1833 Rhode Island gubernatorial election
Appearance
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County results Francis: 50–60% 60–70% 70–80% Arnold: 50–60% | |||||||||||||||||
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Elections in Rhode Island |
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The 1833 Rhode Island gubernatorial election was held on 3 April 1833 in order to elect the governor of Rhode Island. Democratic nominee and former member of the Rhode Island Senate John Brown Francis defeated incumbent National Republican governor Lemuel H. Arnold.[1]
General election
[edit]On election day, 3 April 1833, Democratic nominee John Brown Francis won the election by a margin of 733 votes against his opponent incumbent National Republican governor Lemuel H. Arnold, thereby gaining Democratic control over the office of governor. Francis was sworn in as the 13th governor of Rhode Island on 1 May 1833.[2]
Results
[edit]Party | Candidate | Votes | % | |
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Democratic | John Brown Francis | 4,025 | 54.98 | |
National Republican | Lemuel H. Arnold (incumbent) | 3,292 | 44.97 | |
Scattering | 4 | 0.05 | ||
Total votes | 7,321 | 100.00 | ||
Democratic gain from National Republican |
References
[edit]- ^ "Lemuel Hastings Arnold". National Governors Association. Retrieved 5 April 2024.
- ^ "RI Governor". ourcampaigns.com. 6 June 2005. Retrieved 5 April 2024.
Categories:
- 1833 Rhode Island elections
- Rhode Island gubernatorial elections
- 1833 in Rhode Island
- 1833 United States gubernatorial elections
- April 1833 events
- 1830s in Rhode Island
- 1830s Rhode Island elections
- 1833 elections
- 1833 elections in North America
- 1833 elections in the United States
- Government of Rhode Island
- United States gubernatorial elections in the 1830s