- 2073 Magh · 8 marks
Slip-ring IM starting and running current
Short answer
At standstill, rotor current = 230.9/4.123 = 56.0 A, stator current = 28.0 A. At 1400 rpm (slip 0.0667), rotor current = 15.40/1.035 = 14.9 A, stator current = 7.44 A.
A 4 pole, 50 Hz, 3-phase slip ring induction motor has star connected stator and rotor windings. The rotor winding has impedance of (1+j4) Ω per phase at stand still. The stator to rotor turn ratio is 2. Given that emf induced in rotor circuit is 400 V between two slip rings at stand still. Calculate starting current and running current at 1400 rpm.
Given: 4-pole, 50 Hz, star-connected stator and rotor; rotor impedance at standstill Z₂ = 1 + j4 Ω/phase; stator:rotor turns ratio = 2; rotor open-circuit emf between slip rings = 400 V (standstill).
N_s = 120 × 50 / 4 = 1500 rpm.
Rotor emf per phase (star) E₂ = 400 / √3 = 230.9 V. |Z₂| = √(1² + 4²) = 4.123 Ω.
Starting current (s = 1)
Rotor current I2 = E2 / |Z2| = 230.9 / 4.123 = 56.0 A
Stator current I1 = I2 / K = 56.0 / 2 = 28.0 A
(K = stator turns/rotor turns = 2; the magnetising current is neglected.)
Running current at 1400 rpm
Slip s = (1500 − 1400) / 1500 = 0.0667.
Rotor emf sE2 = 0.0667 × 230.9 = 15.40 V
Rotor reactance sX2 = 0.0667 × 4 = 0.267 ohm
|Z2r| = √(1² + 0.267²) = 1.035 ohm
I2 = 15.40 / 1.035 = 14.88 A
I1 = I2 / 2 = 7.44 A
| Condition | Slip | Rotor current | Stator current |
|---|---|---|---|
| Starting | 1 | 56.0 A | 28.0 A |
| 1400 rpm | 0.0667 | 14.88 A | 7.44 A |
Answer: starting current is 56.0 A in the rotor (28.0 A in the stator); running current at 1400 rpm is 14.9 A in the rotor (7.44 A in the stator).
✓ Answer checked by Claude Opus 5.5 (high effort)
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