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1890 United States House of Representatives election in Wyoming

There were two United States House of Representatives elections in Wyoming in 1890, both held on September 11, 1890. The first was for the 51st United States Congress, with a term expiring on March 4, 1891; this election was necessary due to Wyoming's admittance to the Union as the 44th state earlier in 1890. The second election was for the full term in the 52nd United States Congress beginning on March 4, 1891. Republican lawyer Clarence D. Clark defeated Democratic George T. Beck with about 58% of the vote in both elections and became the first person to represent Wyoming in the House of Representatives.

1890 United States House of Representatives elections in Wyoming
September 11, 1890 (1890-09-11) 1892 →
 
Nominee Clarence D. Clark George T. Beck
Party Republican Democratic
Popular vote 8,751 6,219
Percentage 58.46% 41.54%
Elected U.S. Representative

Clarence D. Clark
Republican

Results

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1890 United States House of Representatives election in Wyoming (51st Congress)[1]
Party Candidate Votes %
Republican Clarence D. Clark 9,087 58.22%
Democratic George T. Beck 6,520 41.78%
Total votes 15,607 100%
1890 United States House of Representatives election in Wyoming (52nd Congress)[1]
Party Candidate Votes %
Republican Clarence D. Clark 8,751 58.46%
Democratic George T. Beck 6,219 41.54%
Total votes 14,970 100%

References

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  1. ^ a b Wyoming Secretary of State (1897). "Appendix B". Report of the Secretary of State of State of Wyoming for the Period Ending Sept. 30 1896. Cheyenne, Wyoming: The S. A. Bristol Company, Printers and Bookbinders. pp. xxiv.


   

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