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The1892 United States presidential election in Rhode Island took place on November 8, 1892, as part of the1892 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.
Rhode Island voted for theRepublican nominee, incumbentPresidentBenjamin Harrison, over theDemocratic nominee, former PresidentGrover Cleveland, who was running for a second, non-consecutive term. Harrison won the state by a narrow margin of 4.96%.
| 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 ofIndiana(incumbent) | Whitelaw Reid ofNew York | 26,975 | 50.71% | 4 | 100.00% | ||
| Democratic | Grover Cleveland ofNew York | Adlai Ewing Stevenson I ofIllinois | 24,336 | 45.75% | 0 | 0.00% | ||
| Prohibition | John Bidwell ofCalifornia | James Cranfill ofTexas | 1,654 | 3.11% | 0 | 0.00% | ||
| Populist | James Baird Weaver ofIowa | James Gaven Field ofVirginia | 228 | 0.43% | 0 | 0.00% | ||
| N/A | Others | Others | 3 | 0.01% | 0 | 0.00% | ||
| Total | 53,196 | 100.00% | 4 | 100.00% | ||||
State and district results of the1892 United States presidential election | ||
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