Current Electricity · Example 3.26

Equivalent Resistance: A–B and A–D

In the given network of resistors, find the equivalent resistance between points A and B, and between A and D.

Square D-C-B-A with edges 6/2/3/1 Ω, plus the 5 Ω diagonal A–C.

Given (edge list)

D–C: 6 Ω (top)
A–D: 2 Ω (left)
C–B: 3 Ω (right)
A–B: 1 Ω (bottom)
A–C: 5 Ω (diagonal)

Asked

(a) R_AB and (b) R_AD.

Formula & Key Idea

Series: R₁+R₂ · Parallel: R_aR_b/(R_a+R_b).

The trick: a corner that is not a terminal connects to only two other nodes, so it is just a series link — collapse it and the bridge becomes a plain series–parallel ladder.

a Resistance between A and B

Solution — R_AB

STEP 1 — D is the "free" corner ⇒ collapse it

D touches only A (2 Ω) and C (6 Ω), so the path A–D–C is series: \( 2 + 6 = 8\ \Omega \) between A and C.

STEP 2 — That 8 Ω is parallel to the 5 Ω diagonal (both join A–C)

\[ R_{AC} = \frac{5 \times 8}{5 + 8} = \frac{40}{13}\ \Omega \]

STEP 3 — Path A→C→B in series, then parallel with direct A–B (1 Ω)

\[ R_{A\to C\to B} = \frac{40}{13} + 3 = \frac{79}{13}\ \Omega \] \[ R_{AB} = \frac{1 \times \tfrac{79}{13}}{1 + \tfrac{79}{13}} = \frac{79}{92}\ \Omega \]

R_AB 79⁄92 ≈ 0.86 Ω

b Resistance between A and D

Solution — R_AD

STEP 1 — Now B is the "free" corner ⇒ collapse it

B touches only A (1 Ω) and C (3 Ω), so the path A–B–C is series: \( 1 + 3 = 4\ \Omega \) between A and C.

STEP 2 — That 4 Ω is parallel to the 5 Ω diagonal (both join A–C)

\[ R_{AC} = \frac{5 \times 4}{5 + 4} = \frac{20}{9}\ \Omega \]

STEP 3 — Path A→C→D in series, then parallel with direct A–D (2 Ω)

\[ R_{A\to C\to D} = \frac{20}{9} + 6 = \frac{74}{9}\ \Omega \] \[ R_{AD} = \frac{2 \times \tfrac{74}{9}}{2 + \tfrac{74}{9}} = \frac{37}{23}\ \Omega \]

R_AD 37⁄23 ≈ 1.61 Ω

Short Cut

Pick the terminals → the leftover corner is always a pure series link. For A–B, D is free (2+6 = 8 Ω); for A–D, B is free (1+3 = 4 Ω). Merge it, leaving the diagonal 5 Ω parallel to that series sum across A–C, then finish with one more series + parallel step. No star–delta, no Kirchhoff loops needed.

Why it works: a degree-2 node carries the same current in and out, so it can never cause a balanced/bridge complication — it is electrically just a resistor in line.

Given → Asked → Formula → Solution → Short Cut · verified by node-voltage analysis