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Slip ring motor: rotor current at 4% slip

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 Ω0.3\ \Omega and standstill leakage reactance per phase of 1.5 Ω1.5\ \Omega. Calculate the current/phase when running short-circuited with 4% slip.

Given: star-connected rotor, emf between slip rings at standstill =120= 120 V, R2=0.3 ΩR_2 = 0.3\ \Omega, X2=1.5 ΩX_2 = 1.5\ \Omega per phase, slip s=0.04s = 0.04, rotor short-circuited.

Standstill emf per phase:

E2=1203=69.28 VE_2 = \frac{120}{\sqrt3} = 69.28\ \text{V}

Rotor emf at 4 % slip:

E2s=sE2=0.04×69.28=2.771 VE_{2s} = sE_2 = 0.04\times 69.28 = 2.771\ \text{V}

Rotor impedance per phase at slip ss:

Z2s=R22+(sX2)2=0.32+(0.04×1.5)2=0.09+0.0036=0.3059 ΩZ_{2s} = \sqrt{R_2^2 + (sX_2)^2} = \sqrt{0.3^2 + (0.04\times1.5)^2} = \sqrt{0.09+0.0036}=0.3059\ \Omega

Rotor current per phase:

I2=E2sZ2s=2.7710.3059=9.06 AI_2 = \frac{E_{2s}}{Z_{2s}} = \frac{2.771}{0.3059} = 9.06\ \text{A}

Answer: rotor current =9.06= 9.06 A per phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

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