1792 United States presidential election in Delaware
(Redirected from 1792 US presidential election in Delaware)
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The 1792 United States presidential election in Delaware took place 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]| United States presidential election in Delaware, 1792 | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Independent | George Washington (incumbent) | 100.00% | 3 | ||
| Totals | 100.00% | 3 | |||
See also
[edit]References
[edit]- ^ 1792 Presidential Electoral Vote Count Dave Leip's Atlas of U.S. Presidential Elections.
| Candidates |  | |
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| General articles | ||
| Local results | ||
| Other 1792 elections | ||