Single-Phase vs Three-Phase Supply: Star and Delta Relations Worked Out
Most Indian homes receive a single-phase 230 V supply, while workshops, apartment blocks and factories get a three-phase 415 V supply. Both come from the same distribution transformer: a single-phase connection is simply one phase wire and the neutral of a three-phase system. Understanding how the 230 V and 415 V figures relate, and how loads connect in star or delta, is essential for wiring, motor selection and load balancing.
This article explains why three-phase power is used, derives the star and delta relations, and works through a balanced load calculation in both connections.
Why Generation and Transmission Use Three Phases
A three-phase alternator has three windings placed 120 electrical degrees apart. It produces three voltages of equal magnitude, each displaced by 120 degrees. In India the phases are labelled R, Y and B (red, yellow, blue), and the normal phase sequence is R-Y-B.
vR = Vm sin(ωt) vY = Vm sin(ωt − 120°) vB = Vm sin(ωt − 240°) For a balanced system: vR + vY + vB = 0 at every instant
| Feature | Single-phase | Three-phase |
|---|---|---|
| Typical Indian LV voltage | 230 V phase to neutral | 415 V line to line (400 V per IS 12360), 230 V line to neutral |
| Power delivery | Pulsates at twice supply frequency, 100 Hz | Total instantaneous power is constant for a balanced load |
| Motors | Need a capacitor or auxiliary winding to start | Self-starting rotating magnetic field |
| Conductor material for same power and losses | Reference (2 wires) | About 75% of single-phase for 3-wire transmission |
| Typical use | Homes, small shops, lighting | Industry, pumps above a few kW, large buildings |
The constant total power is the main reason three-phase motors run smoothly with less vibration than single-phase motors. The copper saving is why every transmission line you see has conductors in sets of three.
Star (Wye) and Delta Connections
Three windings, whether in a generator, transformer or load, can be connected in two ways. In star, one end of each winding joins a common point called the neutral, and the free ends go to the three lines. In delta, the windings are joined end to end in a closed loop and the lines are taken from the three junctions.
STAR DELTA R R | / \ Zph Zph Zph | / \ N ---- Zph ---- Y Y --Zph-- B | Zph | B
In star, the line voltage between two lines is the phasor difference of two phase voltages that are 120 degrees apart. Working out the phasor geometry gives a magnitude √3 times the phase voltage, leading by 30 degrees. The same current flows through each line and its phase winding. In delta, each winding sits directly across two lines, so line and phase voltages are equal, but each line current is the phasor difference of two phase currents and is √3 times larger.
STAR: VL = √3 × Vph IL = Iph DELTA: VL = Vph IL = √3 × Iph Power (either connection, balanced load): P = 3 Vph Iph cos φ = √3 VL IL cos φ (W) Q = √3 VL IL sin φ (VAR) S = √3 VL IL (VA)
Check with Indian values: 230 × 1.732 = 398 V, which is why the nominal 400 V and the older 415 V figure (240 × 1.732) are both quoted for the same system.
Worked Example: Same Load in Star and Delta
Three identical impedances, each 20 Ω at 0.8 power factor lagging, are connected to a 415 V, 50 Hz three-phase supply. Calculate the phase current, line current and total power for star and for delta connection.
Star connection
- Phase voltage: Vph = 415/√3 = 239.6 V.
- Phase current: Iph = 239.6/20 = 11.98 A.
- Line current: IL = Iph = 11.98 A.
- Power: P = √3 × 415 × 11.98 × 0.8 = 6889 W, about 6.89 kW.
Delta connection
- Phase voltage: Vph = VL = 415 V.
- Phase current: Iph = 415/20 = 20.75 A.
- Line current: IL = √3 × 20.75 = 35.94 A.
- Power: P = √3 × 415 × 35.94 × 0.8 = 20667 W, about 20.67 kW.
| Quantity | Star | Delta | Ratio delta/star |
|---|---|---|---|
| Phase voltage | 239.6 V | 415 V | √3 |
| Phase current | 11.98 A | 20.75 A | √3 |
| Line current | 11.98 A | 35.94 A | 3 |
| Total power | 6.89 kW | 20.67 kW | 3 |
The same three impedances take exactly three times the line current and three times the power when switched from star to delta. This ratio is the basis of the star-delta motor starter, which starts a motor in star to cut the starting current to one third and then switches to delta for normal running.
Single-Phase Loads on a Three-Phase Supply
In a building with three-phase supply, single-phase circuits are connected between each phase and neutral. If the loads on R, Y and B are unequal, current flows in the neutral, equal to the phasor sum of the three line currents. A practical goal is to distribute circuits so that each phase carries roughly the same load. For example, if a flat has 4 kW on R, 1 kW on Y and 1 kW on B, moving some circuits from R to the others reduces neutral current, voltage drop and transformer heating.
Common Mistakes
- Applying √3 in the wrong place: in star, multiply voltage; in delta, multiply current.
- Using 3 × VL × IL × cos φ. The correct form is √3 × VL × IL × cos φ, or 3 × Vph × Iph × cos φ.
- Connecting a motor rated 230/400 V in delta on a 415 V supply. Its delta rating is 230 V, so it must run in star.
- Assuming the neutral carries no current in a building. Only perfectly balanced linear loads give zero neutral current; harmonics from LED drivers and computers add to it.
- Reversing phase sequence when reconnecting a motor, which reverses its direction of rotation.
Summary
Three-phase supply gives constant power, self-starting motors and less conductor material, which is why it dominates generation and distribution. In star, line voltage is √3 times phase voltage; in delta, line current is √3 times phase current. Total power is √3 VL IL cos φ in both cases, and the same load draws three times the power in delta compared with star.