1800 United States presidential election in Delaware

Election in Delaware

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% -
Elections in Delaware
Presidential elections
Presidential primaries
Democratic
2004
2008
2016
2020
Republican
2008
2012
2016
2020
2024
U.S. Senate elections
U.S. House of Representatives elections
Special elections
Senate
1795
1796
1798
1799
1802
1804
1810
1813
1822
1824
1827
1830
1836
1837
1841
1849
1857
1864
1869
1885
1897
1899
1903
1906
1922
1930
2010
House
1805
1806
1807
1822
1827
1863
1900
Mayoral elections
  • v
  • t
  • e

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.

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

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.