1792 United States presidential election in Kentucky
Election in Kentucky
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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]
Results
| 1792 United States presidential election in Kentucky[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | – | – | 4 | |
| Totals | – | – | 4 | ||
See also
References
- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788-1860: The Official Results by County and State. Jefferson: McFarland & Company. ISBN 9780786410170.
- ^ "1792 Presidential General Election Results". U.S. Election Atlas. Retrieved July 14, 2023.
- ^ "1792 Presidential Election". 270towin.com. Retrieved July 14, 2023.
