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The ladder network represents a commonly used circuit style that is configured purely on the basis of series and parallel connections. A ladder is a network in which the product of its series and shunt impedances is independent of frequency within the range of interest. We can derive the equivalent resistance of the ladder circuit by successive applications of the series and parallel reduction formulas.
A three-section ladder network appears in [Fig. 1]. The reason for the terminology is quite obvious for the repetitive structure. Basically two approaches are used to solve ladder networks.

#### 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. 2: An alternative approach for ladder networks.
The assigned notation for the current through the final branch is $I_6$:
$$\begin{split} I_6 &= {V_4 \over R_5 + R_6} \\ &= {V_4 \over 1 Ω + 2 Ω}\\ &= {V_4 \over 3 Ω}\\ \end{split}$$
or
$$V_4 = (3 Ω)I_6$$
so that
$$I_4 = {V_4 \over R_4} = (3 Ω)I_6 6 Ω = 0.5 I_6$$
and
$$I_3 = I_4 + I_6 = 0.5 I_6 + I_6 = 1.5 I_6$$
$$V_3 = I_3 R_3 = (1.5 I_6)(4 Ω) = (6 Ω)I_6$$
Also,
$$V_2 = V_3 + V_4 = (6 Ω)I_6 + (3 Ω)I_6 = (9 Ω)I_6$$
so that
$$I_2 ={V_2 \over R_2} = {(9 Ω)I_6 \over 6 Ω }= 1.5 I_6$$
and
$$I_S = I_2 + I_3 = 1.5 I_6 + 1.5 I_6 = 3 I_6$$
with
$$V_1 = I_1 R_1 = I_S R_1 = (5 Ω) I_s$$
so that
$$\begin{split} E &= V_1 + V_2 = (5 Ω)I_s + (9 Ω)I_6\\ &= (5 Ω)(3 I_6) + (9 Ω)I_6 \\ &= (24 Ω)I_6\\ \end{split}$$
and
$$I_6 = {E \over 24 Ω} = {240 V \over 24 Ω} = 10 A$$
with
$$V_6 = I_6 R_6 = (10 A)(2 Ω) = 20 V$$

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