--- title: "T-S characteristics; why linear portion operation" subject: "Electrical Machines" chapter: 5 key: "semester_4/electrical_machines:ts-linear-portion-operation" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ts-linear-portion-operation site: BCT Engineering --- # T-S characteristics; why linear portion operation Electrical Machines · Chapter 5: Three Phase Induction Machines - Asked in: 2071 Bhadra, 8 marks - Video: [Torque speed Characteristics of three phase induction Motor](https://www.youtube.com/watch?v=i-eagFqa5UQ) — SuccessMantra - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ts-linear-portion-operation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5 ## Question Explain torque slip characteristics of 3-phase induction motor. Why the induction motor operates only in linear portion of torque-slip characteristics? ## Short answer At small slip, sX2 is much less than R2, so T is proportional to slip and the curve is linear. This region is stable because a load increase raises slip and torque until equilibrium; beyond maximum torque, torque falls with slip, causing stalling, and rotor copper loss and efficiency also favor low-slip operation. ## Answer #### Torque-slip characteristics The torque produced by a three-phase induction motor at slip $s$ is $$ T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2} $$ The curve starts at $T_{st}$ for $s=1$, rises to $T_{max}$ at $s_m = R_2/X_2$, then drops to zero at $s=0$. ```text T | Tmax | _.-*-._ | Tst _-' `-._ | *-' `-. | `-. +--+--------+--------------*--> s s=1 s_m linear s=0 (unstable)(stable) ``` - Between $s=1$ and $s_m$ the torque falls as slip rises ($T \propto 1/s$): **unstable**. - Between $s_m$ and $0$ the torque is nearly proportional to slip ($T \propto s$): **stable and almost linear**. - The maximum torque is $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$. #### Why the motor operates only in the linear portion 1. For small slip, $sX_2 \ll R_2$, so $T \approx \frac{3E_2^2}{\omega_s R_2}s \propto s$. The curve is a straight line. 2. In this region, if the load rises, the speed falls slightly, the slip rises and the torque increases in proportion. A new equilibrium is reached quickly, so the operation is **stable**. 3. Beyond $s_m$, a rise in load makes the speed fall, which reduces the torque, so the speed falls further and the motor stalls. The operation is **unstable**. 4. In the linear region the rotor resistance dominates the rotor reactance ($R_2 \gg sX_2$), so the rotor power factor is nearly unity. The slip is small (2 to 5 %), so the rotor copper loss ($= sP_g$) is small and the **efficiency is high**. 5. The speed remains almost constant (close to $N_s$) from no load to full load, which is the behaviour most drives need. So the motor is run in the low-slip region of the curve, with full-load slip well below $s_m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the linear region torque is proportional to slip because: A. sX_2 is much smaller than R_2 B. sX_2 is much larger than R_2 C. R_2 = 0 D. Voltage falls with slip Answer: A. Small slip makes the denominator ≈ R_2^2. (Key point: Linear approximation) 2. Why is the linear portion stable? A. Load rise slows the rotor, slip rises and torque rises to match B. Load rise increases speed C. Torque falls with slip D. Torque is constant Answer: A. Torque increases with slip so a new equilibrium is reached. (Key point: Stability) 3. What happens if the load pushes operation beyond s_m? A. Speed rises B. Torque rises, motor stabilises C. Torque falls as speed falls, so the motor stalls D. Nothing changes Answer: C. Falling torque with falling speed is unstable. (Key point: Beyond s_m) 4. Why is efficiency high in the linear region? A. Slip is small so rotor copper loss s P_g is small B. Slip is large C. Rotor loss is zero at all slips D. The stator loss vanishes Answer: A. Rotor copper loss = s P_g. 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