1864 United States presidential election in Delaware
The 1864 United States presidential election in Delaware took place on November 8, 1864, as part of the 1864 United States presidential election. Delaware voters chose 3 representatives, or electors, to the Electoral College, who voted for president and vice president.[1]
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| Elections in Delaware | ||||||||||||
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Delaware was won by the 4th Commanding General of the United States Army George B. McClellan (D–Pennsylvania), running with Representative George H. Pendleton, with 51.81% of the popular vote against the incumbent President Abraham Lincoln (R-Illinois), running with former Senator and Military Governor of Tennessee Andrew Johnson, with 48.19% of the popular vote.[1]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | George B. McClellan | 8,767 | 51.81% | |
| National Union | Abraham Lincoln | 8,155 | 48.19% | |
| Total votes | 16,922 | 100.00% | ||
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References
State results of the 1864 U.S. presidential election | ||
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