1812 United States elections
The 1812 United States elections elected the members of the 13th United States Congress. The election took place during the First Party System, and shortly after the start of the War of 1812. The Federalist Party made a relatively strong showing, winning seats in both chambers while supporting a competitive challenge to the incumbent Democratic-Republican President. However, the Democratic-Republican Party continued its control of the Presidency and both houses of Congress.
In the Presidential election, incumbent Democratic-Republican President James Madison defeated New York Lieutenant Governor and New York City Mayor DeWitt Clinton.[1] Clinton was a member of the Democratic-Republican Party, but his presidential bid received the support of both anti-Madison Democratic-Republicans and many Federalists.[2] Although Madison won, the Presidential election was the closest since the 1800 election, as Clinton won New England and three mid-Atlantic states.
Following the 1810 census, 39 seats were added to the House. Federalists won major gains, but Democratic-Republicans continued to dominate the chamber.[3]
In the Senate, Federalists picked up a small number of seats, but Democratic-Republicans retained a dominant majority.[4]
See also
- United States presidential election, 1812
- United States House of Representatives elections, 1812
- United States Senate elections, 1812 and 1813
References
- ^ "1812 Presidential Election". The American Presidency Project. Retrieved 25 June 2014.
- ^ History of American Presidential Elections, Volume I 1789-1844; Arthur M. Schlesinger Jr.; Pgs 249-272
- ^ "Party Divisions of the House of Representatives". United States House of Representatives. Retrieved 25 June 2014.
- ^ "Party Division in the Senate, 1789-Present". United States Senate. Retrieved 25 June 2014.
  | This American elections-related article is a stub. You can help Wikipedia by expanding it. |