1792 United States presidential election in Georgia

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1792 United States presidential election in Georgia
Flag of the State of Georgia (non-official).svg
  1788–89 November 2 – December 5, 1792 1796  
  George Washington Portrait (3x4 cropped).jpg George Clinton by Ezra Ames (3x4 cropped).jpg
Nominee George Washington George Clinton
Party Independent Democratic-Republican
Home state Virginia New York
Electoral vote44

The 1792 United States presidential election in Georgia took place between November 2 and December 5, 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]

Contents

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]

Results

1792 United States presidential election in Georgia [3]
PartyCandidateVotesPercentageElectoral votes
Independent George Washington (incumbent)4
Totals4

See also

References

  1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson: McFarland & Company. ISBN   9780786410170.
  2. "1792 Presidential General Election Results". U.S. Election Atlas. Retrieved July 10, 2023.
  3. "1792 Presidential Election". 270towin.com. Retrieved July 10, 2023.