1940 United States presidential election in Rhode Island

1940 United States presidential election in Rhode Island
	November 5, 1940
Roosevelt 50–60% 60–70% 70–80%	Willkie 50–60% 60–70% 70–80% 80–90%
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%. Rhode Island was one of five states that swung more Democratic compared to 1936, alongside New Hampshire, Maine, Vermont, and North Carolina.