Find the equivalent resistance between A and B for the 3-rung ladder shown.
Top & bottom rails (teal) connected by three vertical rungs (orange): 9 Ω, 5 Ω, 1 Ω.
Given
Top rail: 4 Ω – 2 Ω – 1 Ω
Bottom rail: 4 Ω – 3 Ω – 1 Ω
Left rung: 9 Ω
Middle rung: 5 Ω
Right rung: 1 Ω
Asked
Equivalent resistance RAB between A (top-left) and B (bottom-left).
Formula & Strategy
Series R₁+R₂ · Parallel
RaRb/(Ra+Rb).
Collapse the ladder from the open right end inward — each rung becomes parallel with
"the rest of the ladder to its right."
Solution — fold from the right
STEP 1 — Right rung + the two 1 Ω rail bits
The right arm (top 1 Ω + right rung 1 Ω + bottom 1 Ω) bridges the middle junctions:
\( 1 + 1 + 1 = 3\ \Omega \). This sits parallel to the 5 Ω middle rung:
\[ R_{\text{mid}} = \frac{5 \times 3}{5 + 3} = \frac{15}{8}\ \Omega \]
STEP 2 — Walk left to the 9 Ω rung
Going through the middle section: \( 2 + \tfrac{15}{8} + 3 = \tfrac{55}{8}\ \Omega \).
Parallel with the 9 Ω left rung:
\[ R_{\text{left}} = \frac{9 \times \tfrac{55}{8}}{9 + \tfrac{55}{8}}
= \frac{495/8}{127/8} = \frac{495}{127}\ \Omega \approx 3.90\ \Omega \]
STEP 3 — Add the two outer 4 Ω in series (A→ and →B)
Always fold a ladder from the dead-end side. The rightmost rung
has nothing beyond it, so start there: combine its rung with the rail pieces around it, then repeat one rung
at a time moving toward the source. Each move is just one "series-then-parallel."
Checkpoints (match your book's margin): middle stage → 55/8 Ω,
after the 9 Ω rung → 495/127 Ω, then +4 Ω +4 Ω gives the answer.
The two end 4 Ω resistors carry the whole current, so they are pure series — never parallel.
Given → Asked → Formula → Solution → Short Cut · verified by full nodal analysis (R = 1511/127 Ω)