Open main menu

1892 United States presidential election in Wisconsin

  (Redirected from United States presidential election in Wisconsin, 1892)

The 1892 United States presidential election in Wisconsin was held on November 8, 1892. Wisconsin voters chose twelve electors to the Electoral College, who voted for president and vice president.

United States presidential election, 1892

← 1888 November 8, 1892 1896 →
  Grover Cleveland - NARA - 518139.tif Pach Brothers - Benjamin Harrison.jpg
Nominee Grover Cleveland Benjamin Harrison
Party Democratic Republican
Home state New York Indiana
Running mate Adlai Stevenson I Whitelaw Reid
Electoral vote 12 0
Popular vote 177,325 171,101
Percentage 47.72% 46.05%

President before election

Benjamin Harrison
Republican

Elected President

Grover Cleveland
Democratic

Democratic Party candidate Grover Cleveland won the state with 48% of the popular vote, winning Wisconsin's twelve electoral votes.[1] This marked the first time since 1852 where the Democratic candidate carried Wisconsin.

ResultsEdit

United States presidential election in Wisconsin, 1892
Party Candidate Votes Percentage Electoral votes
Democratic Grover Cleveland 177,325 47.72% 12
Republican Benjamin Harrison (incumbent) 171,101 46.05% 0
Prohibition John Bidwell 13,136 3.54% 0
People's James Weaver 10,019 2.7% 0
Totals 371,581 100.0% 12

ReferencesEdit

  1. ^ "1892 Presidential General Election Results - Wisconsin". Retrieved August 18, 2016.