NEC Article 220 Load Calculations: Exam Guide

Load calculations are where the electrician exam gets mathematical, and Article 220 is the framework for nearly all of it. Whether you're sizing a service for a new home, calculating feeder loads for a commercial building, or determining how many amps a range demands, Article 220 is the article you'll be working from.

This guide walks through the standard calculation method for dwelling units step by step, covers the optional method, and includes the key tables and formulas you need to have down cold for the exam.

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Article 220 Structure

Article 220 is divided into five parts:

• Part I: General (220.1–220.5)
• Part II: Branch Circuit Load Calculations (220.10–220.18)
• Part III: Feeder and Service Load Calculations, Standard Method (220.40–220.61)
• Part IV: Optional Feeder and Service Load Calculations (220.82–220.87)
• Part V: Farm Loads (220.100–220.103)

The exam focuses on Parts II, III, and IV, branch circuit, feeder, and service loads.

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Part II: Branch Circuit Loads

Table 220.12, General Lighting Loads

Lighting loads for branch circuit and feeder calculations are based on floor area using the unit loads from Table 220.12.

Floor area calculation: Use outside dimensions of the building (or dwelling unit). Do not include open porches, unfinished spaces not adaptable for future use, or garages.

220.14, Other Outlets

Outlets other than general lighting are calculated at the larger of the actual load or 180 VA per outlet, for receptacle outlets where the load is not known.

220.18, Fixed Electric Space Heating

Fixed electric space heating loads are calculated at 100% of the total connected load.

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Part III: Standard Method for Feeders and Services

Step-by-Step Dwelling Unit Calculation (Standard Method)

Here is the complete standard method for sizing a dwelling unit service or feeder:

Step 1: General Lighting Load Multiply the floor area (sq ft) by 3 VA/sq ft from Table 220.12.

Step 2: Small Appliance Branch Circuit Loads Add 1,500 VA for each small appliance branch circuit. Minimum two circuits required (220.52(A)), so minimum 3,000 VA.

Step 3: Laundry Branch Circuit Load Add 1,500 VA for the laundry circuit.

Step 4: Apply Lighting Demand Factors (Table 220.42)

Apply these factors to the combined total of Steps 1 + 2 + 3.

Step 5: Appliance Loads Add the nameplate ratings of all fixed appliances (dishwasher, garbage disposal, water heater, etc.) at 100%. If four or more fixed appliances (excluding A/C, heating, dryers, ranges) are served by the same feeder, a 75% demand factor applies.

Step 6: Dryer Load Use the nameplate rating or 5,000 VA, whichever is larger.

Step 7: Range/Cooking Appliance Load Use Table 220.55. For one range rated 12 kW or less, the demand is 8 kW. See Table 220.55 details below.

Step 8: Heating or Air Conditioning Load Add 100% of the largest of: electric heating, heat pumps, or air conditioning. Do not add both heating and A/C (noncoincident loads per 220.60).

Step 9: Total and Convert to Amperes Add all loads, then divide by voltage to get amperes.

Amps = Total VA ÷ 240V (for single-phase 120/240V service)

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Complete Standard Method Worked Example

This is the type of multi-step problem that shows up on every journeyman exam. Work through it before checking the answer.

Given: A 1,800 sq ft single-family dwelling with a 120/240V, single-phase service. It has: 2 small appliance circuits, 1 laundry circuit, a 12 kW range, a 5.5 kW dryer, a 4.5 kW water heater, a 1,200 VA dishwasher, and a 10 kW electric heating system. No A/C. Size the minimum service.

Step 1, General lighting load: 1,800 sq ft × 3 VA = 5,400 VA

Step 2, Small appliance loads: 2 circuits × 1,500 VA = 3,000 VA

Step 3, Laundry circuit: 1 circuit × 1,500 VA = 1,500 VA

Step 4, Apply Table 220.42 demand factors: Subtotal of Steps 1 + 2 + 3 = 9,900 VA

Step 5, Fixed appliances:

• Dishwasher: 1,200 VA
• Water heater: 4,500 VA
• Subtotal: 5,700 VA (fewer than 4 fixed appliances, so no demand factor, 100%) = 5,700 VA

Step 6, Dryer: Nameplate is 5,500 VA. Minimum is 5,000 VA. Use the larger: 5,500 VA

Step 7, Range: 12 kW range, rated 12 kW or less. Table 220.55 Column C: 8,000 VA

Step 8, Heating vs. A/C (noncoincident loads): Electric heat: 10,000 VA. No A/C. Use 10,000 VA at 100%.

Step 9, Total and convert to amperes:

Service amperes = 34,615 ÷ 240 = 144A → minimum 150A service

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Table 220.42 Demand Factors in Practice

Demand factor only example: A 1,800 sq ft dwelling has 2 small appliance circuits and 1 laundry circuit.

• Lighting: 1,800 × 3 = 5,400 VA
• Small appliances: 2 × 1,500 = 3,000 VA
• Laundry: 1,500 VA
• Subtotal: 9,900 VA

Apply Table 220.42:

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

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Table 220.55, Electric Ranges and Cooking Appliances

Table 220.55 gives demand loads for household ranges. Key rules:

One range, 12 kW or less (Column C): 8 kW demand

One range over 12 kW, not over 27 kW (Column B): Take the Column C value (8 kW) and increase by 5% for each kW or fraction thereof over 12 kW.

Example: One 14 kW range 14 kW − 12 kW = 2 kW over → 2 × 5% = 10% increase 8 kW × 1.10 = 8.8 kW demand

Multiple ranges: Table 220.55 lists demand values for 2 through 61+ ranges. For example, 5 ranges rated 12 kW or less = 20 kW demand (Column C).

Split ranges (separate top + oven): Treat as a single appliance. Two units, one rated at 3 kW or less and another at 6 kW or less, are added together and compared to Table 220.55.

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220.54, Electric Clothes Dryers

Use the nameplate rating or 5,000 VA (5 kW), whichever is larger, for each dryer.

For multiple dryers (such as in a multifamily building), Table 220.54 demand factors apply:

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220.60, Noncoincident Loads

Where two or more loads will not be used simultaneously, only the largest load need be included in the calculation.

The most common exam application: electric heating and air conditioning. You calculate both but only use the larger one in the total load.

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220.61, Neutral Load

The neutral conductor need only carry the maximum unbalanced load between the neutral and any ungrounded conductor.

For feeders or services over 200 amperes: Only 70% of the demand load above 200 amperes is required for the neutral.

For ranges, ovens, and dryers: The neutral is sized at 70% of the demand load from Tables 220.54 and 220.55.

For 3-wire DC or single-phase AC: Line-to-neutral loads only.

Exam note: Electric ranges and dryers operate on 240V but have a 120V neutral for lights, timers, and controls, the neutral load is 70% of the appliance demand.

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Part IV: Optional Calculation Method (220.82)

The optional method is available for dwelling units with a 100-ampere or larger single-phase 120/240V or 208Y/120V service. It simplifies the calculation by combining most loads into two categories.

Optional Method Loads

General loads (all loads except heating/cooling):

• 100% of the first 10 kVA
• 40% of the remainder

Include: 1,500 VA per small appliance circuit, 1,500 VA for laundry, nameplate ratings for all other appliances and loads.

Heating and cooling loads (use the larger of heating or cooling, not both):

• 100% of the selected load

Optional Method Worked Example

A 2,000 sq ft dwelling unit with: 2 small appliance circuits, 1 laundry circuit, 12 kW range, 5 kW dryer, 6 kW water heater, 4 kW dishwasher, 1,200 VA disposal, 10 kW heat, 5 kW A/C.

General loads:

• Lighting: 2,000 × 3 = 6,000 VA
• Small appliances: 3,000 VA
• Laundry: 1,500 VA
• Range: 12,000 VA
• Dryer: 5,000 VA
• Water heater: 6,000 VA
• Dishwasher: 4,000 VA
• Disposal: 1,200 VA
• Subtotal: 38,700 VA

Apply demand: 10,000 × 100% + 28,700 × 40% = 10,000 + 11,480 = 21,480 VA

Largest heating/cooling: Heat at 10,000 VA (100%) = 10,000 VA

Total: 21,480 + 10,000 = 31,480 VA Service amps: 31,480 ÷ 240 = 131A → size for 150A service

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Key Formulas for the Exam

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Related Reading

• NEC Article 230, Services
• NEC Article 250, Grounding and Bonding
• NEC Article 430, Motors
• Electrical Math Study Guide
• Start Your Practice Test