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Transformación estrella / triángulo

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Ejemplo Ra×Rb + Rb×Rc + Rc×Ra = 12.76 (kΩ)² — every Δ arm is that one number divided by the Y arm facing it, so the arm facing the smallest Y resistor comes out largest.

Ejemplo Unbalanced, so the shortcut RΔ = 3 RY does not apply and each arm needs its own line above.

Some networks — a Wheatstone bridge with a load across the middle is the standard one — have no two resistors in plain series or plain parallel, so the usual reduction never starts. Swapping one star for its delta (or the reverse) breaks that deadlock: the network becomes reducible without changing what any terminal sees.

Y → Δ: $R_{\mathrm{ab}}=\frac{R_a\times R_b+R_b\times R_c+R_c\times R_a}{R_c}$ — the numerator is the same for all three arms, only the divisor changes. Δ → Y: $R_a=\frac{R_{\mathrm{ab}}\times R_{\mathrm{ca}}}{R_{\mathrm{ab}}+R_{\mathrm{bc}}+R_{\mathrm{ca}}}$. Balanced networks collapse to $R_{\Delta }=3\times R_Y$.