Current Electricity · Example 3.27

Effective Resistance Between A and B

Find the effective resistance between points A and B for the network shown.

C D E F A B 2Ω 2Ω 4Ω 2Ω 4Ω 2Ω 4Ω 2Ω 2Ω
Red = 4 Ω direct spokes from A · Teal = 2 Ω chain links. Find R across A–B.
Given (edge list)
  • C–D: 2 Ω
  • C–A: 2 Ω
  • D–A: 4 Ω
  • D–E: 2 Ω
  • A–E: 4 Ω
  • E–F: 2 Ω
  • A–F: 4 Ω
  • F–B: 2 Ω
  • A–B: 2 Ω
Asked

Effective resistance RAB between A and B.

Formula & Pattern

Series R₁+R₂; Parallel RaRb/(Ra+Rb). The network repeats a unit: each node X has a chain step (prev→X = 2 Ω) and a direct spoke (A→X = 4 Ω).

\[ R_{A\text{-}X} = \big(R_{\text{prev}} + 2\big)\ \parallel\ 4 \]
Solution — peel A outward, node by node
STEP 1 — Node D (eliminate C)
A–C–D in series = \(2+2 = 4\,\Omega\), parallel with the direct A–D = 4 Ω: \[ R_{AD} = \frac{4\times4}{4+4} = 2\ \Omega \]
STEP 2 — Node E
\( (R_{AD}+2)\parallel 4 = (2+2)\parallel 4 = \dfrac{4\times4}{8} = 2\ \Omega \)
STEP 3 — Node F
\( (R_{AE}+2)\parallel 4 = (2+2)\parallel 4 = 2\ \Omega \)
STEP 4 — Node B (last spoke is 2 Ω, not 4 Ω)
\( (R_{AF}+2)\parallel 2 = (2+2)\parallel 2 = \dfrac{4\times2}{4+2} = \dfrac{8}{6} = \dfrac{4}{3}\ \Omega \)
Stage(prev + 2)∥ direct spokeR to A
D2+2 = 4∥ 42 Ω
E2+2 = 4∥ 42 Ω
F2+2 = 4∥ 42 Ω
B2+2 = 4∥ 24/3 Ω
Effective Resistance RAB = 4⁄3 ≈ 1.33 Ω
Short Cut

Spot the fixed point. Whenever the running resistance to A is 2 Ω and the next unit is "+2 Ω in series, then ∥ 4 Ω," you get (4 ∥ 4) = 2 Ω again — it never changes. So D, E, F are all just 2 Ω from A without any arithmetic. Only the last spoke differs (2 Ω instead of 4 Ω), and that single step gives the answer: (2+2) ∥ 2 = 4/3 Ω.

Reading tip: in a fan-from-one-node network, always reduce starting at the node farthest from your terminals and walk inward — each "triangle" collapses to a single series-then-parallel move.

Given → Asked → Formula → Solution → Short Cut · verified by full nodal analysis (R = 4/3 Ω)