Castleport Test Prep

Electrical Load Calculation Practice Problems

24 original, unofficial electrical load calculation practice problems with worked answers, from basic VA and three-phase math to dwelling calculation exercises. Code choices use the 2023 NEC unless an item says otherwise; changed 2026 NEC answers appear in the worked solutions.

Electrical math (problems 1–4)

1. Convert a single-phase load to amps

LC-01 · Any edition

A single-phase load draws 5.4 kVA at 240 V. What is the load current, to the nearest tenth of an ampere?

  • A. 13.0 A
  • B. 18.0 A
  • C. 22.5 A
  • D. 45.0 A
Show worked answer

Answer: C — 22.5 A

  1. Convert kVA to VA: 5.4 kVA = 5,400 VA.
  2. Single-phase current: I = VA ÷ V = 5,400 ÷ 240 = 22.5 A.

Why not the others: A divides by 240 × √3, which is the three-phase formula. B multiplies by 0.8 first; this load is already given in VA, so there's no power factor to apply. D uses 120 V.

Source: DOE Electrical Science Handbook, Vol. 3, Module ES-09 (pp. ES-09-2, -4, -20, -21)

2. Watts are not volt-amperes

LC-02 · Any edition

A single-phase load uses 3,600 W of real power at a power factor of 0.80 on a 240 V supply. What current does it draw?

  • A. 12.0 A
  • B. 15.0 A
  • C. 18.75 A
  • D. 23.4 A
Show worked answer

Answer: C — 18.75 A

  1. Apparent power: VA = W ÷ PF = 3,600 ÷ 0.80 = 4,500 VA.
  2. Current: 4,500 ÷ 240 = 18.75 A.

Why not the others: B divides watts by volts and ignores the power factor. A multiplies by 0.80 instead of dividing. D adds a 125% factor the question never asked for.

Source: DOE Electrical Science Handbook, Vol. 3, Module ES-09 (pp. ES-09-2, -4, -20, -21)

3. Three-phase current

LC-03 · Any edition

A 208Y/120 V, 3-phase service has a balanced calculated load of 150,000 VA. What is the line current, to the nearest ampere?

  • A. 416 A
  • B. 625 A
  • C. 721 A
  • D. 1,250 A
Show worked answer

Answer: A — 416 A

  1. Balanced three-phase current: I = VA ÷ (V × √3), using line-to-line voltage.
  2. 150,000 ÷ (208 × √3) = 416.36… → 416 A.

Why not the others: C divides by 208 without √3 (a single-phase shortcut). B and D divide by 240 V and 120 V, which aren't this system's line-to-line voltage.

Source: DOE Electrical Science Handbook, Vol. 3, Module ES-09 (pp. ES-09-2, -4, -20, -21)

4. Add loads at different voltages

LC-04 · Any edition

Three resistive loads (power factor 1.0) run at the same time: 18 A at 240 V, 12 A at 120 V, and 8 A at 120 V. What is their total load in volt-amperes?

  • A. 4,560 VA
  • B. 6,720 VA
  • C. 9,120 VA
  • D. 2,400 VA
Show worked answer

Answer: B — 6,720 VA

  1. Convert each load to VA: 240 × 18 = 4,320; 120 × 12 = 1,440; 120 × 8 = 960.
  2. Total: 4,320 + 1,440 + 960 = 6,720 VA.

Why not the others: A and C add the amps first (18 + 12 + 8 = 38 A) and multiply by one voltage. Amps at different voltages can't be added that way; convert each load to VA first. D counts only the two 120 V loads.

Source: DOE Electrical Science Handbook, Vol. 3, Module ES-09 (pp. ES-09-2, -4, -20, -21)

Dwelling standard method (problems 5–17)

5. General lighting demand

LC-05 · 2023 key; 2026 answer shown

A 2,200 ft² one-family dwelling has the two required small-appliance circuits and one laundry circuit. Using the 2023 NEC, what is the minimum demand load for general lighting, small-appliance, and laundry loads combined?

  • A. 3,885 VA
  • B. 5,835 VA
  • C. 8,100 VA
  • D. 11,100 VA
Show worked answer

Answer: B — 5,835 VA

  1. Lighting: 2,200 ft² × 3 VA/ft² = 6,600 VA (220.41).
  2. Small-appliance and laundry: 3 × 1,500 = 4,500 VA (220.52).
  3. Total 11,100 VA. Table 220.45: first 3,000 at 100% = 3,000; remaining 8,100 × 35% = 2,835.
  4. 3,000 + 2,835 = 5,835 VA.

Why not the others: D skips the demand factor. C is only the part above 3,000 VA. A applies 35% to everything, including the first 3,000 VA.

2026 NEC: Lighting is 2,200 ft² × 2 VA/ft² = 4,400 VA (120.41). Total 8,900 → 3,000 + 5,900 × 35% = 5,065 VA; this value is not one of the 2023 choices above.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

6. Count the circuit allowances

LC-06 · 2023 and 2026 NEC

A dwelling has three small-appliance branch circuits and one laundry branch circuit, all 2-wire circuits covered by 220.52 (2023) or 120.52 (2026). What load do these circuits add before demand factors?

  • A. 4,500 VA
  • B. 6,000 VA
  • C. 7,200 VA
  • D. 9,600 VA
Show worked answer

Answer: B — 6,000 VA

  1. Each small-appliance and laundry circuit counts 1,500 VA (220.52(A) and (B) in 2023; 120.52(A) and (B) in 2026).
  2. 4 circuits × 1,500 = 6,000 VA.

Why not the others: A uses the minimum of two small-appliance circuits instead of the three actually installed. C and D multiply breaker amps by voltage (15 A or 20 A × 120 V); the Code assigns 1,500 VA per circuit instead.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

7. Fastened-in-place appliances

LC-07 · 2023 and 2026 NEC

A dwelling service supplies these fastened-in-place appliances: 4,500 W resistive water heater, 1,200 VA dishwasher, 900 VA disposal, and 1,500 VA trash compactor. Each appliance is rated at least ¼ hp or 500 W and qualifies for the appliance demand factor. What is their minimum permitted demand load?

  • A. 4,050 VA
  • B. 5,670 VA
  • C. 6,075 VA
  • D. 8,100 VA
Show worked answer

Answer: C — 6,075 VA

  1. There are four qualifying appliances served by the same service, so the permitted 75% factor applies (220.53 in 2023; 120.53 in 2026).
  2. 4,500 + 1,200 + 900 + 1,500 = 8,100 × 0.75 = 6,075 VA.

Why not the others: D skips the permitted factor. B uses 70%, which isn't the Code's number. A uses 50%.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

8. A dryer doesn't make four

LC-08 · 2023 and 2026 NEC

A dwelling service supplies a 4,500 W resistive water heater, 1,200 VA dishwasher, 900 VA disposal, and a 5,000 W clothes dryer. The first three are qualifying fastened-in-place appliances, each rated at least ¼ hp or 500 W. What load goes on the fastened-in-place appliance line?

  • A. 4,950 VA
  • B. 6,600 VA
  • C. 8,700 VA
  • D. 11,600 VA
Show worked answer

Answer: B — 6,600 VA

  1. Clothes dryers are excluded from the appliance group in 220.53 (2023) and 120.53 (2026).
  2. That leaves three appliances, so no 75% factor: 4,500 + 1,200 + 900 = 6,600 VA. The dryer goes in separately under 220.54 (2023) or 120.54 (2026).

Why not the others: C takes 75% of all four, counting the dryer. A takes 75% of three appliances. D adds everything at 100% on one line.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

9. A small fan doesn't make four either

LC-09 · 2023 and 2026 NEC

A dwelling service supplies three qualifying fastened-in-place appliances rated 4,800 VA, 1,400 VA, and 900 VA. It also has a bathroom exhaust fan rated 60 W and 1/20 hp, already covered by the general lighting load. What goes on the fastened-in-place appliance line?

  • A. 5,325 VA
  • B. 5,370 VA
  • C. 7,100 VA
  • D. 7,160 VA
Show worked answer

Answer: C — 7,100 VA

  1. 220.53 (2023) and 120.53 (2026) count only appliances rated ¼ hp or more, or 500 W or more. The fan meets neither threshold.
  2. Three qualifying appliances means no 75% factor: 4,800 + 1,400 + 900 = 7,100 VA. The fan isn't added again.

Why not the others: B counts the fan as the fourth appliance and takes 75%. A applies 75% to only three appliances. D adds the fan's 60 W a second time.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

10. Range rated over 12 kW

LC-10 · 2023 and 2026 NEC

Using Column C and Note 1 of the standard-method range table, what is the minimum permitted feeder demand load for one 15.6 kW household range?

  • A. 8.0 kW
  • B. 9.2 kW
  • C. 9.6 kW
  • D. 15.6 kW
Show worked answer

Answer: C — 9.6 kW

  1. Table 220.55 (2023) or Table 120.55 (2026), Column C, one range: 8 kW.
  2. For this range rated over 12 kW and not over 27 kW, Note 1 adds 5% for each kW, or major fraction of a kW, above 12 kW. 15.6 − 12 = 3.6 kW, which counts as 4.
  3. 8 × (1 + 4 × 0.05) = 8 × 1.20 = 9.6 kW.

Why not the others: B counts 3.6 kW as 3. A forgets Note 1. D uses the nameplate.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

11. The dryer minimum

LC-11 · 2023 and 2026 NEC

One household electric dryer has a 4,800 W nameplate. Using the standard method, what minimum dryer load goes into the calculation?

  • A. 3,840 VA
  • B. 4,800 VA
  • C. 5,000 VA
  • D. 6,000 VA
Show worked answer

Answer: C — 5,000 VA

  1. 220.54 (2023) or 120.54 (2026): use the nameplate or 5,000 W (VA), whichever is larger.
  2. 5,000 > 4,800, so 5,000 VA.

Why not the others: B uses the nameplate even though it's below the minimum. A takes 80% of the nameplate. D multiplies the 4,800 W nameplate by 125%, rather than applying the dryer-load rule asked for here.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

12. Four dryers on one feeder

LC-12 · 2023 key; 2026 answer differs

A 120/240 V single-phase feeder supplies four household dryers rated 4,800 W, 5,200 W, 5,800 W, and 6,200 W. Using the 2023 NEC dryer table, what is the minimum permitted dryer demand load?

  • A. 17,760 VA
  • B. 18,870 VA
  • C. 22,000 VA
  • D. 22,200 VA
Show worked answer

Answer: D — 22,200 VA

  1. Apply the 5,000 W minimum to each dryer first: 5,000 + 5,200 + 5,800 + 6,200 = 22,200 VA.
  2. 2023 Table 220.54: 1 to 4 dryers = 100%. Demand = 22,200 VA.

Why not the others: C skips the 5,000 W minimum on the first dryer. B uses 85%, the 2023 factor for five dryers. A is the 2026 answer.

2026 NEC: Table 120.54 changed: 3 to 5 dryers = 80%. 22,200 × 0.80 = 17,760 VA (choice A).

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

13. Thirteen dryers

LC-13 · 2023 and 2026 NEC

A single-phase feeder supplies thirteen household dryers, each rated 5,500 W. Using the dryer demand table, what is the minimum permitted dryer demand load?

  • A. 24,310 VA
  • B. 32,175 VA
  • C. 33,605 VA
  • D. 71,500 VA
Show worked answer

Answer: B — 32,175 VA

  1. Connected load: 13 × 5,500 = 71,500 VA.
  2. The 12–23 dryer row: 47% minus 1 percentage point for each dryer over 11. Thirteen is 2 over, so 47 − 2 = 45%.
  3. 71,500 × 0.45 = 32,175 VA.

Why not the others: C uses 47% without the reduction. A subtracts 13 points (34%) instead of 2. D applies no demand factor.

Source: NFPA 2026 NEC Handbook replacement page: 120.52–120.54 and corrected Table 120.54; NFPA 70, 2023 NEC (free-access edition; section locators above)

14. Dryer neutral load

LC-14 · 2023 and 2026 NEC

A 5.5 kW, 120/240 V dryer is the only dryer in the dwelling. Using the permitted 70% neutral reduction, what minimum dryer neutral load does it add to the standard-method service calculation?

  • A. 3,500 VA
  • B. 3,850 VA
  • C. 5,000 VA
  • D. 5,500 VA
Show worked answer

Answer: B — 3,850 VA

  1. Line load: the larger of 5,000 W or nameplate = 5,500 VA (220.54 in 2023; 120.54 in 2026).
  2. The neutral may be taken at 70% of that load (220.61(B)(1) in 2023; 120.61(B)(1) in 2026): 5,500 × 0.70 = 3,850 VA.

Why not the others: A takes 70% of the 5,000 W minimum instead of the larger nameplate. D is the line load. C is the minimum with no neutral reduction.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

15. Heat or air conditioning

LC-15 · 2023 and 2026 NEC

A dwelling has 10 kW of resistive electric baseboard heat and a 5,000 VA air-conditioning comparison load, including any applicable motor allowance. Controls prevent the heating and cooling loads from operating simultaneously. Using the standard method, what heating/cooling load goes into the service calculation?

  • A. 5,000 VA
  • B. 6,500 VA
  • C. 10,000 VA
  • D. 15,000 VA
Show worked answer

Answer: C — 10,000 VA

  1. These stated heating and cooling loads cannot run at the same time, so only the larger is used (220.60 in 2023; the noncoincident-load provisions in Article 120 in 2026).
  2. Fixed electric space heat counts at 100% (220.51 in 2023; 120.51 in 2026): 10,000 VA.

Why not the others: D adds both. B applies a 65% factor from the optional method, which doesn't belong in the standard method. A picks the smaller load.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

16. Standard-method subtotal to amps

LC-16 · 2023 key; 2026 answer shown

A 1,800 ft² one-family dwelling, 120/240 V single-phase, has two small-appliance circuits and one laundry circuit; a 12 kW range; a 5 kW dryer; fastened-in-place appliances of 4,500 W resistive water heater, 1,400 VA dishwasher, 800 VA disposal, and 1,100 VA trash compactor; a 6,000 VA air conditioner; and an 8 kW resistive electric furnace. The four fastened-in-place appliances each qualify for the 75% factor. Controls prevent the furnace and A/C from operating simultaneously. Treat all loads as given, with no separate motor adder. This exercise isolates demand factors and does not determine final service or conductor size. Using the 2023 NEC factors, what is this calculated subtotal in amperes, rounded to the nearest ampere?

  • A. 134 A
  • B. 143 A
  • C. 151 A
  • D. 159 A
Show worked answer

Answer: A — 134 A

  1. Lighting 1,800 ft² × 3 VA/ft² = 5,400 + small-appliance/laundry 4,500 = 9,900 VA → 3,000 + 6,900 × 35% = 5,415 VA (220.41, 220.52, Table 220.45).
  2. Fastened-in-place appliances: 7,800 × 75% = 5,850 VA (220.53).
  3. Range, Column C, one 12 kW range: 8,000 VA (Table 220.55).
  4. Dryer: 5,000 VA (220.54).
  5. Larger of heat 8,000 or A/C 6,000: 8,000 VA (220.60).
  6. Total: 5,415 + 5,850 + 8,000 + 5,000 + 8,000 = 32,265 VA.
  7. 32,265 ÷ 240 = 134.4375 → 134 A, rounded as the question asks. Keep the unrounded value for any subsequent calculation.

Why not the others: B skips the 75% appliance factor. C uses the range nameplate (12 kW) instead of Column C. D adds the heat and A/C together.

2026 NEC: Lighting is 1,800 ft² × 2 VA/ft² = 3,600 VA, so the first line becomes 4,785 VA. Total 31,635 VA ÷ 240 = 131.8125 → 132 A. This is the same bounded subtotal, not a final equipment-size selection; 132 A is not one of the 2023 choices above.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

17. EV charger load (2023 NEC)

LC-17 · 2023 only

Under the 2023 NEC standard method, what load value does 220.57 assign to an electric vehicle supply equipment (EVSE) unit with a 5,800 VA nameplate? No energy-management reduction applies. This asks only for the 220.57 load value, not the separate branch-circuit, conductor, or overcurrent-device sizing calculation.

  • A. 5,800 VA
  • B. 7,200 VA
  • C. 7,250 VA
  • D. 9,000 VA
Show worked answer

Answer: B — 7,200 VA

  1. 220.57: use 7,200 W (VA) or the nameplate, whichever is larger.
  2. 7,200 > 5,800, so 7,200 VA.

Why not the others: A ignores the 7,200 VA floor. C multiplies the 5,800 VA nameplate by 125%; D multiplies the 7,200 VA load value by 125%. Neither is the load value requested under 220.57.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above)

One-family dwelling: optional method (problems 18–23)

18. Optional-method demand on the general loads

LC-18 · 2023 key; 2026 answer shown

A dwelling qualifies for the optional method. Its general loads under 220.82(B) total 24,600 VA (heating, air conditioning, and any EVSE are not included in that number). Using the 2023 NEC tiers, what is the demand load on those general loads?

  • A. 9,840 VA
  • B. 14,640 VA
  • C. 15,840 VA
  • D. 24,600 VA
Show worked answer

Answer: C — 15,840 VA

  1. 220.82(B): first 10,000 VA at 100%, the rest at 40%.
  2. 10,000 + (24,600 − 10,000) × 0.40 = 10,000 + 5,840 = 15,840 VA.

Why not the others: A applies 40% to everything. D applies no demand factor. B is the 2026 answer.

2026 NEC: For the same stated 24,600 VA subtotal, the first tier is 8,000 VA: 8,000 + 16,600 × 0.40 = 14,640 VA (choice B).

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

19. A dryer in the optional method

LC-19 · 2023 and 2026 NEC

A dwelling using the optional method has a dryer with a 4,600 W nameplate; use power factor 1.0 for this exercise. What value goes into the 220.82(B) (2023) or 120.82(B) (2026) general loads for this dryer, before the demand tiers are applied?

  • A. 3,680 VA
  • B. 4,600 VA
  • C. 5,000 VA
  • D. Nothing; dryers are added separately
Show worked answer

Answer: B — 4,600 VA

  1. The optional method collects the nameplate rating of appliances, dryers included, in the general loads (220.82(B)(3) in 2023; 120.82(B)(3) in 2026).
  2. So the dryer enters at 4,600 VA, then rides through the same 100%/40% tiers as everything else.

Why not the others: C is the standard-method dryer rule (220.54), which the optional method doesn't use. A applies an extra factor before the tiers. D confuses the optional method with the standard method's separate dryer line.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

20. Don't discount twice

LC-20 · 2023 key; 2026 answer shown

In an optional-method calculation, the general loads include 12,000 VA of lighting, small-appliance, and laundry loads plus 10,000 VA of fastened-in-place appliance nameplates. There are no heating, A/C, or EVSE loads. Using the 2023 NEC tiers, what is the correct demand load?

  • A. 13,600 VA
  • B. 13,800 VA
  • C. 14,800 VA
  • D. 22,000 VA
Show worked answer

Answer: C — 14,800 VA

  1. Add the nameplates at 100%: 12,000 + 10,000 = 22,000 VA.
  2. 10,000 + 12,000 × 0.40 = 14,800 VA.

Why not the others: B first cuts the appliances to 75% (7,500 VA) and then applies the tiers: 10,000 + 9,500 × 0.40 = 13,800 VA. The 75% factor is a standard-method rule, so that understates the load by 1,000 VA. D applies no demand factor. A is the 2026 answer.

2026 NEC: For the same stated 22,000 VA subtotal: 8,000 + 14,000 × 0.40 = 13,600 VA (choice A). The double-discount mistake gives 12,600 VA, again 1,000 VA low.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

21. Optional-method calculation to amps

LC-21 · 2023 key; 2026 answer shown

Use the same appliance scenario as LC-16: a 1,800 ft² dwelling supplied by one 120/240 V, single-phase, 3-wire service rated at least 100 A; two small-appliance circuits and one laundry circuit; a 12 kW range; a 5 kW dryer; a 4,500 W resistive water heater, 1,400 VA dishwasher, 800 VA disposal, and 1,100 VA trash compactor; a 6,000 VA air conditioner; and one 8 kW central electric furnace. Use power factor 1.0 for the watt-rated nameplates in this exercise. The stated HVAC values include the complete equipment loads; there is no heat pump, thermal-storage heating, EVSE, or other load. The dwelling qualifies for the optional method (220.82). Using the 2023 NEC, what is the calculated load current, rounded to the nearest ampere?

  • A. 105 A
  • B. 108 A
  • C. 130 A
  • D. 170 A
Show worked answer

Answer: B — 108 A

  1. General loads at nameplate: lighting 5,400 + small-appliance/laundry 4,500 + range 12,000 + dryer 5,000 + water heater 4,500 + dishwasher 1,400 + disposal 800 + compactor 1,100 = 34,700 VA.
  2. Tiers: 10,000 + 24,700 × 0.40 = 19,880 VA.
  3. Heating/cooling, larger of: A/C 6,000 × 100% = 6,000, or the one central electric heating unit 8,000 × 65% = 5,200 → 6,000 VA.
  4. Total 25,880 VA ÷ 240 = 107.833… → 108 A, rounded as the question asks; this does not select a service rating or conductor size.

Why not the others: A uses the heat (5,200 VA) instead of the larger A/C. C adds both. D applies no demand factor. In this example, the optional-method result is lower than the bounded standard-method subtotal in LC-16.

2026 NEC: Lighting 3,600 → general loads 32,900 → 8,000 + 24,900 × 0.40 = 17,960 → + 6,000 = 23,960 VA ÷ 240 = 99.833… → 100 A, rounded as requested. This value is not one of the 2023 choices above.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

22. Heat with four or more units

LC-22 · 2023 and 2026 NEC

A dwelling using the optional method has six separately controlled 1,500 W resistive baseboard heaters and a 5,000 VA air conditioner. There is no heat pump or thermal-storage heating. What heating/cooling value goes into the calculation?

  • A. 3,600 VA
  • B. 5,000 VA
  • C. 5,850 VA
  • D. 9,000 VA
Show worked answer

Answer: B — 5,000 VA

  1. Four or more separately controlled heating units: 40% of 9,000 = 3,600 VA.
  2. Air conditioning: 100% = 5,000 VA.
  3. Use the larger: 5,000 VA.

Why not the others: A is the heat value, which is smaller. C uses 65%, the factor for fewer than four units. D uses the heat at 100%.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above); NFPA 70, 2026 NEC (free-access edition; section locators above)

23. EV charger in the optional method (2026 NEC only)

LC-23 · 2026 only

Using the 2026 NEC optional method, a qualifying dwelling's 120.82(B) general loads (lighting already at 2 VA/ft²) total 30,000 VA, excluding HVAC and EVSE. It has a 5,000 VA air conditioner, the larger heating/cooling selection, and an EVSE with an available 11,520 VA nameplate. No energy-management reduction applies. What is the calculated load current at 240 V, rounded to the nearest ampere?

  • A. 110 A
  • B. 139 A
  • C. 144 A
  • D. 194 A
Show worked answer

Answer: B — 139 A

  1. General loads: 8,000 + 22,000 × 0.40 = 16,800 VA (120.82(B)).
  2. A/C: 5,000 VA (120.82(C)).
  3. EVSE at 100% of nameplate, outside the tiers: 11,520 VA (120.82(D)).
  4. Total 33,320 ÷ 240 = 138.8 → 139 A.

Why not the others: A puts the EVSE inside the 40% tier. C uses the old 10,000 VA first tier. D applies no demand factor.

Source: NFPA 70, 2026 NEC (free-access edition; section locators above)

Feeder basics (problem 24)

24. Continuous and noncontinuous feeder load

LC-24 · 2023 NEC

A feeder supplies 60 A of continuous load and 40 A of noncontinuous load. The assembly is not listed for operation at 100% of its rating, and no other exception applies. Under 2023 NEC 215.2(A)(1), what minimum ampacity must its ungrounded conductors have for this load calculation, before comparing adjustment or correction requirements?

  • A. 100 A
  • B. 115 A
  • C. 125 A
  • D. 140 A
Show worked answer

Answer: B — 115 A

  1. Feeder conductors: noncontinuous load at 100% plus continuous load at 125% (215.2(A)(1) in the 2023 NEC).
  2. 60 × 1.25 = 75; 75 + 40 = 115 A.

Why not the others: C applies 125% to the whole load. A ignores the continuous-load rule. D counts the 40 A noncontinuous load twice instead of adding 25% of the 60 A continuous load.

Source: NFPA 70, 2023 NEC (free-access edition; section locators above)

Review the step you missed

Rework the problem without peeking, then try its neighbor. A right total reached the wrong way is still worth another pass.

Review the step you missed
If you missed…The skillGo back to
1–4kVA to amps, power factor, three-phase currentFormulas below
5–6Building the general lighting group and its demand tiersProblem 5, problem 16
7–9Which appliances count toward the 75% factorProblem 7, problem 8, problem 9
10–14Range and dryer rules, including the neutralProblems 10–14
15Heat vs. air conditioningProblem 15, problem 21, problem 22
16Putting a standard-method subtotal togetherProblem 16
17, 23EV chargers and edition-specific calculation rulesProblem 17: 2023, problem 23: 2026
18–22Keeping the optional method separate from the standard methodProblem 19, problem 20
24Continuous vs. noncontinuous feeder loadProblem 24

These 24 problems are a focused practice set, not a full-length licensing exam. A count of correct answers describes your work on these items—not an official score or a prediction of passing. The service examples calculate stated loads; they do not select conductors, breakers, or a complete installation design.

Which NEC edition should you practice?

The one your exam names. An exam can switch editions on a different date than the installation code does.

Two real examples, checked September 29, 2026:

  • Texas: TDLR's electrician exam page names the 2026 NEC in its Reference Materials section (TDLR exam information).
  • Minnesota: the 2026 NEC took effect for permits on Aug. 17, 2026, and licensing exams began moving to the 2026 NEC on Sept. 8, 2026 (Minnesota DLI).

Check your candidate bulletin, not the year printed on a practice book. This page labels its 2023 answers as 2023; it does not present them as the current answer for every jurisdiction. LC-17 is a 2023-only exercise and LC-23 is a 2026-only exercise.

What changed in the 2026 NEC

These are the edition differences used on this page:

What changed in the 2026 NEC
Topic2023 NEC2026 NEC
Main load-calculation articleArticle 220Article 120; individual section numbers still need to be read in that edition
Dwelling feeder/service unit load3 VA/ft² (220.41)2 VA/ft² (120.41)
Unit load for counting branch circuits3 VA/ft²3 VA/ft² under 120.13—not the 2 VA/ft² feeder/service unit load
Dryer demand, 1–5 dryers1–4 dryers 100%; 5 dryers 85%1–2 dryers 100%; 3–5 dryers 80%
Optional method first tierFirst 10,000 VA at 100%First 8,000 VA at 100%
EVSE in the optional methodNo separate EVSE subsection in 220.82EVSE is added separately under 120.82(D), outside the general-load tiers; LC-23 uses the stated nameplate at 100%

Code references: NFPA 70, 2023 edition, Article 220; NFPA 70, 2026 edition, Article 120. The 2026 dryer percentages are shown in NFPA's March 27, 2026 replacement page, printed page 103.

Code values used in these problems

Code values used in these problems
RuleValue used hereCode locator
General lighting and receptacles, dwelling3 VA/ft² (2023) · 2 VA/ft² (2026)220.41 / 120.41
Lighting demand, dwellingFirst 3,000 VA at 100%; the portion over 3,000 VA through 120,000 VA at 35%. All examples here stay within those two tiersTable 220.45 / Table 120.45
Small-appliance and laundry circuits1,500 VA for each qualifying 2-wire circuit220.52 / 120.52
Fastened-in-place appliancesA permitted 75% factor for four or more qualifying appliances, each at least ¼ hp or 500 W, served by the same feeder or service. Do not count a dryer, heating load, or A/C toward these four220.53 / 120.53
DryersNameplate or 5,000 W (VA), whichever is larger, for each dryer; then the applicable dryer-table factor. For 12–23 dryers, subtract one percentage point from 47% for each dryer over 11220.54 / 120.54
RangesFor the one 12 kW range in LC-16, Column C gives 8 kW. For the 15.6 kW range in LC-10, Note 1 increases that by 5% per kW or major fraction over 12 kW. That note's rating range is over 12 through 27 kWTable 220.55 / Table 120.55, Column C and Note 1
Heat vs. A/C (standard method)Compare the applicable calculated loads when they are noncoincident; use the larger. The ordinary resistive fixed electric heat in these examples is included at 100%220.51 and 220.60 (2023); Article 120 heating and noncoincident-load provisions (2026)
Dryer neutralThe permitted reduction used in LC-14 is 70% of the dryer load calculated under the dryer table—not 70% of the whole service220.61(B)(1) / 120.61(B)(1)
Optional method, one dwellingGeneral loads: first tier at 100%, remainder at 40%. Compare the applicable heating/cooling selections separately. These examples use A/C at 100%, one central electric heating unit at 65%, or four or more separately controlled ordinary electric heating units at 40%; these are not universal heat-pump or thermal-storage factors220.82 / 120.82

Sources for the section locators: 2023 NEC and 2026 NEC. NFPA's 2026 replacement page reproduces 120.52–120.54, including their conditions and exclusions.

The optional method used here applies to a single dwelling unit supplied by one 3-wire 120/240 V or 208Y/120 V set of service or feeder conductors rated 100 A or more; see 220.82(A) or 120.82(A). The examples expressly assume that eligibility condition.

Where a question asks for rounding, round only its final displayed current. A rounded practice answer is not an instruction to round an installation's required equipment rating down.

Formulas

Formulas
FindSingle-phaseBalanced three-phase
Current from VAI = VA ÷ VI = VA ÷ (V × √3)
VA from wattsVA = W ÷ power factorVA = W ÷ power factor

In three-phase formulas, V is the line-to-line voltage (208 V on a 208Y/120 V system). Keep √3 in your calculator until the final step. Source: DOE Electrical Science Handbook, Module ES-09, pages 2, 4, 20–21.

Standard or optional method: which do you use?

Whichever the question asks for. When the optional method's eligibility conditions are met, it is an alternative calculation—not a rule to calculate both methods and pick the larger. In the supplied scenario, the bounded standard-method subtotal in problem 16 is 134 A and the optional-method result in problem 21 is 108 A under the 2023 NEC; neither number selects the final service equipment.

The one mistake to avoid: mixing them. The optional method takes appliances, ranges, and dryers at nameplate and applies only its own tiers, so the 75% appliance factor, the dryer minimum, and the range table don't belong inside it (problems 19 and 20). See 2023 NEC 220.82 or 2026 NEC 120.82.

Sources and verification

NFPA's code-reading service may require a free account; the complete practice questions and worked answers remain on this page.

Last verified: September 29, 2026, for the Texas and Minnesota exam-edition statements and NFPA's replacement page covering 120.52–120.54. The electrical-math formulas were checked against the DOE handbook, and the displayed answers were recalculated from the stated inputs. These checks are not a professional review of an electrical installation.

By Castleport Test Prep Editorial Team

Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by NFPA or any licensing board or testing vendor. Code, exam, and credential names identify their subjects; trademarks belong to their respective owners. These are original, unofficial practice problems, not actual exam questions, and they don't guarantee an exam result or license.