1817 Georgia gubernatorial election
Appearance
November 10, 1817
| |||||||||||||||||
| |||||||||||||||||
| |||||||||||||||||
| Elections in Georgia |
|---|
The 1817 Georgia gubernatorial election was held on November 10, 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]
General election
[edit]On election day, November 10, 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 November 10, 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 2023-12-08.
- ^ "GA Governor". ourcampaigns.com. July 18, 2022. Retrieved 2023-12-08.
- ^ Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson: McFarland & Company. ISBN 9780786414390.
