Election to the United States Senate in 1944

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On November 7, 1944, one-third of the members of the United States Senate were elected in the United States . The election was part of the general election to the 79th United States Congress that year, which also elected all members of the House of Representatives. At the same time the presidential election of 1944 took place, which was won for the fourth time in a row by Democrat Franklin D. Roosevelt .

Since the adoption of the 17th Amendment to the United States Constitution in 1913, all US senators in their respective states have been directly elected by the people of their state. Each state has 2 senators. Under the United States Constitution , US Senators are elected for six years. However, all members of the Senate are never elected at the same time. The election follows a scheme according to which a third of the senators are elected every two years at the same time as the election to the US House of Representatives. For this purpose, the Senate is divided into three classes , which determine the election year of the Senators. In 1944, Class III senators stood for election. At that time, the United States consisted of 48 states. This results in a total of 96 senators, 37 of which were available for election. The Democrats had to cede a mandate to the Republicans, but retained a clear absolute Senate majority. Like the elections of 1942, these elections were overshadowed by the events of World War II .

Senate composition after the election

Total: 96

The results of the last elections on November 3, 1942 are in brackets. Changes in the course of the legislative period that do not affect the elections themselves are not included in these figures. But are noted in the article on the 79th Congress in the section on the members of the Senate by the corresponding names of the senators.

See also