1892 United States presidential election in Rhode Island
Appearance
(Redirected from United States presidential election in Rhode Island, 1892)
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County Results
Harrison 40-50% 50-60%
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Elections in Rhode Island |
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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%.
Results
[edit]1892 United States presidential election in Rhode Island[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Benjamin Harrison of Indiana (incumbent) | Whitelaw Reid of New York | 26,975 | 50.71% | 4 | 100.00% | ||
Democratic | Grover Cleveland of New York | Adlai Ewing Stevenson I of Illinois | 24,336 | 45.75% | 0 | 0.00% | ||
Prohibition | John Bidwell of California | James Cranfill of Texas | 1,654 | 3.11% | 0 | 0.00% | ||
Populist | James Baird Weaver of Iowa | James Gaven Field of Virginia | 228 | 0.43% | 0 | 0.00% | ||
N/A | Others | Others | 3 | 0.01% | 0 | 0.00% | ||
Total | 53,196 | 100.00% | 4 | 100.00% |
See also
[edit]References
[edit]- ^ "1892 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.