See main article: 1792 United States presidential election.
| Election Name: | 1792 United States presidential election in Kentucky |
| Type: | presidential |
| Ongoing: | no |
| Next Election: | 1796 United States presidential election in Kentucky |
| 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 Kentucky took place between 2 November and 5 December 1792, as part of the 1792 United States presidential election. The voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President. It was the first presidential election Kentucky participated in since being admitted to the Union on 1 June 1792.[1]
Kentucky 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 Thomas Jefferson from Virginia, the only state to do so. The state was divided into four electoral districts with one elector each, whereupon each district's voters chose the electors.[2]
| 1792 United States presidential election in Kentucky[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | – | – | 4 | |
| Totals | – | – | 4 | ||