1812 United States presidential election in Rhode Island
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| Elections in Rhode Island |
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The 1812 United States presidential election in Rhode Island took place as part of the 1812 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.
Rhode Island voted for the Federalist candidate, DeWitt Clinton, over the Democratic-Republican candidate, James Madison. Clinton won Rhode Island by a margin of 65.93%. With Madison's second loss (the first one was 1808), this marked the first time that a candidate or an incumbent president lost two presidential elections in Rhode Island.
Results
[edit]| 1812 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Federalist | DeWitt Clinton | 4,032 | 65.93% | 4 | |
| Democratic-Republican | James Madison | 2,084 | 34.07% | – | |
| Totals | 6,116 | 100.00% | 4 | ||
See also
[edit]Notes
[edit]- ^ While commonly labeled as the Federalist candidate, Clinton technically ran as a Democratic-Republican and was not nominated by the Federalist party itself, the latter simply deciding not to field a candidate. This did not prevent endorsements from state Federalist parties (such as in Pennsylvania), but he received the endorsement from the New York state Democratic-Republicans as well.
References
[edit]- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-08-31.
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