1866 Rhode Island gubernatorial election
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 County results Burnside: 60–70% 70–80% 80–90% | |||||||||||||||||
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The 1866 Rhode Island gubernatorial election was held on 4 April 1866 in order to elect the governor of Rhode Island. Republican nominee and former Union Army Major General Ambrose Burnside defeated Democratic nominee Lyman Pierce.[1]
General election
[edit]On election day, 4 April 1866, Republican nominee Ambrose Burnside won the election by a margin of 5,381 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Burnside was sworn in as the 30th governor of Rhode Island on 1 May 1866.[2]
Results
[edit]| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Ambrose Burnside | 8,197 | 73.36 | |
| Democratic | Lyman Pierce | 2,816 | 25.20 | |
| Scattering | 160 | 1.44 | ||
| Total votes | 11,221 | 100.00 | ||
| Republican hold | ||||
References
[edit]- ^ "Ambrose Burnside". National Governors Association. Retrieved 7 April 2024.
- ^ "RI Governor". ourcampaigns.com. 26 July 2005. Retrieved 7 April 2024.
Categories:
- Rhode Island gubernatorial elections
- 1866 United States gubernatorial elections
- April 1866 events
- 1860s in Rhode Island
- 1860s Rhode Island elections
- 1866 elections
- 1866 elections in North America
- 1866 elections in the United States
- United States gubernatorial elections in the 1860s
- Government of Rhode Island