1908 United States presidential election in Delaware

Main article: 1908 United States presidential election

The 1908 United States presidential election in Delaware took place on November 3, 1908. All contemporary 46 states were part of the 1908 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.

1908 United States presidential election in Delaware

← 1904 November 3, 1908 1912 →

Nominee William Howard Taft William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate James S. Sherman John W. Kern Electoral vote 3 0 Popular vote 25,014 22,055 Percentage 52.10% 45.94%

County Results Taft   40–50%   50–60%

President before election Theodore Roosevelt Republican Elected President William Howard Taft Republican

Delaware was won by the Republican nominees, former Secretary of War William Howard Taft of Ohio and his running mate James S. Sherman of New York. They defeated the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate John W. Kern. Taft won the state by a margin of 6.16%.

Bryan had previously lost Delaware to William McKinley in both 1896 and 1900.

Results

1908 United States presidential election in Delaware[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	William Howard Taft	25,014	52.10%	3
	Democratic	William Jennings Bryan	22,055	45.94%	0
	Prohibition	Eugene W. Chafin	670	1.40%	0
	Social Democratic	Eugene V. Debs	239	0.50%	0
	Independence	Thomas L. Hisgen	29	0.06%	0
Totals			48,007	100.00%	3
Voter turnout					—

Results by county

County	William Howard Taft Republican		William Jennings Bryan Democrat		Various candidates Others		Total votes
	%	#	%	#	%	#
Kent	49.75%	4,158	49.00%	4,095	1.25%	104	8,357
New Castle	52.29%	14,979	45.26%	12,964	2.45%	701	28,644
Sussex	53.25%	5,870	45.48%	5,013	1.27%	140	11,023

See also

• United States presidential elections in Delaware

Notes