1. Box fill
NEC 314.16 — the box-fill table and the counting rules above it
Pure procedure. Every wrong answer on a box-fill question is a counting mistake, not a math mistake.
Exam-style problem
A device box contains four 12 AWG current-carrying conductors spliced through, one 12 AWG equipment grounding conductor, and one duplex receptacle on 12 AWG. Minimum box volume?
- 1Count each conductor that originates outside and is spliced or terminated inside: 4 conductors = 4 allowances.
- 2ALL equipment grounding conductors together count as ONE allowance (largest EGC size): +1.
- 3Each strap/yoke (the receptacle) counts as TWO allowances at the largest conductor connected to it: +2.
- 4Total allowances: 4 + 1 + 2 = 7, all at the 12 AWG volume (2.25 in³ each in the 314.16 table).
- 5Multiply: 7 × 2.25 in³.
Answer
15.75 in³ minimum box volume
The trap
The receptacle counting as two, and all grounds together counting as one — the exam writes distractors for candidates who count them per-wire.
2. Conduit fill
NEC Chapter 9, Table 1 (the 40% rule) with Tables 4 (conduit areas) and 5 (wire areas)
Three-table lookups against a clock. The procedure is fixed; speed comes from knowing which table answers which question.
Exam-style problem
How many 10 AWG THHN conductors are permitted in ¾-inch EMT?
- 1Three or more conductors → the fill limit is 40% of the conduit's internal area (Chapter 9, Table 1).
- 2Table 4 (EMT column): ¾-inch EMT total area ≈ 0.533 in²; its 40% column already shows ≈ 0.213 in².
- 3Table 5: one 10 AWG THHN ≈ 0.0211 in².
- 4Divide: 0.213 ÷ 0.0211 ≈ 10.09.
- 5Fractions are truncated when the count is 3 or more: 10 conductors.
Answer
10 × 10 AWG THHN in ¾" EMT
The trap
Rounding UP. 10.09 becomes 10, never 11 — and the exam always offers the rounded-up number as a distractor. (Annex C pre-computes common combinations; use it when all conductors are the same type.)
3. Ampacity & derating
NEC 310.16 base ampacities; 310.15(C) bundling adjustment; 310.15(B) temperature correction
The Texas data says this family, plus the ones below, is where the 21% pass rate is made. The procedure has a strict ORDER, and the order is what candidates lose.
Exam-style problem
Nine 12 AWG THHN current-carrying conductors share one raceway; ambient is 96°F. What is the adjusted ampacity of each?
- 1Start from the conductor's 90°C column ampacity (THHN is a 90°C insulation): 12 AWG THHN = 30 A base.
- 2Bundling: 7–9 current-carrying conductors → 70% adjustment (310.15(C)): 30 × 0.70 = 21 A.
- 3Ambient: 96°F falls in the 87–95°F… next band; take the 90°C-column correction for 96°F (≈ 0.91): 21 × 0.91 ≈ 19.1 A.
- 4Termination check: the final protected ampacity can never exceed the 75°C (or 60°C) termination rating — compare and take the lower.
Answer
≈ 19 A adjusted ampacity per conductor
The trap
Starting from the 75°C column. Derating STARTS at the insulation's 90°C ampacity — that is the entire reason THHN exists — and only the END result is capped by the termination column.
4. Voltage drop
Not a mandatory NEC rule for most circuits — informational notes recommend 3% branch / 5% total. The formula is the exam content.
The one family that is algebra rather than lookup. Two formulas, single-phase and three-phase, and the exam tests whether you picked the right one.
Exam-style problem
A single-phase 240 V circuit runs 150 ft one-way on copper conductors with a 40 A load. Using K = 12.9, what size keeps voltage drop within 3%?
- 1Allowed drop: 240 × 0.03 = 7.2 V.
- 2Single-phase formula: CM = (2 × K × I × D) ÷ VD.
- 3Plug in: CM = (2 × 12.9 × 40 × 150) ÷ 7.2 = 154,800 ÷ 7.2 = 21,500 circular mils.
- 4Table 8 (Chapter 9): next size UP from 21,500 CM → 6 AWG (26,240 CM). 8 AWG (16,510 CM) is too small.
The trap
The 2 (single-phase) vs 1.732 (three-phase) multiplier, and using one-way distance — the 2 already accounts for the return path. Doubling the distance AND using 2 is the classic self-inflicted wrong answer.
5. Motor calculations
NEC Article 430: Table 430.248/250 for FLC, 430.22 (conductors), 430.52 (branch protection)
The family with the most counter-intuitive rule set — nameplate current is IGNORED for most of it, and the protection multipliers feel illegally large the first time.
Exam-style problem
A 5 HP, 230 V single-phase motor: size the branch-circuit conductors and the inverse-time breaker.
- 1FLC comes from TABLE 430.248, not the nameplate: 5 HP @ 230 V single-phase = 28 A.
- 2Conductors (430.22): 125% of FLC → 28 × 1.25 = 35 A minimum conductor ampacity.
- 3Branch protection (430.52, inverse-time breaker): up to 250% of FLC → 28 × 2.5 = 70 A.
- 4No standard 70 A issue here (70 A is standard); when the calc lands between sizes, 430.52 permits the NEXT size up.
Answer
35 A minimum conductor ampacity; 70 A inverse-time breaker maximum
The trap
A 70 A breaker on conductors rated 35 A looks wrong and is right — motor branch protection covers short-circuit/ground-fault only; the OVERLOAD device protects the wire. The exam preys on the instinct to match breaker to conductor.
Worked with commonly used table values for teaching; your exam runs on the NEC edition named in YOUR state's bulletin, and table values shift between editions — always pull the number from your own code book, which is the open-book skill the exam is actually testing.
Which NEC edition is YOUR exam on?
Not the edition your state enforces on job sites — the one the exam bulletin names. They differ in most states, and every row below was read from the official bulletin on the date shown. Texas switched to the 2026 NEC on September 1, 2026; Tennessee still tests on 2017.
Oregon and Montana are deliberately absent: Oregon mails each candidate a private approved-materials letter and publishes no edition, and Montana's board page could not be read directly at verification time. When a bulletin does not state the edition, ask the board — do not infer it from the state building code.
