# Answer to Question #230235 in Electrical Engineering for KHANYA

Question #230235

A 150 KW ,3000 v, 50 Hz, 6-pole-star-connected induction motor has a start connected slip-ring with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.2 Ω/phase and its per phase/leakage reactance is 14.7 Ω/phase. The stator impedance may be neglected. Find the starting current and starting torque at rated voltage with short circuited slip–ring.

1
2021-08-31T01:42:53-0400

"X_2= 2 \\pi *50*5.61*10^{-3}=1.13 \\Omega\\\\\nK= 1\/3.6, R_2'= R_2\/K^2 = 3.6^2*0.1 = 1.3 \\Omega\\\\\nX_2= 2 \\pi *50*3.61*10^{-3}=1.13 \\Omega ; X_2' =3.6^2*1.13 = 14.7 \\Omega\\\\\nSo, I_{st}= \\frac{V}{(R_2')^2+(X_2')^2}\\\\\n I_{st}= \\frac{3000\/\\sqrt{3}}{\\sqrt{(1.3)^2+(14.7)^2}}\\\\\nI_{st}=117.4A\\\\\nN_s= 120*50\/6=1000 rpm=50\/3 rps\\\\\nT= \\frac{3}{2 \\pi N_s}* \\frac{V^2R_2'}{(R_2')^2+(X_2')^2}\\\\\nT=\\frac{3}{2 \\pi* 50\/3}* \\frac{(3000\/\\sqrt3)^2*1.3}{(1.3)^2+(14.7)^2}\\\\\nT=513 Nm"

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