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]
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 County results Francis: 50–60% 60–70% 70–80% Arnold: 50–60% | |||||||||||||||||
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General election
editOn 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 | % | |
|---|---|---|---|---|
| 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.