United States presidential election in Rhode Island, 1992
|Elections in Rhode Island|
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. Clinton ultimately won the national vote, defeating incumbent President Bush.
|United States presidential election in Rhode Island, 1992|
|Republican||George H.W. Bush (incumbent)||131,601||29.02%||0|
|New Alliance||Lenora Fulani||1,878||0.41%||0|
|Ind. For LaRouche||Lyndon LaRouche||494||0.11%||0|
|Natural Law||John Hagelin||262||0.06%||0|
|U.S. Taxpayers'||Howard Phillips||215||0.05%||0|
|Write-in||James Bo Gritz||3||0.00%||0|
|Write-in||J. Quinn Brisben||2||0.00%||0|
- "1992 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 9 June 2012.
- "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 9 June 2012.