Ladder network — equivalent resistance and node voltage: 22-Elec-A1 Circuits, December 2019, Q1

Question 1: Ladder network — equivalent resistance and node voltage [10 + 10]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Figure‑1 — redrawn resistive ladder. Terminal A is the top‑left node; B is the bottom rail. Series links (top): 12 Ω, 5 Ω, 10 Ω. Shunt legs (to B): 10 Ω, 20 Ω ($V_1$ across it), 10 Ω, 25 Ω. The 10 &#. See the official exam paper.]

Given. A four‑section resistive ladder between terminals A (top) and B (bottom rail). The node A = N₁ feeds successive nodes N₂, N₃, N₄ through the series resistors, and each node has a shunt resistor to B. An open switch sits in the 10 Ω link between N₃ and N₄. For part (b) an ideal 50 V source is applied directly across A–B, and $V_1$ is the voltage across the 20 Ω shunt at N₂.

Section	Series link	Shunt to B
N₁ (= A)	—	$10\,\Omega$
N₁→N₂	$12\,\Omega$	$20\,\Omega$ (across it: $V_1$)
N₂→N₃	$5\,\Omega$	$10\,\Omega$
N₃→N₄	$10\,\Omega$ + open switch	$25\,\Omega$

Find. (a) $R_{AB}$; (b) $V_1$ with 50 V applied.

Check — switch state. The reconstructed figure shows the 10 Ω + 25 Ω final section isolated by an open switch, so it carries no current and is excluded from both parts. If your copy shows that switch closed, include the last section ($R_{AB}=6.64\,\Omega$, $V_1=19.7$ V instead).

Approach. With the switch open the ladder truncates at N₃; collapse it from the far end back to A by alternating series additions and parallel (shunt) combinations, then apply a voltage divider for $V_1$.

Right end (N₃). The open switch removes the 10 Ω link and the 25 Ω leg, so the only resistance at N₃ is its own shunt: $R_{N_3}=10\,\Omega$.
Back through the 5 Ω link and N₂ shunt. Add the series 5 Ω, then parallel the 20 Ω shunt: $$20\;\|\;(5+10)=20\;\|\;15=\frac{20\cdot15}{35}=\frac{60}{7}=8.571\,\Omega.$$
Back through the 12 Ω link and N₁ shunt. Add the series 12 Ω, then parallel the 10 Ω shunt at A: $$R_{AB}=10\;\|\;\Big(12+\tfrac{60}{7}\Big)=10\;\|\;\tfrac{144}{7}=\frac{10\cdot\frac{144}{7}}{10+\frac{144}{7}}=\boxed{6.73\,\Omega.}$$
Apply 50 V for $V_1$. The ideal source fixes $V_A=50$ V; the 10 Ω shunt at A draws current but cannot change $V_A$. Looking right of N₂, the load on that node is $\tfrac{60}{7}=8.571\,\Omega$, reached through the 12 Ω link, so $V_1$ is a divider of $V_A$: $$V_1=V_A\cdot\frac{R_{N_2}}{12+R_{N_2}}=50\cdot\frac{60/7}{12+60/7}=50\cdot\frac{60}{144}=\boxed{20.83\,\text{V}.}$$

Quantity	Result
Equivalent resistance $R_{AB}$	$6.73\,\Omega$
Node voltage $V_1$ (with 50 V applied)	$20.83\,\text{V}\;(=250/12)$

Concept & Theory

Topic: ladder‑network reduction and the voltage divider.

Key relations: series $R_s=\sum R_i$; parallel $R_p=\dfrac{R_aR_b}{R_a+R_b}$; voltage divider $V_x=V\dfrac{R_x}{R_{\text{series}}+R_x}$.

Why this works: a ladder has no bridging element, so every node reduces by a strict alternation of one series step (the link resistor) and one parallel step (the shunt). Working from the open end guarantees each combination is a clean two‑terminal reduction. An ideal source clamps its node, so any resistor directly across it is irrelevant to downstream voltages — only to the source current.

Common pitfall: forgetting that an open switch deletes everything beyond it, or trying to include the 10 Ω shunt at A in the $V_1$ divider — it sits across the ideal source and drops out.

Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., §2.5–2.6.

Question 1 of 6: Ladder network — equivalent resistance and node voltage

Question 1: Ladder network — equivalent resistance and node voltage [10 + 10]