Service and Feeder Load Calculations for the Texas Journeyman Exam: How to Break Down the Problem, With Worked Examples (2026 NEC)

Tsai Electro is an independent learning site. Not affiliated with TDLR, PSI or NFPA. We explain the method in our own words and do not reproduce NEC text. Check every rule and table value in your own 2026 NEC.

What changed for 2026

Texas adopted the 2026 NEC effective September 1, 2026, and the TDLR exams are referenced to the 2026 NEC from the same date (TDLR rule adoption notice, PSI Candidate Information Bulletin). The biggest change for calculation questions is where the rules live: the City of Denison’s summary of 2026 changes lists a new Article 120 for branch-circuit, feeder and service load calculations, previously located in Article 220 (Denison 2026 NEC summary). The same summary notes a clarification on how the 125% rule for continuous loads applies in service and feeder calculations. If your tabs and notes still say 220, retab before exam day.

How many exam items this covers

Texas Journeyman Calculations area (PSI outline)	Scored items
Electrical Services, Service Equipment and Separately Derived Systems	4
Electrical Feeders	3
Branch Circuit Calculations and Conductors	4
Motors and Generators	2
All other areas	11
Total scored items	24

The Calculations part has 26 items including 2 that are not scored, in 110 minutes. That is about 4 minutes per question, which is enough only if you have a fixed routine.

The four-step routine

1. List the loads in volt-amperes. Convert everything first: amps × volts, kilowatts × 1,000, square feet × the unit load.
2. Apply demand factors. Only the ones the method allows, and only to the loads they cover.
3. Add the extras. 125% of the largest motor where required, and any minimum values the method sets. Continuous loads are not multiplied by 125% inside the Article 120 load total. When the question asks for minimum feeder conductor ampacity, apply 125% to the continuous loads under the feeder rules in Article 215.
4. Convert to amps and size. For a 120/240 V single-phase service, divide by 240. For a balanced three-phase load, divide total volt-amperes by the line-to-line voltage times 1.732 (√3); at 208 V that denominator is 360.3. Then apply the conductor and overcurrent-protection rules that fit the job.

Worked example 1: dwelling service, optional method

A 2,400 square-foot single-family house at 120/240 V has these loads: two small-appliance circuits, one laundry circuit, a 1,200 VA dishwasher, a 900 VA disposal, a 4,500 W water heater, a 12 kW range, a 5,000 W dryer, a 7,680 VA EV charger (32 A at 240 V), 4,800 VA of central air conditioning and a 15 kW central electric furnace.

The optional method for dwellings (120.82) takes the first 8 kVA of general load at 100% and the rest at 40%, then adds the larger of the heating or cooling load at its own percentage. An EV charger does not go in the 40% group: it is added separately at 100% of its nameplate rating (120.82(D)). In the 2026 NEC, general lighting for service and feeder calculations is 2 VA per square foot (120.41); the 3 VA per square foot figure now applies only when you count branch circuits (120.13). The 2023 edition used the first 10 kVA and 3 VA per square foot, so older practice books give different answers.

1. General lighting: 2,400 × 2 VA = 4,800 VA.
2. Small-appliance and laundry circuits: 3 × 1,500 = 4,500 VA.
3. Appliances at nameplate, without the EV charger: 1,200 + 900 + 4,500 + 12,000 + 5,000 = 23,600 VA.
4. General load total: 4,800 + 4,500 + 23,600 = 32,900 VA.
5. First 8,000 VA at 100%: 8,000 VA. Remaining 24,900 VA at 40%: 9,960 VA. Subtotal: 17,960 VA.
6. Heating or cooling: air conditioning at 100% is 4,800 VA. Central electric heat at 65% is 15,000 × 0.65 = 9,750 VA. Use the larger: 9,750 VA.
7. EV charger, separately at 100% of nameplate: 7,680 VA.
8. Total: 17,960 + 9,750 + 7,680 = 35,390 VA.
9. Amps: 35,390 ÷ 240 = 147.46 A, about 147.5 A. The next standard rating at or above that is 150 A. The load calculation sets the minimum; the service equipment, conductors and overcurrent protection still have to meet their own NEC rules.

Check: rebuild the subtotal another way. 32,900 × 0.40 = 13,160, plus the 60% of the first 8,000 that stays in (4,800), gives 17,960. Same subtotal. Add the other two lines in a different order: 7,680 + 9,750 = 17,430, plus 17,960 gives 35,390 VA. Same total.

Worked example 2: service conductors for that house

For a 120/240 V single-phase dwelling service, the code allows service and main feeder conductors with an ampacity of at least 83% of the service rating (310.12(A)).

1. Service rating from example 1: 150 A.
2. Minimum conductor ampacity: 150 × 0.83 = 124.5 A.
3. Choose a conductor from the ampacity table with at least 124.5 A at the terminal temperature rating.

Check: 124.5 ÷ 0.83 = 150. The 83% rule applies only to the conductors that carry the whole dwelling load, not to feeders for a subpanel serving part of it.

Worked example 3: commercial feeder with continuous loads and a motor

A 208Y/120 V three-phase feeder supplies a small retail space. The question asks for the minimum feeder conductor ampacity:

• 18,000 VA of lighting that runs more than 3 hours (continuous)
• 9,000 VA of general receptacles (noncontinuous)
• one 10 hp, 208 V, three-phase motor with a table full-load current of 30.8 A

1. Lighting at 125% for conductor sizing: 18,000 × 1.25 = 22,500 VA. The 125% comes from the feeder conductor rules in Article 215; the Article 120 load total itself counts the lighting at 18,000 VA.
2. Receptacles: 9,000 VA. Receptacle loads up to 10 kVA stay at 100% (120.14(I) and Table 120.47).
3. Non-motor total: 22,500 + 9,000 = 31,500 VA.
4. Non-motor amps: 31,500 ÷ 360.3 = 87.4 A.
5. Motor at 125% (it is the largest and only motor): 30.8 × 1.25 = 38.5 A.
6. Minimum feeder conductor ampacity: 87.4 + 38.5 = 125.9 A.

Check: compute the non-motor amps another way. 31,500 ÷ 208 ÷ 1.732 = 87.44 A. Add 38.5 and you get 125.94 A, so 125.9 A is right.

Where candidates lose points

• Applying demand factors to the wrong loads. The 35% and 25% steps in the standard method apply to the lighting, small-appliance and laundry group only.
• Forgetting the furnace percentage. In the optional method, central electric heat goes in at 65%, not 100%.
• Putting the EV charger in the 40% group. In the 2026 optional method it is added separately at 100% of nameplate.
• Mixing methods. Do not take demand factors from the standard method and apply them inside the optional method.
• Using 240 for three-phase. A 208 V three-phase load divides by 360.3.
• Missing the 125%. When sizing feeder conductors, continuous loads and the largest motor both add 25%. The Article 120 load total itself does not multiply continuous loads.

Show the method and supplied factors before the total

Write the requested result before calculating: connected load, calculated load, current, conductor selection, or protective-device selection. Then list each supplied input with units. Do not combine methods or apply a factor twice because two worksheets use different layouts.

For this original arithmetic exercise only, the problem supplies a 120 V resistive load with 1,800 W in one group and 1,200 W in another. It explicitly directs a factor of 1.25 on the second group and 1.00 on the first. The calculated total is 1,800 + (1,200 × 1.25) = 3,300 W. At the supplied 120 V, that is 27.5 A.

Those factors are given exercise data, not a statement that every service or feeder uses that treatment. A real 2026 NEC problem needs its applicable method, eligible loads, demand factors, and all relevant equipment and conductor checks. Keep the code reference beside each factor you select.

Load-method arithmetic practice

1. TX-160: What is the total using the factors expressly supplied by the exercise?
   Answer: 3,300 W.
2. TX-160: What is the resulting current at the supplied 120 V for this resistive example?
   Answer: 27.5 A: 3,300 W divided by 120 V.
3. TX-160: Does 27.5 A by itself select a real feeder conductor or breaker?
   Answer: No. The applicable code method and selection requirements remain separate.

Official references checked October 16, 2026

• U.S. Department of Energy: Electrical Science, Volume 1
• NFPA 70 official standard page

Check yourself

Our own practice questions. Answers below each.

1. General load total 30,000 VA under the optional method, no heating or cooling. Demand load? Answer: 8,000 + (22,000 × 0.40) = 16,800 VA.
2. A 200 A dwelling service under the 83% rule. Minimum conductor ampacity? Answer: 200 × 0.83 = 166 A.
3. 14,400 VA of continuous lighting at 208 V three-phase. Minimum feeder conductor ampacity for that load? Answer: 14,400 × 1.25 = 18,000 VA for conductor sizing; 18,000 ÷ 360.3 = 49.96 A, about 50.0 A. The load itself is 14,400 ÷ 360.3 = 39.97 A; the 125% applies only when you size the conductors.

FAQ

Sources checked October 2026

• TDLR rule adoption notice, 2026 NEC: https://www.tdlr.texas.gov/news/rulemaking/2026/09/01/commission-adopts-rules-12/
• PSI Candidate Information Bulletin for TDLR (updated 9/15/2026): https://test-takers.psiexams.com/api/content/bulletin/1682
• City of Denison, 2026 NEC significant changes: https://denisontx.gov/DocumentCenter/View/4131/2026-National-Electrical-Code-Significant-Changes
• IAEI Magazine, NEC 2026 significant code changes (Article 220 renumbered as 120): https://iaeimagazine.org/standards/nec-2026-significant-code-changes/
• NFPA 70 (NEC), 2026 edition, Articles 120, 310 and 430, read in your own copy: https://www.nfpa.org/codes-and-standards/nfpa-70-standard-development/70