1940 United States presidential election in Delaware

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 3[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1940 United States presidential election

United States presidential election in Delaware, 1940

November 5, 1940[1]
All 3 Delaware votes to the Electoral College
Nominee Franklin D. Roosevelt Wendell Willkie Party Democratic Republican Home state New York Indiana Running mate Henry A. Wallace Charles L. McNary Electoral vote 3 0 Popular vote 74,599 61,440 Percentage 54.7% 45.1%
County Results Roosevelt   50-60%
President before election Franklin D. Roosevelt Democratic Elected President Franklin D. Roosevelt Democratic

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–Indiana), running with Minority Leader Charles L. McNary, with 45.05% of the popular vote.[3][4]

Results

United States presidential election in Delaware, 1940

Party		Candidate	Votes	%
	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%

County	Dem	Rep	Oth
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

References

1. "United States Presidential election of 1940 - Encyclopædia Britannica". Retrieved August 20, 2018.
2. "1940 Election for the Thirty-ninth Term (1941-45)". Retrieved August 20, 2018.
3. "1940 Presidential General Election Results - Delaware". Retrieved August 20, 2018.
4. "The American Presidency Project - Election of 1940". Retrieved August 20, 2018.