Dwelling Unit Service Load Calculations – Standard Method

Standard and Optional Methods

Electricians use two different methods to calculate the size of electrical service for a Dwelling. One is called the Standard Method, which has many steps but produces arguably more accurate figures while the other, the Optional Method, has fewer steps but yields slightly inflated figures. Now before we just dive into examples, let’s take a look at Article 220 in the National Electrical Code and talk about a few things that will help us know how these calculations work.

Article 220 is something that electricians at any level seem to either overthink, make more complicated in their own minds, or maybe just misunderstand altogether. Like all articles in the codebook, it is separated into parts, which are identified with Roman numerals which help to separate different topics within the same article. Sometimes the NEC will reference the same information in different parts so it is always important to check to see which part you’re in while gathering information.  

An example of this is when looking up Show windows. Show windows can be found in both Part II and Part III of Article 220. 220.14(G) is in Part II, which covers “Branch Circuits”, and 220.43(A) is in Part III which covers “Feeder and Service Load Calculations” also known as the Standard Method. Knowing which part you are in is vital to your understanding of Article 220. Also note that Part IV covers “Feeders and Services” and is considered the Optional Method, which we will not discuss here, but will dive into in a later lesson. Just note that the major difference between the two is that each load in Part III has its own demand factor and in Part IV the Watts or Volt-Amps from each load are added up and have a demand factor for all loads added together. 

Demand Factors

You might be asking yourself, “What is a demand factor?”. I could write the definition found in article 100 of the codebook but it doesn’t make sense to the average person. In my experience teaching the code, my students usually draw a blank on that one. Think of it as a multiplier that is used to account for variations or special circumstances, in different types of loads. For example, a continuous load is one that is expected to be running for three hours or more. Typically you would just size the conductors at 100% of the current rating of the circuit, but because this is a continuous load, more heat will build up in the conductor over time. In this situation, the ampere rating of the load must be multiplied by 125% per article 215.2(A)(1)(a) as well as 210.19(A)(1) of the NEC. This is to account for this heat buildup, so the insulation around the conductors does not overheat. This puts the actual load at only 80% of the circuit rating, which is less stress on the conductor insulation. 

Not all demand factors increase the circuit to the load. In Part III of Article 220, we use demand factors to size the feeder and the services for a dwelling unit. If feeders and services are sized without the demand factors in article 220, the breakers and feeder conductors would be oversized for the actual demand. This is absolutely ok in all technical senses, but it may be wasteful and therefore unnecessary if the load will never reach those levels. In this instance, we could apply demand factors to decrease the size of the expensive copper conductors and still be within safe code speculations. 

The reason we need to understand demand factors is that part of both the Standard Method and Optional Method use percentage multipliers to account for these varying demand loads. So we apply demand factors to the calculation to help give us extra allowances.

The Standard Method

So let’s get started! As previously stated in this article, Part III (Standard Method) has demand factors for each type of load. I recommend getting out your 2020 National Electrical Code book and following along. I’m going to use Example D1(a) found in Annex D. 

To find the general lighting load, we have to go to 220.14(J) “Dwelling Units”. This tells us to use three VA per square foot of the dwelling. Going to Annex D1(a) we see that this dwelling is 1,500 square feet. So what we read in 220.14(J) tells us to multiply 1,500 by 3 VA per square foot, which gives us 4,500 VA

Before we add a demand factor to the general lighting load, we need to add the Small-Appliance and Laundry loads. A lot of people miss this step. For this, we go to 220.52(A) and (B). These two references state that we can use them with the demand factors found in Table 220.42, which is the same demand factor we use for the general lighting load. 220.52(A) tells us that the two Small-Appliance branch circuits that are required by 210.11(C)(1) must be calculated at 1,500 VA each. 220.52(B) tells us that the laundry branch circuit also must be calculated at 1,500 VA. 

So we would add up the three loads (lighting, laundry, and small appliance) and get 4,500 VA + 1,500 VA + 3,000 VA = 9,000 VA. Now that we have this information added together, we can do the demand factor calculation that is in Table 220.42.

Table 220.42 of the NEC, Lighting Load Demand Factors – Source: link.nfpa.org

Table 220.42 says to take the first 3,000 VA at 100%, and the entire remainder up to 120,000 VA at 35%. So let’s add that in:

General Lighting4,500 VA
Small-Appliance            3,000 VA
Laundry1,500 VA
Total Lighting & Small Appliance Load9,000 VA
3,000 VA at 100% 3,000 VA
Remainder 6,000 VA at 35%                  2,100 VA
Total Demand Load5,100 VA
Rather than using 9,000 VA, code lets us derate based on realistic usage down to 5,100 VA

In some standard method calculations, we will have another demand factor we can apply based on the number of appliances fastened in place. 220.53 doesn’t apply to this dwelling found in the Annex D example, because it doesn’t have any appliances other than the “Small-Appliance” that serves the countertops. But if it did, we would be allowed a demand factor of 75% if we had four or more appliances fastened in place, per 220.53.

The next step is 220.54, electric clothes dryer. In the Annex D example, we have a 5,500 VA dryer. 220.54 says that we can’t go under 5000 Watts, but if the nameplate rating of the dryer is higher than 5000 Watts then we would have to use that value. Therefore we know that 5500 Watts must be the nameplate rating. There is also a table in 220.54 for dryers, however, we don’t get a demand factor until we have over four. Look at that table for more info. Our dryer load, therefore, equals 5,500 VA.

Next is 220.55, electric cooking equipment. In this Annex D example, we have a unit that’s rated at 12,000 Watts. We have to go to Table 220.55 for this to get our demand factor. We can see in Table 220.55 Column C which is over 8¾ kW but not over 12 kW the demand factor is 8 kW or 8,000 Watts = 8,000 VA.

So let’s put this all together. 

General Lighting load4,500 VA
Small-Appliance3,000 VA
Laundry1,500 VA
Total Lighting Load9,000 VA
Next, apply demand factors to that 9,000 VA
First 3,000 VA at 100%3,000 VA
6,000 VA remainder at 35%2,100 VA
Net Load5,100 VA
Finally, add net load to Appliances for total calculated load
Range8,000 VA
Dyer5,500 VA
Total Calculated Load18,600 VA
After demand factors are applied, add appliances to the net load, or demand load

We have to remember what we are doing when doing a load calculation. We are sizing the service entrance or feeder conductors, and the main breaker for an occupancy – in this case, a dwelling unit. Like sizing most loads, we use the amperage of the load to size the breaker and conductors. That might leave you thinking, “Why have we been using VA and Watts up to this point?” It’s as simple as basic Ohm’s law, P/E=I. Our net load is 18,600 VA. The voltage in most occupancies is 240 volts. 

18,600 VA / 240 V = 78 A

Service Entrance Conductor or Feeder Size

There are a couple of codes that don’t allow us to put anything less than a 100 Amp service on a dwelling unit. The code reference for this claim can be found in 230.42(B) and 230.79; feel free to read into them at your leisure. 

So the next step is to go to Table 310.12 to size the service conductor or feeder. Table 310.12 tells us 4-AWG copper or 2-AWG aluminum is the conductor size for 100 Amps or less. So we’re going to use 4-AWG copper in this example. Now, we don’t have to use Table 310.12, we can also use Table 310.16 and manually look through the tables to figure out what we need, and add ambient correction factors, to get a very precise answer. But Table 310.12 is a kind of “shortcut” table if you will. As long as no ambient temperature correction factors or adjustment factors are being calculated, T310.12 is just fine to use. If you do have adjustments and/or corrections to factor in, you MUST use Table 310.16, rather than Table 310.12.

Breaker Size 

To size the breaker we have to see if there is such a thing as a 100 Amp breaker. Most people in the field already know there is, but for educational purposes let’s look at Table 240.6(A). This table gives us a list of the standard sizes for breakers, and 100 Amp is on the list. Now, there are other things to consider with more advanced dwellings, as well as larger services and the ability to add more or fewer demand factors. Not every dwelling is the same. We will cover more lessons on more advanced dwelling calculations in other articles.


When prepping for an exam you should study Article 220 Part III. Your questions probably won’t ask you to size a full house, but more than likely you will be asked to calculate the individual loads such as a dryer or a range. 

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