1792 United States presidential election in Vermont

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]

References

^ Coolidge, Austin Jacobs; Mansfield, John Brainard (1859). A History and Description of New England, General and Local: Maine, New Hampshire, and Vermont. A.J. Coolidge. p. 990.
^ State, Vermont Office of Secretary of (1886). Vermont Legislative Directory. Rand, Avery. p. 72.
^ Michigan Official Directory and Legislative Manual for the Years. State of Michigan. 1905. p. 429.

[1] Coolidge, Austin Jacobs; Mansfield, John Brainard (1859). A History and Description of New England, General and Local: Maine, New Hampshire, and Vermont. A.J. Coolidge. p. 990.

[2] State, Vermont Office of Secretary of (1886). Vermont Legislative Directory. Rand, Avery. p. 72.

[3] Michigan Official Directory and Legislative Manual for the Years. State of Michigan. 1905. p. 429.

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v t e State results of the 1792 U.S. presidential election
Candidates	George Washington
General articles	Election timeline
Local results	Connecticut Delaware Georgia Kentucky Maryland Massachusetts New Hampshire New Jersey New York North Carolina Pennsylvania Rhode Island South Carolina Vermont Virginia
Other 1792 elections	House Senate

See also

References