1860 United States presidential election in Delaware

The 1860 United States presidential election in Delaware took place on November 6, 1860, as part of the 1860 United States presidential election. Delaware voters chose 3 representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1860 United States presidential election
1860 United States presidential election in Delaware
November 6, 1860
 
Nominee John C. Breckinridge John Bell
Party Southern Democratic Constitutional Union
Home state Kentucky Tennessee
Running mate Joseph Lane Edward Everett
Electoral vote 3 0
Popular vote 7,339 3,888
Percentage 45.54% 24.13%
 
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote 0 0
Popular vote 3,822 1,066
Percentage 23.72% 6.61%
President before election

James Buchanan
Democratic

 Elected President

Abraham Lincoln
Republican

Elections in Delaware

Delaware was won by the 14th Vice President of the United States John C. Breckinridge (SD–Kentucky), running with Senator Joseph Lane, with 45.54% of the popular vote, against Senator John Bell (CU–Tennessee), running with the Governor of Massachusetts Edward Everett, with 24.13% of the popular vote, Illinois Representative Abraham Lincoln (R–Kentucky), running with Senator Hannibal Hamlin, with 23.72% of the popular vote and the 15th Senator Stephen A. Douglas (D–Vermont), running with 41st Governor of Georgia Herschel V. Johnson, with 6.61% of the popular vote.

Results
1860 United States presidential election in Delaware[1]
Party Candidate Votes %
Southern Democratic John C. Breckinridge 7,339 45.54%
Constitutional Union John Bell 3,888 24.13%
Republican Abraham Lincoln 3,822 23.72%
Democratic Stephen A. Douglas 1,066 6.61%
Total votes 16,115 100.00%
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References
  1. "1860 Presidential Election Results Delaware".
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