1792 United States presidential election in Georgia

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1792 United States presidential election in Georgia
Flag of Georgia (U.S. state).svg
  1788-89 2 November - 5 December 1792 1796  
  Gilbert Stuart Williamstown Portrait of George Washington.jpg
Nominee George Washington
Party Independent
Home state Virginia
Running mateNone
Electoral vote4

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]

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

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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.