1819 Georgia gubernatorial election
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The 1819 Georgia gubernatorial election was held on 5 November 1819 in order to elect the Governor of Georgia. Democratic-Republican candidate John Clark defeated fellow Democratic-Republican candidate and former United States Senator for Georgia George Troup in a Georgia General Assembly vote.[1]
General election
[edit]On election day, 5 November 1819, Democratic-Republican candidate John Clark won the election against his opponent fellow Democratic-Republican candidate George Troup. Clark was sworn in as the 31st Governor of Georgia on 5 November 1819.[2]
Results
[edit]| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic-Republican | John Clark | 73 | 54.89 | |
| Democratic-Republican | George Troup | 60 | 45.11 | |
| Total votes | 133 | 100.00 | ||
| Democratic-Republican hold | ||||
References
[edit]- ^ "Georgia 1819 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.
Categories:
- Georgia (U.S. state) gubernatorial elections
- 1819 United States gubernatorial elections
- 1819 Georgia (U.S. state) elections
- Indirect elections
- 1819 in Georgia (U.S. state)
- 1810s in Georgia (U.S. state)
- 1810s Georgia (U.S. state) elections
- 1819 elections
- 1819 elections in North America
- 1819 elections in the United States
- United States gubernatorial elections in the 1810s
- Government of Georgia (U.S. state)