1808 United States presidential election in New Jersey explained

See main article: 1808 United States presidential election.

Election Name:1808 United States presidential election in New Jersey
Country:New Jersey
Type:presidential
Ongoing:no
Previous Election:1804 United States presidential election in New Jersey
Previous Year:1804
Next Election:1812 United States presidential election in New Jersey
Next Year:1812
Election Date:November 4 - December 7, 1808
Image1:James Madison.jpg
Percentage1:55.97%
Nominee1:James Madison
Party1:Democratic-Republican Party
Home State1:Virginia
Running Mate1:George Clinton
Electoral Vote1:8
Popular Vote1:18,670
Nominee2:Charles C. Pinckney
Percentage2:44.03%
Popular Vote2:14,687
Electoral Vote2:0
Running Mate2:Rufus King
Home State2:South Carolina
Party2:Federalist Party (United States)
Map Size:230px
President
Before Election:Thomas Jefferson
Before Party:Democratic-Republican Party
After Election:James Madison
After Party:Democratic-Republican Party

The 1808 United States presidential election in New Jersey took place between November 4 and December 7, 1808, as part of the 1808 United States presidential election. The state chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, New Jersey cast its eight electoral votes to Democratic Republican candidate and Secretary of State James Madison.[1]

See also

Notes and References

  1. Book: Dubin, Michael J. . United States Presidential Elections, 1788-1860: The Official Results by County and State . McFarland & Company . 2002 . 9780786410170 . Jefferson . 15.