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1812 United States presidential election in Rhode Island
The 1812 United States presidential election in Rhode Island took place as part of the 1812 United States presidential election. Voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.
The 1812 United States presidential election in Rhode Island took place as part of the 1812 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 Democratic-Republican and Federalist supported candidate, DeWitt Clinton, over normal Democratic-Republican candidate, incumbent President 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.
| 1812 United States presidential election in Rhode Island | ||||
|---|---|---|---|---|
| Democratic-Republican/Federalist | DeWitt Clinton | 4,032 | 65.93% | 4 |
| Democratic-Republican | James Madison | 2,084 | 34.07% | – |
| Totals | 6,116 | 100.00% | 4 |
- United States presidential elections in Rhode Island
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