United States presidential election in Iowa, 1992

United States presidential election in Iowa, 1992

← 1988 November 3, 1992 1996 →
Nominee Bill Clinton George H.W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle Electoral vote 7 0 0 Popular vote 586,353 504,891 253,468 Percentage 43.3% 37.3% 18.7%
County Results   Clinton—50-60%   Clinton—40-50%   Clinton—<40%   Bush—<40%   Bush—40-50%   Bush—50-60%   Bush—70-80%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

George H. W. Bush
Republican

Bill Clinton
Democratic

The 1992 United States presidential election in Iowa took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Iowa was won by Democratic Governor Bill Clinton of Arkansas with 43.29% of the popular vote over incumbent Republican President George H.W. Bush's 37.27%, a victory margin of 6.01%. Independent businessman Ross Perot finished in third with 18.71% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.

In the 1992 Iowa Democratic caucuses, most candidates for the Democratic nomination did not campaign in Iowa because of the presence of Tom Harkin, a longtime Senator representing Iowa.

Harkin was running for President in 1992 as a populist with labor union support who criticized George H.W. Bush for being out of touch with working class Americans. Harkin was an early favorite in a small field of five candidates. Harkin easily won the caucus with 76% of the vote, uncommited got second place with 11%, Senator Paul Tsongas came in third with 4%, and Governor Bill Clinton finished fourth with 2%.