United States presidential election in Rhode Island, 1996

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United States presidential election in Rhode Island, 1996
Rhode Island
← 1992 November 5, 1996 2000 →
  Bill Clinton.jpg Bob Dole, PCCWW photo portrait.JPG Ross Perot.jpg
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 59.7% 26.8% 11.2%

Rhode Island Presidential Election Results by County, 2004.svg
County Results

President before election

Bill Clinton

Elected President

Bill Clinton

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 4 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 .[1]


United States presidential election in Rhode Island, 1996
Party Candidate Running mate Votes Percentage Electoral votes
Democratic Bill Clinton (incumbent) Al Gore 233,050 59.71% 4
Republican Bob Dole Jack Kemp 104,683 26.82% 0
Reform Ross Perot Patrick Choate 43,723 11.20% 0
Green Ralph Nader Winona LaDuke 6,040 1.55% 0
Libertarian Harry Browne Jo Jorgensen 1,109 0.28% 0
U.S. Taxpayers' Party Howard Phillips Herbert Titus 1,021 0.26% 0
Natural Law Dr. John Hagelin Dr. V. Tompkins 435 0.11% 0
Workers World Party Monica Moorehead Gloria La Riva 186 0.05% 0
No party Write-in 37 0.01% 0


See also[edit]