The Delta-Wye (Δ-Y) Transformation
A star arm is one-third of a balanced triangle side
A balanced triangle (Δ) with three equal sides R_Δ = 9 Ω and a star (Y) with three equal arms R_Y share the same three terminals A, B, C. For the two networks to be indistinguishable from outside, the resistance between any two terminals must match. The triangle is fixed at 6 Ω between terminals (= 2/3·R_Δ); the star is 2·R_Y. Slide R_Y to find the value where the star looks identical to the triangle.
A circuit series-parallel cannot crack
In a bridge circuit, no resistor is in pure series or parallel with another. When a Wheatstone bridge is off balance, the middle branch stalls any series-parallel reduction. Replacing a triangle (Δ) cluster that shares three terminals with a star (Y) cluster, or the reverse, reopens the blocked series-parallel path. As long as the two shapes behave the same at the three outer terminals, they are completely interchangeable parts to the rest of the circuit.
Match what the terminals see
Measure the resistance between two terminals of the balanced triangle. One side R_Δ is in parallel with the other two in series, 2R_Δ, so R_Δ ∥ 2R_Δ = (R_Δ·2R_Δ)/(3R_Δ) = (2/3)R_Δ. In the balanced star, two terminals are just two arms in series, 2R_Y. For the networks to look the same, 2R_Y = (2/3)R_Δ, that is R_Y = R_Δ/3. So a 9 Ω triangle is indistinguishable from a 3 Ω star at the terminals. The star arm is smaller because from a terminal it travels only one leg to the center.
The general formula, and going back
A triangle with unequal sides converts to a star by the same principle. Each star arm is the product of the two triangle sides touching that terminal, divided by the sum of all three sides: R_a = R_ab·R_ca/(R_ab+R_bc+R_ca). With three equal sides the product is R_Δ² and the sum is 3R_Δ, returning R_Δ/3. To go back, a star unfolds to a triangle by taking the sum of the pairwise products divided by the opposite arm. Unfold the middle triangle of an unbalanced Wheatstone bridge into a star once with this transform, and the rest solves cleanly by series-parallel.
Back to the first screen
As you slid the star arm R_Y, the star’s terminal resistance 2R_Y grew, and at R_Y = 3 Ω it matched the triangle’s 6 Ω exactly, so the two networks became indistinguishable at the terminals. The 9 Ω side of the balanced triangle folded into a 3 Ω arm, one-third, in the star. The reason was simple: between terminals the triangle is one side parallel with two, (2/3)R_Δ, while the star is two arms in series, 2R_Y; setting them equal gives R_Y = R_Δ/3. That single swap is the key that cracks a bridge series-parallel could not.