1864 United States presidential election in Rhode Island
Jump to navigation
Jump to search
![]() | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
![]() County Results
Lincoln 60-70%
| ||||||||||||||||||||||||||
|
Elections in Rhode Island |
---|
![]() |
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, 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 |
See also
References
Categories:
- Articles with short description
- Short description matches Wikidata
- Use mdy dates from September 2023
- Elections using electoral votes
- 1864 United States presidential election by state
- United States presidential elections in Rhode Island
- 1864 Rhode Island elections
- All stub articles
- Rhode Island election stubs