Quick answer: An electrical load is a device or part of a circuit that consumes electric power (Wikipedia, electrical load). To estimate a home's connected load, add the input wattages of its appliances; use their operating hours separately to calculate energy use.
Connected load, peak demand and electricity bills answer different questions. This guide explains the arithmetic for each, with assumed appliance examples, and shows why a running-cost estimate does not establish the service capacity your home needs.
What should you calculate: watts, amps or kWh?
Use watts for power, amperes for current and kilowatt-hours for energy over time. A kilowatt-hour is the energy delivered by a kilowatt over an hour. (Wikipedia, kilowatt-hour)
| Your question | Quantity to calculate | Calculation or distinction |
|---|---|---|
| What is the combined listed load? | Connected power, W | Add each appliance's input wattage, including quantity |
| How much current does a resistive load draw? | Current, A | Divide power by voltage: I = P ÷ V |
| How much energy does it use? | Energy, kWh | Multiply watts by operating hours, then divide by 1,000 |
| What will that energy cost? | Cost in your billing currency | Multiply kWh by your utility rate per kWh |
The energy and cost calculations follow the DOE appliance energy-use guide. Its annual method multiplies daily energy by days used per year, then by the electricity rate.
Keep the time period attached to the result. Watts describe the load while it operates; kWh describes how much energy it uses during the chosen period. An annual energy total alone cannot tell you the highest simultaneous load.
How do watts translate into circuit current?
For a resistive load, divide watts by volts to estimate amperes. For an AC load more generally, include power factor: real power = voltage × current × power factor. According to Wikipedia's power-factor explanation, power factor is real power divided by apparent power; voltage and current are in phase in a purely resistive circuit.
This also distinguishes load types. According to the same power-factor source, reactive components such as inductors and capacitors store energy and shift the timing between voltage and current. A calculation that treats every AC appliance as purely resistive leaves that distinction out.
Worked example: a resistive heater
Assumed inputs: a purely resistive heater drawing 1,800 W from a 120 V supply, with power factor 1.
- Formula: I = P ÷ (V × power factor).
- Denominator = 120 × 1 = 120.
- Current = 1,800 ÷ 120 = 15 A.
- Arithmetic check: P = V × I = 120 × 15 = 1,800 W.
That result describes the assumed heater's current. It is not a recommendation for a breaker or conductor size. For another AC appliance, use its stated current or include its power factor in the calculation rather than transferring the heater's assumption.
How much energy does a lighting load use?
Multiply total wattage by operating time, then convert watt-hours to kWh. The DOE energy-use formula is daily kWh = wattage × hours per day ÷ 1,000.
The electricity cost calculator also includes quantity. It calculates monthly energy from daily energy and days per month, then annualizes that monthly result.
Worked example: a group of LED bulbs
Assumed inputs: 6 LED bulbs drawing 9 W each, used for 5 hours per day. Use the calculator's planning figure of 30 days per month. Let r be your actual electricity rate per kWh; no rate is assumed here.
- Total power = quantity × bulb wattage = 6 × 9 = 54 W.
- Daily energy = total power × hours ÷ 1,000 = 54 × 5 ÷ 1,000 = 270 ÷ 1,000 = 0.27 kWh.
- Monthly energy = daily energy × days = 0.27 × 30 = 8.1 kWh.
- Annualized energy = monthly energy × 12 = 8.1 × 12 = 97.2 kWh.
- Daily cost = 0.27 × r; monthly cost = 8.1 × r; annualized cost = 97.2 × r.
That annual estimate represents 30 × 12 = 360 operating days. With an alternative assumed schedule of 365 operating days, annual energy = 0.27 × 365 = 98.55 kWh, and energy cost = 98.55 × r. Use the same schedule when comparing estimates.
For a replacement comparison, use the LED savings calculator. A lower input wattage reduces calculated energy use when quantity and operating hours stay the same; compare the intended lighting output separately from the power calculation.
Why is peak demand different from connected load?
Connected load adds the listed loads; peak demand describes their highest combined use. According to Wikipedia's demand-factor definition, demand factor is the fraction of maximum possible demand actually used and is at most 1.
Worked example: connected load versus measured peak
Assumed inputs: appliances with a combined connected load of 20 kW and a measured peak demand of 12 kW over the observation period.
- Demand factor = peak demand ÷ connected load.
- Demand factor = 12 ÷ 20 = 0.6.
- Percentage = 0.6 × 100 = 60%.
- Difference between connected load and observed peak = 20 − 12 = 8 kW.
The difference is a result for that assumed observation period, not permission to add that much equipment. Do not apply the example's demand factor to a different home as a service-sizing allowance.
Diversity factor asks a related but different question. According to Wikipedia's diversity-factor definition, it divides the sum of individual non-coincident maximum loads by the maximum demand of the complete system and is at least 1. It compares separate peaks with the combined peak, so it is not automatically the reciprocal of a demand factor based on connected nameplate load.
How should you estimate a cycling appliance?
Use operating time at the stated power, rather than assuming it draws that power throughout every plugged-in hour. According to the DOE appliance guide, nameplate wattage represents maximum consumption; its refrigerator estimate divides plugged-in time by 3 to approximate time at maximum wattage.
Treat that divisor as DOE's refrigerator estimation method, not a universal duty cycle for every appliance. For your estimate:
- Separate the input power from the number of hours the appliance is connected.
- Choose operating hours that correspond to the power figure used.
- Calculate energy as watts × those hours ÷ 1,000.
- Use that energy result for cost; retain the full operating load when considering simultaneous use.
Can these estimates size a home's electrical service?
These calculations describe load and energy; they do not establish a safe service, breaker or wire size. According to Wikipedia's ampacity definition, ampacity is the maximum current a conductor can carry continuously under its conditions of use without exceeding its temperature rating.
A current calculation therefore answers only part of the capacity question. The assumed heater's 15 A result does not assess the conductor's conditions of use, and the lighting example's annual kWh does not specify a peak current.
Use a qualified electrician for mains wiring and service-capacity decisions. Bring the appliance input ratings, which loads may run together and any measured demand data. Those inputs are more useful than selecting a service rating from floor area alone.
FAQ
What counts as an electrical load?
A component or part of a circuit that consumes electric power is a load, according to Wikipedia's electrical-load definition. Calculate a group of identical loads by multiplying each unit's wattage by the quantity.
Can I calculate amps by dividing watts by volts?
Yes for the resistive assumption used above. For an AC load generally, use I = P ÷ (V × power factor), consistent with Wikipedia's definition of power factor as real power divided by apparent power.
Is kWh the same as kW?
No: kW measures power, while kWh measures energy delivered over time. A kilowatt delivered for an hour is a kilowatt-hour. (Wikipedia, kilowatt-hour)
Should I add every appliance's nameplate wattage?
Do that to estimate connected load, but do not call the result measured peak demand. According to Wikipedia's demand-factor explanation, the fraction actually used can be less than the maximum possible demand.
Does reducing operating hours reduce the load while a light is on?
With wattage held constant, it reduces energy consumption rather than operating power. In the assumed bulb example, halving daily use gives 5 ÷ 2 = 2.5 hours, so daily energy = 54 × 2.5 ÷ 1,000 = 135 ÷ 1,000 = 0.135 kWh; the operating load remains 54 W.
Jack Shi
Founder & editor, LEDaskJack Shi builds and writes LEDask, an independent LED-lighting tools project operated by clooms. He designs the calculators, checks their formulas and reference values against published engineering data, and writes the guides across the site.
