Texas journeyman calculations: the formulas the exam uses, with worked examples
The Calculations portion of the Texas journeyman exam has 26 questions (24 scored) in 110 minutes, and you need 70% to pass it. That is about 4 minutes per question. Most questions use the same short list of formulas. If you know them cold and know where each value sits in the 2026 NEC, the rest is careful arithmetic.
The formula sheet
K is the resistance of a circular-mil-foot of conductor: 12.9 for copper and 21.2 for aluminum. L is the one-way length in feet. Load calculations moved from Article 220 to Article 120 in the 2026 NEC; see our guide on 2026 NEC changes for the full list.
| Topic | Formula | Where to look (2026 NEC) |
|---|---|---|
| Single-phase current | I = P ÷ E | - |
| Three-phase current | I = P ÷ (E × 1.732) | - |
| Continuous load | Overcurrent device ≥ 125% of continuous load + 100% of noncontinuous load | 215.5 (feeders) |
| Conductor ampacity | Table ampacity × temperature correction × adjustment for more than 3 current-carrying conductors | Table 310.16, Table 310.15(B)(1)(1), Table 310.15(C)(1) |
| Small conductors | 14 Cu on 15 A, 12 Cu on 20 A, 10 Cu on 30 A | 240.4(D) |
| Voltage drop, single-phase | VD = 2 × K × I × L ÷ CM | Chapter 9 for conductor area (CM) |
| Voltage drop, three-phase | VD = 1.732 × K × I × L ÷ CM | Chapter 9 for conductor area (CM) |
| Motor branch-circuit conductors | 125% of table full-load current | 430.22, Table 430.250 |
| Motor short-circuit protection | Inverse time breaker up to 250% of table FLC, next higher standard size allowed | Table 430.52(C)(1), 240.6(A) |
| Transformer, primary-only protection | 125% of primary current, next higher standard size allowed | Table 450.5(B) |
| Dwelling general lighting | 2 VA per ft², then demand factors | 120.41, Table 120.45 |
| Box fill | Count conductor volumes × volume per conductor | 314.16 |
Worked example 1: three-phase current
A 30 kW resistive load on a 480 V three-phase system. I = 30,000 ÷ (480 × 1.732) = 36.09 A.
Common wrong answers: 62.5 A (used the single-phase formula) and 20.83 A (divided by 3 instead of 1.732).
Worked example 2: continuous load
A feeder serves 40 A of continuous load and 20 A of noncontinuous load. Minimum overcurrent device: 40 × 1.25 + 20 = 70 A. 70 A is a standard size in 240.6(A), so the answer is 70 A.
Common wrong answers: 60 A (no 125%) and 75 A (125% applied to the whole 60 A).
Worked example 3: ampacity with correction and adjustment
Six current-carrying 12 AWG THHN copper conductors in one raceway, ambient 43 °C. Start from the 90 °C column of Table 310.16: 30 A. Temperature correction for 41–45 °C in the 90 °C column: 0.87. Adjustment for 4–6 conductors: 80%.
30 × 0.87 × 0.80 = 20.88 A. The overcurrent device is still limited to 20 A by 240.4(D).
Two rules: use the 90 °C column only as the starting point for correction and adjustment, and never round an ampacity up.
Worked example 4: voltage drop
A 240 V single-phase load draws 30 A at 150 ft (one way) on 10 AWG copper (10,380 CM).
VD = 2 × 12.9 × 30 × 150 ÷ 10,380 = 11.18 V, which is 4.66% of 240 V.
To keep the drop at 3% (7.2 V): CM = 2 × 12.9 × 30 × 150 ÷ 7.2 = 16,125 CM. The next size with at least that area is 8 AWG (16,510 CM).
Voltage drop only sets a minimum size. The conductor must also meet the ampacity rules. Our free voltage drop tool shows both steps.
Worked example 5: motor branch circuit
A 5 hp, 230 V, three-phase motor. Use the table full-load current, not the nameplate (430.6(A)(1)): Table 430.250 gives 15.2 A.
Branch-circuit conductors: 15.2 × 1.25 = 19 A minimum ampacity.
Inverse time breaker: 15.2 × 2.5 = 38 A. 38 A is not a standard size, so the next higher standard size, 40 A, is allowed.
Overload protection is different: it uses the nameplate current.
Worked example 6: transformer primary protection
A 45 kVA three-phase transformer with a 480 V primary, protected on the primary only. Primary current: 45,000 ÷ (480 × 1.732) = 54.13 A. At 125%: 67.66 A.
Table 450.5(B) lets you go to the next higher standard size, so the answer is 70 A.
Worked example 7: box fill
A box with four 12 AWG conductors, one receptacle on a single yoke and two equipment grounding conductors.
Count: 4 conductors + 2 for the yoke + 1 for all the equipment grounding conductors (up to four) = 7 volumes. Each 12 AWG volume is 2.25 in³, so the box needs at least 7 × 2.25 = 15.75 in³.
Habits that save points
- Write the formula first, then the numbers. Most wrong options come from one skipped factor: 1.732, 125% or a correction factor.
- Round up only where the NEC allows it: the next standard breaker or fuse size. Never round an ampacity or a voltage drop result in your favor.
- For motors, table FLC for conductors and short-circuit protection, nameplate for overloads.
- For dwelling loads on the 2026 NEC, use 2 VA/ft², not the 3 VA/ft² in older books.
- Watch the clock: about 4 minutes per question. Skip a long one and come back.
Check yourself in 30 minutes
Our free 20-question diagnostic is built on the 2026 NEC and scores each exam portion separately. No sign-up.
Take the free diagnostic See the full Texas Journeyman Electrician packSources
- IAEI: Voltage drop formulas (K = 12.9 copper, 21.2 aluminum)
- OpenExamPrep: Texas exam two-part format (2026 NEC, 26 calculations questions, 110 minutes)
- OpenExamPrep: Texas conductor insulations and ampacity (correction and adjustment method)
- OpenExamPrep: Texas motor overcurrent and short-circuit protection
- EC&M Magazine: What's new in Chapter 4 of the 2026 NEC (15 Sep 2026)
- PSI Candidate Information Bulletin, Texas electrician examinations
Updated 2026-10-15. Found something wrong? Tell us and we will fix it and log the fix on our quality page.