See main article: 1800 United States presidential election.
Election Name: | 1800 United States presidential election in Delaware |
Country: | Delaware |
Type: | presidential |
Ongoing: | no |
Previous Election: | 1796 United States presidential election in Delaware |
Previous Year: | 1796 |
Next Election: | 1804 United States presidential election in Delaware |
Next Year: | 1804 |
Election Date: | 31 October – 3 December 1800 |
Image1: | Gilbert Stuart, John Adams, c. 1800-1815, NGA 42933.jpg |
Nominee1: | John Adams |
Party1: | Federalist Party (United States) |
Home State1: | Massachusetts |
Running Mate1: | Charles C. Pinckney |
Electoral Vote1: | 3 |
Percentage1: | 100.00% |
Nominee2: | Thomas Jefferson |
Party2: | Democratic-Republican Party (United States) |
Home State2: | Virginia |
Running Mate2: | Aaron Burr |
Electoral Vote2: | 0 |
Percentage2: | - |
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]
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 | ||