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]
 | |||||||||||||||||||||||||
| |||||||||||||||||||||||||
| |||||||||||||||||||||||||
| |||||||||||||||||||||||||
.jpg.webp)
| Elections in Connecticut |
|---|
 |
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
Samuel Huntington, Richard Law, Matthew Griswold, Erastus Wolcott, Thaddeus Burr, Jedediah Huntington, and Oliver Wolcott served as electors.[2]
References
- "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
- 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.
State and district results of the 1788–89 United States presidential election | ||
|---|---|---|
 | ||
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.