Jefferson, Ohio | |
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![]() Ashtabula County Courthouse | |
![]() Location of Jefferson, Ohio | |
![]() Location of Jefferson in Ashtabula County | |
Coordinates: 41°44′19″N 80°45′25″W / 41.73861°N 80.75694°W | |
Country | United States |
State | Ohio |
County | Ashtabula |
Area | |
• Total | 2.54 sq mi (6.57 km2) |
• Land | 2.54 sq mi (6.57 km2) |
• Water | 0.00 sq mi (0.00 km2) |
Elevation | 961 ft (293 m) |
Population | |
• Total | 3,226 |
• Density | 1,271.58/sq mi (491.04/km2) |
Time zone | UTC-5 (Eastern (EST)) |
• Summer (DST) | UTC-4 (EDT) |
ZIP code | 44047 |
Area code | 440 |
FIPS code | 39-38500[4] |
GNIS feature ID | 1042061[2] |
Website | https://jeffersonohio.gov/ |
Jefferson is a village in Ashtabula County, Ohio, United States. The population was 3,226 at the 2020 census.[3] It is the county seat of Ashtabula County.[5]
Municipalities and communities of Ashtabula County, Ohio, United States | ||
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Footnotes | ‡This populated place also has portions in an adjacent county or counties |