#### Thank you for visiting one of our most popular classic articles. If you’d like to see updated information on this topic, please check out this recently published article, **Load Calculations — Part 1**.

*Note: This article is based on the 2008 NEC.*

A dwelling unit is a single structure that provides complete and independent living facilities, according to the NEC definition found in Art. 100 (**Fig. 1** ).

Dwelling units have special requirements for load calculations. Although most of the actual load calculation requirements are in Art. 220, others are scattered throughout the Code and still come into play when making certain calculations (see **SIDEBAR:****Where to Find Dwelling Unit Code Requirements Outside Art. 220**at end of article). Keep the following considerations in mind when making dwelling unit calculations:

**Voltages**. Unless other voltages are specified, calculate branch-circuit, feeder, and service loads using the nominal system voltage [220.5(A)]. For a single-family dwelling unit, the nominal voltage is typically 120/240V.**Motor VA**. Use motor table voltage and current values, such as 115V, 230V, or 460V — not 120V, 240V, or 480V [430.248 and 430.250]. A much more accurate VA rating is obtained by using the motor’s rated voltage and current, which were used in developing the Code Tables.**Rounding**. Where calculations result in a fraction of less than 0.50A, you can drop the fraction [220.5(B)].**Receptacles**. You can use 15A or 20A receptacles on 20A circuits as long as there is more than one receptacle on the circuit. For these purposes, a duplex receptacle is considered to be two receptacles [210.21(B)(3)].**Continuous loads**. A continuous load is one in which the maximum current is expected to continue for 3 hr or more, according to the Art. 100 definition. Fixed electric heating is one example of a continuous load [424.3(B)]. When sizing branch circuit conductors and overcurrent devices for a continuous load, multiply the load by 125% [210.19(A)(1) and 210.20(A)].**Laundry rooms**. A laundry area receptacle is required [210.52(F)], at least one of which must be within 6 ft of a washing machine [210.50(C)]. Any receptacle within 6 ft of the outside edge of a laundry sink must be GFCI protected [210.8(A)(7)].

### Required circuits

In addition to the circuits required for dedicated appliances and those needed to serve the general lighting and receptacle load, a dwelling unit must have the following circuits:

- A minimum of two 20A, 120V small-appliance branch circuits for receptacles in the kitchen, dining room, breakfast room, pantry, or similar dining areas [220.11(C)(1)]. These circuits must not be used to serve other outlets, such as lighting outlets or receptacles from other areas [210.52(B)(2) Ex]. These circuits are included in the feeder/service calculation at 1,500VA for each circuit [220.52(A)].
- One 20A, 120V branch circuit for the laundry receptacle(s). It can’t serve any other outlet(s), such as lighting, and can serve only receptacle outlets in the laundry area [210.52(F) and 210.11(C)(2)]. In your feeder/service load calculation, include 1,500VA for the 20A laundry receptacle circuit [220.52(B)], as shown in
**Fig. 2**

## Feeder and service calculations

Occupants don’t use all loads simultaneously under normal living conditions, so “demand factors” can be applied to many of the dwelling unit loads in order to size the service. Some demand factors provided in the Code are intended for use in dwellings only; others are allowed only in non-dwellings. Therefore, be careful to apply demand factors only as allowed by the NEC.

The NEC provides two dwelling service load calculation methods: the standard method and the optional method.

### Standard method for feeder and service load calculations

The standard method consists of three calculation steps:

**General lighting VA load**. When calculating branch circuits and feeder/service loads for dwellings, include a minimum 3VA per sq ft for general lighting and general-use receptacles [220.12]. When determining the area, use the outside dimensions of the dwelling. Don’t include open porches, garages, or spaces not adaptable for future use.**Small appliance and laundry circuits**. The 3VA per sq ft rule includes general lighting and all 15A and 20A, 125V general-use receptacles, but doesn’t include small-appliance or laundry circuit receptacles. Therefore, you must calculate those at 1,500VA per circuit. See 220.14(J) for details.**Number of branch circuits**. Determine the number of branch circuits required for general lighting and general-use receptacles from the general lighting load and rating of the circuits [210.11(A)]. Although this is explained in Annex D, Example D1(a) of the NEC, let’s look at an another example.

**Question: **What’s the general lighting and receptacle load for a 2,000-sq-ft dwelling unit that has 34 convenience receptacles and 12 luminaires rated 100W each (**Fig. 3)?**

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:

- General loads [220.82(B)]
- Heating and air-conditioning load [220.82(C)]
- 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:

- General lighting and receptacles: 3VA per sq ft
- Small-appliance and laundry branch circuits: 1,500VA for each 20A, 120V small-appliance and laundry branch circuit specified in 220.52.
- 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):

- Air-conditioning equipment: 100%
- Heat-pump compressor without supplemental heating: 100%
- 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.
- Space-heating units (three or fewer separately controlled units): 65%.
- Space-heating units (four or more separately controlled units): 40%.
- 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."

## FAQs

### What is the unit load per square meter for dwelling units? ›

Unit loads are provided in both volt-amperes per square foot and volt-amperes per square meter. Unit loads range from ¼ to 3½ volt-amperes per square foot.

### How do you fill out a load calculation? ›

**Here is one:**

- Start by adding the wattage of all lighting branch circuits.
- Add in the wattage rating of all plug-in outlets.
- Add in the wattage of all permanent appliances, such as washer/dryer, electric range, or water heaters.
- Subtract 10,000 & multiply this number by . ...
- Add 10,000.

### How do you calculate kW electrical load? ›

The conversion of watts to kilowatts is just as straight-forward as you may have guessed. We find the power in kilowatts P(kW) by dividing the power in watts P(W) by 1,000. Here's the Formula for Converting Watts Into Kilowatts: **P(kW) = P(W) / 1,000**.

### What is the purpose of dwelling calculations? ›

A single dwelling calculation is **used for a stand alone building or within a multi-unit building**. When determining the service for a multi-unit building, the designer is allowed to first calculate each individual dwelling, then use the multi-dwelling calculation for each feeder for a floor(s).

### What's the general lighting load for a dwelling unit with a total area of 2800 square feet? ›

General lighting and receptacle loads

Table 220.12 in the National Electrical Code considers a residence a listed occupancy at 3 VA per square foot; therefore, the general lighting load is determined by multiplying the square footage. For example, 2,800 square feet times 3 VA is **8,400 VA**.

### How do you calculate kw in a house? ›

- Hi there,
- To calculate the electrical load of a house, you need to understand the number of appliances that run in your house. ...
- The amount of electricity required to run:
- Fans - 2 nos. - ...
- Lights - 3 nos. - ...
- TV - 1 no. ...
- Add these individual loads:
- 500 Wh + 480 Wh + 300 Wh = 1280 Wh or 1.28 kWh.

### How many amps can you put on a 100-amp panel? ›

For example, a 100-amp service panel could have circuit breakers that add up to **more than 200 amps**.

### How do you calculate square Metres per load? ›

**Multiply the load per unit area or length by the total area or length**. For the rectangle, you compute 10 kN per square meter multiplied by 24 square meters to get 240 kN. For the beam, you calculate 10 kN per meter multiplied by 5 meters to get 50 kN.

### When calculating the total load on a dwelling What is the minimum VA that must be added for the two required small-appliance circuits? ›

In each dwelling unit, the load shall be calculated at **1500 volt-amperes** for each 2-wire small-appliance branch circuit as covered by 210.11(C)(1).

### What is a typical residential floor dead load? ›

Generally, the customary floor dead load is **10-12 PSF (pounds per square foot) for floors, 12-15 PSF for roof rafters and 20 PSF for roof trusses**. However, these may increase when a heavy finish material, such as brick veneer walls or tile floors/roofs, is specified.

### What is typical roof dead load? ›

Dead Loads

Your roof then needs to be built to support itself. Normally, the dead load of a typical asphalt shingle roofing system with wood frames is **15 pounds per square foot**. A clay-tiled roof, on the other hand, has a dead load of 27 pounds per square foot.