1946 Iowa Senate election

1946 Iowa Senate election

← 1944 November 5, 1946 1948 →

23 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Party Republican Democratic
Last election 45 5
Seats after 46 4
Seat change Increase1 Decrease1

The 1946 Iowa State Senate elections took place as part of the biennial 1946 United States elections. Iowa voters elected state senators in 23 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 1946 elections is provided by the Iowa General Assembly here.

The primary election on June 3, 1946, determined which candidates appeared on the November 5, 1946 general election ballot.[1][2]

Following the previous election, Republicans had control of the Iowa state Senate with 45 seats to Democrats' 5 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 21 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1946 general election with the balance of power shifting to Republicans holding 46 seats and Democrats having 4 seats (a net gain of 1 seat for Republicans).

  1. ^ "Primary Election 1946 For State Senator" (PDF). Iowa Official Register. Retrieved June 15, 2020.
  2. ^ "General Election 1946 For State Senator" (PDF). Iowa Official Register. Retrieved June 15, 2020.