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Sync motor back emf at 0.5 pf leading

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).

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.

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)

 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)

 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)

              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°.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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