1832 United States presidential election in Delaware

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

November 2 – December 5, 1832


Nominee	Henry Clay	Andrew Jackson
Party	National Republican	Democratic
Home state	Kentucky	Tennessee
Running mate	John Sergeant	Martin Van Buren
Electoral vote	3	0
Popular vote	4,276	4,110
Percentage	50.99%	49.01%

Federal government

U.S. President
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Presidential primaries
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U.S. Senate
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U.S. House
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Special elections
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State government

General elections
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Gubernatorial elections
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Lieutenant gubernatorial elections
2004 2008 2012 2016 2020
Senate elections
2018 2020
House of Representatives elections
2008 2010 2012 2014 2016 2018 2020
Special elections
10th Senate district, 2017

Delaware voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won the state by a narrow margin of 0.98%. This was the first election in which Delaware voted by popular vote for President.

Results

1832 United States presidential election in Delaware[1]
Party		Candidate	Votes	Percentage	Electoral votes
	National Republican	Henry Clay	4,276	50.99%	3
	Democratic	Andrew Jackson	4,110	49.01%	0
Totals			8,386	100.00%	3

References

"1832 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 12 April 2013.

State and district results of the 1832 United States presidential election
Alabama Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia

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1832 United States presidential election in Delaware

Results

See also

References