The 1804 United States presidential election in Rhode Island took place as part of the 1804 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
.png)
Rhode Island voted for the Democratic-Republican candidate, Thomas Jefferson, over the Federalist candidate, Charles C. Pinckney. Jefferson won Rhode Island by a margin of 100.00%, the largest margin and landslide victory ever since in the state. Because Pinckney was not on the ballot.
Results
edit| 1804 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Thomas Jefferson (incumbent) | 1,312 | 100.00% | 4 | |
| Federalist | Charles C. Pinckney (not on ballot) |
— | — | 0 | |
| Totals | 1,312 | 100.0% | 4 | ||
See also
editReferences
edit- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-08-31.