1792 United States presidential election in Rhode Island
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The 1792 United States presidential election in Rhode Island took place as part of the 1792 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for President and Vice President.
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Rhode Island unanimously voted for the incumbent Independent President George Washington.
Results
[edit]| 1792 United States presidential election in Rhode Island[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | — | — | 4 | |
| Federalist | John Adams | — | — | 4 | |
| Totals | — | — | 8 | ||
See also
[edit]References
[edit]- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-07-19.
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| Other 1792 elections | ||