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1864 United States presidential election in Delaware information


Main article: 1864 United States presidential election
1864 United States presidential election in Delaware
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
President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

Elections in Delaware
1864 United States presidential election in Delaware
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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]

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 margin of 3.62%.

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

  1. ^ a b "1864 Presidential Election Results Delaware".
  2. ^ "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.

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