Dwelling Unit Calculations (2024)

The calculation is pretty simple.

2,000 sq ft x 3VA = 6,000VA.

No additional load is required for general-use receptacles and lighting outlets because they are included in the 3VA per sq ft load specified by Table 220.12 for dwelling units. See 220.14(J).

Now let’s work through an example to determine the number of circuits required.

Question: How many 15A circuits are required for a 2,000-sq-ft dwelling unit?

Step 1: General lighting VA = 2,000 sq ft x 3VA = 6,000VA

Step 2: General lighting amperes:
I = VA ÷ E
I = 6,000VA ÷ 120V*
I = 50A
*Use 120V, single-phase unless specified otherwise.

Step 3: Determine the number of circuits:
Number of circuits = General lighting amperes ÷ circuit amperes
Number of circuits = 50A ÷ 15A
Number of circuits = 3.30, or 4 circuits. Any fraction of a circuit must be rounded up.

Optional method for feeder and service load calculations

You can use the optional method [Art. 220, Part IV] only for dwelling units served by a single 120/240V or 120/208V 3-wire set of service or feeder conductors with an ampacity of 100A or larger [220.82]. The optional method consists of three calculation steps:

1. General loads [220.82(B)]
2. Heating and air-conditioning load [220.82(C)]
3. Feeder/service conductors [310.15(B)(6)]

Step 1: General loads [220.82(B)]

The general calculated load must be at least 100% for the first 10kVA, plus 40% of the remainder of the following loads:

1. General lighting and receptacles: 3VA per sq ft
2. Small-appliance and laundry branch circuits: 1,500VA for each 20A, 120V small-appliance and laundry branch circuit specified in 220.52.
3. Appliances: The nameplate VA rating of all appliances and motors that are fastened in place (permanently connected) or located on a specific circuit, not including heating or air-conditioning.

Be sure to calculate the range and dryer at their nameplate ratings.

Step 2: Heating and air-conditioning load [220.82(C)]

Include the larger of (1) through (6):

1. Air-conditioning equipment: 100%
2. Heat-pump compressor without supplemental heating: 100%
3. Heat-pump compressor and supplemental heating: 100% of the nameplate rating of the heat-pump compressor and 65% of the supplemental electric heating for central electric space-heating systems. If the control circuit is designed so that the heat-pump compressor can’t run at the same time as the supplementary heat, omit the compressor from the calculation.
4. Space-heating units (three or fewer separately controlled units): 65%.
5. Space-heating units (four or more separately controlled units): 40%.
6. Thermal storage heating: 100%.

Step 3: Feeder/service conductors [310.15(B)(6)]

• 400A and less. For individual dwelling units of one-family, two-family, and multi-family dwellings, use Table 310.15(B)(6) to size 3-wire, single-phase, 120/240V service or feeder conductors (including neutral conductors) that serve as the main power feeder. Feeder conductors aren’t required to have an ampacity rating greater than the service conductors [215.2(A)(3)]. Size the neutral conductor to carry the unbalanced load per Table 310.15(B)(6). Table 310.15(B)(6) can’t be used for sizing the feeder or service conductors that supply more than a single dwelling unit.
• Over 400A. Size ungrounded conductors and the neutral conductor using Table 310.16 for feeder/services over 400A and those that do not fill all of the requirements for using Table 310.15(B)(6). Let’s try a calculation example.

Question: What size service conductor is required for a 1,500-sq-ft dwelling unit containing the following loads?
Cooktop: 6,000VA
Disposal: 900VA
Dishwasher: 1,200VA
Dryer: 4,000VA
Ovens (two each): 3,000VA
Water heater: 4,500VA
A/C: 17A, 230V
Electric heating (one control unit): 10kVA

Step 1: General loads [220.82(B)]
General lighting: 1,500 sq ft x 3VA = 4,500VA
Small-appliance circuits: 1,500VA x 2 circuits = 3,000VA
Laundry circuit: 1,500VA
Appliances (nameplate):
Cooktop: 6,000VA
Disposal: 900VA
Dishwasher: 1,200VA
Dryer: 4,000VA
Ovens (each 3 kW): 6,000VA
Water heater: 4,500VA

Total connected load: 31,600VA

First 10kW at 100%: 10,000VA x 1.00 = 10,000VA

Remainder at 40%: 21,600VA x 0.40 = 8,640VA

Calculated general load: 10,000VA + 8,640VA

Calculated general load: 18,640VA

Step 2: Air-Conditioning versus heat [220.82(C)]

Air-conditioning at 100% [220.82(C)(1)] vs. electric space heating at 65% [220.82(C)(4)]

Air conditioner [Table 430.248]:
A/C VA = V x A
A/C VA = 230V x 17A
A/C VA = 3,910VA (omit)

Electric space heat: 10,000VA x 0.65 = 6,500VA

Step 3: Feeder/service conductors [310.15(B)(6)]

Calculated general load (Step 1): 18,640VA

Heat calculated load (Step 2): 6,500VA

Total calculated load = 18,640VA + 6,500VA = 25,140VA

I = VA ÷ E

I = 25,140VA ÷ 240V = 105A

Therefore, the feeder/service ungrounded conductor is sized to 110A, 3 AWG [310.15(B)(6)].

The NECdoesn’t explain how demand factors were derived, and it’s not essential that you understand this in order to apply them correctly. Be sure to work on some practice calculations so you understand how to apply the various demand factors to a dwelling unit calculation.

The standard calculation and the optional calculation methods were both discussed in this article. These are two distinctly different calculation methods, so be careful not to mix them. Remember that the standard method is in Part III of Art. 220, and the optional method is contained in Part IV. When you are evaluating the necessary loads in either type of calculation method, follow the requirements for specific loads covered in other Articles outside of Art. 220. Which method is better to use? On an exam, you’ll likely be told which method to use on a specific question. However, if the question doesn’t specify a method, use the standard calculation. The optional method is usually faster and easier to apply, so it has a natural advantage for daily use on the job.

SIDEBAR: Where to Find Dwelling Unit Code Requirements Outside Art. 220

Branch circuits — Art. 210

Areas supplied by small appliance circuits — 210.52(B)(1)

Feeders — Art. 215

Services — Art. 230

Overcurrent protection — Art. 240

Wiring methods — Art. 300

Conductors — Art. 310

Appliances — Art. 422

Electric space-heating equipment — Art. 424

Motors — Art. 430

Air-conditioning equipment — Art. 440

For more information, read "Load Calculations -- Part 1."