1832 United States presidential election in Rhode Island
Election in Rhode Island
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 County Results
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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 four representatives, or electors to the Electoral College, who voted for President and Vice President.
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% | 4 | |
| Democratic | Andrew Jackson (incumbent) | 2,126 | 43.07% | 0 | |
| Totals | 4,936 | 100.0% | 4 | ||
See also
References
- ^ "1832 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved April 12, 2013.
State and district results of the 1832 United States presidential election | ||
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