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Slip-ring IM rotor current at 4% slip

Short answer

Rotor standstill emf per phase = 120/√3 = 69.28 V. At 4% slip, rotor emf = 2.771 V, reactance = 0.06 Ω, impedance = √(0.3²+0.06²) = 0.3059 Ω. Rotor current = 2.771/0.3059 = 9.06 A per phase.

A 3-phase, slip-ring, induction motor with star-connected rotor has an induced e.m.f. of 120 volts between slip-rings at standstill with normal voltage applied to the stator. The rotor winding has resistance per phase of 0.3 Ohm and standstill leakage reactance per phase of 1.5 Ohm. Calculate the current/phase when running short-circuited with 4% slip.

Given: star-connected rotor; standstill emf between slip rings = 120 V; R₂ = 0.3 Ω/phase; standstill reactance X₂ = 1.5 Ω/phase; slip s = 0.04; rotor short-circuited.

Step 1: Rotor emf per phase at standstill

E₂ = 120/√3 = 69.28 V

Step 2: Quantities at 4 % slip

 Rotor emf per phase  sE2 = 0.04 × 69.28 = 2.771 V
 Rotor reactance      sX2 = 0.04 × 1.5   = 0.06 ohm
 Rotor impedance |Z2r| = √(0.3² + 0.06²) = √0.0936 = 0.3059 ohm

Step 3: Rotor current

I₂ = sE₂ / |Z₂ᵣ| = 2.771 / 0.3059 = 9.06 A per phase

Rotor power factor = R₂ / |Z₂ᵣ| = 0.3/0.3059 = 0.981 lagging.

Answer: rotor current ≈ 9.06 A per phase (rotor frequency 0.04 × 50 = 2 Hz if the supply is 50 Hz).

✓ Answer checked by Claude Opus 5.5 (high effort)

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