See main article: 1792 United States presidential election.
| Election Name: | 1792 United States presidential election in Vermont |
| Country: | Vermont |
| Flag Image: | Flag of the Vermont Republic.svg |
| Type: | presidential |
| Ongoing: | no |
| Next Election: | 1796 United States presidential election in Vermont |
| Next Year: | 1796 |
| Election Date: | November 2 - December 5, 1792 |
| Image1: | Gilbert Stuart Williamstown Portrait of George Washington.jpg |
| Nominee1: | George Washington |
| Party1: | Independent (politician) |
| Home State1: | Virginia |
| Electoral Vote1: | 3 |
| Percentage1: | 100.00% |
| Nominee2: | John Adams |
| Party2: | Federalist Party (United States) |
| Home State2: | Massachusetts |
| Electoral Vote2: | 3 |
| Percentage2: | – |
| President | |
| Before Election: | George Washington |
| Before Party: | Independent (politician) |
| After Election: | George Washington |
| After Party: | Independent (politician) |
The 1792 United States presidential election in Vermont took place between November 2 and December 5, 1792, as part of the 1792 United States presidential election. The state legislature chose four members of the Electoral College, each of whom, under the provisions of the Constitution prior to the passage of the Twelfth Amendment, cast two votes for president.
Vermont participated in its first ever presidential election, having become the 14th state on March 4, 1791. The state cast three electoral votes for incumbent George Washington and three for the incumbent vice president John Adams;[1] [2] one elector did not cast his votes.[3]