1833 Rhode Island gubernatorial election
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The 1833 Rhode Island gubernatorial election was held on April 3, 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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April 3, 1833
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 County results Francis: 50â60% 60â70% 70â80% Arnold: 50â60% | |||||||||||||||||
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General election
On election day, April 3, 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 May 1, 1833.[2]
Results
| 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 | ||||