Delta and Y connections

Every three-phase machine you will ever wire — a generator, a transformer winding, an induction motor, a load bank — has exactly three coils or three branches, and there are only two sane ways to join their six terminals into three output lines: star (Y), where one end of each coil meets at a common neutral point, and delta (Δ), where the coils are chained end-to-end into a closed triangle. That single wiring choice fixes a fixed, memorisable relationship between what a line carries and what one winding carries: star scales up the voltage by $\sqrt3$ , delta scales up the current by $\sqrt3$ , and both effects trace back to the same piece of phasor geometry — subtracting two equal vectors $120^\circ$ apart. This article builds that geometry from scratch, applies it to the 230 V and 415 V numbers you already know from Indian supply mains, and flags the traps that cost the most marks.

The core insight

In a balanced star connection the line voltage is $\sqrt3$ times the phase voltage while line current equals phase current, and in a balanced delta connection the line current is $\sqrt3$ times the phase current while line voltage equals phase voltage — both $\sqrt3$ factors come from subtracting two equal-magnitude phasors $120^\circ$ apart.

The core idea

A three-phase source or load is built from three identical windings (or branches), each carrying its own phase voltage $V_{ph}$ and phase current $I_{ph}$ . The three windings are excited $120^\circ$ apart in time, which is what makes them "three-phase" in the first place. Once you bring three wires out to the external world, two new quantities appear: the line voltage $V_L$ , measured between any pair of the three output terminals, and the line current $I_L$ , the current actually flowing in one line conductor. How $V_L,I_L$ relate to $V_{ph},I_{ph}$ depends entirely on which of the two standard interconnections was used.

Star (Y) connection. One end of every winding is tied together at a common point, the neutral; the free ends become the line terminals R, Y, B. Because each line wire is in series with exactly one winding, the current in the line and the current in that winding are the same physical current:

I_L = I_{ph}

The line voltage, however, is not one winding's voltage but the phasor difference between two of them, e.g. $\vec V_{RY} = \vec V_R - \vec V_Y$ . Since $V_R$ and $V_Y$ have equal magnitude and are $120^\circ$ apart, the subtraction produces a resultant $\sqrt3$ times either one:

V_L = \sqrt3\,V_{ph}

where: $V_L$ is the line-to-line voltage, $V_{ph}$ is the voltage across one winding, and the resulting line-voltage phasor leads its corresponding phase voltage by $30^\circ$ .

Delta (Δ) connection. The three windings are joined end-to-end into a closed mesh, and the line terminals are taken from the three junction points. Now a pair of line terminals sits directly across one winding, so:

V_L = V_{ph}

But the current in a line conductor is the phasor difference of the two winding currents meeting at that junction, e.g. $\vec I_R = \vec I_{RY} - \vec I_{BR}$ , and the same $120^\circ$ subtraction gives:

I_L = \sqrt3\,I_{ph}

with the line current lagging its corresponding phase (winding) current by $30^\circ$ . Both $\sqrt3$ results are the same piece of vector algebra applied to voltages in one case and currents in the other:

|\vec A - \vec B| = \sqrt3\,|\vec A| \quad \text{when } |\vec A| = |\vec B| \text{ and the angle between } \vec A,\vec B \text{ is } 120^\circ

This is worth internalising as a single geometric fact rather than two separate formulas to memorise, because it also tells you why one relation involves voltage and the other current: in star it is two phase voltages that get subtracted to make a line voltage, and in delta it is two phase currents that get subtracted to make a line current — the connection topology decides which quantity does the subtracting.

Figure

Star connection with three windings meeting at a neutral point beside a delta connection with three windings forming a closed triangle, each labelled with phase and line quantities

Derivation ADeriving the sqrt3 factor from two phasors 120 degrees apart

The goal is to show, using only vector subtraction and basic trigonometry, that whenever you subtract two phasors of equal magnitude that are $120^\circ$ apart, the result always has magnitude exactly $\sqrt3$ times the original magnitude and is displaced by $30^\circ$ from each — the single geometric fact that produces both the star voltage relation and the delta current relation, so it only needs proving once.

Step 1. Set up the two phasors. Take $\vec A$ and $\vec B$ , both of magnitude $M$ , with $\vec B$ rotated $120^\circ$ ahead of $\vec A$ . Place $\vec A$ along the reference axis: $\vec A = M\angle 0^\circ$ and $\vec B = M\angle 120^\circ$ . In component form, $\vec A = (M, 0)$ and $\vec B = (M\cos120^\circ,\, M\sin120^\circ) = (-M/2,\, M\sqrt3/2)$ .

Step 2. Subtract the phasors component-wise. $\vec A - \vec B = \left(M - \left(-\dfrac{M}{2}\right),\; 0 - \dfrac{M\sqrt3}{2}\right) = \left(\dfrac{3M}{2},\; -\dfrac{M\sqrt3}{2}\right)$ .

Step 3. Find the magnitude of the result. $|\vec A - \vec B| = \sqrt{\left(\dfrac{3M}{2}\right)^2 + \left(\dfrac{M\sqrt3}{2}\right)^2} = \sqrt{\dfrac{9M^2}{4} + \dfrac{3M^2}{4}} = \sqrt{3M^2} = M\sqrt3$ . This confirms $|\vec A - \vec B| = \sqrt3\,M$ , independent of which two of the three $120^\circ$ -spaced phasors were chosen, since the geometry is identical for any such pair.

Step 4. Find the phase displacement. The angle of the resultant is $\tan^{-1}\!\left(\dfrac{-M\sqrt3/2}{3M/2}\right) = \tan^{-1}\!\left(\dfrac{-1}{\sqrt3}\right) = -30^\circ$ , i.e. the resultant sits $30^\circ$ from $\vec A$ (and, by symmetry, $30^\circ$ from $\vec B$ on the other side). Applying this to star voltages, $\vec V_{RY} = \vec V_R - \vec V_Y$ makes $V_L = \sqrt3\,V_{ph}$ with the line voltage leading the phase voltage by $30^\circ$ ; applying it to delta currents, $\vec I_R = \vec I_{RY} - \vec I_{BR}$ makes $I_L = \sqrt3\,I_{ph}$ with the line current lagging the phase current by $30^\circ$ — the same algebra, two different physical quantities.

Worked example

A balanced star-connected alternator has phase voltage $V_{ph} = 230\ \text{V}$ and supplies a balanced delta-connected load whose phase impedance is $Z_{ph} = 20\ \Omega$ . Find (a) the line voltage delivered by the source, (b) the phase current in the delta load, and (c) the line current drawn from the supply.

Step 1 — Line voltage from the star source. Since the source is star-connected, $V_L = \sqrt3\,V_{ph} = 1.732\times230 = 398.4\ \text{V} \approx 400\ \text{V}$ , matching the familiar 230 V/400 V figures used across Indian mains supply.

Step 2 — Voltage seen by the delta load. In a delta connection the line voltage feeds each winding directly, so the load's phase voltage equals the line voltage supplied to it: $V_{ph,\text{load}} = V_L = 400\ \text{V}$ (using the rounded value; $398.4\ \text{V}$ if carried unrounded).

Step 3 — Phase current in the delta load. $I_{ph} = \dfrac{V_{ph,\text{load}}}{Z_{ph}} = \dfrac{398.4}{20} = 19.92\ \text{A}$ .

Step 4 — Line current from the delta relation. $I_L = \sqrt3\,I_{ph} = 1.732\times19.92 = 34.5\ \text{A}$ .

The alternator delivers $V_L \approx 400\ \text{V}$ , each delta winding carries $I_{ph} \approx 19.9\ \text{A}$ , and the supply lines carry $I_L \approx 34.5\ \text{A}$ — a direct illustration of how a star source and a delta load are mixed in the same balanced system, with the star relation fixing the line voltage and the delta relation fixing the line current.

Common traps

Applying $V_L=\sqrt3\,V_{ph}$ or $I_L=\sqrt3\,I_{ph}$ without first checking which connection — star or delta — is actually being asked about; the $\sqrt3$ multiplies voltage in star but current in delta, never both.
Forgetting the "silent" equalities: in star $I_L = I_{ph}$ (no $\sqrt3$ ), and in delta $V_L = V_{ph}$ (no $\sqrt3$ ) — each connection has one direct equality and one $\sqrt3$ -scaled relation, not two scaled relations.
Mixing up the $30^\circ$ phase-shift direction: line voltage leads phase voltage in star, but line current lags phase current in delta.
Assuming source and load use the same connection; a star source can feed a delta load (as in the worked example) and the two connections' relations must be applied separately to source and load.

Ask the tutor: want to work through the full phasor-subtraction diagram for a delta-source-star-load combination, or see how the star-delta motor starter uses these same relations to limit starting current?

Why does star give sqrt3 times the phase voltage but delta gives sqrt3 times the phase current, not both?

It comes down to what each connection makes you subtract. In star, each line wire carries exactly one winding's current, so line current equals phase current directly — there is nothing to subtract for current. But line voltage is measured between two separate winding terminals, so it is the phasor difference of two phase voltages, which gives the sqrt3 factor. In delta it is the reverse: each pair of line terminals sits directly across one winding, so line voltage equals phase voltage directly, while line current at a junction is the phasor difference of two winding currents, producing the sqrt3 factor there instead.

Why is star preferred for distribution and delta for some motor windings?

Star gives each winding only the phase voltage (230 V rather than 400 V for the same line voltage), which reduces insulation stress, and it provides a neutral point so a four-wire system can also supply single-phase loads at 230 V. Delta has no neutral but is useful where windings must carry the full line voltage by design, and the star-delta arrangement is exploited in motor starters: starting a motor in star limits the starting current, then switching to delta for running restores full-rated torque and voltage across each winding.

If I know a load is star connected, can I still use the sqrt3 formulas for current?

No — in a star connection line current equals phase current exactly, with no sqrt3 involved; the sqrt3 factor there applies only to voltage. You would need the sqrt3 relation for current only if that same load were rewired as delta. Always match the formula to the connection: identify whether you are looking at a star or a delta first, then pick the one relation that carries the sqrt3 factor for that connection (voltage for star, current for delta) and use the direct equality for the other quantity.

Delta and Y connections

The core idea

Worked example

Common traps

Before this

Where this leads