1812 United States presidential election in Georgia
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Elections in Georgia |
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The 1812 United States presidential election in Georgia took place between 30 October and 2 December 1812, as part of the 1812 United States presidential election. The state legislature chose eight representatives, or electors to the Electoral College, who voted for President and Vice President. Georgia had gained two additional electors compared to the previous election in 1808.
Georgia cast eight electoral votes for the Democratic-Republican candidate and incumbent President James Madison over the Federalist candidate DeWitt Clinton. The electoral votes for Vice president were cast for Madison's running mate Elbridge Gerry from Massachusetts. These electors were elected by the Georgia General Assembly, the state legislature, rather than by popular vote.[1]
Results
1812 United States presidential election in Georgia[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | James Madison (incumbent) | – | – | 8 | |
Federalist | DeWitt Clinton | – | – | 0 | |
Totals | – | – | 8 |
See also
References
- ^ "1812 Presidential General Election Results". U.S. Election Atlas. Retrieved July 10, 2023.
- ^ "1812 Presidential Election". 270towin.com. Retrieved July 10, 2023.
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