1864 United States presidential election in Rhode Island

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1864 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1860 November 8, 1864 1868  
  Abraham Lincoln November 1863.jpg GeorgeMcClellan2 (cropped).jpg
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois New Jersey
Running mate Andrew Johnson George H. Pendleton
Electoral vote40
Popular vote13,9628,470
Percentage62.24%37.76%

Rhode Island Presidential Election Results 1864.svg
County Results
Lincoln
  60–70%


President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Contents

Rhode Island voted for the National Union candidate, incumbent Republican Party President Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic candidate, George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a margin of 24.48%.

Results

1864 United States presidential election in Rhode Island [1]
PartyCandidateVotesPercentageElectoral votes
National Union Abraham Lincoln (incumbent)13,96262.24%4
Democratic George B. McClellan 8,47037.76%0
Totals22,432100.0%4

See also

References

  1. "1864 Presidential General Election Results - Rhode Island".