Current Electricity · Symmetry Method

Reducing a Symmetric Network (R between A and B)

A network symmetric about the perpendicular bisector is simplified by deleting the resistors between equal-potential points, leaving three parallel branches. Find RAB.

Notation: the page mixes the symbols R and r — they denote the same base resistance, so 2r = 2R, 4r = 4R, etc. The card uses R throughout.
A B 4R 4R 2R 2R 4R 4R = 8R = 4R = 8R
After symmetry removal: three parallel branches of 8R, 4R, and 8R between A and B.
Given (after symmetry reduction)
Asked

Equivalent resistance RAB.

Key Idea + Formula

Symmetry: points at equal potential carry no current between them, so the connecting resistors are redundant and removed. Then combine in parallel:

\[ \frac{1}{R_{AB}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \]
Solution
STEP 1 — Each branch is a series pair
\( 4R+4R = 8R,\quad 2R+2R = 4R,\quad 4R+4R = 8R. \)
STEP 2 — Three branches in parallel
\[ \frac{1}{R_{AB}} = \frac{1}{8R}+\frac{1}{4R}+\frac{1}{8R} = \frac{1+2+1}{8R} = \frac{4}{8R} = \frac{1}{2R} \]
STEP 3 — Invert
\[ R_{AB} = 2R \]
Equivalent Resistance RAB = 2R
Short Cut

Find the mirror line first. If a network is symmetric about a line through A and B, the points it reflects onto each other are at the same potential — any resistor joining such a pair carries zero current and can be erased before you compute anything. That usually leaves a clean set of parallel (or series) branches.

Fast finish here: 8R ∥ 8R = 4R, then 4R ∥ 4R = 2R. Two equal resistors in parallel always halve, so no LCM needed.

Given → Asked → Formula → Solution → Short Cut · symmetry (equipotential) method · R = r