1926 Iowa Senate election

1926 Iowa Senate election

← 1924 November 2, 1926 1928 →

21 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party Third party
 
Party Republican Democratic Independent
Last election 45 4 1
Seats after 49 1 0
Seat change Increase4 Decrease3 Decrease1

Results      Republican gain
     Republican hold

The 1926 Iowa State Senate elections took place as part of the biennial 1926 United States elections. Iowa voters elected state senators in 21 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 1926 elections is provided by the Iowa General Assembly here.

The primary election on June 7, 1926, determined which candidates appeared on the November 2, 1926 general election ballot.[1][2]

Following the previous election, Republicans had control of the Iowa state Senate with 45 seats, the Democrats had 4 seats, and 1 seat was held by an independent.

Republicans maintained control of the Iowa State Senate following the 1926 general election with the balance of power shifting to Republicans holding 49 seats and Democrats having 1 seat (a net gain of 4 seats for Republicans). Like 6 years prior, Democrats lost every district up for election, including four held by them. After the election, Lloyd Ellis of the 3rd Senate District was the only Democrat in the chamber.

  1. ^ "Primary Election 1926 For State Senator" (PDF). Iowa Official Register. Retrieved January 18, 2021.
  2. ^ "General Election 1926 For State Senator" (PDF). Iowa Official Register. Retrieved January 18, 2021.