1948 Iowa Senate election

1948 Iowa Senate election

← 1946 November 5, 1948 1950 →

29 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Party Republican Democratic
Last election 46 4
Seats after 43 7
Seat change Decrease3 Increase3

The 1948 Iowa State Senate elections took place as part of the biennial 1948 United States elections. Iowa voters elected state senators in 29 of the state senate's 50 districts. State senators serve four-year terms in the Iowa State Senate.

A statewide map of the 50 state Senate districts in the year 1948 is provided by the Iowa General Assembly here.

The primary election on June 3, 1948, determined which candidates appeared on the November 5, 1948 general election ballot.[1][2]

Following the previous election, Republicans had control of the Iowa state Senate with 46 seats to Democrats' 4 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 22 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1948 general election with the balance of power shifting to Republicans holding 43 seats and Democrats having 7 seats (a net gain of 3 seats for Democrats).

  1. ^ "Primary Election 1948 For State Senator" (PDF). Iowa General Assembly. Retrieved June 15, 2020.
  2. ^ "General Election 1948 For State Senator" (PDF). Iowa General Assembly. Retrieved June 15, 2020.

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