1788–89 United States presidential election in Connecticut
(Redirected from United States presidential election in Connecticut, 1788–1789)
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]
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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
editSamuel Huntington, Richard Law, Matthew Griswold, Erastus Wolcott, Thaddeus Burr, Jedediah Huntington, and Oliver Wolcott served as electors.[2]
See also
editReferences
edit- ^ "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.
- ^ Jensen & Becker 1976, p. xxvii.
Works cited
edit- Jensen, Merrill; Becker, Robert, eds. (1976). The First Federal Elections 1788-1790: Congress, South Carolina, Pennsylvania, Massachusetts, New Hampshire. Vol. 1. University of Wisconsin Press. ISBN 0299066908.
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