See main article: 1812 United States presidential election.
| Election Name: | 1812 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1808 United States presidential election in Rhode Island |
| Previous Year: | 1808 |
| Next Election: | 1816 United States presidential election in Rhode Island |
| Next Year: | 1816 |
| Election Date: | October 30 – December 2, 1812 |
| Color1: | EA9978 |
| Nominee1: | DeWitt Clinton |
| Party1: | Democratic-Republican |
| Alliance1: | Federalist Party |
| Running Mate1: | Jared Ingersoll |
| Electoral Vote1: | 4 |
| Popular Vote1: | 4,032 |
| Percentage1: | 65.93% |
| Nominee2: | James Madison |
| Party2: | Democratic-Republican Party |
| Running Mate2: | Elbridge Gerry |
| Electoral Vote2: | 0 |
| Popular Vote2: | 2,084 |
| Percentage2: | 34.07% |
| President | |
| Before Election: | James Madison |
| Before Party: | Democratic-Republican Party |
| After Election: | James Madison |
| After Party: | Democratic-Republican Party |
| Flag Year: | 1812 |
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.
| 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 | |||