1788–89 United States presidential election in New Hampshire
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.
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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
edit| 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
editReferences
edit- ^ a b "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 16, 2024.
- ^ 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.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 11, 2024.
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