1788–89 United States presidential election in Connecticut

The 1788–89 United States presidential election in Connecticut took place on January 7, 1789, as part of the 1789 United States presidential election. The state legislature chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.[1]

1788–1789 United States presidential election in Connecticut

January 7, 1789 1792 →
 
Nominee George Washington John Adams Samuel Huntington
Party Independent Federalist Federalist
Home state Virginia Massachusetts Connecticut
Electoral vote 7 5 2
Percentage 100.00%

President before election

George Washington
Independent

Elected President

George Washington
Independent

Connecticut, which had become the 5th state on January 9, 1788, unanimously cast its seven electoral votes for George Washington during its first presidential election.

Electors

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Samuel Huntington, Richard Law, Matthew Griswold, Erastus Wolcott, Thaddeus Burr, Jedediah Huntington, and Oliver Wolcott served as electors.[2]

See also

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References

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  1. ^ "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.
  2. ^ Jensen & Becker 1976, p. xxvii.

Works cited

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