United States presidential election in Georgia, 1824
Main article: United States presidential election, 1824
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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 nine 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 four different candidates from this party sought the Presidency. Georgia cast nine electoral votes for native 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.
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