1992 United States presidential election in Rhode Island

1992 United States presidential election in Rhode Island

← 1988 November 3, 1992 1996 →
Turnout	76.6%[1] 6.4 pp
Nominee Bill Clinton George H. W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle James Stockdale Electoral vote 4 0 0 Popular vote 213,299 131,601 105,045 Percentage 47.04% 29.02% 23.16%
County Results Municipality Results Clinton   30–40%   40–50%   50–60%   60–70% Bush   30–40%   40–50%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

Elections in Rhode Island
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Ballot measures 2020 Question 1
Providence Mayoral elections 1998 2002 2006 2010 2014 2018 2022
v t e

The 1992 United States presidential election in Rhode Island took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island was won by Governor Bill Clinton (D-Arkansas) with 47.04% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 29.02%. Businessman Ross Perot (I-Texas) finished in third, with 23.16% of the popular vote.^[2] Clinton ultimately won the national vote, defeating incumbent President Bush.^[3]