1808 United States presidential election in Rhode Island

Main article: 1808 United States presidential election

1808 United States presidential election in Rhode Island

←  1804 November 4 – December 7, 1808 1812  →

Nominee Charles Cotesworth Pinckney James Madison Party Federalist Democratic-Republican Home state South Carolina Virginia Running mate Rufus King George Clinton Electoral vote 4 0 Popular vote 3,072 2,692 Percentage 53.30% 46.70%

President before election Thomas Jefferson Democratic-Republican Elected President James Madison Democratic-Republican

The 1808 United States presidential election in Rhode Island took place as part of the 1808 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.

Rhode Island voted for the Federalist candidate, Charles Cotesworth Pinckney, over the Democratic-Republican candidate, James Madison. Pinckney won Rhode Island by a margin of 53.30%.

Results

1808 United States presidential election in Rhode Island [1]
Party		Candidate	Votes	Percentage	Electoral votes
	Federalist	Charles Cotesworth Pinckney	3,072	53.30%	4
	Democratic-Republican	James Madison	2,692	46.70%	–
Totals			5,764	100.00%	4

See also

• United States presidential elections in Rhode Island

References

1. "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 31, 2024.