Current Electricity · Example 3.29

Equivalent Resistance Between A and B

Find the equivalent resistance between A and B in the two cases shown.

i Chain with two jumper wires
A B M N 6R 9R 12R plain wire: A ≡ N plain wire: M ≡ B
A–6R–M–9R–N–12R–B, with a top wire joining A to N and a bottom wire joining M to B.
Given
Asked

Equivalent resistance RAB.

Key Idea + Formula

A plain wire has zero resistance, so its two ends are the same node. Merge them, then combine in parallel:

\[ \frac{1}{R_{AB}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \]
Solution — RAB
STEP 1 — Collapse the two jumpers
Top wire ⇒ A ≡ N.   Bottom wire ⇒ M ≡ B. So there are only two distinct nodes: {A, N} and {M, B}.
STEP 2 — Re-label each resistor by its end nodes
6R: A–M → A–B.   9R: M–N → B–A.   12R: N–B → A–B.
All three resistors connect A to B ⇒ pure parallel.
STEP 3 — Add the reciprocals
\[ \frac{1}{R_{AB}} = \frac{1}{6R}+\frac{1}{9R}+\frac{1}{12R} = \frac{6+4+3}{36R} = \frac{13}{36R} \] \[ R_{AB} = \frac{36R}{13} \approx 2.77\,R \]
Case (i) RAB = 36R⁄13 ≈ 2.77 R
ii Spokes inside a conducting ring
A B 2R 2R 2R 2R ring = zero-resistance wire ⇒ all rim points are one node
Four 2R spokes from centre A to a bare conducting ring; B is a point on the ring.
Given
Asked

Equivalent resistance RAB between centre A and rim point B.

Key Idea + Formula

A bare conducting loop shorts all the points it touches into one node. Equal resistors in parallel:

\[ R_{\text{parallel}} = \frac{r}{n} \quad (n \text{ equal resistors of } r) \]
Solution — RAB
STEP 1 — The whole rim is a single node
Every spoke's outer end lands on the same node. Since B is on that rim, the rim node is B.
STEP 2 — All four spokes run A → B in parallel
\[ R_{AB} = \frac{2R}{4} = \frac{R}{2} \]
Case (ii) RAB = R⁄2 = 0.5 R
Short Cut (both cases)

First merge every node joined by a bare wire, then look again. A resistor-free wire means "same point." Networks that look like long chains or rings usually collapse into a handful of nodes — and very often into a simple parallel bundle.

(i) Two jumpers ⇒ two nodes ⇒ 6R ∥ 9R ∥ 12R = 36R/13.   (ii) Conducting ring ⇒ one rim node ⇒ n equal spokes give r/n → 2R/4 = R/2.

Given → Asked → Formula → Solution → Short Cut · merge wire-shorted nodes first