See main article: 1788–89 United States presidential election.
| Election Name: | 1788–1789 United States presidential election in Connecticut |
| Country: | Connecticut |
| Type: | presidential |
| Ongoing: | no |
| Next Election: | 1792 United States presidential election in Connecticut |
| Next Year: | 1792 |
| Election Date: | January 7, 1789 |
| Image1: | Gilbert Stuart Williamstown Portrait of George Washington.jpg |
| Nominee1: | George Washington |
| Party1: | Independent (politician) |
| Home State1: | Virginia |
| Electoral Vote1: | 7 |
| Percentage1: | 100.00% |
| Nominee2: | John Adams |
| Party2: | Federalist Party |
| Home State2: | Massachusetts |
| Electoral Vote2: | 5 |
| Percentage2: | – |
| Image3: | Samuel Huntington - Charles Willson Peale.jpg |
| Nominee3: | Samuel Huntington |
| Party3: | Federalist Party |
| Home State3: | Connecticut |
| Electoral Vote3: | 2 |
| Percentage3: | – |
| President | |
| Before Election: | George Washington |
| Before Party: | Independent (politician) |
| After Election: | George Washington |
| After Party: | Independent (politician) |
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]
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.
Samuel Huntington, Richard Law, Matthew Griswold, Erastus Wolcott, Thaddeus Burr, Jedediah Huntington, and Oliver Wolcott served as electors.