1792 United States presidential election in Delaware explained

See main article: 1792 United States presidential election.

Election Name:1792 United States presidential election in Delaware
Country:Delaware
Type:presidential
Ongoing:no
Previous Election:1788–89 United States presidential election in Delaware
Previous Year:1788–89
Next Election:1796 United States presidential election in Delaware
Next Year:1796
Election Date:November 2 – December 5, 1792
Image1:Gilbert Stuart Williamstown Portrait of George Washington.jpg
Nominee1:George Washington
Party1:Independent (politician)
Home State1:Virginia
Electoral Vote1:3
Percentage1:100.00%
Nominee2:John Adams
Party2:Federalist Party
Home State2:Massachusetts
Electoral Vote2:3
Percentage2:
President
Before Election:George Washington
After Election:George Washington
Before Party:Independent (politician)
After Party:Independent (politician)

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

United States presidential election in Delaware, 1792
PartyCandidateVotesPercentageElectoral votes
IndependentGeorge Washington (incumbent)100.00%3
Totals100.00%3

See also

Notes and References

  1. https://uselectionatlas.org/RESULTS/ 1792 Presidential Electoral Vote Count