1892 United States presidential election in Rhode Island explained

See main article: 1892 United States presidential election.

Election Name:1892 United States presidential election in Rhode Island
Country:Rhode Island
Flag Year:1882
Type:presidential
Ongoing:no
Previous Election:1888 United States presidential election in Rhode Island
Previous Year:1888
Next Election:1896 United States presidential election in Rhode Island
Next Year:1896
Election Date:November 8, 1892
Image1:Benjamin Harrison 1896.jpg
Nominee1:Benjamin Harrison
Party1:Republican Party (United States)
Home State1:Indiana
Running Mate1:Whitelaw Reid
Electoral Vote1:4
Popular Vote1:26,975
Percentage1:50.71%
Nominee2:Grover Cleveland
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Adlai Stevenson I
Electoral Vote2:0
Popular Vote2:24,336
Percentage2:45.75%
Map Size:250px
President
Before Election:Benjamin Harrison
Before Party:Republican Party (United States)
After Election:Grover Cleveland
After Party:Democratic Party (United States)

The 1892 United States presidential election in Rhode Island took place on November 8, 1892, as part of the 1892 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Republican nominee, incumbent President Benjamin Harrison, over the Democratic nominee, former President Grover Cleveland, who was running for a second, non-consecutive term. Harrison won the state by a narrow margin of 4.96%.

See also