Question
A wire of resistance is bent into a regular hexagon. A battery of emf and negligible internal resistance is connected across two diametrically opposite vertices. Find the equivalent resistance between those vertices, the current drawn from the battery, and the current through any one of the six sides. Each side of the hexagon has resistance .
Solution — Step by Step
Two diametrically opposite vertices split the hexagon into two parallel paths. Each path consists of three sides in series.
Each path: .
Two paths of each in parallel:
By symmetry, the current splits equally between the two paths. Current through one side .
Final answer: , total current , current per side .
Why This Works
When you connect a battery across two symmetric points of a regular network, current divides equally along symmetric branches. The hexagon has two-fold symmetry across the diameter, so the two halves carry equal currents.
This is a classic JEE Advanced setup — they often replace “hexagon” with cube, tetrahedron, or pentagon, but the symmetry-then-parallel trick is identical.
Alternative Method
Apply Kirchhoff’s laws directly with three loop equations. You will get three equations for three loop currents. Set up the matrix, solve. Same answer, four times the effort.
For polygon-with-battery problems: count the number of independent paths between the battery terminals, find each path’s resistance, combine in parallel. This handles 90% of polygon networks.
Students sometimes treat all six sides as a single parallel combination of six resistors. That gives , which is wrong because the sides are not all in parallel — they are arranged into series-parallel paths.