1940 United States presidential election in Delaware
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All 3 Delaware votes to the Electoral College | ||||||||||||||||||||||||||
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County Results
Roosevelt 50-60%
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Elections in Delaware |
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The 1940 United States presidential election in Delaware took place on November 5, 1940, as part of the 1940 United States presidential election. Delaware voters chose three[2] representatives, or electors, to the Electoral College, who voted for president and vice president.
Delaware was won by incumbent President Franklin D. Roosevelt (D–New York), running with Secretary Henry A. Wallace, with 54.70% of the popular vote, against Wendell Willkie (R–New York), running with Minority Leader Charles L. McNary, with 45.05% of the popular vote.[3][4] Delaware was one of six states that swung more Democratic compared to 1936, alongside Maine, New Hampshire, Rhode Island, Vermont, and North Carolina.
Results
Party | Candidate | Votes | % | |
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Democratic | Franklin D. Roosevelt (inc.) | 74,599 | 54.70% | |
Republican | Wendell Willkie | 61,440 | 45.05% | |
Write-in | 335 | 0.25% | ||
Total votes | 136,334 | 100% |
Results by county
County | Dem | Rep | Oth |
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Kent | 9,226 | 8,079 | 48 |
New Castle | 52,167 | 41,508 | 261 |
Sussex | 13,206 | 11,853 | 26 |
STATEWIDE | 74,599 | 61,440 | 335 |
See also
References
- ^ "United States Presidential election of 1940 - Encyclopædia Britannica". Retrieved August 20, 2018.
- ^ "1940 Election for the Thirty-ninth Term (1941-45)". Retrieved August 20, 2018.
- ^ "1940 Presidential General Election Results - Delaware". Retrieved August 20, 2018.
- ^ "The American Presidency Project - Election of 1940". Retrieved August 20, 2018.