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 3 representatives, or electors to the Electoral College, who voted for President and Vice President.

United States presidential election in Delaware, 1832

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%

Elections in Delaware

Federal government

U.S. President
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U.S. Senate
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U.S. House
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Special elections
Senate, 2010

State government

General elections
2010 2012 2014 2016 2018
Gubernatorial elections
1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020
Lieutenant gubernatorial elections
2004 2008 2012 2016 2020
Senate elections
2018
House of Representatives elections
2008 2010 2012 2014 2016 2018
Special elections
10th Senate district, 2017

Wilmington

Mayoral elections
2012 2016 2020

Delaware voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Delaware by a margin of 0.98%.

Results

United States presidential election in Delaware, 1832[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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