Three phase systems

A three-phase system uses three voltages of equal magnitude and frequency displaced by 120° from each other. It transmits more power per kilogram of conductor than single-phase, gives constant instantaneous power to balanced loads, and produces a rotating magnetic field for induction motors. Nepal's distribution is 400 V line-to-line (230 V line-to-neutral), 50 Hz, star-connected with neutral. This topic covers star and delta connections, line and phase relationships, the neutral, unbalanced loads and power in three-phase circuits.

Key concepts

• Generation: three identical windings displaced by 120° electrical. $v_R = V_m\sin\omega t$, $v_Y = V_m\sin(\omega t - 120^\circ)$, $v_B = V_m\sin(\omega t - 240^\circ)$. Phase sequence R-Y-B (positive); reversing any two lines reverses the sequence and the direction of motor rotation. In a balanced set the phasor sum is zero.
• Star (Y) connection: one end of each phase joined to a common neutral point.
  • $V_L = \sqrt3V_{ph}$; line voltage leads the corresponding phase voltage by 30°.
  • $I_L = I_{ph}$.
  • Allows a neutral wire, giving two voltages (400 V and 230 V) — used for LV distribution and generators.
• Delta (Δ) connection: phases connected end to end in a closed loop.
  • $V_L = V_{ph}$.
  • $I_L = \sqrt3I_{ph}$; line current lags the corresponding phase current by 30°.
  • No neutral; used for motors and transformer windings; delta windings allow circulation of third-harmonic currents.
• Neutral concept: the neutral is the star point; ideally at earth potential. In a balanced load, neutral current = 0 (phase currents sum to zero), so the neutral could be omitted. With unbalanced loads the neutral carries the phasor sum of line currents and keeps each phase voltage at 230 V. In most LV systems the neutral is earthed at the distribution transformer. Triplen harmonics (3rd, 9th…) from non-linear loads such as LED drivers and computers add in the neutral and can overload it.
• Effect of unbalanced load:
  • 4-wire star with neutral: phase voltages remain balanced; phase currents differ; neutral current $\bar I_N = \bar I_R + \bar I_Y + \bar I_B$ flows.
  • 3-wire star without neutral (or broken neutral): the load star point shifts (neutral shifting); some phase voltages rise above 230 V and others fall — lightly loaded phases get overvoltage, which can damage appliances. This is why a loose neutral in houses is dangerous.
  • Unbalance causes negative-sequence currents, extra heating in motors and generators, unequal voltage drops and higher losses. Single-phase loads should be distributed evenly across phases.
• Active and reactive power (balanced): per phase $P_{ph} = V_{ph}I_{ph}\cos\phi$, total $P = 3V_{ph}I_{ph}\cos\phi = \sqrt3V_LI_L\cos\phi$ for both star and delta. φ is the angle between phase voltage and phase current. $Q = \sqrt3V_LI_L\sin\phi$, $S = \sqrt3V_LI_L$.
• Star vs delta load with same impedance: a delta-connected load draws three times the line current and three times the power of the same impedances in star (this is the basis of star-delta motor starting).
• Power measurement: three-wattmeter method (4-wire) and two-wattmeter method (3-wire, balanced or unbalanced): $P = W_1 + W_2$; for balanced loads $\tan\phi = \sqrt3\frac{W_1 - W_2}{W_1 + W_2}$. At pf 0.5 one wattmeter reads zero; below 0.5 one reads negative.
• Advantages of three-phase: constant power to balanced loads (smooth motor torque), about 75 % of the conductor material of single-phase for the same power and losses, self-starting induction motors, smaller and cheaper machines.

High-yield points

• Star: $V_L = \sqrt3V_{ph}$, $I_L = I_{ph}$; delta: $V_L = V_{ph}$, $I_L = \sqrt3I_{ph}$.
• $P = \sqrt3V_LI_L\cos\phi$, φ = load impedance angle (phase V to phase I).
• Same impedances in delta draw 3 × the line current and 3 × the power of star.
• Balanced load: zero neutral current; triplen harmonics add in the neutral.
• Two-wattmeter: $\tan\phi = \sqrt3(W_1 - W_2)/(W_1 + W_2)$; one reading zero at pf 0.5.
• Broken neutral with unbalanced load overvoltages the lightly loaded phase.

Formulas

$\text{Star: } V_L = \sqrt3V_{ph},\ I_L = I_{ph};\qquad \text{Delta: } V_L = V_{ph},\ I_L = \sqrt3I_{ph}$

$P = \sqrt3V_LI_L\cos\phi,\quad Q = \sqrt3V_LI_L\sin\phi,\quad S = \sqrt3V_LI_L$

Worked example

Three identical impedances of $Z = 8 + j6\ \Omega$ are star-connected to a 400 V, 3-phase supply. Find line current, power factor and total power. Repeat for delta.

1. |Z| = √(64 + 36) = 10 Ω; pf = 8/10 = 0.8 lagging.
2. Star: $V_{ph} = 400/\sqrt3 = 230.9$ V; $I_L = I_{ph} = 23.09$ A.
3. $P = \sqrt3\times400\times23.09\times0.8 = 12.8$ kW (check: $3I^2R = 3\times533.2\times8 = 12.8$ kW ✓).
4. Delta: $I_{ph} = 400/10 = 40$ A, $I_L = 69.3$ A, P = 3 × 40² × 8 = 38.4 kW — three times the star value.

Exam tips

• Star: line voltage = √3 × phase voltage; delta: line current = √3 × phase current.
• In P = √3 V_L I_L cos φ, φ is the phase angle of the load impedance, not the angle between line quantities.
• Balanced load → zero neutral current; broken neutral with unbalanced load → overvoltage on lightly loaded phase.
• Two-wattmeter method: one reading is zero at pf = 0.5, negative below 0.5.
• Same impedances in delta take 3× the power of star.

Exam-style questions

From Calmon's question bank, which has 23 questions on this topic.

1. A balanced three-phase, 400 V load consumes 30 kW at 0.75 power factor lagging. The line current is about:

• A. 75 A
• B. 43.3 A
• C. 100 A
• D. 57.7 A

Answer: D. $I_L = \dfrac{P}{\sqrt3\,V_L\cos\phi} = \dfrac{30000}{1.732\times400\times0.75} \approx 57.7$ A.

2. A balanced delta-connected load with impedance $(8 + j6)\,\Omega$ per phase is connected to a 400 V, three-phase supply. The total reactive power is:

• A. 38.4 kvar
• B. 28.8 kvar
• C. 9.6 kvar
• D. 16.6 kvar

Answer: B. $I_{ph} = 400/10 = 40$ A; $Q = 3I_{ph}^2X = 3\times1600\times6 = 28.8$ kvar.

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