Residential Calculations: Estimating Elements of the Electrical System

By James G. Stallcup | Apr 15, 2009
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What method should be used to calculate the volt-amperes (VA) of a residence?

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An estimator asked what method to use to calculate the volt-amperes (VA) of a residence. He hoped to use the results as a guide to estimate the elements of the electrical system and wanted the standard method applied. Below, I explain my interpretation of this method to determine the VA and serve as an estimating tool.

Grouping loads

The standard calculation requires the loads to be divided as follows:

  • Load 1: General lighting and receptacle and small appliance loads
  • Load 2: Cooking equipment loads
  • Load 3: Special appliance loads
  • Load 4: Dryer load
  • Load 5: Heating load
  • Load 6: Largest motor

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. Use this total to determine the number of outlets for lighting and general-purpose receptacles. Installing a 20-amp circuit supplied by 120 volts, the 8,400 VA divided by 2,400 VA (20 A × 120 V = 2,400 VA) is 3.5, when rounded up requires four 20-amp circuits. Five 15-amp circuits would be the minimum required for a 15-amp circuit.

Small appliance loads

At least two small appliance circuits per 210.52(A) must be installed to supply receptacle outlets in the kitchen, breakfast room, pantry and dining room. One is needed for the laundry room per 210.52(B). Small appliance circuits are calculated at 1,500 VA each. Therefore, 4,500 VA is added to the general lighting load. These outlets are not to be connected to the circuits supplied by the general-purpose or special appliance circuits.

A demand factor, as permitted in Table 220.42, can be applied to these loads. Depending on the VA, the first 3,000 VA can be calculated at 100% and the remaining VA at 35%.

Special appliance load

Directed circuits usually supply special appliance circuits, which are not connected to the general-purpose or small appliance circuits. Such loads are water heaters, heating units, ranges, air conditioning units, cooking equipment, motors, etc. For example, 10 kW is transposed to 10,000 VA and used in the calculation to determine the total load in VA.

Fixed appliance loads, such as dishwashers, disposals, water heaters, compactors, etc., are permitted to have a 75% demand factor applied to their total VA.

Appliances that are not included when using this demand factor are heating units, air conditioning units, dryers or cooking equipment. When these appliances are removed from the calculation, all other appliances are considered fixed and are eligible for the 75% demand.

Demand factor

As mentioned, a term constantly used in modern design is “demand factor,” the ratio of the maximum demand of a system (or part of a system) to the connected load on the system (or part of the system). It is always less than 1.

Application of demand factors


Tables 220.12 and 220.42 may be applied as follows:

General-purpose lighting and receptacle loads—2,800 sq. ft. × 3 VA = 8,400 VA

Small appliance and laundry loads—1,500 VA × 3 = 4,500 VA

ANSWER: 8,400 VA + 4,500 VA = 12,900 VA

Application of demand factors

First 3,000 VA × 100% = 3,000 VA

Next 9,900 VA × 35% = 3,465 VA

ANSWER: 3,000 VA + 3,465 VA = 6,465 VA


Table 220.55, Col. B (65%) may be applied as follows:

8,500-VA cooktop and 8,000-VA oven

Application of demand factors

ANSWER: 8,500 VA + 8,000 VA × 65% = 10,725 VA


Section 220.53 (75% rule) may be applied as follows:

The fixed appliance load of 13,200 VA consists of a water heater, water pump, disposal, compactor, dishwasher, microwave and blower motor.

Application of demand factors

ANSWER: 13,200 VA × 75% = 9,900 VA


Table 220.54 permits a 5,000-VA dryer to be calculated at 5,000 VA.


Section 220.60 permits the largest between a 10,000-VA heating unit and 5,500-VA air conditioning unit with the smaller load dropped.


Section 220.50 requires 25% for the largest motor (2,600 VA water pump) to be added to the calculation at 650 VA (25% of 2,600 VA = 650 VA).


Add the VA of 6,465; 10,725; 9,900; 5,000; 10,000; and 650 together, and 42,740 VA is derived. Total amps is 178 (42,740A/240V = 178A)

This procedure can be used to derive the VA to determine the amps for selecting the service elements and aid in estimating wiring methods and equipment.

About The Author

James G. Stallcup is the CEO of Grayboy Inc., which develops and authors publications for the electrical industry and specializes in classroom training on the NEC and OSHA, as well as other standards. Contact him at 817.581.2206.





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