1800 United States presidential election in Delaware
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A presidential election was held in Delaware between October 31 and December 3, 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.
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October 31 â December 3, 1800
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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) | â | 100.00% | 3 | |
| Democratic-Republican | Thomas Jefferson | â | â | 0 | |
| Totals | â | 100.00% | 3 | ||