- 2069 Poush
6-pole slip-ring IM no-load current
Short answer
At standstill, rotor current = 230.9/4.079 = 56.6 A, stator current = 14.15 A. At no-load (960 rpm, slip 0.04), rotor current = 9.24/0.816 = 11.3 A, stator current = 2.83 A.
A 6 pole, 50 Hz, 3-ph, slip ring induction motor has star connected stator and rotor windings. The rotor windings has impedance of 0.8 + j4 Ω/phase at standstill. The induced emf between slip rings at standstill is 400 V. The stator to rotor turn ratio is 4. The motor runs at 960 rpm at no load. Calculate current drawn by motor at standstill and no load.
Given: 6-pole, 50 Hz, 3-phase slip-ring motor, star stator and rotor; Z₂ = 0.8 + j4 Ω/phase at standstill; emf between slip rings at standstill = 400 V; stator:rotor turns ratio = 4; no-load speed = 960 rpm.
Assumption: the magnetising (no-load) current of the stator is neglected, so the current drawn is the rotor current reflected to the stator by the turns ratio. Stator current I₁ = I₂ / 4.
N_s = 120 × 50 / 6 = 1000 rpm.
Rotor emf per phase E₂ = 400/√3 = 230.9 V. |Z₂| at standstill = √(0.8² + 4²) = 4.079 Ω.
At standstill (s = 1)
I2 = 230.9 / 4.079 = 56.6 A
I1 = 56.6 / 4 = 14.15 A
At no load (960 rpm)
Slip s = (1000 − 960)/1000 = 0.04.
sE2 = 0.04 × 230.9 = 9.24 V
sX2 = 0.04 × 4 = 0.16 ohm
|Z2r| = √(0.8² + 0.16²) = 0.816 ohm
I2 = 9.24 / 0.816 = 11.32 A
I1 = 11.32 / 4 = 2.83 A
| Condition | Rotor current | Stator (line) current |
|---|---|---|
| Standstill | 56.6 A | 14.15 A |
| No load, 960 rpm | 11.32 A | 2.83 A |
Answer: the motor draws about 14.2 A at standstill and about 2.83 A at no load (rotor current 56.6 A and 11.3 A respectively).
✓ Answer checked by Claude Opus 5.5 (high effort)
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