SolarSizer

What Size Solar System Do I Need for an Average Home?

Square footage is not in the EIA figures. The average home here is 10,791 kWh a year, and the states at the ends of that table are part of the answer.

Full breakdown: what size solar system.

Panel-count context for a whole house: how many solar panels.

Quick answer: An average U.S. home, at 10,791 kWh a year, needs about 6.1 kW on the Phoenix yield: 10,791 ÷ 1,755 = 6.15 kW. The same yield puts Louisiana's 14,774 kWh at 8.42 kW and Hawaii's 6,178 kWh at 3.52 kW. A generic 4 peak sun hour estimate on the monthly 899 kWh is 8.71 kW. Run the house, not the national average, in PVWatts or the solar size calculator.
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National average and the two end states

CaseAnnual kWh÷ 1,755 kWh/kW400 W panels, rounded upSource
U.S. residential average10,7916.15 kW16EIA; PVWatts
Louisiana (highest)14,7748.42 kW22EIA; PVWatts
Hawaii (lowest)6,1783.52 kW9EIA; PVWatts
Generic, 4 peak sun hours and 14% losses899 kWh/month8.71 kW22EIA; Battle Born; PVWatts v5
Generic, 5 peak sun hours and 14% losses899 kWh/month6.97 kW18EIA; Battle Born; PVWatts v5

About 6.1 kW is 10,791 ÷ 1,755 rounded to one decimal (6.15 kW before that rounding). Panel counts use the unrounded quotient ÷ 400, rounded up. 400 W is an illustrative divisor.

How to size it

EIA's FAQ is an average of electricity sold, not a floor plan. The highest state in the same note is Louisiana and the lowest is Hawaii, so a single national kilowatt is already a range before solar resource changes.

  1. Phoenix path for the U.S. average: 10,791 ÷ 1,755 = 6.15 kW, about 6.1 kW. The yield is the Phoenix 1 kW run (tilt 20°, azimuth 180°, 14% losses).
  2. Same yield, other annual totals: 14,774 ÷ 1,755 = 8.42 kW, and 6,178 ÷ 1,755 = 3.52 kW. This step moves the bill and holds the weather still, so you can see the EIA spread by itself.
  3. Generic path from the monthly figure. 899 ÷ 30 = 29.97 kWh/day. At 4 peak sun hours: (29.97 ÷ 4) ÷ (1 − 0.14) = 8.71 kW. At 5: 6.97 kW.
  4. Illustrative panels: the 6.15 kW Phoenix case rounds up to 16 modules of 400 W. The 4-hour generic case rounds up to 22.

The generic 4-hour result (8.71 kW) is higher than the Phoenix year (6.15 kW) because 4 peak sun hours is a cautious baseline and the loss term enlarges the array. Phoenix's own year produces 1,755 kWh per kW, with annual plane-of-array sun of 6.54 kWh/m²/day and a December low of 4.75. Use the Phoenix column as a sunny-climate year, and the 4-hour column when you only have Battle Born's baseline. Your address belongs in PVWatts. Daily loads belong in the solar size calculator.

The off-grid reading of this same month — 899 ÷ 30 ≈ 30 kWh/day, then watt-hours ÷ peak sun hours — is a different page, because that method adds 20 to 30 percent and does not use the 1,755 annual yield.

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FAQ

What size solar system does an average U.S. home need?

EIA reports 10,791 kWh a year. Divided by Phoenix PVWatts' 1,755 kWh per kW, that is 6.15 kW, about 6.1 kW.

How wide is the state range on the same yield?

Louisiana's 14,774 kWh ÷ 1,755 = 8.42 kW. Hawaii's 6,178 kWh ÷ 1,755 = 3.52 kW. The weather assumption stays the Phoenix run so the EIA totals are the only thing moving.

What is the 4 peak sun hour estimate?

899 kWh ÷ 30 = 29.97 kWh/day. (29.97 ÷ 4) ÷ (1 − 0.14) = 8.71 kW. At 5 peak sun hours the result is 6.97 kW.

Why is the 4-hour result larger than 6.1 kW?

About 6.1 kW uses a full Phoenix year at 1,755 kWh per kW. The 8.71 kW figure pushes one day through 4 peak sun hours and divides by (1 − 0.14). Phoenix's annual plane-of-array sun is 6.54 and its December is 4.75, so the national 4-hour baseline is the more conservative of the two.

Related questions

All of these questions are listed on the topic map.

Sources