Sunday, 22 February 2026

 

📘 Single Phase Induction Motor – Numerical Problems

Numerical problems are mainly based on double revolving field theory. Forward and backward rotating fields must be considered.


🔹 Problem 1 – Slip Calculation

Given: Frequency = 50 Hz Number of poles = 4 Rotor speed = 1440 rpm Find slip. 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 or 4%


🔹 Problem 2 – Forward and Backward Slip

Given slip s = 0.04 Forward field slip = s = 0.04 Backward field slip: s_b = 2 − s = 2 − 0.04

Backward slip = 1.96

Important concept: Backward slip always close to 2.

🔹 Problem 3 – Air Gap Power

Given: Rotor input power = 1000 W Slip = 0.05 Mechanical power developed: P_mech = Rotor input × (1 − s) = 1000 × (1 − 0.05) = 1000 × 0.95

P_mech = 950 W

Rotor copper loss: = s × Rotor input = 0.05 × 1000

Rotor copper loss = 50 W


🔹 Problem 4 – Starting Torque Ratio

If starting torque = 1.5 times full load torque, And full load torque = 20 Nm

Starting torque = 30 Nm

Conceptual understanding question.

🔹 Problem 5 – Output Power

Given: Mechanical power developed = 950 W Mechanical losses = 50 W Output power: = 950 − 50

Output power = 900 W


🔹 Important Concepts to Remember

  • Single phase IM behaves like two induction motors
  • Forward slip = s
  • Backward slip = 2 − s
  • Net torque = Forward torque − Backward torque
  • At starting, net torque = 0

🎯 GATE Important Points

  • Backward slip formula very important
  • Rotor copper loss = s × Rotor input
  • Mechanical power = (1 − s) × Rotor input
  • Forward & backward torque concept common question

Single Phase IM = Double Revolving Field Theory

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