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1812 United States presidential election in Rhode Island

← 1808 October 30 – December 2, 1812 1816 →

Nominee DeWitt Clinton James Madison Party Democratic-Republican Democratic-Republican Alliance Federalist  – Running mate Jared Ingersoll Elbridge Gerry Electoral vote 4 0 Popular vote 4,032 2,084 Percentage 65.93% 34.07%

President before election James Madison Democratic-Republican Elected President James Madison Democratic-Republican

Elections in Rhode Island
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v t e

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

See also

• United States presidential elections in Rhode Island

Notes

1. 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

1. "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 31, 2024.

v t e State results of the 1812 United States presidential election
Connecticut Delaware Georgia Kentucky Louisiana Maryland Massachusetts New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia

• 1812 United States presidential election by state
• United States presidential elections in Rhode Island
• 1812 Rhode Island elections