Three-phase circuits

Why it matters

Almost every industrial motor, CNC machine, compressor and VFD in a plant runs from a three-phase supply (400 V line-to-line, 50 Hz in India). Three-phase systems deliver constant instantaneous power, need less conductor material than single-phase for the same power, and produce the rotating magnetic field that makes induction motors self-starting. Reading a motor nameplate, sizing a cable or wiring a star-delta starter all need the relations in this topic.

Key ideas

Generating three phases. Three identical windings spaced 120° apart on a generator produce three EMFs of equal magnitude displaced by 120° in time. With phase sequence R-Y-B (also called A-B-C or positive sequence): V_R = V∠0°, V_Y = V∠−120°, V_B = V∠+120°. In a balanced set the three phasors sum to zero at every instant. Reversing any two supply lines reverses the sequence, which reverses the direction of rotation of a three-phase motor.

Star (wye, Y) connection. One end of each phase is joined at a neutral point N.

• Phase voltage V_ph is measured from line to neutral; line voltage V_L is between two lines.
• V_L = √3·V_ph, and each line voltage leads the corresponding phase voltage by 30° (for positive sequence).
• Line current equals phase current: I_L = I_ph.
• A neutral wire (4-wire system) carries the phasor sum of the line currents: zero for a balanced load, non-zero for an unbalanced one. Domestic single-phase 230 V supplies are one phase plus neutral of a 400 V star system.

Delta (mesh, Δ) connection. The three phases are connected end to end in a closed loop; there is no neutral.

• V_L = V_ph.
• I_L = √3·I_ph, and each line current lags the corresponding phase current by 30° (positive sequence).

Balanced loads. When all three phase impedances are equal, analyse one phase (the "per-phase equivalent") and multiply power by three. A delta load can be converted to an equivalent star with Z_Y = Z_Δ/3. The phase angle φ in the power formulas is always the angle between phase voltage and phase current (the angle of the load impedance), not between line quantities.

Power. Total power is the same expression for star and delta: P = √3·V_L·I_L·cos φ. For a balanced system the instantaneous total power is constant (no pulsation at 2f), which gives smooth motor torque.

Star versus delta for the same impedance. Connecting the same three impedances in delta instead of star, on the same line voltage, triples the power and the line current. A star-delta starter uses this: it starts the motor in star (one-third of the delta starting current and torque) and switches to delta to run.

Measuring power: two-wattmeter method. In a three-wire system (balanced or unbalanced), two wattmeters with current coils in two lines and pressure coils from those lines to the third line read a total equal to the three-phase power. For a balanced load W1 = V_L·I_L·cos(30° − φ) and W2 = V_L·I_L·cos(30° + φ) (lagging φ). Consequences:

• At unity pf both read equal.
• At pf = 0.5 (φ = 60°) one wattmeter reads zero.
• Below pf 0.5 one reading goes negative (reverse its current coil and subtract).

Unbalanced loads. With an unbalanced star load and no neutral, the star point shifts away from the supply neutral (neutral shift), and the phase voltages are no longer equal; solve with nodal analysis (Millman's theorem). This is why single-phase loads in a building must be spread across the three phases.

Formulas

V_L = √3 · V_ph; I_L = I_ph (star) V_L = V_ph; I_L = √3 · I_ph (delta)

• V_L: line-to-line RMS voltage (V); V_ph: phase RMS voltage (V); I_L, I_ph: RMS currents (A).

I_ph = V_ph / |Z_ph|; cos φ = R_ph / |Z_ph|

• Z_ph: per-phase impedance (Ω); φ: angle of Z_ph.

P = 3·V_ph·I_ph·cos φ = √3·V_L·I_L·cos φ (W) Q = √3·V_L·I_L·sin φ (VAR) S = √3·V_L·I_L (VA)

• Valid for balanced loads, star or delta.

Z_Y = Z_Δ / 3 (balanced delta-to-star conversion)

P = W1 + W2; Q = √3·(W1 − W2); tan φ = √3·(W1 − W2) / (W1 + W2)

• Two-wattmeter method; the Q and tan φ relations hold for a balanced load and sinusoidal supply; W1 is the larger reading for a lagging load.

Worked examples

Example 1 (standard: star and delta load). Three identical impedances Z = 8 + j6 Ω are connected to a 400 V, 50 Hz three-phase supply. Find line current and total power (a) in star, (b) in delta.

1. |Z| = √(8² + 6²) = 10 Ω; cos φ = 8/10 = 0.8 lagging.
2. (a) Star: V_ph = V_L/√3 = 400/1.732 = 230.9 V.
3. I_ph = I_L = 230.9/10 = 23.09 A.
4. P = 3·I_ph²·R = 3 × 23.09² × 8 = 12 800 W. Check: √3 × 400 × 23.09 × 0.8 = 12 800 W.
5. (b) Delta: V_ph = 400 V, I_ph = 400/10 = 40 A, I_L = √3 × 40 = 69.28 A.
6. P = 3 × 40² × 8 = 38 400 W.

Star: I_L = 23.1 A, P = 12.8 kW. Delta: I_L = 69.3 A, P = 38.4 kW (three times the star value).

Example 2 (GATE level: two-wattmeter method). A balanced three-phase induction motor on a 400 V supply is tested with two wattmeters reading W1 = 8 kW and W2 = 4 kW. Find the total power, power factor, reactive power and line current.

1. P = W1 + W2 = 8 + 4 = 12 kW.
2. tan φ = √3·(W1 − W2)/(W1 + W2) = 1.732 × 4/12 = 0.5774, so φ = 30.0°.
3. pf = cos 30° = 0.866 lagging (motor load).
4. Q = √3·(W1 − W2) = 1.732 × 4 = 6.93 kVAR.
5. I_L = P/(√3·V_L·cos φ) = 12 000/(1.732 × 400 × 0.866) = 20.0 A.

P = 12 kW, pf = 0.866 lagging, Q = 6.93 kVAR, I_L = 20.0 A.

Common mistakes

• Using line voltage directly as phase voltage in a star load (forgetting the √3).
• Taking φ as the angle between line voltage and line current; it is the phase-impedance angle.
• Writing P = 3·V_L·I_L·cos φ (it is √3 with line values, 3 with phase values).
• Expecting the neutral current to be non-zero in a balanced load.
• Discarding a negative wattmeter reading instead of subtracting it when pf < 0.5.
• Assuming phase voltages stay equal in an unbalanced star load without a neutral.

For GATE ME

Questions are usually short numericals: line and phase currents for balanced star or delta loads, total power, star-delta conversion, effect of changing connection on power or starting current, and pf from two wattmeter readings. Practise the per-phase approach and the two-wattmeter formulas until they are automatic.

Quick check

1. A star-connected supply has 230 V phase voltage. What is the line voltage?
2. In a delta load the phase current is 10 A. What is the line current?
3. Two wattmeters read equal values on a balanced load. What is the power factor?
4. One wattmeter reads zero on a balanced load. What is the power factor?
5. The same three resistors are reconnected from star to delta on the same supply. By what factor does the power change?

Answers: 1. 398 V (nominally 400 V). 2. 17.3 A. 3. Unity. 4. 0.5. 5. It triples.