The 1817 Georgia gubernatorial election was held on 10 November 1817 in order to elect the Governor of Georgia. Democratic-Republican candidate and incumbent acting Governor William Rabun defeated fellow Democratic-Republican candidate John Clark in a Georgia General Assembly vote.[1]
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General election
editOn election day, 10 November 1817, Democratic-Republican candidate William Rabun won the election against his opponent fellow Democratic-Republican candidate John Clark. Rabun was officially sworn in as the 29th Governor of Georgia on 10 November 1817.[2]
Results
edit| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic-Republican | William Rabun | 62 | 52.10 | |
| Democratic-Republican | John Clark | 57 | 47.90 | |
| Total votes | 119 | 100.00 | ||
| Democratic-Republican hold | ||||
References
edit- ^ "Georgia 1817 Governor". Tufts Digital Collations and Archives. A New Nation Votes: American Election Returns 1787–1825. Tufts University. Retrieved 8 December 2023.
- ^ "GA Governor". ourcampaigns.com. 18 July 2022. Retrieved 8 December 2023.
- ^ Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson: McFarland & Company. ISBN 9780786414390.