.svg)
1792 United States presidential election in Rhode Island
From Wikipedia, the free encyclopedia
A presidential election was held in Rhode Island 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.
 | ||||||||||||||||||||
November 2 â December 5, 1792
| ||||||||||||||||||||
| ||||||||||||||||||||
| ||||||||||||||||||||
.jpg)
_(farther_3x4_cropped)_(cropped).jpg)
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) | â | â | 4 | |
| Federalist | John Adams | â | â | 4 | |
| Totals | â | â | 8 | ||