Wye-delta transforms

YOU WILL BE ABLE TO

Recognise networks that are neither series nor parallel.
Convert a delta (Δ) network to an equivalent wye (Y), and back.
Use the balanced-network shortcut R_Δ = 3R_Y.
Apply a transform to unlock a bridge network.

Some three-terminal resistor arrangements defeat series/parallel reduction entirely — the classic being the bridge, where a middle resistor spans between two dividers. The escape is to swap one three-terminal pattern for its electrically identical twin: the delta (Δ), three resistors in a triangle, and the wye (Y), three resistors meeting at a central node. Seen only from their three terminals, a correctly chosen Δ and Y are indistinguishable.

The wye (Y): three resistors meeting at a central node, seen from terminals a, b, c.

The delta (Δ), drawn squared-off: one resistor between each pair of terminals.

Converting Δ to Y: each Y resistor is the product of the two Δ resistors adjacent to its terminal, divided by the sum of all three.

R_1 = \frac{R_b R_c}{R_a + R_b + R_c}

Δ → Y, one of three symmetric formulas

Converting Y to Δ: each Δ resistor is the sum of all pairwise products of the Y resistors, divided by the Y resistor at the opposite terminal.

R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1}

Y → Δ, one of three symmetric formulas

The subscript pattern is what to internalise, not the letters: going to Y, adjacent pair over total; going to Δ, all pairs over the opposite one. Δ resistances are always larger than their Y counterparts.

R_\Delta = 3\,R_Y

balanced case

In practice you'll reach for this tool rarely but decisively: spot the Δ (or Y) that is jamming the reduction, transform it, and watch the rest of the network fall to ordinary series/parallel steps. Bridge circuits — which reappear in ME-6 as sensor front-ends — are the canonical customer.

The bridge: R_5 spans two dividers. No pair is in series or parallel — transform to unlock it.

WORKED EXAMPLE

Δ → Y, worked

A Δ has R_ab = 6 Ω, R_bc = 3 Ω, R_ca = 3 Ω. Sum = 12 Ω.
Y resistor at terminal a (adjacent to R_ab and R_ca): (6 × 3)/12 = 1.5 Ω.
At terminal b (adjacent to R_ab and R_bc): (6 × 3)/12 = 1.5 Ω.
At terminal c (adjacent to R_bc and R_ca): (3 × 3)/12 = 0.75 Ω.

R_a = 1.5 Ω, R_b = 1.5 Ω, R_c = 0.75 Ω

WORKED EXAMPLE

Balanced shortcut

A Δ of three 30 Ω resistors.
Balanced: R_Y = R_Δ/3 = 10 Ω each.

An equivalent Y of three 10 Ω resistors

PRACTICE

Convert a balanced Δ of 90 Ω resistors to its equivalent Y.

A Y has R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 20 Ω. Find the equivalent Δ.

A bridge network has a resistor spanning between the midpoints of two dividers. Explain, in one or two sentences, why plain series/parallel reduction fails and how a Δ→Y transform rescues it.

Using the general Δ→Y formulas with all three Δ resistors equal to R, show that each Y resistor is R/3.