1832 United States presidential election in Rhode Island

1832 United States presidential election in Rhode Island
	November 2 – December 5, 1832
Clay 40–50% 50–60%	Jackson 50–60%

The 1832 United States presidential election in Rhode Island took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Rhode Island by a margin of 13.86%.

Jackson remains the only Democrat to win two consecutive terms without carrying Rhode Island either time.

Results

1832 United States presidential election in Rhode Island^[1]
Party		Candidate	Votes	Percentage	Electoral votes
	National Republican	Henry Clay	2,810	56.93%	4
	Democratic	Andrew Jackson (incumbent)	2,126	43.07%	0
Totals			4,936	100.0%	4

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