Port Jefferson, Ohio
Port Jefferson, Ohio | |
|---|---|
Main Street downtown | |
Location of Port Jefferson, Ohio | |
Location of Port Jefferson in Shelby County | |
| Coordinates: 40°19′49″N 84°05′33″W / 40.33028°N 84.09250°W | |
| Country | United States |
| State | Ohio |
| County | Shelby |
| Area | |
• Total | 0.15 sq mi (0.40 km2) |
| • Land | 0.15 sq mi (0.39 km2) |
| • Water | 0 sq mi (0.00 km2) |
| Elevation | 971 ft (296 m) |
| Population (2020) | |
• Total | 308 |
| • Density | 2,000/sq mi (780/km2) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| ZIP code | 45360 |
| Area codes | 937, 326 |
| FIPS code | 39-64262 |
| GNIS feature ID | 2399004 |
Port Jefferson is a village in Salem Township, Shelby County, Ohio, United States. The population was 308 at the 2020 census.