1800 United States presidential election in Delaware

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1800 United States presidential election in Delaware
Flag of Delaware.svg
  1796 31 October – 3 December 1800 1804  
  Gilbert Stuart, John Adams, c. 1800-1815, NGA 42933.jpg Thomas Jefferson by Rembrandt Peale, 1800.jpg
Nominee John Adams Thomas Jefferson
Party Federalist Democratic-Republican
Home state Massachusetts Virginia
Running mate Charles C. Pinckney Aaron Burr
Electoral vote30
Percentage100.00%-

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.

Contents

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]
PartyCandidateVotesPercentageElectoral votes
Federalist John Adams (incumbent)100.00%3
Democratic-Republican Thomas Jefferson 0
Totals100.00%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.