United States presidential election in Nebraska, 1892
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Elections in Nebraska |
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The 1892 United States presidential election in Nebraska took place on November 8, 1892. All contemporary 44 states were part of the 1892 United States presidential election. Nebraska voters chose eight electors to the Electoral College, which selected the president and vice president.
Nebraska was won by the Populist nominees, James B. Weaver of Iowa and his running mate James G. Field of Virginia. Weaver and Field defeated the Republican nominees, incumbent President Benjamin Harrison of Indiana and his running mate Whitelaw Reid of New York.
Results
United States presidential election in Nebraska, 1892[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | Benjamin Harrison (incumbent) | 87,213 | 43.56% | 8 | |
People's | James Weaver | 83,134 | 41.53% | 0 | |
Democratic | Grover Cleveland | 24,943 | 12.46% | 0 | |
Prohibition | John Bidwell | 4,902 | 2.45% | 0 | |
Totals | 200,192 | 100.00% | 8 | ||
Voter turnout | — |
References
- ↑ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Nebraska
Notes
This article is issued from
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