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1788–89 United States presidential election in New Hampshire

1788–89 United States presidential election in New Hampshire

December 15, 1788 – January 7, 1789 (1788-12-15 – 1789-01-07) 1792 →
 
Nominee George Washington John Adams
Party Independent Federalist
Home state Virginia Massachusetts
Electoral vote 5 5
Popular vote 1,759
Percentage 100.00%

President before election

Office established

Elected President

George Washington
Independent

The 1788–89 United States presidential election in New Hampshire took place on January 7, 1789, as part of the 1788–89 United States presidential election to elect the first President. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

New Hampshire unanimously voted for independent candidate and commander-in-chief of the Continental Army, George Washington. The total vote was composed of 1,759 for Federalist electors, all of whom were supportive of Washington.[1] Several candidates of unknown affiliation also received votes.

Voters voted for electors on December 15, 1788, through a general ticket. As no elector candidate received the "requisite number for a choice," the election went to the legislature,[1][2] which selected the five best-performing elector candidates from the top ten on January 7, 1789.[3][2]

Results

1788-1789 United States presidential election in New Hampshire
Party Candidate Votes Percentage Electoral votes
Independent George Washington 1,759 100.00% 5
Totals 1,759 100.00% 5

See also

References

  1. ^ a b "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 16, 2024.
  2. ^ a b "The Electoral Count for the Presidential Election of 1789". The Papers of George Washington. Archived from the original on September 14, 2013. Retrieved May 4, 2005.
  3. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 11, 2024.


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