Answer to Question #230236 in Electrical Engineering for KHANYA

Question #230236

A 1100 V ,50 Hz, delta connected induction motor has a star connected slip ring rotor with a phase transformation ratio of 3.8. The rotor resistance and standstill leakage reactance are 0.012 Ω/phase and 0.25 Ω/phase respectively. Neglecting the stator impedance and magnetising current determine the rotor current at 4 % slip with slip rings shorted


1
Expert's answer
2021-08-31T23:56:13-0400

"X_r=sX_s\\\\\nX_s=0.25\\Omega\\\\\ns=0.04\\\\\nX_r=0.04\\times{0.25}=0.01\\Omega\\\\\nZ_r=\\sqrt{0.012^2+0.01^2}=0.0156\\Omega\\\\\nk=\\dfrac{1}{3.8}=0.26\\\\\nRotor\\ phase\\ voltage\\ at\\ standstill\\\\\nE_2=1100\\times{k}=1100\\times{0.26}=289.5V\\\\\nE_r=sE_2=0.04\\times{289.5}=11.58V\\\\\nI_r=\\dfrac{E_r}{Z_r}=\\dfrac{11.58}{0.0156}=742.3A"Rotor current at 4% slip with slip rings shorted is 742.3A


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