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Free Advanced NEC Calculations Practice — NEC 2023 practice questions (2026)

What is the format? These ten original multiple-choice questions introduce Advanced NEC Calculations Practice — NEC 2023 study topics. They are practice material, not actual exam items. Official assessments may use different formats; check the current candidate guide for your credential.

What does this prepare me for? General code practice. Not aligned to any specific state exam blueprint. Check your licensing authority’s current bulletin. CodeReady does not confer a license.

What should I study? Start with load calculations, transformers and motors, grounding and conductors. This bank uses NEC 2023. Check the current exam objectives, adopted editions, and permitted references before planning your study.

How should I practice? Choose an answer before opening the explanation. Read the cited reference and revisit unfamiliar topics. These samples are general preparation, not jurisdiction-specific advice.

Explore full Advanced NEC Calculations Practice — NEC 2023 →

1 / Load calculations

Under the standard dwelling calculation, the general lighting, small-appliance, and laundry loads total 12,000 VA before demand factors. What is their demand load using 100% of the first 3,000 VA and 35% of the remainder?

  1. 4,200 VA
  2. 7,200 VA
  3. 12,000 VA
  4. 6,150 VA
Show answer & explanation

6,150 VA

Apply the dwelling demand factor to the combined eligible load: 3,000 + (9,000 × 0.35) = 6,150 VA. This does not apply the same factor to every load in the dwelling.

NEC 2023 Table 220.45, Dwelling units row; 220.52(A), (B)

2 / Load calculations

A dwelling's undemanded lighting/small-appliance/laundry subtotal is 13,200 VA. It includes three small-appliance circuits and one laundry circuit at 1,500 VA each. At 3 VA/ft², what calculated dwelling floor area does the worksheet imply?

  1. 2,900 ft²
  2. 2,400 ft²
  3. 3,400 ft²
  4. 4,400 ft²
Show answer & explanation

2,400 ft²

Remove circuit allowances: 13,200 − (3 + 1) × 1,500 = 7,200 VA of lighting. Floor area = 7,200/3 = 2,400 ft². This reverse check can detect a missing circuit allowance.

NEC 2023 220.41; 220.52(A), (B)

3 / Load calculations

A dwelling has a 4.2 kW dryer and a 6.2 kW dryer. For two dryers use a 100% demand factor after applying the per-dryer minimum. What total dryer load enters the standard calculation?

  1. 10.4 kW
  2. 11.2 kW
  3. 10 kW
  4. 8.32 kW
Show answer & explanation

11.2 kW

The individual bases are max(4.2, 5) = 5 kW and max(6.2, 5) = 6.2 kW. Total = (5 + 6.2) × 1.00 = 11.2 kW.

NEC 2023 220.54; Table 220.54

4 / Load calculations

A sign outlet is required, but no sign is selected. Using the minimum 1,200 VA per required sign branch circuit, what load is reserved for two such circuits?

  1. 1,200 VA
  2. 3,000 VA
  3. 2,400 VA
  4. 600 VA
Show answer & explanation

2,400 VA

Two required circuits contribute 2 × 1,200 = 2,400 VA before any additional continuous-load sizing treatment.

NEC 2023 220.14(F)

5 / Load calculations

A feeder's adjusted conductor ampacity is exactly 100 A. Its calculated noncontinuous load is 98 A. With no special equipment exception, may the ordinary next-size rule support a 110 A breaker?

  1. Yes; every conductor may use one higher rating
  2. No; 100 A is already a standard rating
  3. Yes; the load is below 100 A
  4. Yes; all ratings below 800 A can be increased
Show answer & explanation

No; 100 A is already a standard rating

The load fits: 98 ≤ 100 A. However, 100 A is a standard size, so the nonstandard-ampacity condition of 240.4(B) is not met. The stated facts do not authorize 110 A protection.

NEC 2023 240.4(B); 240.6(A); Table 240.6(A)

6 / Transformers and motors

A 150 kVA transformer supplies 96 kW at a measured 0.80 power factor. What percentage of its kVA rating is being used?

  1. 64%
  2. 96%
  3. 125%
  4. 80%
Show answer & explanation

80%

Load apparent power = 96/0.80 = 120 kVA. Loading = 120/150 × 100 = 80%. Compare kVA with kVA, not kW with kVA.

Trade math — kVA loading; NEC 2023 450.3 context only

7 / Transformers and motors

A continuous-duty 5 hp, 230 V single-phase motor has a Table 430.248 current of 28 A. Its 90°C branch conductors will be subject to a combined 0.70 adjustment/correction factor. What minimum 90°C base ampacity is needed to retain the required motor ampacity?

  1. 35 A
  2. 40 A
  3. 50 A
  4. 62.5 A
Show answer & explanation

50 A

Motor requirement = 28 × 1.25 = 35 A. Required insulation-column base ampacity = 35/0.70 = 50 A. Termination limits must also allow at least 35 A.

NEC 2023 430.22 introductory paragraph; Table 430.248; 310.15

8 / Transformers and motors

A motor feeder serves two continuous-duty motors. Motor A has table FLC 54 A and nameplate current 50 A; motor B has table FLC 40 A and nameplate current 57 A. What minimum motor-feeder conductor ampacity follows from the correct largest-current comparison?

  1. 104 A
  2. 117.5 A
  3. 107.5 A
  4. 94 A
Show answer & explanation

107.5 A

Compare applicable table currents, not the nameplate currents used for overloads. The largest table current is 54 A: 54 × 1.25 + 40 = 107.5 A. Motor B's larger nameplate current does not determine the feeder's largest-motor addition.

NEC 2023 430.24

9 / Grounding and conductors

A 400 A feeder's copper EGC is initially 3 AWG, 52,620 circular mils. Voltage-drop upsizing doubles the phase-conductor circular-mil area, with no exception. Supplied areas are 1 AWG = 83,690 and 1/0 = 105,600 circular mils. What is the minimum of those EGC sizes?

  1. 1 AWG copper
  2. 1/0 AWG copper
  3. 2 AWG copper
  4. 3 AWG copper
Show answer & explanation

1/0 AWG copper

Required EGC area = 52,620 × 2 = 105,240 circular mils. 1 AWG's 83,690 is insufficient; 1/0's 105,600 meets the area requirement.

NEC 2023 250.122(B); Table 250.122; Chapter 9, Table 8

10 / Grounding and conductors

Three parallel phase-conductor sets each provide 180 A final allowable ampacity after all corrections. The continuous service load is 400 A, with ordinary equipment. What ampacity margin remains after the required 125% load calculation?

  1. 140 A
  2. 40 A
  3. 100 A
  4. 180 A
Show answer & explanation

40 A

Combined conductor ampacity = 3 × 180 = 540 A. Required service ampacity = 400 × 1.25 = 500 A. Margin = 540 − 500 = 40 A. Proper parallel installation is assumed.

NEC 2023 230.42(A)(1); 310.10(G)

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