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.
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 $$V_6 = I_6 R6 = (10 A)(2 Ω) = 20 V$$