1792 United States presidential election in Rhode Island
November 2 – December 5, 1792
| ||||||||||||||||||||
| ||||||||||||||||||||
| ||||||||||||||||||||
| Elections in Rhode Island |
|---|
The 1792 United States presidential election in Rhode Island took place as part of the 1792 United States presidential election. Voters chose four representatives, or electors to the Electoral College who voted for President and Vice President.
Rhode Island unanimously voted for the incumbent Independent President George Washington.
Results
| 1792 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | — | — | 4 | |
| Federalist | John Adams | — | — | 4 | |
| Totals | — | — | 8 | ||
See also
References
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 19, 2024.