Electrical Load Calculations Practice

Dwelling service loads, demand factors, range demand and the optional method — the single biggest block of marks on the master calculations portion.

Practise Load Calcs

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What it is, and why electricians need it

A load calculation adds up every load a service must supply, applies the demand factors that recognise not everything runs at once, and converts the result into amperes to size the service. It is the most heavily weighted calculation on the exam: Services and Service Equipment accounts for 8 of 30 scored items on the Texas master calculations portion, more than a quarter of it.

The reason demand factors exist is worth internalising rather than memorising. Four 12 kW ranges in an apartment building represent 48 kW of connected load, but the calculated demand is 17 kW, because the chance of all four running flat out simultaneously is negligible. Without that diversity allowance every service in the country would be two or three times larger than it needs to be.

The discipline that makes these calculations reliable is keeping the categories separate. General lighting, small-appliance circuits and the laundry circuit are added together and then put through the lighting demand factors. Ranges go through their own table. Dryers have their own floor value. Fixed appliances may get a 75 percent factor but only if there are four or more of them. Heating and air conditioning are handled separately again, and you take the larger of the two rather than both. Mixing a range into the lighting subtotal is the most common way to get the whole calculation wrong.

The optional method is a genuine shortcut where it applies: 100 percent of the first 10 kVA of general loads plus 40 percent of the remainder. But it is conditional on the dwelling being served by a single three-wire 120/240-volt or 208Y/120-volt service or feeder of 100 amperes or greater — and the Texas bulletin instructs candidates not to use optional methods unless a question directs them to.

Formulas and principles

  • General lighting = floor area × 3 VA per ft² (dwellings)

    Floor area from outside dimensions, excluding open porches, garages and unused space. Covers all general lighting and general-use receptacles.

    Commercial occupancies differ: stores and schools 3, offices and banks 3.5, hospitals and hotels 2, warehouses 0.25.

  • Small appliance 1,500 VA per circuit; laundry 1,500 VA

    Minimum two small-appliance circuits plus one laundry circuit in a dwelling.

    Per circuit actually installed, so three small-appliance circuits means 4,500 VA.

  • Lighting demand: 3,000 VA at 100%, 3,001–120,000 at 35%, remainder at 25%

    Applies to the general lighting, small-appliance and laundry subtotal only.

    Not to ranges, dryers, fixed appliances, heating or air conditioning.

  • Range demand, Column C: 1→8 kW, 2→11, 3→14, 4→17, 5→20, 6→21

    Household ranges rated not over 12 kW each.

    Over 12 kW through 27 kW: add 5 percent per kilowatt above 12 kW.

  • Dryer = 5,000 VA or nameplate, whichever is larger

    Each household electric clothes dryer in a dwelling.

  • Appliance demand = nameplate × 75% for four or more fastened in place

    Excluding ranges, dryers, space heating and air conditioning from the count.

  • Optional method general load = 10,000 VA + 40% of the remainder

    Where the dwelling has a single 120/240 V or 208Y/120 V three-wire service of 100 A or greater.

    Heating and air conditioning are added separately afterwards.

  • Service current = calculated VA ÷ service voltage

    Divide by 1.732 × voltage as well for a three-phase service.

Now try it on 20 questions

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Start the Load Calcs drill

Worked examples

Scenario: A 2,000 square foot dwelling has two small-appliance circuits, one laundry circuit, a 12 kW range and a 4.8 kW dryer, on a 240-volt single-phase service. Find the calculated load and minimum service size by the standard method.

  1. General lighting: 2,000 × 3 = 6,000 VA.
  2. Small appliance: 2 × 1,500 = 3,000 VA.
  3. Laundry: 1,500 VA.
  4. Subtotal subject to demand factors: 10,500 VA.
  5. Demand factors: 3,000 at 100% plus 7,500 at 35% = 3,000 + 2,625 = 5,625 VA.
  6. Range, Column C for one appliance not over 12 kW: 8,000 VA.
  7. Dryer: nameplate 4,800 VA, but the floor is 5,000 VA, so use 5,000 VA.
  8. Total: 5,625 + 8,000 + 5,000 = 18,625 VA.
  9. Service current: 18,625 ÷ 240 = 77.6 A, so a 100-ampere service.

Answer: 18,625 VA, or 77.6 A — a 100-ampere service

Scenario: A single household range is rated 16 kW. What demand load is used?

  1. Column C for one appliance not over 12 kW is 8 kW.
  2. The range is 16 kW, so it is 4 kW over the 12 kW threshold.
  3. Note 1: increase Column C by 5 percent per kilowatt above 12 kW.
  4. Increase = 5% × 4 = 20 percent.
  5. Demand = 8 × 1.20 = 9.6 kW.
  6. Note this is 5 percent per kilowatt, not 5 percent once — that error would give 8.4 kW.

Answer: 9.6 kW

Scenario: Under the optional method, a dwelling's general loads total 25,000 VA. Find the calculated general load.

  1. First 10,000 VA at 100 percent: 10,000 VA.
  2. Remainder: 25,000 − 10,000 = 15,000 VA.
  3. That remainder at 40 percent: 15,000 × 0.40 = 6,000 VA.
  4. General load = 10,000 + 6,000 = 16,000 VA.
  5. Air conditioning and heating would then be added separately, taking the larger of the two.

Answer: 16,000 VA of general load

Common mistakes

  • Putting the range or dryer through the lighting demand factors. Those factors apply only to the general lighting, small-appliance and laundry subtotal. Ranges, dryers, fixed appliances, heating and air conditioning each have their own rule and are added afterwards.
  • Adding 5 percent once for a range over 12 kW. It is 5 percent for each additional kilowatt above 12 kW. A 16 kW range is 4 kW over, so 20 percent, giving 9.6 kW rather than 8.4 kW.
  • Using the dryer nameplate when it is below 5,000 VA. The load is 5,000 VA or the nameplate, whichever is larger. A 4,800 VA dryer is calculated at 5,000 VA.
  • Applying the 75 percent appliance factor to fewer than four appliances. It needs four or more fastened in place on the same feeder or service, and ranges, dryers, heating and air conditioning do not count toward the four.
  • Including both heating and air conditioning at full value. They cannot run simultaneously, so you take the larger of the two rather than adding both.
  • Counting receptacles separately in a dwelling. The 3 VA per square foot figure already covers all general lighting outlets and general-use receptacles in a dwelling. Commercial occupancies are the opposite — there receptacles are calculated separately.
  • Using the optional method when the question has not invited it. The Texas bulletin tells candidates not to consider optional calculation methods unless the question specifically directs them to. Default to the standard method.

Related practice topics

Topics that come up alongside this one on the exam.

Questions about this topic

Why do load calculations use demand factors at all?

Because no building ever has every load running at full power simultaneously. A dwelling with four 12 kW ranges has 48 kW of connected range load but a calculated demand of only 17 kW, because the probability of all four running flat out at the same moment is negligible. Demand factors let services be sized for realistic peak load rather than for an impossible worst case, which keeps services affordable without making them inadequate.

What is the difference between the standard and optional methods?

The standard method works through each load category with its own demand rule. The optional method is a shortcut: 100 percent of the first 10 kVA of general loads plus 40 percent of the remainder, with heating and air conditioning added separately. The optional method may only be used where the dwelling is served by a single 120/240-volt or 208Y/120-volt three-wire service or feeder with an ampacity of 100 amperes or greater.

Which loads do the dwelling lighting demand factors apply to?

Only the general lighting load, the small-appliance circuits and the laundry circuit. Those three are added together and then put through the factors: the first 3,000 VA at 100 percent, from 3,001 to 120,000 VA at 35 percent, and the remainder at 25 percent. Ranges, dryers, fixed appliances, heating and air conditioning each have their own rules and are added afterwards at full calculated value.

How do I handle a range rated over 12 kW?

Note 1 to the cooking appliance table applies to ranges over 12 kW but not over 27 kW, all of the same rating: increase the Column C figure by 5 percent for each additional kilowatt above 12 kW. A single 16 kW range is 4 kW over, so the increase is 20 percent and the demand is 8 × 1.20 = 9.6 kW. Candidates often add 5 percent once instead of 5 percent per kilowatt.

When does the 75 percent appliance demand factor apply?

Only where four or more appliances fastened in place are served by the same feeder or service in a dwelling, and electric ranges, clothes dryers, space-heating equipment and air-conditioning equipment are expressly excluded from the count. Check the count before applying it — with only three qualifying appliances you include the full nameplate total, so adding a fourth can actually reduce the calculated load.

Why is a 4,800 VA dryer calculated at 5,000 VA?

The dryer load is 5,000 watts or the nameplate rating, whichever is larger. The 5,000 VA floor exists so the calculation is not undermined when a small dryer is later replaced with a bigger one. A 5,500 VA dryer is calculated at 5,500 VA, since the nameplate is then the larger figure.

How do commercial load calculations differ from dwelling ones?

The unit lighting loads differ by occupancy — stores and schools are 3 VA per square foot, offices and banks 3.5, hospitals and hotels 2, warehouses only 0.25 — and, crucially, receptacle outlets are calculated separately rather than being included in the general lighting load as they are in a dwelling. That single difference catches people who have only ever done residential calculations.

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