Delta-Wye (Δ-Y) Transformation

Intuition

A triangle exchanged for a star

Reduction by series and parallel steps needs a pair of resistors to work on, and some networks offer none. The standard case has four resistances in two chains across the supply, with a fifth joining the two midpoints. Every junction in it carries three components, so no two of the resistances have a junction to themselves and no two run between the same points. Redrawing the same connections a different way leaves the situation exactly where it was.

What that network does contain is a triangle: three resistances joining three points. A triangle can be lifted out and replaced by three different resistances arranged as a star, one from each of those points to a new junction in the middle. Three components go in and three come out. The three points meeting the rest of the circuit stay where they are, and everything outside carries on as it did.

Both names come from the drawing. Three resistances closed into a triangle look like the Greek capital delta, and a three-armed star looks like a letter Y. Star is the everyday word for the Y arrangement, and star-delta is the commoner name for the swap outside textbooks.

The smallest case can be checked by hand. A delta of three 90 Ω resistors has a star equivalent whose arms are each 30 Ω, a third of the delta value. Put an ohmmeter across two of the terminals. On the delta the meter meets one resistor in parallel with the other two in series, 60 Ω; on the star it meets two arms end to end, 60 Ω. The readings match, and so do those from the other two pairs. That agreement at the terminals is what "equivalent" means here.

Put that swap into the stalled network and pairs appear where there were none, and the reduction runs to the end.

Practitioner

Which product over which sum

Both directions need a labelling convention before the expressions mean anything, and one is standard. Number the three terminals 1, 2 and 3. In the delta, name each resistance after the terminal it does not touch: R_a joins terminals 2 and 3, R_b joins 3 and 1, R_c joins 1 and 2. In the star, name each arm after the terminal it reaches, so R_1 runs from terminal 1 to the centre.

With those names, the arm at a terminal is the product of the two delta resistances meeting at that terminal, over the sum of all three:

The library entry gives one arm. Advance the labels 1 → 2 → 3 → 1 for the other two: the arm at terminal 2 is R_c R_a over the same sum, and the arm at terminal 3 is R_a R_b over it. The divisor never changes, so the three arms are quick to write out together.

The reverse direction divides one common numerator by each star arm in turn:

R_1 R_2 + R_2 R_3 + R_3 R_1 is the same quantity for all three, so R_b is that quantity over R_2 and R_c is that quantity over R_3. A star of small arms therefore becomes a delta of large resistances, and a delta arm can never come out smaller than either of the two star arms it spans.

Either of a bridge's two triangles — each of them two arms plus the bridging element — can go to a star, and each of the two corners where three elements meet is already a star that can go to a triangle. Any one of those four moves unlocks the reduction; pick whichever leaves the tidiest arithmetic.

The tests themselves and the two combining rules are unchanged:

What that produces deserves a caution. The star arms are three numbers and the star point is a junction between them — contrived to behave at the three terminals the way the original triangle behaved — and none of it is a place on the board. A current calculated in a star arm belongs to the star.

Engineer

Matching the network at every pair of terminals

Interchangeability means that nothing outside the two networks can tell them apart, and there is only so much an outside circuit can do to a three-terminal resistive network: drive current between one pair of its terminals at a time. So the test is the resistance seen between terminals 1 and 2, then between 2 and 3, then between 3 and 1, with the third terminal left open each time.

For the star those readings are plain sums: R_1 + R_2, R_2 + R_3, R_3 + R_1. On the delta each is one resistance in parallel with the other two in series, so between terminals 1 and 2 it is R_c against R_a + R_b.

Set the pairs equal and there are three equations in the three star arms, linear in those arms and therefore solvable outright. Adding all three star readings gives twice R_1 + R_2 + R_3, so half the sum of the three delta readings is the sum of the arms; subtract any one reading from that and a single arm is left on its own. Grinding the delta expressions through that subtraction produces the product-over-sum form. The argument treats all three terminals alike, so a single line covers all three arms and the labels can simply be advanced.

Equal resistances make the collapse visible. With all three delta values the same, every arm becomes R² over 3R, one third of the delta value — the 30 Ω against 90 Ω met earlier — and the reverse direction multiplies by three. Anyone unsure of a conversion can put three equal numbers through it in a few seconds and see whether the machinery is pointing the right way.

Getting the interior back

The equivalence holds at the three terminals and stops there. Inside, the two networks are simply different circuits: the delta carries no central node, the star carries no element between C and D, and each arm is a number in its own right rather than a stand-in for any one triangle resistance. So a current read off a star arm describes the star and not the triangle.

The interior is recoverable all the same, because two of the three terminals are interior corners of the bridge. Carry the worked example forward. The supply current 12.0 mA passes through the arm to A and drops 6.48 V across it, leaving the star point at 5.52 V. That potential drives both chains: 6.0 mA down the chain through C and 6.0 mA down the chain through D, equal here only because the two chains came out equal. Those currents fall across the original lower resistances, which puts C at 3.72 V and D at 2.82 V.

Both are real terminals of the triangle that was removed, so both potentials transfer straight back to the original drawing. Restore the triangle and its three elements can be read off: 0.90 V across the bridging element drives 0.9 mA through it, the A–C resistance carries 6.9 mA, and the A–D resistance carries 5.1 mA. Those last two add to the supply current, as they must.

The star point is the one thing that does not come back. 5.52 V is a potential in a network drawn to make the arithmetic work, and a probe moved over the real board will find no conductor holding it. Information crosses in either direction only at the three shared terminals.

The round trip and its small print

Running the transformation backwards returns the original, and doing it once settles whether anything leaks away on the journey. Feed the three arms — 540 Ω, 300 Ω, 450 Ω — into the reverse expression and the resistance opposite the A arm comes back as 1.0 kΩ, the one opposite the C arm as 1.8 kΩ, and the one opposite the D arm as 1.2 kΩ. Every positive-valued delta has exactly one positive-valued star and the reverse, so the pair of transformations is a clean round trip.

The derivation took several things for granted, and a real circuit can withdraw any of them. Every element must be a resistance that holds its value, because the terminal-pair equations treat all three as constants: a lamp, a diode or a thermistor in one arm rules the method out at any operating point where its value moves. The triangle must also contain nothing but those three resistances. A source between two corners, or a fourth component across one of them, belongs to the network being replaced and has to be combined in or handled separately, which is where Thévenin's theorem takes over. The equations are linear, on the other hand, so the result holds at any current and any voltage, and it extends unchanged to impedances in AC work with the same expressions run on complex quantities.

Professional

What the transformation does not carry across

Dissipation lands somewhere else

An equivalent network draws the same current at the same terminal voltages, so it absorbs the same total power. What the swap does change is how that total divides between three elements.

In the worked example's original triangle, the A–C resistance takes 57.1 mW, the A–D resistance 46.8 mW, and the bridging element, carrying under a milliamp, 0.81 mW, which come to 104.8 mW between them. The star that stood in for it comes to the same 104.8 mW, split as 77.8 mW in the arm to A, 10.8 mW in the arm to C and 16.2 mW in the arm to D. Set the two lists side by side and no figure in one is near any in the other: the busiest star arm outruns every element of the triangle, and the element that runs coolest in the real circuit has no counterpart at all. Power ratings belong to parts that get soldered down, so the dissipation pass runs on the original network once the currents have come back out of the reduction.

Star and delta on three-phase equipment

Three-phase loads, motor windings and transformer windings are connected in one arrangement or the other, and star and delta are the ordinary names for them there. A balanced three-phase load connected in delta looks, from its three line terminals, like a star-connected load of one third the per-phase impedance — the symmetric case above, with the arms equal. Unbalanced loads take the general expressions the same way a bridge does. What follows from each connection about line and phase quantities, and why a machine might be started in one and run in the other, belongs to three-phase supplies and to AC analysis generally.

The same algebra under other names

A T-pad and a Pi-pad are a star and a delta of one three-terminal network, so the expressions in attenuators that turn one topology into the other are this transformation with the symbols renamed; the T-to-Pi conversions in filter tables are the same again. A Wheatstone bridge read away from balance — a strain gauge under load, a resistance thermometer off its set point — has a source resistance at one diagonal and a detector at the other, and converting one triangle turns the whole thing into a reduction that gives the detector current directly. The same substitution simplifies meshed power distribution networks before anyone solves them.

When to reach for something else

There is a case for the transformation only where a triangle or a star sits somewhere its removal exposes a pair. Networks with several interlocked bridges can absorb two or three conversions and still not open up, and every conversion is a sum, three products and three divisions done by hand, with a fresh drawing to keep straight afterwards. Mesh analysis and nodal analysis need no reducible topology at all, at the price of simultaneous equations; past a dozen elements a simulator beats both. Set against those, this method keeps one thing they discard — an expression for the equivalent resistance in terms of the component values, which is what answers the question of how the result moves when one resistor is changed or drifts.

Tolerance comes through the conversion differently from the way it comes through a plain reduction. Each star arm depends on all three delta values, so one resistor at the edge of its band shifts all three arms together, and their deviations move as a set rather than independently. Combining them in quadrature, the way tolerance is usually carried through a series-parallel reduction of independent parts, then understates the spread. Where the number matters, walk the original component values to their corners and re-run the whole reduction from there.