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📘 Induction Motor – Numerical Problems (GATE Level)

Important numerical problems on slip, torque, power flow and maximum torque. Very high weightage topic.

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🔹 Problem 1 – Slip & Rotor Speed

Given: f = 50 Hz P = 4 poles Rotor speed = 1440 rpm Step 1: Synchronous speed Ns = 120f / P = 120 × 50 / 4 = 1500 rpm Step 2: Slip s = (Ns − Nr) / Ns = (1500 − 1440) / 1500 = 60 / 1500

Slip = 0.04 (4%)

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🔹 Problem 2 – Rotor Copper Loss

Given: Rotor input power = 5000 W Slip = 0.05 Rotor copper loss: = s × Rotor input = 0.05 × 5000

Rotor copper loss = 250 W

Mechanical power developed: = (1 − s) × Rotor input = 0.95 × 5000

Mechanical power = 4750 W

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🔹 Problem 3 – Torque Calculation

Given: Output power = 10 kW Speed = 1440 rpm Torque formula: T = (9550 × P(kW)) / N = (9550 × 10) / 1440

T ≈ 66.3 Nm

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🔹 Problem 4 – Maximum Torque Condition

Condition for maximum torque:

R2 = sX2

Slip at maximum torque: s_max = R2 / X2 Given: R2 = 0.4 Ω X2 = 2 Ω

s_max = 0.4 / 2 = 0.2

Slip = 20%

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🔹 Problem 5 – Starting Torque Ratio

Starting torque: s = 1 Full load slip = 0.05 Using torque proportional relation: T ∝ s / (R2² + (sX2)²) At starting: Denominator large due to X2 Conclusion: Starting torque less than maximum torque. Conceptual question frequently asked.

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🔹 Problem 6 – Efficiency

Given: Input power = 12 kW Output power = 10 kW Efficiency: η = Output / Input = 10 / 12

η = 83.3%

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🔹 Important Formulas to Remember

• Ns = 120f / P
• s = (Ns − Nr) / Ns
• Rotor copper loss = s × Rotor input
• Mechanical power = (1 − s) × Rotor input
• T = (9550P) / N
• s_max = R2 / X2

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🎯 GATE Important Areas

• Slip based power distribution
• Maximum torque condition
• Rotor copper loss questions
• Torque-speed relation

Induction Motor Numericals = Direct Scoring Area