1864 United States presidential election in Rhode Island

Main article: 1864 United States presidential election

November 8, 1864


Nominee	Abraham Lincoln	George B. McClellan
Party	National Union	Democratic
Home state	Illinois	New Jersey
Running mate	Andrew Johnson	George H. Pendleton
Electoral vote	4	0
Popular vote	13,962	8,470
Percentage	62.24%	37.76%

County Results Lincoln 60-70%

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Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

Elections in Rhode Island

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Question 1

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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.

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
Party		Candidate	Votes	Percentage	Electoral votes
	National Union	Abraham Lincoln (incumbent)	13,962	62.24%	4
	Democratic	George B. McClellan	8,470	37.76%	0
Totals			22,432	100.0%	4

References

"1864 Presidential General Election Results - Rhode Island".

v t e State results of the 1864 U.S. presidential election
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