1864 United States presidential election in Delaware

Main article: 1864 United States presidential election

The 1864 United States presidential election in Delaware took place on November 8, 1864, as part of the 1864 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president.^[1]

1864 United States presidential election in Delaware

← 1860 November 8, 1864 1868 →

Nominee George B. McClellan Abraham Lincoln Party Democratic National Union Home state New Jersey Illinois Running mate George H. Pendleton Andrew Johnson Electoral vote 3 0 Popular vote 8,767 8,155 Percentage 51.81% 48.19%

County Results McClellan   50–60% Lincoln   50–60%

President before election Abraham Lincoln Republican Elected President Abraham Lincoln National Union

Delaware was won by the Democratic nominee, 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton. They defeated the National Union nominee, incumbent President Abraham Lincoln of Illinois and his running mate Senator and Military Governor of Tennessee Andrew Johnson.^[1] McClellan won the state by a narrow margin of 3.62%.

With 51.81% of the popular vote, Delaware would prove to be McClellan's third strongest state after Kentucky and New Jersey, his only two other winning states.^[2]

Results

1864 United States presidential election in Delaware^[1]

Party		Candidate	Votes	%
	Democratic	George B. McClellan	8,767	51.81%
	National Union	Abraham Lincoln (incumbent)	8,155	48.19%
Total votes			16,922	100.00%

See also

• United States presidential elections in Delaware