2 EMT?
- A. 7
- B. 9
- C. 10
- D. 11
Show table values
1-1/2 EMT at 40% = 0.814 in². 4 AWG THHN = 0.0824 in².
Show answer and explanation
Answer: C. 10
Work: 0.814 ÷ 0.0824 ≈ 9.88 → 10
The division gives 9.88. The decimal meets the 0.8 threshold, so the answer is 10.
Why not the others:
- A: 7 is the 1-1/4 EMT answer (0.598 ÷ 0.0824 ≈ 7.26).
- B: 9 drops the decimal.
- D: 11 is the 1-1/2 IMC answer (0.890 ÷ 0.0824 ≈ 10.80). Check the raceway type.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Minimum raceway size with mixed wires
Question 16 of 30 · CF-16 · Don't forget the ground
Three 8 AWG THHN conductors and one 10 AWG THHN equipment grounding conductor go in EMT. What is the minimum trade size?
- A. 1/2
- B. 3/4
- C. 1
- D. 1-1/4
Show table values
8 AWG THHN = 0.0366 in². 10 AWG THHN = 0.0211 in². EMT at 40%: 1/2 = 0.122, 3/4 = 0.213 in².
Show answer and explanation
Answer: B. 3/4
Work: (3 × 0.0366) + 0.0211 = 0.1309 → 0.122 < 0.1309 ≤ 0.213 → 3/4
Add every conductor, including the ground. The total is 0.1309 in². That's more than 1/2 EMT allows (0.122) and within 3/4 EMT (0.213).
Why not the others:
- A: 1/2 works only if you forget the grounding conductor (0.1098 in²).
- C: 1 works but isn't the minimum.
- D: 1-1/4 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 17 of 30 · CF-17 · Bare grounding conductor
Three 8 AWG THHN conductors and one bare solid 10 AWG copper equipment grounding conductor go in EMT where the bare grounding conductor is permitted. What is the minimum trade size?
- A. 1/2
- B. 3/4
- C. 1
- D. 1-1/4
Show table values
Bare solid 10 AWG = 0.008 in² (Table 8). 8 AWG THHN = 0.0366 in². 1/2 EMT at 40% = 0.122 in².
Show answer and explanation
Answer: A. 1/2
Work: (3 × 0.0366) + 0.008 = 0.1178 ≤ 0.122 → 1/2
A bare conductor may use its Table 8 area. The total drops to 0.1178 in², which fits within 1/2 EMT's 0.122 in². It still has to be counted (Note 3), just at its real size.
Why not the others:
- B: 3/4 is what you get by treating the bare wire as insulated.
- C: 1 works but isn't the minimum.
- D: 1-1/4 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017); Table 8 (2017).
Question 18 of 30 · CF-18 · Three sizes in one raceway
What is the minimum EMT trade size for three 6 AWG THWN-2, one 8 AWG THWN-2 equipment grounding conductor, and four 12 AWG THHN?
- A. 1/2
- B. 3/4
- C. 1
- D. 1-1/4
Show table values
6 AWG = 0.0507, 8 AWG = 0.0366, 12 AWG = 0.0133 in² (THHN/THWN/THWN-2 row). EMT at 40%: 3/4 = 0.213, 1 = 0.346 in².
Show answer and explanation
Answer: C. 1
Work: (3 × 0.0507) + 0.0366 + (4 × 0.0133) = 0.2419 → 1
THWN and THWN-2 share the THHN row in Table 5. The total is 0.2419 in². That's more than 3/4 EMT allows (0.213) and within 1 EMT (0.346).
Why not the others:
- A: 1/2 is far too small.
- B: 3/4 fits only if you leave out the four 12 AWG conductors (0.1887 in²).
- D: 1-1/4 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 19 of 30 · CF-19 · PVC Schedule 80 sizing
What is the minimum PVC Schedule 80 trade size for four 2 AWG THHN and one 6 AWG THHN equipment grounding conductor?
- A. 1
- B. 1-1/4
- C. 1-1/2
- D. 2
Show table values
2 AWG THHN = 0.1158 in². 6 AWG THHN = 0.0507 in². PVC Schedule 80 at 40%: 1-1/4 = 0.495, 1-1/2 = 0.684 in².
Show answer and explanation
Answer: C. 1-1/2
Work: (4 × 0.1158) + 0.0507 = 0.5139 → 1-1/2
The total is 0.5139 in². Schedule 80 1-1/4 allows 0.495, which is too small. 1-1/2 allows 0.684.
Why not the others:
- A: 1 is far too small.
- B: 1-1/4 would work in Schedule 40 (0.581 in²), but this is Schedule 80.
- D: 2 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 20 of 30 · CF-20 · Close to the fill limit still fits
What is the minimum RMC trade size for four 500 kcmil THHN and one 3 AWG THHN equipment grounding conductor?
- A. 2
- B. 2-1/2
- C. 3
- D. 3-1/2
Show table values
500 kcmil THHN = 0.7073 in². 3 AWG THHN = 0.0973 in². RMC at 40%: 2-1/2 = 1.946, 3 = 3.000 in².
Show answer and explanation
Answer: C. 3
Work: (4 × 0.7073) + 0.0973 = 2.9265 ≤ 3.000 → 3
The total is 2.9265 in². 3 RMC allows 3.000 in², so it meets this fill requirement. That uses 97.6% of the allowable space, but the question asks what Table 1 permits, and it's permitted. (Table 1's first informational note says a bigger raceway is worth considering for hard pulls, but that's field judgment, not the minimum.)
Why not the others:
- A: 2 is far too small.
- B: 2-1/2 allows 1.946 in², far too small.
- D: 3-1/2 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 21 of 30 · CF-21 · IMC with a bare stranded ground
What is the minimum intermediate metal conduit (IMC) trade size for six 10 AWG THHN, three 12 AWG THHN, and one bare 7-strand 10 AWG copper equipment grounding conductor, where the bare conductor is permitted?
- A. 3/4
- B. 1
- C. 1-1/4
- D. 1-1/2
Show table values
10 AWG THHN = 0.0211, 12 AWG THHN = 0.0133 in². Bare 7-strand 10 AWG copper = 0.011 in² (Table 8). IMC at 40%: 1/2 = 0.137, 3/4 = 0.235 in².
Show answer and explanation
Answer: A. 3/4
Work: (6 × 0.0211) + (3 × 0.0133) + 0.011 = 0.1775 → 3/4
The total is 0.1775 in². 1/2 IMC allows 0.137, which is too small. 3/4 IMC allows 0.235.
Why not the others:
- B: 1 works but isn't the minimum.
- C: 1-1/4 works but isn't the minimum.
- D: 1-1/2 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017); Table 8 (2017).
Question 22 of 30 · CF-22 · When Annex C doesn't fit
A raceway will hold a mix of 10 AWG and 12 AWG THHN conductors. Which method accounts for each conductor's actual area when finding the minimum trade size?
- A. Add the Annex C maximum counts for each size
- B. Treat every conductor as 10 AWG and use that Annex C count
- C. Treat every conductor as 12 AWG and use that Annex C count
- D. Add up each conductor's area from Table 5 and compare the total with the raceway's Table 4 area
Show answer and explanation
Answer: D. Add up each conductor's area from Table 5 and compare the total with the raceway's Table 4 area
Annex C gives maximum counts for conductors that are all the same size (Note 1). For mixed sizes, Note 6 sends you to Table 5 for conductor areas and Table 4 for the raceway. Add up the areas and compare.
Why not the others:
- A: Annex C counts assume the whole raceway is one size. You can't combine them.
- B: Treating every conductor as 10 AWG overstates the smaller conductors' areas. It can be conservative, but it does not calculate the actual mixed-conductor area requested.
- C: Treating every conductor as 12 AWG understates the larger conductors' areas and can select a raceway that is too small.
References: Table 1 and notes (2017).
Question 23 of 30 · CF-23 · The rounding trap with mixed sizes
What is the minimum EMT trade size for four 600 kcmil THHN and one 2 AWG THHN equipment grounding conductor?
- A. 2-1/2
- B. 3
- C. 3-1/2
- D. 4
Show table values
600 kcmil THHN = 0.8676 in². 2 AWG THHN = 0.1158 in². EMT at 40%: 3 = 3.538, 3-1/2 = 4.618 in².
Show answer and explanation
Answer: C. 3-1/2
Work: (4 × 0.8676) + 0.1158 = 3.5862 → 3.538 < 3.5862 ≤ 4.618 → 3-1/2
The total is 3.5862 in². 3 EMT allows 3.538, which is too small. 3-1/2 EMT allows 4.618.
Why not the others:
- A: 2-1/2 allows 2.343 in², far too small.
- B: This is the trap. 3.538 ÷ 3.5862 ≈ 0.99 looks like it "rounds up," but the 0.8 rule is only for conductors that are all the same size. With mixed sizes, the total can't be more than the allowable area.
- D: 4 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Adding to a raceway that already has wires
Question 24 of 30 · CF-24 · How many more will fit
A 1-1/4 EMT already holds three 6 AWG THHN and one 10 AWG THHN equipment grounding conductor. How many 12 AWG THHN conductors can be added?
- A. 31
- B. 32
- C. 44
- D. 45
Show table values
1-1/4 EMT at 40% = 0.598 in². 6 AWG = 0.0507, 10 AWG = 0.0211, 12 AWG = 0.0133 in² (THHN).
Show answer and explanation
Answer: A. 31
Work: Existing: (3 × 0.0507) + 0.0211 = 0.1732. Left: 0.598 − 0.1732 = 0.4248. 0.4248 ÷ 0.0133 ≈ 31.94 → 31
Subtract what's already there from the allowable area, then divide by the new conductor's area. The result is 31.94, so 31. The raceway holds mixed sizes, so the 0.8 round-up doesn't apply. Checking: 32 added conductors would bring the total to 0.5988 in², more than 0.598.
Why not the others:
- B: 32 uses the 0.8 round-up on a mixed-size raceway.
- C: 44 ignores the existing conductors and drops the decimal from 0.598 ÷ 0.0133 ≈ 44.96. That is not the available-space calculation.
- D: 45 is the all-12-AWG count for an otherwise empty raceway after applying Note 7. This raceway already contains other sizes.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 25 of 30 · CF-25 · Adding a different size
A 3/4 IMC already holds four 12 AWG THHN. How many 10 AWG THHN conductors can be added?
- A. 7
- B. 8
- C. 9
- D. 11
Show table values
3/4 IMC at 40% = 0.235 in². 12 AWG THHN = 0.0133, 10 AWG THHN = 0.0211 in².
Show answer and explanation
Answer: B. 8
Work: 0.235 − (4 × 0.0133) = 0.1818. 0.1818 ÷ 0.0211 ≈ 8.62 → 8
The space left is 0.1818 in², and 0.1818 ÷ 0.0211 ≈ 8.62, so 8. Nine would need 0.1899 in², more than is left.
Why not the others:
- A: 7 leaves usable space unused.
- C: 9 needs more area than remains.
- D: 11 ignores the conductors already in the pipe (0.235 ÷ 0.0211 ≈ 11.14).
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017).
Question 26 of 30 · CF-26 · Percent fill uses the total area
About what percent of the total cross-sectional area of a 1 EMT is taken up by three 6 AWG THHN and one 10 AWG THHN?
- A. 12%
- B. 20%
- C. 40%
- D. 50%
Show table values
1 EMT total area (100%) = 0.864 in². 6 AWG THHN = 0.0507, 10 AWG THHN = 0.0211 in².
Show answer and explanation
Answer: B. 20%
Work: (3 × 0.0507) + 0.0211 = 0.1732. (0.1732 ÷ 0.864) × 100 ≈ 20.0%
Divide the conductor area by the raceway's total (100%) area: (0.1732 ÷ 0.864) × 100 ≈ 20.0%. That's well under the 40% limit.
Why not the others:
- A: 12% divides by the 1-1/4 EMT total area (1.496 in²).
- C: 40% is the limit, not the amount actually used.
- D: 50% divides by the 40% column (0.346). That tells you how much of the allowable space is used, not how full the pipe is.
References: Table 4 (2017); Tables 5 and 5A (2017).
Nipples and cables
Question 27 of 30 · CF-27 · Counting in a nipple
A 12-inch trade size 1/2 EMT nipple (measured without its connectors) connects two enclosures. What is the maximum number of 12 AWG THHN conductors, with no other wires or cables present?
- A. 9
- B. 12
- C. 13
- D. 14
Show table values
1/2 EMT at 60% = 0.182 in². 12 AWG THHN = 0.0133 in².
Show answer and explanation
Answer: C. 13
Work: 0.182 ÷ 0.0133 ≈ 13.68 → 13
Qualifying nipples use 60% of total area. The division gives 13.68. The decimal is below 0.8, so the answer is 13.
Why not the others:
- A: 9 uses the 40% column.
- B: 12 leaves room unused.
- D: 14 rounds at 0.5.
References: Table 1 and notes (2017); Table 4 (2017); Tables 5 and 5A (2017); 2023 Note 4 excerpt.
Question 28 of 30 · CF-28 · Nipples and ampacity adjustment
A 12-inch trade size 1/2 EMT nipple, measured without its connectors, connects two enclosures and carries more than three current-carrying conductors. What about the ampacity adjustment factors in 310.15(C)(1)?
- A. They always apply
- B. They apply only if fill is over 40%
- C. They apply only to aluminum conductors
- D. Note 4 says they don't have to be applied to a nipple that meets its conditions
Show answer and explanation
Answer: D. Note 4 says they don't have to be applied to a nipple that meets its conditions
The same Note 4 that allows 60% fill says the 310.15(C)(1) adjustment factors need not apply to a qualifying nipple (that's the 2023 cross-reference; older editions number it differently). That does not waive other applicable ampacity requirements.
Why not the others:
- A: Note 4 carves out qualifying nipples.
- B: Fill and ampacity adjustment are separate checks.
- C: Conductor material has nothing to do with this rule.
References: 2023 Note 4 excerpt.
Question 29 of 30 · CF-29 · Oval (elliptical) cable
One jacketed cable with an oval cross section runs alone in EMT. Its major (wide) diameter is 0.72 inch and its minor diameter is 0.40 inch. What is the minimum EMT trade size?
- A. 3/4
- B. 1
- C. 1-1/4
- D. 1-1/2
Show table values
EMT at 53%: 3/4 = 0.283, 1 = 0.458 in².
Show answer and explanation
Answer: B. 1
Work: π/4 × 0.72² ≈ 0.407 in² → 0.283 < 0.407 ≤ 0.458 → 1
Note 9 says to figure an elliptical cable's area as a circle using the major diameter: π/4 × 0.72² ≈ 0.407 in². One cable counts as one conductor, so use 53%. 3/4 EMT allows 0.283 in², which is too small. 1 EMT allows 0.458 in².
Why not the others:
- A: 3/4 fits only if you use the true oval area (π/4 × 0.72 × 0.40 ≈ 0.226 in²). The NEC tells you to use the major diameter instead.
- C: 1-1/4 is what you get using the 40% column (1 EMT at 40% = 0.346 in², too small).
- D: 1-1/2 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017).
Question 30 of 30 · CF-30 · Two cables in one raceway
Two round jacketed cables, each 0.45 inch in outside diameter, run together in EMT. What is the minimum trade size?
- A. 3/4
- B. 1
- C. 1-1/4
- D. 1-1/2
Show table values
EMT at 31%: 3/4 = 0.165, 1 = 0.268, 1-1/4 = 0.464 in².
Show answer and explanation
Answer: C. 1-1/4
Work: 2 × (π/4 × 0.45²) ≈ 0.318 in² → 0.268 < 0.318 ≤ 0.464 → 1-1/4
Each cable counts as one conductor (Note 9), so two cables use the 31% column. Each cable's area is π/4 × 0.45² ≈ 0.159 in², so 0.318 in² together. 1 EMT allows 0.268 in² at 31%, which is too small. 1-1/4 EMT allows 0.464 in².
Why not the others:
- A: 3/4 EMT allows 0.165 in² under the applicable 31% column, less than the two cables need.
- B: 1 EMT would fit if you wrongly used 40% (0.346 in²), but two cables in this complete run require the 31% column.
- D: 1-1/2 works but isn't the minimum.
References: Table 1 and notes (2017); Table 4 (2017).
The method for these conduit fill questions
- Count everything in the raceway. That includes hots, neutrals, and equipment grounding conductors, whether they're insulated or bare. A jacketed multiconductor cable counts as one.
- Pick the Table 1 percentage. One conductor is 53%, two is 31%, and three or more is 40%. A nipple 24 inches or shorter between enclosures (measured without its connectors in the 2023 wording) can use 60%.
- Look up each conductor's area. Use Table 5 for insulated wire, where THWN and THWN-2 share the THHN row. Use Table 5A for compact-stranded wire, Table 8 for bare wire, and the actual outside diameter for cables. For an oval cable, use the major (wide) diameter as a circle.
- Find the raceway row in Table 4. Use the row for that raceway type and schedule, and read the column that matches your percentage.
- Compare or divide.
- What size with mixed wires or cables? Add up the areas. The total must be at or below the column value. There's no rounding with mixed sizes. For a uniform group of conductors, use the count rule below; Question 14 shows why a straight area comparison can miss its exception.
- How many? Divide the column value by one conductor's area. Round up only when every conductor has the same total cross-sectional area, including insulation, and the unrounded quotient has a fractional part of 0.8 or more (Note 7).
A shortcut for mixed-size sizing. Divide the total conductor area by the fill fraction to estimate the required total (100%) area. Compare that with Table 4's total-area column, then confirm the size against the applicable percentage column; the printed table entries are rounded. For question 18, that's 0.2419 ÷ 0.40 = 0.60475 in² (about 0.605 in²). 3/4 EMT's total area is 0.533 in², which is too small. 1 EMT's is 0.864 in², which works. It's the same answer the 40% column gives.
Worked example (question 16). Three 8 AWG THHN plus one 10 AWG THHN ground come to (3 × 0.0366) + 0.0211 = 0.1309 in². 1/2 EMT allows 0.122 in² at 40%, so it's too small. 3/4 EMT allows 0.213 in², so 3/4 is the answer. Leave out the ground and you'd wrongly pick 1/2.
References: Table 1 and its notes (2017 text); 2023 Note 4 wording.
Where each value lives in Chapter 9
| What you need | Where to find it |
|---|---|
| The fill percentage (53, 31, 40) | Table 1 |
| The rules on grounds, nipples, rounding, cables, and bare wire | Notes to Tables, printed with Table 1 |
| Raceway areas at 100%, 60%, 53%, 31%, and 40% | Table 4, by raceway type |
| Insulated conductor areas | Table 5 |
| Compact-stranded conductor areas | Table 5A |
| Bare conductor areas | Table 8 |
| Ready-made counts for conductors all the same size and insulation type | Informative Annex C |
Table 4 separates raceway types and schedules. EMT, IMC, RMC, PVC Schedule 40, and PVC Schedule 80 of the same trade size all have different areas, and several questions above test exactly that.
Reference: Chapter 9 tables (2017 text). AWG sizes and kcmil sizes identify conductors; the fill calculations use the separate area in square inches (in²).
Review your misses by mistake
Count only answers you chose before revealing the explanation. Your result is the number correct out of the number attempted, not a pass prediction or an official exam score; an answer revealed first is review, not a scored attempt.
| If you missed it because you… | Redo these |
|---|---|
| Used the wrong percentage | 1, 2, 5, 6, 29, 30 |
| Forgot a grounding conductor or counted it wrong | 4, 16, 17 |
| Used the wrong table or insulated area for bare wire | 7, 17, 21, 22 |
| Read the wrong raceway row or PVC schedule | 10, 13, 15, 19 |
| Used the wrong insulation type | 11, 12, 18 |
| Always rounded down, or rounded at 0.5 | 8, 9, 11, 12, 14, 15, 27 |
| Rounded up with mixed sizes | 23, 24, 25 |
| Divided by the wrong area for percent fill | 26 |
| Missed a nipple rule | 3, 27, 28 |
| Handled a cable like separate wires | 6, 29, 30 |
Quick answers
Do ground wires count toward conduit fill?
Yes. Equipment grounding and bonding conductors count at their actual size (Note 3). A bare one can use the smaller bare area from Table 8 (Note 8). This is the opposite of ampacity adjustment, where grounds don't count as current-carrying conductors.
References: Notes 3 and 8 (2017 text); Minnesota DLI, ampacity calculations.
When can I round up the number of conductors?
For the multi-conductor counts in this set, every conductor must have the same total cross-sectional area, including insulation, and the unrounded quotient must have a fractional part of 0.8 or more (Note 7). So 21.96 becomes 22, but 13.70 stays 13. With mixed sizes, the total area has to fit; do not apply this count-rounding exception. Note 7 also has a companion sentence for sizing a raceway around a single conductor, which uses the same 0.8 test.
Reference: Note 7 (2017 text).
Should I use Annex C or do the math?
Use Annex C when every conductor is the same size and type. It's a fast cross-check for same-size problems. For mixed sizes, Annex C doesn't apply, so add up areas from Table 5 and compare with Table 4 (Note 6). Annex C is informative, not a requirement; the Chapter 9 tables and notes are the rule.
Reference: Notes 1 and 6 (2017 text).
Is conduit fill the same as derating or box fill?
No. These are three separate checks:
- Conduit fill is about physical space. Every conductor counts.
- Ampacity adjustment (derating) uses the applicable current-carrying conductor count and adjustment rules to determine allowable current.
- Box fill is a volume calculation for boxes, in Article 314, not Chapter 9.
Passing conduit fill doesn't mean the conductors have enough ampacity, or that the whole installation complies.
Reference: Minnesota DLI examination guide, conduit fill, ampacity, and box-fill sections.
Which NEC edition should I study for this?
Use the edition named in your own exam bulletin. States move to new NEC editions on their own schedules, and an exam can switch editions on a different date than the installation code does. Look up any value you're unsure of in that edition.
For an official example of separate installation-code and examination transition dates, see Minnesota DLI’s code and exam notices; those dates apply to Minnesota, not every jurisdiction.
Keep going
- Journeyman Electrician Exam Prep: a study plan and practice for the rest of the journeyman exam.
- Free Journeyman Electrician Practice Test: 50 mixed questions once you're comfortable with fill.
- Free Master Electrician Practice Test: broader calculation practice at the master level.
Sources
By Castleport Test Prep Editorial Team
- NFPA 70, 2017 Chapter 9 — historical reference text: Table 1 and Notes to Tables on PDF page 1; Table 4 on pages 2–5; Tables 5 and 5A on pages 6–10; Table 8 on page 11. These are the inspected sources for the supplied reference values, not a claim that every later edition is identical.
- 2023 NEC wording of Chapter 9, Note 4: the section headed “2023 Code Language” reproduces the nipple-length measurement, 60% allowance, and 310.15(C)(1) exception.
- Minnesota DLI, Electrical License Examination Guide: the conduit/raceway-fill, ampacity, and box-fill calculation sections provide a free official explanation of the separate methods.
- NFPA’s 2023 NEC read-only access: the issuing organization’s edition-specific reader; free access requires an account. Use the edition specified by your own exam authority rather than assuming this edition applies to your test.
Source and calculation checks — September 29, 2026: The supplied table values and Chapter 9 rules were checked in the 2017 NFPA text and page images, and the 2023 wording of Note 4 was checked in the identified excerpt. The numerical answers were recalculated from the supplied inputs. This check does not establish that all values and notes match the final 2023 or 2026 edition.
This resource was developed and edited with AI assistance. Source and calculation checks are not credentialed professional review. See our methodology for the distinction.
Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by NFPA or any electrician licensing board or testing company. NEC and National Electrical Code are used only to identify the subject; trademarks belong to their respective owners. These are original practice questions, not official or recalled exam questions. A practice score doesn't guarantee an exam result, a license, or installation approval.