See main article: 1800 United States presidential election.
| Election Name: | 1800 United States presidential election in Kentucky |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1796 United States presidential election in Kentucky |
| Previous Year: | 1796 |
| Next Election: | 1804 United States presidential election in Kentucky |
| Next Year: | 1804 |
| Election Date: | 31 October – 11 December 1800 |
| Image1: | Thomas Jefferson by Rembrandt Peale, 1800.jpg |
| Nominee1: | Thomas Jefferson |
| Party1: | Democratic-Republican Party (United States) |
| Home State1: | Virginia |
| Running Mate1: | Aaron Burr |
| Electoral Vote1: | 4 |
| Popular Vote1: | 119 |
| Percentage1: | 100.00% |
| Nominee2: | John Adams |
| Party2: | Federalist Party (United States) |
| Home State2: | Massachusetts |
| Running Mate2: | Charles C. Pinckney |
| Electoral Vote2: | 0 |
| Popular Vote2: | 0 |
| Percentage2: | 0.00% |
The 1800 United States presidential election in Kentucky took place between 31 October and 3 December 1800, as part of the 1800 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.[1]
Kentucky cast four electoral votes for the Democratic-Republican candidate and incumbent Vice President Thomas Jefferson over the Federalist candidate and incumbent President John Adams. The electoral votes for Vice president were cast for Jefferson's running mate Aaron Burr from New York. The state was divided into two electoral districts with two electors each, whereupon each district's voters chose the electors.[2]
| 1800 United States presidential election in Kentucky[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Thomas Jefferson | 119 | 100.00% | 4 | |
| Federalist | John Adams (incumbent) | 0 | 0.00% | 0 | |
| Totals | 119 | 100.00% | 4 | ||