1792 United States presidential election in Georgia
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Elections in Georgia |
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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]
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] | |||||
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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 10, 2023.
- ^ "1792 Presidential Election". 270towin.com. Retrieved July 10, 2023.
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- 1792 United States presidential election by state
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- 1792 Georgia (U.S. state) elections
- 1792 elections
- 1792 elections in North America
- 1792 elections in the United States
- 1792 in Georgia (U.S. state)