1832 United States presidential election in Rhode Island
The 1832 United States presidential election in Rhode Island took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Rhode Island | ||||||
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Rhode Island voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Rhode Island by a margin of 13.86%.
Results
1832 United States presidential election in Rhode Island[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
National Republican | Henry Clay | 2,810 | 56.93% | 21 | |
Democratic | Andrew Jackson | 2,126 | 43.07% | 0 | |
Totals | 4,936 | 100.0% | 4 | ||
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References
- "1832 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 12 April 2013.
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