1792 United States presidential election in Virginia explained

Election Name:1792 United States presidential election in Virginia
Country:Virginia
Type:presidential
Ongoing:no
Previous Election:1788–89 United States presidential election in Virginia
Previous Year:1788–89
Next Election:1796 United States presidential election in Virginia
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 Vote1:21
Popular Vote1:962
President
Before Election:George Washington
After Election:George Washington
Before Party:Independent (politician)
After Party:Independent (politician)
Percentage1:100%

The 1792 United States presidential election in Virginia took place between November 2 and December 5, 1792, as part of the 1792 United States presidential election. Virginia's 21 electors each cast one vote for the incumbent, George Washington, and one vote for John Adams, the incumbent Vice President.[1]

Virginia unanimously voted for independent candidate and incumbent president, George Washington. The total vote is composed of 962 for Democratic-Republican electors, all of whom were supportive of Washington and George Clinton.[2] The totals for Virginia appears to be incomplete.

Results

Virginia was divided into 21 electoral districts.

1792 United States presidential election in Virginia
PartyCandidateVotesPercentageElectoral votes
IndependentGeorge Washington962100%21
Totals962100%21

Notes and References

  1. Web site: ELECTIONS FROM 1789 TO 1828 . dead . 2024-05-01 . Virginia Museum of History & Culture . 2023-12-09 . https://web.archive.org/web/20231209181833/https://virginiahistory.org/learn/getting-message-out-presidential-campaign-memorabilia-collection-allen-frey/elections-1789-1828 .
  2. Web site: A New Nation Votes . 2024-07-16 . elections.lib.tufts.edu.