1996 United States presidential election in Rhode Island

1996 United States presidential election in Rhode Island

← 1992 November 5, 1996 2000 →
Turnout	64.8%[1] 11.8 pp
Nominee Bill Clinton Bob Dole Ross Perot Party Democratic Republican Reform Home state Arkansas Kansas Texas Running mate Al Gore Jack Kemp Patrick Choate Electoral vote 4 0 0 Popular vote 233,050 104,683 43,723 Percentage 59.71% 26.82% 11.20%
County Results Municipality Results Clinton   40–50%   50–60%   60–70%   70–80% Dole   40–50%
President before election Bill Clinton Democratic 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 1996 United States presidential election in Rhode Island took place on November 5, 1996, as part of the 1996 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 President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 59.71% to 26.82% by a margin of 32.89%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third, with 11.20% of the popular vote.^[2]

As of 2024, this was the most recent presidential election in which the town of Scituate voted for a Democrat.