Balanced Three-Phase Fundamentals
Phase Sequence and Phasors (Positive/ABC Sequence)
V_a = V_p\angle 0°, \quad V_b = V_p\angle -120°, \quad V_c = V_p\angle -240°
| Symbol | Description | Unit |
|---|---|---|
| V_p | Phase voltage magnitude (RMS) | V |
| V_a, V_b, V_c | Phase voltage phasors | V |
Worked example
A balanced 3-phase supply has Vp=230V (RMS), positive sequence. Write all three phase voltages.
Given: Vp=230V, positive sequence (ABC)
- Va = 230∠0° V
- Vb = 230∠-120° V
- Vc = 230∠-240° = 230∠+120° V
- Sum check: Va + Vb + Vc = 0 for balanced system ✓
Answer: Va=230∠0°V, Vb=230∠-120°V, Vc=230∠120°V
Star (Wye) Connection Relations
V_L = \sqrt{3} V_P, \quad I_L = I_P
| Symbol | Description | Unit |
|---|---|---|
| V_L | Line-to-line voltage (RMS) | V |
| V_P | Phase voltage (RMS) | V |
| I_L | Line current | A |
| I_P | Phase current (= line current in star) | A |
Worked example
A star-connected source has phase voltage 230V. Find line voltage.
Given: V_P = 230V
- V_L = sqrt(3) * V_P
- = 1.732 * 230
- = 398.4V ≈ 400V
Answer: V_L = 400V (standard 400V three-phase supply)
Delta Connection Relations
V_L = V_P, \quad I_L = \sqrt{3} I_P
| Symbol | Description | Unit |
|---|---|---|
| I_P | Phase current in delta branch | A |
| I_L | Line current | A |
Worked example
A delta-connected load draws phase current 10A per branch. Find line current.
Given: I_P = 10A
- I_L = sqrt(3) * I_P
- = 1.732 * 10
- = 17.32A
Answer: I_L = 17.32A
Balanced Load Analysis
Line Current in Balanced Star Load
I_L = \frac{V_P}{Z_P} = \frac{V_L}{\sqrt{3} Z_P}
| Symbol | Description | Unit |
|---|---|---|
| Z_P | Impedance per phase | Ω |
| V_P | Phase voltage | V |
Worked example
Balanced star load, ZP=10+j10 Ω per phase, connected to 400V (line) supply. Find IL.
Given: V_L=400V, Z_P=10+j10 Ω
- |Z_P| = sqrt(10^2 + 10^2) = sqrt(200) = 14.14Ω
- V_P = V_L/sqrt(3) = 400/1.732 = 231V
- I_L = I_P = V_P/|Z_P| = 231/14.14 = 16.3A
Answer: I_L = 16.3A
Phase Current in Balanced Delta Load
I_P = \frac{V_L}{Z_P}, \quad I_L = \sqrt{3} I_P
| Symbol | Description | Unit |
|---|---|---|
| V_L | Line voltage = Phase voltage in delta | V |
Worked example
Balanced delta load, ZP=20∠30°Ω per phase, 400V line supply. Find phase and line currents.
Given: V_L=400V, |Z_P|=20Ω
- I_P = V_L/|Z_P| = 400/20 = 20A
- I_L = sqrt(3) * I_P = 1.732 * 20 = 34.64A
Answer: I_P = 20A, I_L = 34.64A
Delta-Star Load Impedance Conversion
Z_Y = \frac{Z_\Delta}{3}, \quad Z_\Delta = 3 Z_Y
| Symbol | Description | Unit |
|---|---|---|
| Z_Y | Equivalent star impedance per phase | Ω |
| Z_Δ | Delta impedance per phase | Ω |
Worked example
A delta load has ZΔ=60∠30° Ω per phase. Convert to equivalent star.
Given: Z_Δ = 60∠30° Ω
- Z_Y = Z_Δ / 3 = 60∠30° / 3
- = 20∠30° Ω
- = 17.32 + j10 Ω
Answer: Z_Y = 20∠30° Ω = 17.32 + j10 Ω
Three-Phase Power
Total Three-Phase Active Power
P = 3 V_P I_P \cos\phi = \sqrt{3} V_L I_L \cos\phi
| Symbol | Description | Unit |
|---|---|---|
| P | Total active power | W |
| cos φ | Power factor (per phase) | — |
| V_L | Line voltage | V |
| I_L | Line current | A |
Worked example
A 3-phase motor draws 20A line current from a 415V supply at pf=0.85 lag. Find active power.
Given: V_L=415V, I_L=20A, cosφ=0.85
- P = sqrt(3) * V_L * I_L * cos_phi
- = 1.732 * 415 * 20 * 0.85
- = 1.732 * 415 * 17
- = 1.732 * 7055
- = 12,219W ≈ 12.22kW
Answer: P ≈ 12.22kW
Total Three-Phase Reactive Power
Q = 3 V_P I_P \sin\phi = \sqrt{3} V_L I_L \sin\phi
| Symbol | Description | Unit |
|---|---|---|
| Q | Total reactive power | VAR |
| sin φ | Reactive factor | — |
Worked example
Same motor: 415V, 20A, pf=0.85. Find Q.
Given: V_L=415V, I_L=20A, cosφ=0.85, sinφ=0.527
- sin φ = sqrt(1 - 0.85^2) = sqrt(1-0.7225) = sqrt(0.2775) = 0.527
- Q = sqrt(3) * 415 * 20 * 0.527
- = 1.732 * 415 * 10.54
- = 7,575 VAR ≈ 7.58kVAR
Answer: Q ≈ 7.58 kVAR (lagging)
Apparent Power and Power Triangle
S = \sqrt{3} V_L I_L = \sqrt{P^2 + Q^2}, \quad \cos\phi = P/S
| Symbol | Description | Unit |
|---|---|---|
| S | Total apparent power | VA |
| P | Active power | W |
| Q | Reactive power | VAR |
Worked example
Find S and verify pf for the same motor (P=12.22kW, Q=7.58kVAR).
Given: P=12220W, Q=7580VAR
- S = sqrt(P^2 + Q^2) = sqrt(12220^2 + 7580^2)
- = sqrt(149,328,400 + 57,456,400)
- = sqrt(206,784,800) = 14,380 VA ≈ 14.38 kVA
- pf = P/S = 12220/14380 = 0.85 ✓
Answer: S = 14.38 kVA, pf = 0.85 ✓
Two-Wattmeter Method
Total Power (Two-Wattmeter Method)
P = W_1 + W_2
| Symbol | Description | Unit |
|---|---|---|
| W1 | Reading of wattmeter 1 | W |
| W2 | Reading of wattmeter 2 | W |
Worked example
Two wattmeters read W1=8kW and W2=3kW. Find total power P.
Given: W1=8000W, W2=3000W
- P = W1 + W2 = 8000 + 3000 = 11000W = 11kW
Answer: P = 11kW
Power Factor from Two-Wattmeter Method
\tan\phi = \sqrt{3} \cdot \frac{W_1 - W_2}{W_1 + W_2}
| Symbol | Description | Unit |
|---|---|---|
| φ | Load power factor angle | ° |
Worked example
W1=8kW and W2=3kW. Find pf using two-wattmeter method.
Given: W1=8000W, W2=3000W
- tan φ = sqrt(3) * (W1-W2)/(W1+W2)
- = 1.732 * (8000-3000)/(8000+3000)
- = 1.732 * 5000/11000
- = 1.732 * 0.4545 = 0.787
- φ = arctan(0.787) = 38.2°
- pf = cos(38.2°) = 0.785
Answer: pf = 0.785 lagging
Reactive Power from Two-Wattmeter Method
Q = \sqrt{3}(W_1 - W_2)
| Symbol | Description | Unit |
|---|---|---|
| W1-W2 | Difference of wattmeter readings | W |
Worked example
W1=8kW, W2=3kW. Find reactive power Q.
Given: W1=8000W, W2=3000W
- Q = sqrt(3) * (W1 - W2)
- = 1.732 * (8000 - 3000)
- = 1.732 * 5000
- = 8660 VAR = 8.66 kVAR
Answer: Q = 8.66 kVAR
Power Factor Correction
Capacitor Bank for PF Correction
Q_C = P(\tan\phi_1 - \tan\phi_2), \quad C = \frac{Q_C}{3 \omega V_P^2}
| Symbol | Description | Unit |
|---|---|---|
| Q_C | Reactive power to be compensated | VAR |
| φ1 | Original power factor angle | ° |
| φ2 | Target power factor angle | ° |
| C | Capacitance per phase (star) | F |
| ω | Angular frequency | rad/s |
Worked example
A 3-phase load: P=15kW, pf=0.707 lag. Correct pf to 0.95 lag using star-connected capacitors. Supply: 415V, 50Hz.
Given: P=15000W, cosφ1=0.707 (φ1=45°), cosφ2=0.95 (φ2=18.2°), VL=415V, f=50Hz
- tanφ1 = tan(45°) = 1.0, tanφ2 = tan(18.2°) = 0.329
- QC = P(tanφ1 - tanφ2) = 15000*(1.0 - 0.329) = 15000*0.671 = 10065 VAR
- VP = 415/sqrt(3) = 240V
- ω = 2π*50 = 314.16 rad/s
- C = QC/(3 * ω * VP^2) = 10065/(3 * 314.16 * 240^2)
- = 10065/(3 * 314.16 * 57600)
- = 10065 / 54,288,691 = 185.4μF
Answer: C = 185.4 μF per phase (star-connected)
Unbalanced Three-Phase Systems
Neutral Shift Voltage (Unbalanced Star Load)
V_{nN} = \frac{V_{an}/Z_a + V_{bn}/Z_b + V_{cn}/Z_c}{1/Z_a + 1/Z_b + 1/Z_c}
| Symbol | Description | Unit |
|---|---|---|
| V_nN | Voltage between load neutral n and source neutral N | V |
| Za, Zb, Zc | Impedances of phases A, B, C | Ω |
| Van, Vbn, Vcn | Source phase voltages | V |
Worked example
Unbalanced star: Za=10Ω, Zb=j10Ω, Zc=-j10Ω. Source: balanced 230V. Find VnN.
Given: Va=230∠0°, Vb=230∠-120°, Vc=230∠120°, Za=10, Zb=j10, Zc=-j10
- Ya=1/10=0.1, Yb=1/j10=-j0.1, Yc=1/(-j10)=j0.1
- Numerator = Va*Ya + Vb*Yb + Vc*Yc
- = 230*0.1 + 230∠-120°*(-j0.1) + 230∠120°*(j0.1)
- = 23 + 23∠(-120°-90°) + 23∠(120°+90°)
- = 23 + 23∠-210° + 23∠210° = 23 + 0 = 23 (the j terms cancel)
- Denominator = 0.1 - j0.1 + j0.1 = 0.1
- VnN = 23/0.1 = 230V
Answer: V_nN = 230V (significant neutral displacement due to reactive loads)
Symmetrical Components (Positive Sequence)
V_a^{(1)} = \frac{1}{3}(V_a + aV_b + a^2 V_c), \quad a = 1\angle 120°
| Symbol | Description | Unit |
|---|---|---|
| a | Operator a = 1∠120° = -0.5+j0.866 | — |
| Va^(1) | Positive sequence component of Va | V |
Worked example
Unbalanced phasors: Va=100∠0°, Vb=80∠-100°, Vc=120∠130°V. Find positive sequence Va^(1).
Given: Va=100∠0°, Vb=80∠-100°, Vc=120∠130°
- a = 1∠120°, a^2 = 1∠240°
- aVb = 80∠(-100°+120°) = 80∠20°
- a^2*Vc = 120∠(130°+240°) = 120∠370° = 120∠10°
- Va + aVb + a^2*Vc = 100∠0° + 80∠20° + 120∠10°
- ≈ 100 + (75.2+j27.4) + (118.2+j20.8) = 293.4 + j48.2
- = 297.3∠9.3°
- Va^(1) = 297.3∠9.3° / 3 = 99.1∠9.3°V
Answer: V_a^(1) ≈ 99.1∠9.3° V
Quick reference
| Formula | Expression |
|---|---|
| Star: V_L vs V_P | V_L = √3 · V_P |
| Star: I_L vs I_P | I_L = I_P |
| Delta: V_L vs V_P | V_L = V_P |
| Delta: I_L vs I_P | I_L = √3 · I_P |
| Active Power (3-phase) | P = √3·V_L·I_L·cosφ |
| Reactive Power | Q = √3·V_L·I_L·sinφ |
| Apparent Power | S = √3·V_L·I_L |
| Two-Wattmeter Total P | P = W1 + W2 |
| Two-Wattmeter pf | tanφ = √3·(W1-W2)/(W1+W2) |
| Two-Wattmeter Q | Q = √3·(W1-W2) |
| Delta-Star conversion | Z_Y = Z_Δ/3 |
| PF correction capacitor | QC = P(tanφ1 - tanφ2) |
| Phase operator a | a = 1∠120°, a² = 1∠240° |
| Positive sequence | Va^(1) = (Va + aVb + a²Vc)/3 |
Exam tips
- GATE papers regularly test the two-wattmeter method — both the formula P = W1+W2 and the pf formula tan φ = √3(W1-W2)/(W1+W2) must be memorised, and the case where one wattmeter reads negative (pf < 0.5) must be handled carefully.
- In balanced three-phase problems, always convert delta loads to equivalent star using Z_Y = Z_Δ/3 before applying nodal or mesh analysis.
- The distinction between phase voltage and line voltage is tested constantly — for star connection VP = VL/√3 ≈ 231V for a 400V supply; confusing these leads to answers off by a factor of √3.
- Power factor correction questions require computing QC = P(tanφ1-tanφ2); always find tan values from the given pf (cos φ) rather than assuming angles.
- Symmetrical components (zero, positive, negative sequence) appear in fault analysis questions — know the operator a = 1∠120° and use it correctly in the decomposition formulas.