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1892 United States presidential election in Rhode Island

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1892 United States presidential election in Rhode Island

1888 November 8, 1892 1896
Nominee Benjamin Harrison Grover Cleveland
Party Republican Democratic
Home state Indiana New York
Running mate Whitelaw Reid Adlai Stevenson I
Electoral vote 4 0
Popular vote 26,975 24,336
Percentage 50.71% 45.75%

County Results
Harrison
40-50%
50-60%


President before election

Benjamin Harrison
Republican

Elected President

Grover Cleveland
Democratic

The1892 United States presidential election in Rhode Islandtook place on November 8, 1892, as part of the1892 United States presidential election.Voters chose four representatives, or electors to theElectoral College,who voted forpresidentandvice president.

Rhode Islandvoted for theRepublicannominee, incumbentPresidentBenjamin Harrison,over theDemocraticnominee, former PresidentGrover 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]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin HarrisonofIndiana(incumbent) Whitelaw ReidofNew York 26,975 50.71% 4 100.00%
Democratic Grover ClevelandofNew York Adlai Ewing Stevenson IofIllinois 24,336 45.75% 0 0.00%
Prohibition John BidwellofCalifornia James CranfillofTexas 1,654 3.11% 0 0.00%
Populist James Baird WeaverofIowa James Gaven FieldofVirginia 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]

  1. ^"1892 Presidential General Election Results - Rhode Island".U.S. Election Atlas.RetrievedDecember 23,2013.