A three-section ladder network appears in Fig. No.1. The reason for the terminology is quite obvious for the repetitive structure. Basically two approaches are used to solve ladder networks.
Fig. No.1: Ladder network.

Method 1

Calculate the total resistance and resulting source current, and then work back through the ladder network until the desired current or voltage is obtained. This way is introduced previously in the reduced and return approach.

Method 2

Assign a letter symbol to the last branch current and work back through the ladder network to the source, maintaining this assigned current or other current of interest. The desired current can then be found directly.
Fig. No.2: An alternative approach for ladder networks.
The assigned notation for the current through the final branch is $I_6$:
$$I_6 = {V_4 \over R5 + R6}$$ $$= {V_4 \over 1 Ω + 2 Ω} = {V_4 \over 3 Ω}$$ or $$V_4 = (3 Ω)I_6$$ so that $$I_4 = {V_4 \over R4} = (3 Ω)I_6 6 Ω = 0.5 I_6$$ and $$I_3 = I_4 + I_6 = 0.5 I_6 + I_6 = 1.5 I_6$$ $$V_3 = I_3 R3 = (1.5 I_6)(4 Ω) = (6 Ω)I_6$$ Also, $$V_2 = V_3 + V_4 = (6 Ω)I_6 + (3 Ω)I_6 = (9 Ω)I_6$$ so that $$I_2 ={V_2 \over R2} = {(9 Ω)I_6 \over 6 Ω }= 1.5 I_6$$ and $$I_S = I_2 + I_3 = 1.5 I_6 + 1.5 I_6 = 3 I_6$$ with $$V_1 = I_1 R1 = I_S R1 = (5 Ω) I_s$$ so that $$E = V_1 + V_2 = (5 Ω)I_s + (9 Ω)I_6$$ $$= (5 Ω)(3 I_6) + (9 Ω)I_6 = (24 Ω)I_6$$ and $$I_6 = {E \over 24 Ω} = {240 V \over 24 Ω} = 10 A$$ with $$V_6 = I_6 R6 = (10 A)(2 Ω) = 20 V$$