Delta to Wye Resistance Transformation

Some networks refuse to be reduced. A Wheatstone bridge has no two resistors in plain series and no two in plain parallel, so the usual tools stall. Arthur Kennelly's 1899 transformation is the way out: swap any triangle of three resistors for an electrically identical star, and the series-parallel structure reappears. The arm of the star at node A is the product of the two triangle legs that touch A, divided by the sum of all three.

The balanced case is worth memorising because it is so clean: three equal 30 Ω legs become three equal 10 Ω arms, $R_Y = R_\Delta/3$. For the unbalanced case take a 10/20/30 Ω delta — the sum is 60, so the arm at A, touching the 10 and the 30, is $10 \times 30/60 = 5$ Ω. Every star arm always comes out smaller than either delta leg it touches, which is a useful sanity check on your own arithmetic.

The trap is bookkeeping: it is dangerously easy to pair an arm with the wrong two legs. Label the nodes on the drawing before you start, and remember the rule in words — the arm at a node uses the two legs meeting at that node, and the leg opposite a node never appears in the numerator. Note also that this is a resistance identity, but it works unchanged for complex impedances, which is how three-phase engineers convert a delta-connected load to its wye equivalent before applying per-phase analysis.