1824 United States presidential election in Georgia
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Elections in Georgia |
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The 1824 United States presidential election in Georgia took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. The state legislature chose 9 representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Georgia cast 9 electoral votes for William H. Crawford.
Results
United States presidential election in Georgia, 1824[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | William H. Crawford | 9 | |||
Democratic-Republican | John Quincy Adams | 0 | |||
Democratic-Republican | Henry Clay | 0 | |||
Democratic-Republican | Andrew Jackson | 0 | |||
Totals | 9 |
References
- ^ "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved 28 February 2013.