1792 United States presidential election in Delaware
November 2 – December 5, 1792
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A presidential election was held in Delaware between November 2 and December 5, 1792, as part of the 1792 United States presidential election. The state legislature chose three members of the Electoral College, each of whom, under the provisions of the Constitution prior to the passage of the Twelfth Amendment, cast two votes for President.
Delaware's three electors each cast one vote for the incumbent, George Washington, and one vote for John Adams, the incumbent Vice President.[1]
Results
[edit]| Party | Candidate | Electoral Vote | % | |
|---|---|---|---|---|
| Independent | George Washington | 3 | 100.00% | |
| Total votes | 3 | 100.00% | ||
Results by elector
[edit]| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Federalist | James Sykes | 26 | 43.33% | |
| Federalist | Gunning Bedford Sr. | 17 | 28.33% | |
| Federalist | William Hill Wells | 17 | 28.33% | |
| Federalist | Nathaniel Mitchell | 0 | 0.00% | |
| Democratic-Republican | Thomas Montgomery[a] | 0 | 0.00% | |
| Total votes | 60 | 100.00% | ||
See also
[edit]Notes
[edit]References
[edit]- ^ 1792 Presidential Electoral Vote Count Dave Leip's Atlas of U.S. Presidential Elections.
- ^ a b "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 12, 2026.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 12, 2026.
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| General articles | ||
| Local results | ||
| Other 1792 elections | ||