1864 United States presidential election in Rhode Island

Main article: 1864 United States presidential election

1864 United States presidential election in Rhode Island

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%

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.

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

See also

• United States presidential elections in Rhode Island