Current Electricity · Reduction Walk-through

Folding the Jumpered Chain into Parallels

Find the equivalent resistance between A and B for the chain A–6R–X–9R–Y–12R–B, where a top wire joins A to Y and a bottom wire joins X to B.

Original network with two resistor-free jumper wires.

⬇ redraw

All three resistors now sit between the same two nodes ⇒ parallel.

Given

Chain resistors: 6R (A–X), 9R (X–Y), 12R (Y–B).
Top jumper (no resistor): A to Y.
Bottom jumper (no resistor): X to B.

Asked

Equivalent resistance R_AB.

Key Idea + Formula

A wire with no resistor ⇒ its ends are one node. After merging:

\[ \frac{1}{R_{AB}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \]

Solution

STEP 1 — Merge the wire-shorted nodes

Top wire ⇒ A ≡ Y; bottom wire ⇒ X ≡ B. Only two nodes remain: {A, Y} and {B, X}.

STEP 2 — Re-read each resistor's endpoints

Resistor	Original ends	After merge
6R	A – X	{A,Y} – {B,X}
9R	X – Y	{B,X} – {A,Y}
12R	Y – B	{A,Y} – {B,X}

All three span the same two nodes ⇒ in parallel.

STEP 3 — Add reciprocals (LCM = 36)

\[ \frac{1}{R_{AB}} = \frac{1}{6R}+\frac{1}{9R}+\frac{1}{12R} = \frac{6+4+3}{36R} = \frac{13}{36R} \] \[ R_{AB} = \frac{36R}{13} \approx 2.77\,R \]

Equivalent Resistance R_AB = 36R⁄13 ≈ 2.77 R

Short Cut

Tag the nodes, then sort resistors by node-pair. The instant you see bare jumper wires, give each junction a letter, merge the ones joined by wire, and group resistors that share the same end-pair — they are automatically parallel. No step-by-step series/parallel chasing required.

Parallel reciprocal trick: with denominators 6, 9, 12, take LCM = 36 → numerators 6+4+3 = 13 → answer flips to 36R/13 in one line.

Given → Asked → Formula → Solution → Short Cut · matches the book's A,Y | B,X redrawing