1792 United States presidential election in New Hampshire explained

See main article: 1792 United States presidential election.

Election Name:1792 United States presidential election in New Hampshire
Country:New Hampshire
Type:presidential
Ongoing:no
Next Election:1796 United States presidential election in New Hampshire
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 Vote1:6
Popular Vote1:2,762
Percentage1:100.00%
President
Before Election:George Washington
Before Party:Independent (politician)
After Election:George Washington
After Party:Independent (United States)

The 1792 United States presidential election in New Hampshire took place between November 2 to December 5, 1792, as part of the 1792 United States presidential election to elect the President. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

New Hampshire voted for incumbent George Washington without any competitor.

Results

1792 United States presidential election in New Hampshire[1]
PartyCandidateVotesPercentageElectoral votes
IndependentGeorge Washington2,762100.00%6
Totals2,762100.00%6

See also

References

  1. Book: Dubin, Michael J. . United States Presidential Elections, 1788-1860: The Official Results by County and State . McFarland & Company . 2002 . 9780786410170 . Jefferson.