1817 Georgia gubernatorial election

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]

1817 Georgia gubernatorial election

← 1815 10 November 1817 1819 →
 
Nominee William Rabun John Clark
Party Democratic-Republican Democratic-Republican
Popular vote 62 57
Percentage 52.10% 47.90%

Governor before election

William Rabun (Acting)
Democratic-Republican

Elected Governor

William Rabun
Democratic-Republican

General election

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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

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Georgia gubernatorial election, 1817[3]
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

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  1. ^ "Georgia 1817 Governor". Tufts Digital Collations and Archives. A New Nation Votes: American Election Returns 1787–1825. Tufts University. Retrieved 8 December 2023.
  2. ^ "GA Governor". ourcampaigns.com. 18 July 2022. Retrieved 8 December 2023.
  3. ^ Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson: McFarland & Company. ISBN 9780786414390.