1864 United States presidential election in Rhode Island
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![]() County Results
Lincoln 60-70%
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Elections in Rhode Island |
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The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%.
Results[]
1864 United States presidential election in Rhode Island[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
National Union | Abraham Lincoln | 13,962 | 62.24% | 4 | |
Democratic | George B. McClellan | 8,470 | 37.76% | 0 | |
Totals | 22,432 | 100.0% | 4 |
See also[]
References[]
Categories:
- 1864 United States presidential election by state
- United States presidential elections in Rhode Island
- 1864 Rhode Island elections
- Rhode Island election stubs