How Many Solar Panels to Power a House Calculator
Work out how many solar panels it takes to power a house from your daily electricity use and local sun.
Your result
Enter your numbers and press the button. Your result appears here.
Formula
Panels = daily kWh × 1,000 ÷ (peak sun hours × panel watts × (1 − losses) × inverter efficiency), rounded up. Area = panels × area of one panel.
Example: 28.4 kWh a day ÷ (4.98 × 400 W × 0.86 × 0.96) × 1,000 = 17.3, so 18 panels (7.2 kW).
Correction to the old formula
The previous version divided by panel efficiency and panel power together, which inflated the answer about fivefold (70 panels for 30 kWh a day). Panel wattage already includes efficiency, so only system losses belong in the formula. 30 kWh a day at 5 sun hours with 300 W panels is about 25 panels.
Panels needed by daily use
At 4.98 peak sun hours (U.S. average), 14% losses and 96% inverter efficiency.
| Daily use | System | 400 W panels | Panel area |
|---|---|---|---|
| 10 kWh | 2.43 kW | 7 | 147 ft² |
| 15 kWh | 3.65 kW | 10 | 210 ft² |
| 20 kWh | 4.86 kW | 13 | 273 ft² |
| 28.4 kWh | 6.91 kW | 18 | 378 ft² |
| 35 kWh | 8.51 kW | 22 | 462 ft² |
| 40 kWh | 9.73 kW | 25 | 525 ft² |
| 50 kWh | 12.16 kW | 31 | 651 ft² |
Common questions
How many solar panels does an average house need?
The average U.S. home used 865 kWh a month in 2024, about 28.4 kWh a day (EIA). At 5 peak sun hours with 400 W panels that is about 18 panels.
Where do I find peak sun hours?
Use the irradiance by ZIP code calculator, or pick your state above.
Sources
- NREL PVWatts V8 API documentation: https://developer.nrel.gov/docs/solar/pvwatts/v8/
- NREL PVWatts v8 state results compiled by TheGreenWatt: https://thegreenwatt.com/average-peak-sun-hours-by-state/