Dwelling Unit Feeder/Service Conductor Calculations
How do you size conductors for residential services and feeders?

By Mike Holt, Extracted from my Electrical Calculations textbook.

Most electrical licensing exams (e.g., journeyman, master electrician, contractor) require you to calculate residential loads and feeders. To do this, you might use the standard method contained in Article 220, Part II. Usually, you can use the optional method in Article 220, Part III. In either case, you are free to exceed the NEC requirements-these are minimum requirements, not design specifications.

Though the standard method involves more steps, many people use it exclusively to avoid using the wrong method. However, most residential construction qualifies for the optional method. So, it’s prudent to understand both methods. The standard method is where we’ll start-it requires six sets of calculations:

  1. General lighting, receptacles, small-appliance, and laundry
  2. Air conditioning versus heat
  3. Appliances
  4. Clothes dryer
  5. Cooking equipment
  6. Conductor size

General lighting and receptacles, small-appliance, and laundry

The NEC recognizes these circuits will not all be on (loaded) simultaneously. Thus, you may apply a demand factor to the total connected load [220.16]. To determine the feeder demand load, refer to Table 220.11 and follow these steps:

  1. Total Connected Load. Determine the total connected load for general lighting, receptacles (3 VA per sq ft), two small-appliance circuits each at 1,500 VA, and one laundry circuit at 1,500 VA.
  2. Demand Factor: Apply Table 220.11 demand factors to the total connected load. First 3,000 VA at 100% demand. Remainder VA at 35% demand

Example: What is the general lighting, small-appliance and laundry demand load for a 2,700 sq ft dwelling unit? (See Figure 9-13).

Graphics are not included in this Newsletter.

(a) 8,100 VA (b) 12,600 VA (c) 2,700 VA (d) 6,360 VA

Answer: (d) 6,360 VA
General lighting/receptacles: 2,700 sq ft x 3 VA = 8,100 VA
Small-appliance circuits: 1,500 VA x 2 = 3,000 VA
Laundry circuit: 1,500 VA x 1 = 1,500 VA
Total Connected Load: 12,600 VA
First 3,000 VA at 100% = 3,000 VA x 1.00 = 3,000 VA
Remainder at 35% = 9,600 VA x 0.35 = 3,360 VA
Total = 3,000 VA + 3,360 VA = 6,360 VA

Air-conditioning versus heat

Because air-conditioning and heating loads are not on simultaneously, you may omit the smaller of the two loads [220.21]. Calculate each of these at 100% [220.15].

Example: What is the service demand load for 5-hp, 230V A/C versus three 3 kW baseboard heaters? (see Figure 9-14).

(a) 6,400 VA (b) 3,000W (c) 8,050 VA (d) 9,000W

Answer: (d) 9,000W
A/C = 230V x 28A = 6,440 VA. Omit this, per 220.21.
Heat = 3,000W x 3 = 9,000W


Per 220.17, you can use a 75% demand factor when four or more “fastened in place” appliances (e.g., dishwasher, waste disposal) are on the same feeder.

Example: What is the demand load for a waste disposal (940 VA), dishwasher (1,250 VA), and water heater (4,500 VA)?

(a) 5,018 VA (b) 6,690 VA (c) 8,363 VA (d) 6,272 VA

Answer: (b) 6,690 VA
Waste disposal: 940 VA
Dishwasher: 1,250 VA
Water heater: 4,500 VA
Total: 6,690 VA

Clothes dryer

Per 220.18, the feeder or service demand load for electric clothes dryers in a dwelling unit shall not be less than 5,000W. Exception: if the nameplate rating exceeds 5,000W, use that rating as the load. You can omit this calculation if the unit has no electric dryer provision-however, it’s common to provide both gas and electric sources. If you see gas on the plans, verify electric will be omitted-do not just assume.

Example: What is the service and feeder demand load for a 4 kW dryer? (see Figure 9-16).

(a) 4,000W (b) 3,000W (c) 5,000W (d) 5,500W

Answer: (c) 5,000W. The dryer load must not be less than 5,000 VA.

Cooking equipment

For household cooking appliances rated over 13/4 kW, you can use the demand factors in 220.19, Table and Notes 1, 2, and 3.

Example: What is the service and feeder demand load for a 13.6 kW range in a dwelling unit?

(a) 8.8 kW (b) 8 kW (c) 9.2 kW (d) 6 kW

Answer: (a) 8.8 kW
13.6 kW exceeds 12 kW by one kW and one major fraction of a kW. Column C value (8 kW) must be increased 5% for each kW or major fraction of a kW (0.5 kW or larger) over 12 kW.
8 kW x 1.1 = 8.8 kW

Feeder and Service Conductor Size.

400A and Less: For 3-wire, 120/240V, 1 systems, size the feeder or service conductors to Table 310.15(B)(6). For all others, use Table 310.16. Size the grounded (neutral) conductor to the maximum unbalanced load [220.22] per Table 310.16.

Over 400A: Size the ungrounded and grounded (neutral) conductors per Table 310.16.

What size THHN feeder or service conductor (120/240V, 1) does the NEC require for a 225A service demand load?
(a) 1/0 AWG (b) 2/0 AWG (c) 3/0 AWG (d) 4/0 AWG

Answer: (c) 3/0 AWG

Optional Method

You can use the easier optional method ( in 220.30) only when the total connected load is served by a single 3-wire, 120/240V or 208Y/120V set of service or feeder conductors with an ampacity of 100A or greater. Because this condition describes the typical residential service, the optional method is likely to apply. Using it can simplify the design process and save you time because you have so many fewer sets of calculations.

General loads.

The calculated load shall not be less than 100% for the first 10 kW, plus 40% of the remainder of the following loads:


Include the largest of the following:

Sizing Service/Feeder Conductors

Now that we’ve seen how to determine residential loads, let’s size the service/feeder conductors.

400A and Less. Size the ungrounded conductors per Table 310.15(B)(6) for 120/240V, 1 systems up to 400A. You must size the grounded (neutral) conductor to carry the unbalanced load per Table 310.16.

Over 400A. Size the ungrounded and grounded (neutral) conductors per Table 310.16.

An example will help illustrate how to do these. What size service conductor does a 1,500 sq ft dwelling unit need, if it contains the following loads?

(a) 6 AWG (b) 4 AWG (c) 3 AWG (d) 2 AWG

Answer: (c) 3 AWG

Step 1: Calculate the general loads [220.30(B)].
Small-appliances = 1,500 VA x 2 circuits = 3,000 VA
General lighting = 1,500 sq ft x 3 VA= 4,500 VA
Laundry = 1,500 VA
Appliances (nameplate) = All the appliance loads summed = 1,200 VA + 4,500 VA + 900 VA + 4,000 VA + 6,000 VA + (3,000 VA x 2 units) + 6,000 VA. Total Connected Load = 31,600 VA

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

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

Demand load = 18,640 VA

Step 2: Air-Conditioning Versus Heat [220.30(C)].
AC = 230V x 28A = 6,440 VA; Electric heat: 10,000 VA x 0.65 = 6,500 VA. Thus, omit the AC in the calculations.
Step 3: Service/Feeder Conductors [310.15(C)(6)].
General Loads = 18,640 VA
Heat = 6,500 VA
Total Demand Load = 25,140 VA
I = VA/V
I = 25,140 VA / 240V = 105A

Note: The feeder/service conductor is sized to 110A, 3 AWG.

Now that we’ve walked through the process of calculating residential feeders, you can see that doing so is fairly easy. You need to calculate the loads, then the feeder size. The NEC provides the requirements in Articles 220 and 230. Doing these calculations correctly can save you money during design and construction, while providing safe homes for the families that occupy them.