📘 Suchenow Academy | TGPSC AEE Electrical
Subject 1 → Chapter 1.6: Three-Phase Circuits — Star, Delta, Power Measurement
| Expected questions | 4–6 every exam — power calculation and star-delta conversion are guaranteed |
| Upgraded features | Live phasor rotation · Star↔Delta converter · Two-wattmeter live calculator · Unbalanced load solver · Current flow animation |
| Target exams | TGPSC AEE, TSGENCO AE, TSTRANSCO AE, TSSPDCL AE, APPSC AEE, GATE EE |
| Syllabus line | Three-phase balanced and unbalanced systems, power measurement |
1. Why Three-Phase? — Physical Meaning First
Single-phase power pulsates (goes to zero twice per cycle). Three-phase power is constant — three sinusoids 120° apart, always summing to the same total. That's why all power generation, transmission and heavy motors use three-phase. It's also 73% more efficient in conductor use than single-phase.
🌾 తెలుగులో: Single-phase లో కరెంట్ పల్సేట్ అవుతుంది (2 సార్లు zero అవుతుంది). Three-phase లో 120° తేడాలో మూడు phases ఉంటాయి — వాటి sum ఎప్పుడూ constant. అందుకే power stations, factories లో three-phase వాడతారు.
⭐ Star (Y) Connection
One end of each winding connected to neutral (N). 4-wire system possible.
V_L = √3 · V_ph
I_L = I_ph
Line voltage leads phase by 30°
Neutral exists ✓
I_L = I_ph
Line voltage leads phase by 30°
Neutral exists ✓
△ Delta Connection
Windings connected end-to-end forming a triangle. No neutral wire.
V_L = V_ph
I_L = √3 · I_ph
Line current lags phase by 30°
No neutral possible ✗
I_L = √3 · I_ph
Line current lags phase by 30°
No neutral possible ✗
🌾 Memory trick: "Star లో Voltage multiply (×√3), Delta లో Current multiply (×√3)" — ఒక్క line గుర్తుంటే direct 2 marks!
⚠️ Most common trap: In Star — V_L = √3·V_ph (voltage multiplies). In Delta — I_L = √3·I_ph (current multiplies). Students often confuse which multiplies in which connection.
🎮 Lab 1 — Live Phasor Rotation (Star System)
Three phase voltages (120° apart) rotate together. Watch V_R, V_Y, V_B and see how line voltage V_RY forms between them:
V_R=0°, V_Y=−120°, V_B=−240°(+120°). Line voltage V_RY=V_R−V_Y is √3 times phase voltage and leads V_R by 30°.
2. Star and Delta — Complete Formula Set
STAR (Y):
V_line = √3 · V_phase (line voltage = √3 × phase voltage)
I_line = I_phase (same current)
V_phase = V_line/√3
Power: P = √3·V_L·I_L·cosφ = 3·V_ph·I_ph·cosφ
DELTA (Δ):
V_line = V_phase (same voltage)
I_line = √3 · I_phase (line current = √3 × phase current)
I_phase = I_line/√3
Power: P = √3·V_L·I_L·cosφ = 3·V_ph·I_ph·cosφ
KEY: Power formula √3·V_L·I_L·cosφ is SAME for both Star AND Delta!
Only V_ph and I_ph differ between the two connections.
🎮 Lab 2 — Star ↔ Delta Live Converter
Enter line voltage and load details — get complete Star and Delta analysis instantly:
3. Power in Three-Phase Systems ⭐
Total Active Power: P = √3·V_L·I_L·cosφ = 3·V_ph²/R (for resistive load)
Total Reactive Power: Q = √3·V_L·I_L·sinφ (VAR)
Total Apparent Power: S = √3·V_L·I_L (VA)
Power Factor: cosφ = P/S = R/Z
Key insight: The formula √3·V_L·I_L·cosφ works for BOTH star and delta — you never need to know which connection it is!
Total Reactive Power: Q = √3·V_L·I_L·sinφ (VAR)
Total Apparent Power: S = √3·V_L·I_L (VA)
Power Factor: cosφ = P/S = R/Z
Key insight: The formula √3·V_L·I_L·cosφ works for BOTH star and delta — you never need to know which connection it is!
🌾 తెలుగులో: Power formula √3·V_L·I_L·cosφ — Star కైనా Delta కైనా ఒక్కటే! V_L మరియు I_L line quantities use చేయండి — phase తెలియకపోయినా సమాధానం వస్తుంది.
🎮 Lab 3 — Two-Wattmeter Method ⭐ (Live Calculator)
The two-wattmeter method measures three-phase power using only 2 wattmeters. Drag the readings — total power, reactive power and PF update live:
TWO-WATTMETER METHOD:
Total power: P = W₁ + W₂
Reactive power: Q = √3·(W₁ − W₂)
Power factor: tanφ = √3·(W₁−W₂)/(W₁+W₂)
cosφ = cos[tan⁻¹(√3·(W₁−W₂)/(W₁+W₂))]
SPECIAL CASES (direct MCQ):
Unity PF (cosφ=1): W₁ = W₂ (both equal)
Zero PF (cosφ=0): W₁ = −W₂ (equal and opposite)
0.5 PF (cosφ=0.5): One wattmeter reads ZERO
W₁ negative → PF < 0.5 (lagging)
Valid for: balanced OR unbalanced 3-phase 3-wire loads
NOT valid for: 4-wire systems (need 3 wattmeters)
⚠️ Two-wattmeter traps: (1) W₁ can be NEGATIVE — don't panic, it means PF<0.5 (2) At PF=0.5, one meter reads ZERO — not a fault (3) Works for 3-wire only — 4-wire needs 3 wattmeters (4) tanφ formula uses (W₁−W₂), not (W₂−W₁) — check sign carefully.
4. Star-Delta Transformation (Impedance)
Star to Delta: Z_Δ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁) / Z_Y (opposite arm)
Delta to Star: Z_Y = (product of adjacent) / (sum of all three)
For balanced (all equal): Z_Δ = 3·Z_Y (delta impedance = 3× star)
Delta to Star: Z_Y = (product of adjacent) / (sum of all three)
For balanced (all equal): Z_Δ = 3·Z_Y (delta impedance = 3× star)
🎮 Lab 4 — Star ↔ Delta Impedance Converter
Enter balanced load impedance in one form — get the other instantly. Also shows equivalent circuit performance:
🎮 Lab 5 — Unbalanced Star Load Solver
Real power systems are rarely perfectly balanced. Enter 3 different phase loads — get line currents, neutral current and total power:
🔴 Phase R
🟡 Phase Y
🔵 Phase B
Enter phase loads and click Solve.
5. Phase Sequence
POSITIVE (RYB / ABC): R leads Y by 120°, Y leads B by 120°
V_R = V∠0°, V_Y = V∠−120°, V_B = V∠−240° (=+120°)
NEGATIVE (RBY / ACB): R leads B by 120°, B leads Y by 120°
V_R = V∠0°, V_B = V∠−120°, V_Y = V∠−240°
Effect: Reverses motor rotation direction
Test: Phase sequence meter, or swap any two wires to reverse
6. "3 Questions That ALWAYS Appear" ⭐
⭐ Guaranteed every TGPSC/TSGENCO exam
Q-TYPE 1: Power calculation (always comes)
Pattern: "3-phase balanced load, V_L=415V, R=10Ω/phase star connected. Find P."
Method: V_ph=V_L/√3=239.6V → I_ph=V_ph/R=23.96A=I_L → P=√3×415×23.96×1=17.23kW
Shortcut: P=3×V_ph²/R=3×239.6²/10=17.23kW (for resistive load)
Q-TYPE 2: Two-wattmeter PF (always comes)
Pattern: "W₁=1200W, W₂=400W. Find PF."
Method: tanφ=√3×(1200−400)/(1200+400)=√3×800/1600=√3×0.5=0.866 → φ=40.9° → cosφ=0.755
Watch for: if W₂ is NEGATIVE → PF<0.5. If W₁=W₂ → PF=1.
Q-TYPE 3: Star↔Delta current/voltage (always comes)
Pattern: "Delta load 30Ω/phase, V_L=415V. Find line current."
Method (Delta): V_ph=V_L=415V → I_ph=V_ph/Z=415/30=13.83A → I_L=√3×I_ph=23.95A
Key: In Delta, V_ph=V_L. In Star, V_ph=V_L/√3. Never mix these up.
7. Solved Problems — Try First! 🎯
P1. Balanced star load: R=15Ω/phase, X=0. V_L=400V. Find I_L and total P.
V_ph=400/√3=231V. I_ph=231/15=15.4A=I_L. P=√3×400×15.4×1=10.67kW. Or P=3×231²/15=10.67kW ✓
P2. Same load reconnected in delta. Find new I_L and P.
V_ph=V_L=400V (delta). I_ph=400/15=26.67A. I_L=√3×26.67=46.19A. P=√3×400×46.19=32kW (= 3× star power). Delta draws 3× power of star for same impedance!
P3 (Key insight). Why does delta draw 3× the power of star?
Star: V_ph=V_L/√3. Delta: V_ph=V_L. Voltage ratio=√3. Power∝V²/R, so power ratio=(√3)²=3. Delta always draws 3× power. This is why motors start in star and run in delta (reduced starting current).
P4. Two-wattmeter: W₁=2000W, W₂=1000W. Find P, Q, cosφ.
P=W₁+W₂=3000W. Q=√3(W₁−W₂)=√3×1000=1732 VAR. tanφ=Q/P=1732/3000=0.577 → φ=30° → cosφ=0.866.
P5. Two-wattmeter: W₁=1500W, W₂=−500W. What does negative W₂ mean? Find cosφ.
Negative wattmeter reading means PF<0.5 (highly inductive load). P=1500+(−500)=1000W. Q=√3×(1500−(−500))=√3×2000=3464 VAR. tanφ=3464/1000=3.464 → φ=73.9° → cosφ=0.277. Very low PF — heavily inductive.
P6. At what PF does one wattmeter read zero in two-wattmeter method?
W₂=0: tanφ=√3×(W₁−0)/(W₁+0)=√3 → φ=60° → cosφ=0.5. At PF=0.5, one wattmeter reads exactly zero. This is a classic exam question.
P7. Balanced delta: Z=6+j8Ω/phase, V_L=400V. Find I_L and total P.
|Z|=√(36+64)=10Ω. cosφ=6/10=0.6. V_ph=V_L=400V (delta). I_ph=400/10=40A. I_L=√3×40=69.28A. P=√3×400×69.28×0.6=28.8kW. Or P=3×V_ph²×R/|Z|²=3×400²×6/100=28.8kW ✓
P8. Star-delta starter: motor runs at V_L=415V. During star start, voltage per phase?
Star start: V_ph=V_L/√3=415/1.732=239.6V. Delta run: V_ph=V_L=415V. Starting torque ∝ V_ph² → star gives (1/3) of delta torque → reduced starting current (1/3).
P9. 3-phase, 4-wire system. How many wattmeters needed to measure total power?
3 wattmeters — one per phase. Two-wattmeter method only works for 3-wire (no neutral). With neutral wire (4th wire), neutral current can flow and 2 meters don't capture full power.
P10. Balanced load: cosφ=0.8 lag. Two-wattmeter readings: W₁>W₂ or W₁<W₂?
For lagging load with cosφ>0.5: both wattmeters read positive and W₁>W₂. For cosφ<0.5: W₂ goes negative. The convention depends on which phase the meters are connected to.
P11. Convert star load (Z_Y=10Ω each) to equivalent delta.
Balanced: Z_Δ=3×Z_Y=3×10=30Ω. For unbalanced: Z_Δ(12)=(Z₁Z₂+Z₂Z₃+Z₃Z₁)/Z₃. General formula uses sum of products divided by opposite arm.
P12 (Full TGPSC pattern). 3-phase balanced load: V_L=440V, P=30kW, cosφ=0.8 lag. Find: (a) I_L (b) Q (c) S (d) Two-wattmeter readings W₁ and W₂.
(a) P=√3·V_L·I_L·cosφ → I_L=30000/(√3×440×0.8)=49.22A
(b) Q=P·tanφ=30k×0.75=22.5 kVAR (sinφ=0.6, tanφ=0.75)
(c) S=P/cosφ=30/0.8=37.5 kVA
(d) φ=cos⁻¹(0.8)=36.87°. W₁+W₂=30kW. W₁−W₂=Q/√3=22500/√3=12990W.
W₁=21.5kW, W₂=8.5kW. Check: W₁>W₂ (cosφ>0.5) ✓
(b) Q=P·tanφ=30k×0.75=22.5 kVAR (sinφ=0.6, tanφ=0.75)
(c) S=P/cosφ=30/0.8=37.5 kVA
(d) φ=cos⁻¹(0.8)=36.87°. W₁+W₂=30kW. W₁−W₂=Q/√3=22500/√3=12990W.
W₁=21.5kW, W₂=8.5kW. Check: W₁>W₂ (cosφ>0.5) ✓
8. PYQ Bank
- [TSGENCO 2015] In star: V_L=√3·V_ph. In delta: I_L=√3·I_ph.
- [TSSPDCL 2018] Two-wattmeter: P=W₁+W₂. Q=√3(W₁−W₂).
- [TSTRANSCO 2018] At PF=0.5, one wattmeter reads zero.
- [APPSC 2016] Star-delta starter reduces starting current by factor 3 (1/3 of DOL).
- [GATE-style] Power formula √3·V_L·I_L·cosφ is same for both star and delta.
- [ESE pattern] Negative wattmeter reading → PF < 0.5 (lagging).
- [TGPSC 2022] Delta draws 3× power of equivalent star (same supply, same impedance).
- [TGPSC 2022] Two-wattmeter valid for 3-wire — NOT for 4-wire (need 3 meters).
9. Traps & Memory Hooks
⚠️ TOP TRAPS:
(1) Star: V multiplies by √3 | Delta: I multiplies by √3 — never reverse
(2) Negative wattmeter = PF<0.5, not instrument fault
(3) Power formula √3·V_L·I_L·cosφ same for both — use LINE quantities
(4) Star-delta: Z_Δ=3Z_Y (balanced only) | Unbalanced needs full formula
(5) Two-wattmeter: 3-wire only — neutral wire needs 3rd meter
(6) Phase sequence reversal reverses motor direction
(1) Star: V multiplies by √3 | Delta: I multiplies by √3 — never reverse
(2) Negative wattmeter = PF<0.5, not instrument fault
(3) Power formula √3·V_L·I_L·cosφ same for both — use LINE quantities
(4) Star-delta: Z_Δ=3Z_Y (balanced only) | Unbalanced needs full formula
(5) Two-wattmeter: 3-wire only — neutral wire needs 3rd meter
(6) Phase sequence reversal reverses motor direction
- "Star Voltage multiplies, Delta Current multiplies" (both by √3)
- "Power = √3 V_L I_L cosφ — always, both" (star and delta same)
- "W equal → PF=1, W zero-one → PF=0.5, W negative → PF<0.5"
- "Delta = 3× Star power" (same impedance, same supply)
- "Star starts, Delta runs" (motor starting)
🌾 Final Telugu revision: "Star లో V×√3, Delta లో I×√3. Power √3·VL·IL·cosφ రెండింటికి same. Wattmeter negative అంటే PF<0.5. Delta = 3×Star power." — ఈ నాలుగు points exam లో ముందు రాయండి!
MASTER CHEAT SHEET:
STAR: V_L=√3·V_ph | I_L=I_ph | V_ph=V_L/√3
DELTA: V_L=V_ph | I_L=√3·I_ph | I_ph=I_L/√3
POWER (both): P=√3·V_L·I_L·cosφ | Q=√3·V_L·I_L·sinφ | S=√3·V_L·I_L
RESISTIVE: P=3V_ph²/R=V_L²/R_delta=3V_L²/R_star (shortcuts)
TWO-WATTMETER:
P=W₁+W₂ | Q=√3(W₁−W₂) | tanφ=√3(W₁−W₂)/(W₁+W₂)
PF=1 → W₁=W₂ | PF=0.5 → one=0 | PF<0.5 → one negative
Valid: 3-wire only | 4-wire → 3 wattmeters
STAR↔DELTA (balanced): Z_Δ=3Z_Y | Delta=3× Star power
Star-delta starter: V_start=V_L/√3 | Starting current=I_DOL/3
🎯 Chapter 1.6 Quiz — 10 Questions
⏱ Exam Timer Drill — 5 Questions · 6 Minutes
TGPSC speed: ~72 sec/question. Timer starts immediately.
6:00
D1. In a balanced star system, V_L=415V. Find V_ph.
a) 415V b) 718V c) 239.6V d) 830VStar: V_ph=V_L/√3=415/1.732=239.6V. In Delta: V_ph=V_L=415V (no division).
D2. Two-wattmeter: W₁=1000W, W₂=1000W. Power factor?
a) 0.5 b) 1.0 (unity) c) 0.866 d) 0.707W₁=W₂ → tanφ=√3×(1000−1000)/2000=0 → φ=0° → cosφ=1.0. Equal readings always means unity PF.
D3. Star load Z=10Ω/phase reconnected in delta. Power ratio (delta/star)?
a) 1 (same) b) √3 c) 3 d) 1/3Delta draws 3× star power for same impedance and supply. In delta V_ph=V_L vs star V_ph=V_L/√3. Power∝V_ph²/R → ratio=(√3)²=3.
D4. At what PF does one wattmeter read ZERO in two-wattmeter method?
a) PF = 1.0 b) PF = 0.866 c) PF = 0.5 d) PF = 0At PF=0.5 (φ=60°): tanφ=√3 → W₁−W₂=W₁+W₂ → W₂=0. Classic direct question.
D5. 3-phase, V_L=400V, P=20kW, cosφ=0.8. Find I_L.
a) 50A b) 36.08A c) 28.87A d) 62.5AP=√3·V_L·I_L·cosφ → I_L=20000/(√3×400×0.8)=20000/554=36.08A.
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