1940 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia
1940 United States presidential election in Rhode Island

← 1936 November 5, 1940 1944 →
  FDRoosevelt1938.png WendellWillkie.jpg
Nominee Franklin D. Roosevelt Wendell Willkie
Party Democratic Republican
Home state New York New York
Running mate Henry A. Wallace Charles L. McNary
Electoral vote 4 0
Popular vote 182,182 138,653
Percentage 56.73% 43.17%

Rhode Island Presidential Election Results 1940.svg
County Results

President before election

Franklin D. Roosevelt
Democratic

Elected President

Franklin D. Roosevelt
Democratic

The 1940 United States presidential election in Rhode Island took place on November 5, 1940. All contemporary 48 states were part of the 1940 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.

Rhode Island was won by incumbent Democratic President Franklin D. Roosevelt of New York, who was running against Republican businessman Wendell Willkie of New York. Roosevelt ran with former Secretary of Agriculture Henry A. Wallace of Iowa as his running mate, and Willkie ran with Senator Charles L. McNary of Oregon.

Roosevelt won Rhode Island by a margin of 13.56%.

Results[]

1940 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Delano Roosevelt of New York Henry Agard Wallace of Iowa 182,182 56.73% 4 100.00%
Republican Wendell Willkie of New York Charles Linza McNary of Oregon 138,653 43.27% 0 0.00%
Communist Earl Russell Browder of Kansas James W. Ford of New York 239 0.07% 0 0.00%
Prohibition Roger Ward Babson of Massachusetts of Illinois 74 0.02% 0 0.00%
Total 321,148 100.00% 4 100.00%

See also[]

References[]

  1. ^ "1940 Presidential General Election Results – Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.


Retrieved from ""