1864 United States presidential election 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.
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November 8, 1864
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 County Results
Lincoln 60–70%
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Rhode Island voted for the National Union candidate, incumbent Republican Party President Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic candidate, George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a margin of 24.48%.
Results
| 1864 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| National Union | Abraham Lincoln (incumbent) | 13,962 | 62.24% | 4 | |
| Democratic | George B. McClellan | 8,470 | 37.76% | 0 | |
| Totals | 22,432 | 100.0% | 4 | ||
