NEC 120 Commercial Load Calculation for the Journeyman Exam: Lighting VA per Square Foot, Receptacles, Show Windows and Kitchens

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How this differs from a dwelling calculation

The dwelling method (general lighting by floor area, small appliance and laundry circuits, the optional method) is on service and feeder load calculations for the Texas Journeyman exam. A commercial building has no small appliance circuits and no dwelling demand table for lighting. Instead, each kind of load has its own line:

Load	How it is counted	Where to look
General lighting	Area × unit load for the occupancy	Table 120.42(A)
General-use receptacles	180 VA per yoke; 90 VA per receptacle on one piece of equipment with four or more	120.14(I)
Receptacle demand	First 10 kVA at 100 percent, remainder at 50 percent	Table 120.47
Show window lighting	200 VA per linear foot of window	120.46(A)
Sign outlet	At least 1,200 VA	120.14(F)
Commercial kitchen equipment	Demand factor by number of units, not less than the two largest	120.56 and Table 120.56
Heating and air conditioning	Use only the larger of two loads that will not run at the same time	120.6

The pieces one by one

• General lighting. Use the outside dimensions of the building and the unit load for its occupancy (NEC Table 120.42(A)). An office is 1.3 VA per square foot and a retail store 1.9 VA per square foot.
• Receptacles. Each single or multiple receptacle on one yoke counts 180 VA. A single piece of equipment made of four or more receptacles counts at least 90 VA per receptacle (NEC 120.14(I)).
• Receptacle demand. Where the non-dwelling receptacle demand table applies, add up the receptacle load, then take the first 10,000 VA at 100 percent and everything above that at 50 percent (NEC Table 120.47). Banks and office buildings have their own receptacle rule with a 1 VA per square foot allowance; read its conditions in the section before you combine it with the demand factor (NEC 120.43).
• Show windows. 200 VA for each linear foot of show window, measured horizontally along its base (NEC 120.46(A)). Track lighting has its own rule in the next subsection (NEC 120.46(B)).
• Sign outlet. At least 1,200 VA for each required sign outlet (NEC 120.14(F)). Where a sign circuit is required is covered with sign circuits on NEC 600, 620 and 670.
• Kitchen equipment. For three or more qualifying units of commercial cooking, dishwasher, water heating and similar kitchen equipment, apply the demand factor for the number of units, after confirming which equipment the table covers: 70 percent for five units and 65 percent for six or more. The result is never less than the sum of the two largest kitchen loads. Space heating, ventilation and air conditioning equipment are not counted in this group (NEC 120.56).
• Noncoincident loads. Where two loads will not run at the same time, such as electric heat and air conditioning, use only the larger one (NEC 120.6).

Exam questions often state that the lighting is a continuous load. In that case, the general conductor sizing rule uses 125 percent of the continuous part plus 100 percent of the rest, subject to the equipment rating provisions and exceptions (NEC 215.4(A)(1)). Keep the calculated load and the conductor sizing step separate, so you do not apply 125 percent twice.

Worked example: small retail store

A retail store is 8,000 square feet. It has 60 general-use duplex receptacles, 30 linear feet of show window, one sign outlet and a 10 kW electric heater that never runs with the 6 kW air conditioner. Find the calculated load in VA using the stated loads and demand factors. Do not apply the 125 percent continuous load multiplier in this step.

1. Lighting. 8,000 × 1.9 = 15,200 VA.
2. Receptacles. 60 × 180 = 10,800 VA. First 10,000 at 100 percent = 10,000. The remaining 800 at 50 percent = 400. Receptacle load = 10,400 VA.
3. Show window. 30 × 200 = 6,000 VA.
4. Sign. 1,200 VA.
5. Heat or air conditioning. The larger is the heater, 10,000 VA. The air conditioner is left out.
6. Total. 15,200 + 10,400 + 6,000 + 1,200 + 10,000 = 42,800 VA.

At 208Y/120 V three-phase, 42,800 ÷ (208 × 1.732) = 42,800 ÷ 360.3 = about 118.8 A before any 125 percent continuous-load step.

Worked example: kitchen equipment demand

A restaurant kitchen has six pieces of equipment that qualify for the commercial kitchen demand table: 12 kW, 10 kW, 6 kW, 5 kW, 3 kW and 2 kW.

1. Connected load. 12 + 10 + 6 + 5 + 3 + 2 = 38 kW.
2. Demand factor for six units. 38 × 0.65 = 24.7 kW.
3. Two largest check. 12 + 10 = 22 kW. 24.7 kW is larger, so the kitchen demand load is 24.7 kW.

If the two largest units had been 15 kW and 12 kW, their sum (27 kW) would be the floor, and you would use it instead of a smaller demand result.

Check yourself

1. What is the general lighting load for a 5,000 square foot office? 5,000 × 1.3 = 6,500 VA.
2. A building has 80 general-use receptacles. What is the receptacle load after the demand factor? 80 × 180 = 14,400 VA. 10,000 + (4,400 × 0.5) = 12,200 VA.
3. Four kitchen units total 20 kW. Which table applies? Table 120.56, for the number of units, but never less than the two largest units.

FAQ

Sources checked October 2026

• PSI Candidate Information Bulletin, Texas electrician examinations (updated September 15, 2026): https://test-takers.psiexams.com/api/content/bulletin/1682
• NFPA 70, National Electrical Code, 2026 Edition, Articles 120 and 215