Wye to Delta Resistance Transformation

Going back the other way, the delta leg between two nodes is the sum of the three pairwise products of the star arms, divided by the arm at the node the leg does not touch. A balanced 10 Ω star becomes a 30 Ω delta, the mirror of $R_\Delta = 3R_Y$. Unbalanced: arms of 5, 10 and 20 Ω give pairwise products 50, 200 and 100, summing to 350, so the leg opposite the 20 Ω arm is $350/20 = 17.5$ Ω.

That opposite-arm divisor is where people go wrong. The numerator is the same for all three legs — compute it once — and only the denominator changes, so the leg opposite the smallest arm comes out largest. Another built-in check: every delta leg must exceed the sum of the two star arms it spans, because the direct path has to imitate both the series path and the roundabout one. Here 17.5 Ω comfortably beats $5 + 10 = 15$ Ω.

Practically this direction shows up when a network has a star that blocks reduction, and in three-phase work when a wye-connected load has to be restated as delta for a source that is delta-connected. It is also the reason a delta-connected motor winding draws three times the current of the same windings in wye at the same line voltage, which is the whole basis of wye-delta starting.