1817 Georgia gubernatorial election
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Elections in Georgia |
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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]
General election
[edit]On 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.
Kategorien:
- Georgia (U.S. state) gubernatorial elections
- 1817 United States gubernatorial elections
- 1817 Georgia (U.S. state) elections
- November 1817 events
- Indirect elections
- 1810s in Georgia (U.S. state)
- 1810s Georgia (U.S. state) elections
- 1817 elections
- 1817 elections in North America
- 1817 elections in the United States
- United States gubernatorial elections in the 1810s
- Government of Georgia (U.S. state)