See main article: 1792 United States presidential election.
| Election Name: | 1792 United States presidential election in Georgia |
| Country: | Georgia (U.S. state) |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1788-89 United States presidential election in Georgia |
| Previous Year: | 1788-89 |
| Next Election: | 1796 United States presidential election in Georgia |
| Next Year: | 1796 |
| Election Date: | 2 November - 5 December 1792 |
| Image1: | Gilbert Stuart Williamstown Portrait of George Washington.jpg |
| Nominee1: | George Washington |
| Party1: | Independent (United States) |
| Home State1: | Virginia |
| Running Mate1: | None |
| Electoral Vote1: | 4 |
The 1792 United States presidential election in Georgia took place between 2 November and 5 December 1792, as part of the 1792 United States presidential election. The state legislature chose four representatives, or electors to the Electoral College, who voted for President and Vice President. Georgia had lost one elector compared to the previous election in 1788-89.[1]
Georgia cast four electoral votes for the Independent candidate and incumbent President George Washington, as he ran effectively unopposed. The electoral votes for Vice president were cast for Democratic-Republican George Clinton from New York. These electors were elected by the Georgia General Assembly, the state legislature, rather than by popular vote.[2]
| 1792 United States presidential election in Georgia[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | – | – | 4 | |
| Totals | – | – | 4 | ||