1940 United States presidential election in Rhode Island
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| Elections in Rhode Island |
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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 4 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 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% | ||||
References
- ^ "1940 Presidential General Election Results – Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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