1832 United States presidential election in Virginia explained

See main article: 1832 United States presidential election.

Election Name:	1832 United States presidential election in Virginia
Country:	Virginia
Flag Year:	1788
Type:	presidential
Ongoing:	no
Previous Election:	1828 United States presidential election in Virginia
Previous Year:	1828
Next Election:	1836 United States presidential election in Virginia
Next Year:	1836
Election Date:	November 2 – December 5, 1832
Image1:	Andrew jackson head.jpg
Nominee1:	Andrew Jackson
Party1:	Democratic Party (United States)
Home State1:	Tennessee
Running Mate1:	Martin Van Buren
Electoral Vote1:	23
Popular Vote1:	34,243
Percentage1:	74.96%
Nominee2:	Henry Clay
Party2:	National Republican Party (United States)
Home State2:	Kentucky
Running Mate2:	John Sergeant
Electoral Vote2:	0
Popular Vote2:	11,436
Percentage2:	25.03%
President
Before Election:	Andrew Jackson
Before Party:	Democratic Party (United States)
After Election:	Andrew Jackson
After Party:	Democratic Party (United States)

The 1832 United States presidential election in Virginia took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose 23 representatives, or electors to the Electoral College, who voted for President and Vice President.

Virginia voted for the Democratic Party candidate, incumbent President Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won Virginia by a margin of 49.93%.

Results

United States presidential election in Virginia, 1832^[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Andrew Jackson (inc.)	34,243	74.96%	23
width: 3px"		National Republican	Henry Clay	11,436	25.03%	0
	Anti-Masonic	William Wirt	3	0.01%	0
Totals			45,682	100.0%	23

Notes and References

1. Web site: 1832 Presidential General Election Results - Virginia. U.S. Election Atlas. 12 April 2013.