1940 United States presidential election in Rhode Island

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.

1940 United States presidential election in Rhode Island

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

County Results

President before election

Franklin D. Roosevelt
Democratic

Elected President

Franklin D. Roosevelt
Democratic

Rhode Island was won by incumbent Democratic President Franklin D. Roosevelt of New York, who was running against Republican businessman Wendell Willkie of Indiana. Roosevelt ran with 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 Indiana 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 Edgar Moorman 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.


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