1836 United States presidential election in Delaware

The 1836 United States presidential election in Delaware took place between November 3 and December 7, 1836, as part of the 1836 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, 1836

November 3 – December 7, 1836


Nominee	William Henry Harrison	Martin Van Buren
Party	Whig	Democratic
Home state	Ohio	New York
Running mate	Francis Granger	Richard Johnson
Electoral vote	3	0
Popular vote	4,736	4,154
Percentage	53.24%	46.70%

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 Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Delaware by a margin of 6.54%.

Results

United States presidential election in Delaware, 1836[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Whig	William Henry Harrison	4,736	53.24%	3
	Democratic	Martin Van Buren	4,154	46.70%	0
	N/A	Other	5	0.06%	0
Totals			8,895	100.00%	3

References

"1836 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 4 August 2012.

State and district results of the 1836 United States presidential election
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