--- title: "Sync motor back emf at 0.5 pf leading" subject: "Electrical Machine" chapter: 6 key: "semester_4/electrical_machine:sync-motor-3-3kv-back-emf" source: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-motor-3-3kv-back-emf site: BCT Engineering --- # Sync motor back emf at 0.5 pf leading Electrical Machine · Chapter 6: Three Phase Synchronous Machines - Asked in: 2071 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-motor-3-3kv-back-emf - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-6 ## Question A 3.3 KV, 3-phase star connected synchronous motor has impedance of 0.2 + j2.2 Ω/phase of the armature winding. The motor is operated at 0.5 pf leading with line current of 100 A. Determine the back emf per phase and also draw phasor diagram. ## Short answer Phase voltage 1905.3 V, current 100∠60° A. Impedance drop IZ = -180.5 + j127.3 V. Back emf per phase E_b = V_ph - IZ = 2090 V, lagging by 3.49° (line 3.62 kV). ## Answer **Given:** 3.3 kV, 3-phase, star-connected synchronous motor; Z_s = 0.2 + j2.2 Ω/phase; I = 100 A at 0.5 pf leading. #### Step 1: Phase voltage and current phasor V_ph = 3300/√3 = 1905.3 V (reference). cos φ = 0.5, so φ = 60° and the current leads V: I = 100 ∠+60° = 50 + j86.60 A. #### Step 2: Impedance drop I·Z_s = (50 + j86.60)(0.2 + j2.2) ```text real part: 50(0.2) - 86.60(2.2) = 10 - 190.5 = -180.5 V imag part: 50(2.2) + 86.60(0.2) = 110 + 17.3 = 127.3 V I Z = -180.5 + j127.3 V (|IZ| = 220.9 V) ``` #### Step 3: Back emf (motor: E_b = V − I Z_s) ```text E_b = 1905.3 - (-180.5 + j127.3) = 2085.8 - j127.3 V |E_b| = √(2085.8² + 127.3²) = 2089.7 V angle = -tan⁻¹(127.3/2085.8) = -3.49° ``` So E_b = 2089.7 V per phase, lagging V by δ = 3.49° (load angle). Line value ≈ √3 × 2089.7 = 3619 V. **Answer: back emf per phase = 2090 V (about 3.62 kV line), at load angle 3.49°. E_b > V, confirming over-excitation for leading pf.** #### Phasor diagram (V as reference) ```text I (100 A, leads V by 60°) / / ---------+-----------------> V = 1905 V (reference) \ \ δ = 3.5° Eb = 2090 V (lags V) Er = V - Eb = I·Zs (= 220.9 V, at angle 145° from V, i.e. I + 85°) I leads V; Eb > V : over-excited motor ``` (The impedance angle is tan⁻¹(2.2/0.2) = 84.8°, so I·Z leads I by 84.8°: 60° + 84.8° = 144.8°.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 3.3 kV star-connected motor, what is the reference phase voltage V_ph? A. 3300 V B. 3300/sqrt(3) = 1905.3 V C. 3300 x sqrt(3) = 5716 V D. 3300/3 = 1100 V Answer: B. In a star connection the phase voltage is the line voltage divided by sqrt(3), so V_ph = 1905.3 V. (Key point: Phase voltage of star connection) 2. At 0.5 pf leading with V as reference, how is the 100 A current written? A. 100 angle -60° = 50 - j86.6 A B. 100 angle 0° = 100 A C. 100 angle +60° = 50 + j86.6 A D. 100 angle +30° = 86.6 + j50 A Answer: C. cos phi = 0.5 gives phi = 60°, and a leading current has a positive angle with respect to V. (Key point: Current phasor at leading pf) 3. Which relation gives the back emf of a synchronous motor per phase? A. E_b = V + I Z_s B. E_b = V - I Z_s C. E_b = V / (I Z_s) D. E_b = I Z_s - V Answer: B. In a motor the supply voltage V supplies the impedance drop and the back emf, so E_b = V - I Z_s. (Key point: Back emf of a motor) 4. What is the result of the calculation and what does it show? A. E_b = 2090 V per phase, larger than V, so the motor is over-excited at leading pf B. E_b = 1905 V per phase, equal to V, so the motor is normally excited C. E_b = 2090 V per phase, larger than V, so the motor is under-excited D. E_b = 1700 V per phase, smaller than V, so the motor is over-excited Answer: A. E_b = 2085.8 - j127.3 V, magnitude about 2090 V at a load angle of 3.49° lagging. E_b > V means over-excitation, which goes with a leading pf. 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