1864 United States presidential election in Vermont
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County Results
Lincoln 60-70% 70-80% 80-90% 90-100%
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Elections in Vermont |
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The 1864 United States presidential election in Vermont took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.
Vermont voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a wide margin of 52.20%.
With 76.10% of the popular vote, Lincoln's victory with in the state would be his second strongest victory in the country in terms of percentage in the popular vote after Kansas.[1]
Results[]
1864 United States presidential election in Vermont[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
National Union | Abraham Lincoln | 42,420 | 76.10% | 5 | |
Democratic | George B. McClellan | 13,322 | 23.90% | 0 | |
Totals | 55,742 | 100.0% | 26 |
See also[]
- United States presidential elections in Vermont
References[]
- ^ "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ^ "1864 Presidential General Election Results - Vermont".
Categories:
- Vermont election stubs
- 1864 United States presidential election by state
- United States presidential elections in Vermont
- 1864 Vermont elections