NEC Load Calculation Practice Problems (10 Examples)

Load calculations are the most calculation-heavy part of the electrician exam, and they appear on nearly every state's journeyman and master test. The key is not memorizing formulas, it's building a repeatable process you can execute quickly with NEC Article 220 open. These 10 worked examples cover the question types you'll actually see on the PSI and Pearson VUE exams.

Key Reference Points Before You Start

• NEC Table 220.12: General lighting load = 3 VA/sq ft for dwelling units
• NEC 220.52(A): Small appliance circuits = 1,500 VA each, minimum two required
• NEC 220.52(B): Laundry circuit = 1,500 VA
• NEC Table 220.42: Demand factors, first 3,000 VA at 100%, remainder at 35%
• NEC 220.54: Clothes dryer = 5,000 VA or nameplate, whichever is larger
• NEC 220.55 / Table 220.55: Electric ranges (see table for demand)
• NEC 220.53: Fixed appliances, 75% demand factor when 4 or more (not including dryer, range, A/C, heat)

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Problem 1: Basic Residential Service (Standard Method)

Given: A 1,800 sq ft single-family home with two 20A small appliance circuits, one laundry circuit, a 5,500W electric dryer, a 12,000W electric range, and a 4,000W central A/C unit. Calculate the minimum service size in amps at 240V single-phase.

Step 1, General lighting, small appliance, and laundry loads:

Step 2, Apply demand factor (Table 220.42):

• First 3,000 VA at 100% = 3,000 VA
• Remaining 6,900 VA at 35% = 2,415 VA
• Demand total = 5,415 VA

Step 3, Add fixed appliances at 100%:

• Dryer: 5,500 W (nameplate), but NEC 220.54 says minimum 5,000W, use 5,500W
• Range: Use Table 220.55. For one range rated 12 kW, Column C = 8,000 VA (demand)
• A/C: 4,000 VA at 100% (NEC 220.60, use larger of A/C or heat, not both)

Step 4, Total load: 5,415 + 5,500 + 8,000 + 4,000 = 22,915 VA

Step 5, Minimum service amps: 22,915 VA ÷ 240V = 95.5A → round up to 100A service (minimum)

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Problem 2: Larger Home with Heat Pump

Given: A 2,400 sq ft home with two small appliance circuits, one laundry circuit, a 5,000W dryer, a 12 kW range, and a 6 kW heat pump (heating dominant, A/C = 5 kW). Calculate minimum service size at 240V.

Lighting, SA, laundry subtotal:

• 2,400 × 3 = 7,200 + 3,000 + 1,500 = 11,700 VA
• Demand: 3,000 @ 100% + 8,700 @ 35% = 3,000 + 3,045 = 6,045 VA

Fixed loads:

• Dryer: 5,000 VA
• Range (12 kW, Column C): 8,000 VA
• Heat pump: 6,000 VA (heat is larger than A/C 5,000 VA, so use 6,000 VA per NEC 220.60)

Total: 6,045 + 5,000 + 8,000 + 6,000 = 25,045 VA

Service: 25,045 ÷ 240 = 104.4A → 125A service

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Problem 3: Demand Factor on Fixed Appliances

Given: A home has a garbage disposal (0.5 kW), dishwasher (1.2 kW), water heater (4.5 kW), trash compactor (1.0 kW), and a 2.5 kW microwave (built-in). What is the calculated load for these fixed appliances?

NEC 220.53: When 4 or more fixed appliances (not including dryer, range, A/C, or heat), apply 75% demand factor.

Total nameplate: 0.5 + 1.2 + 4.5 + 1.0 + 2.5 = 9,700 VA

9,700 VA × 75% = 7,275 VA

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Problem 4: Optional Method Calculation

Given: Same 1,800 sq ft home from Problem 1. Calculate the service size using the optional method (NEC 220.82).

NEC 220.82(B), "All other loads": Total nameplate of all loads:

• General lighting: 5,400 VA
• Small appliance + laundry: 4,500 VA
• Dryer: 5,500 VA
• Range: 12,000 VA
• A/C: 4,000 VA

Total = 31,400 VA

NEC 220.82(B) demand:

• First 10,000 VA at 100% = 10,000 VA
• Remaining 21,400 VA at 40% = 8,560 VA
• Total = 18,560 VA

Service: 18,560 ÷ 240 = 77.3A → 100A service

Note: The optional method gives a lower calculated load in this case (77.3A vs 95.5A), but both result in a 100A minimum service. On larger homes the difference is more significant.

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Problem 5: Multi-Unit Feeder with Demand Factors

Given: A 6-unit apartment building, each unit is 900 sq ft. Each unit has a range (12 kW each), two small appliance circuits, and a laundry circuit. Calculate the total feeder demand load for all six units at 240V.

Per-unit general load:

• Lighting: 900 × 3 = 2,700 VA
• Small appliance: 2 × 1,500 = 3,000 VA
• Laundry: 1,500 VA
• Per-unit subtotal: 7,200 VA

6 units total lighting/SA/laundry: 6 × 7,200 = 43,200 VA

NEC Table 220.42 demand on 43,200 VA:

• First 3,000 VA at 100% = 3,000
• Next 117,000 VA at 35% (we only have 40,200 remaining): 40,200 × 35% = 14,070
• Lighting/SA/laundry demand = 17,070 VA

Ranges (NEC Table 220.55, Column C): 6 ranges at 12 kW each Table 220.55, Column C: 6 ranges = 21 kW demand = 21,000 VA

Total feeder demand: 17,070 + 21,000 = 38,070 VA

Feeder amps: 38,070 ÷ 240 = 158.6A → 175A feeder

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Problem 6: Commercial Lighting Load

Given: A 20,000 sq ft office building. What is the calculated lighting load before demand factors?

NEC Table 220.12, Office buildings: 3.5 VA/sq ft (not 3 VA like residential)

20,000 sq ft × 3.5 VA/sq ft = 70,000 VA

Note: NEC Table 220.12 lists different unit load values by occupancy type. Dwelling units = 3 VA/sq ft. Offices = 3.5 VA/sq ft. Banks = 3.5 VA/sq ft. Schools = 3 VA/sq ft. Hospitals = 2 VA/sq ft (patient care areas use different rules). Know the occupancy types, they appear on exams.

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Problem 7: Clothes Dryer Load for Multifamily

Given: A 12-unit apartment building has one dryer in each unit. Each dryer is rated 5,800W. What is the calculated dryer load for the service?

NEC Table 220.54: Demand factors for clothes dryers in multifamily

12 dryers at 50% demand factor.

Per-dryer: 5,800W (nameplate) vs. 5,000W minimum, use 5,800W (larger)

Total nameplate: 12 × 5,800 = 69,600 VA Demand: 69,600 × 50% = 34,800 VA

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Problem 8: Conductor Sizing for a Dwelling Feeder

Given: The calculated load for a 240V single-phase service is 175A. What is the minimum conductor size for this service using copper THW conductors at 75°C?

From NEC Table 310.16 (75°C column for copper):

• 1/0 AWG: 150A, too small
• 2/0 AWG: 175A, exactly 175A, acceptable
• Minimum: 2/0 AWG copper THW

NEC 230.42(A) requires service conductors to have an ampacity not less than the calculated load. Always round up to the next standard conductor size if the calculation falls between sizes.

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Problem 9: Neutral Conductor Sizing

Given: A service has a calculated line-to-line load of 175A. The calculated maximum unbalanced neutral load is 110A. What is the minimum neutral conductor size in copper at 75°C?

NEC 220.61: The neutral of a feeder or service can be sized based on the maximum unbalanced load, not the full load, with one key exception: large motor loads and non-linear loads require the full neutral size.

For this problem, no motors or non-linear loads, so size the neutral for 110A.

From Table 310.16, 75°C copper: 2 AWG = 95A (too small), 1 AWG = 110A.

Minimum neutral: 1 AWG copper THW

Note: NEC 250.24(C) also requires the neutral be at least the size required by NEC 250.122 for grounding, check both requirements.

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Problem 10: Air Conditioning and Heat Calculation (NEC 220.60)

Given: A house has a 5 kW heat strip and an air conditioner with a 6 kW compressor and a 0.5 kW fan motor. What is the calculated load for the heating and cooling systems combined?

NEC 220.60, Non-coincident loads: Where it is unlikely that two dissimilar loads will be used simultaneously, you can use only the larger load.

Heat strip: 5,000 VA

A/C: 6,000 + 500 = 6,500 VA (compressor + fan, NEC 440.34 requires you include the motor loads)

A/C (6,500 VA) is larger than heat (5,000 VA), so use 6,500 VA.

Note: If the home has an electric heat pump (where the same equipment does both heating and cooling), NEC 220.60 does NOT apply, you cannot use non-coincident loads because the same unit provides both. Use the largest operational load instead.

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Practice Strategy

These 10 problems cover the most common load calculation question types. For exam prep, the goal is not just getting the right answer, it's being able to execute the full process in under 4 minutes per multi-step problem with your NEC open to Article 220.

Work through each problem without looking at the solution first, then compare your work step-by-step. Pay attention to which table you went to at each step. The most common exam errors are using the wrong demand factor table, forgetting to include laundry in the general lighting demand subtotal, and missing the 5,000 VA minimum for dryers.

Related Reading

• Free Dwelling Service Load Calculator (NEC 220.82)
• Free Standard Method Dwelling Load Calculator (NEC 220 Part III)
• NEC Article 220: Load Calculations Deep Dive
• Electrician Exam Math: Complete Formula Guide
• NEC Table 310.16: Wire Sizing and Ampacity Guide
• Calculators on the Electrician Exam: What's Allowed
• NEC Article 230: Services