Answer to Question #230236 in Electrical Engineering for KHANYA

Question

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

Answer 1

$X_r=sX_s\\ X_s=0.25\Omega\\ s=0.04\\ X_r=0.04\times{0.25}=0.01\Omega\\ Z_r=\sqrt{0.012^2+0.01^2}=0.0156\Omega\\ k=\dfrac{1}{3.8}=0.26\\ Rotor\ phase\ voltage\ at\ standstill\\ E_2=1100\times{k}=1100\times{0.26}=289.5V\\ E_r=sE_2=0.04\times{289.5}=11.58V\\ I_r=\dfrac{E_r}{Z_r}=\dfrac{11.58}{0.0156}=742.3A$Rotor current at 4% slip with slip rings shorted is 742.3A