1800 United States presidential election in Delaware

Main article: 1800 United States presidential election

The 1800 United States presidential election in Delaware took place between 31 October and 3 December 1800, as part of the 1800 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.

1800 United States presidential election in Delaware

← 1796 31 October – 3 December 1800 1804 →

Nominee John Adams Thomas Jefferson Party Federalist Democratic-Republican Home state Massachusetts Virginia Running mate Charles C. Pinckney Aaron Burr Electoral vote 3 0 Percentage 100.00% -

Delaware cast three electoral votes for the Federalist candidate and incumbent President John Adams over the Democratic-Republican candidate and incumbent Vice President Thomas Jefferson. These electors were elected by the Delaware General Assembly, the state legislature, rather than by popular vote. The three electoral votes for Vice president were cast for Adams's running mate Charles C. Pinckney from South Carolina.^[1]

Results

1800 United States presidential election in Delaware^[2]
Party		Candidate	Votes	Percentage	Electoral votes
	Federalist	John Adams (incumbent)	–	100.00%	3
	Democratic-Republican	Thomas Jefferson	–	–	0
Totals			–	100.00%	3

See also

• United States presidential elections in Delaware

References

1. "1800 Presidential General Election Results". U.S. Election Atlas. Retrieved July 9, 2023.
2. "1800 Presidential Election". 270towin.com. Retrieved July 9, 2023.