1812 United States presidential election in Rhode Island

1812 United States presidential election in Rhode Island

← 1808 October 30 – December 2, 1812 1816 →
 
Nominee DeWitt Clinton James Madison
Party Democratic-Republican[a] 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

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

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

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 2024-08-31.

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