--- title: "Torque-slip characteristic and linear operating region" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-torque-slip-linear-region" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-torque-slip-linear-region site: BCT Engineering --- # Torque-slip characteristic and linear operating region Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2079 Chaitra, 8 marks - Video: [Electrical Machines | Lecture - 6G | Three-Phase Induction Motors | Torque-Speed Characteristics](https://www.youtube.com/watch?v=Z6GVndgNCOw) — Engineering Devotion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-torque-slip-linear-region - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Explain T-s characteristics of 3-phase induction motor. Why the induction motor operates only in linear portion of T-s characteristics? ## Short answer The T–s curve is linear (T ∝ s) for low slip and hyperbolic (T ∝ 1/s) beyond pull‑out. The motor operates only in the linear region because there dT/ds > 0, ensuring stable operation, high efficiency, low current, and near‑constant speed. ## Answer The **T–s characteristic** shows how the torque developed by a 3-phase induction motor varies with slip (speed) at constant voltage and frequency. It follows from $$ T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} $$ ```text T | Tmax | .--. | .' '. | / B '. | / '-._ | / A '-- Tst (C) | / | | /stable| unstable | / | |/________|______________ s 0 sm 1 (Ns) (N=0) ``` #### Regions of the curve 1. **Low-slip region ($0 < s < s_m$):** $sX_2$ is small compared with $R_2$, so $$ T \approx \frac{K E_2^2}{R_2}\, s \quad\Rightarrow\quad T \propto s $$ The curve is almost a straight line through the origin (region OA–B). 2. **Maximum torque (B):** at $s_m = R_2/X_2$, $T_{max} = KE_2^2/(2X_2)$ — pull-out torque, usually 2–3 times full-load torque. 3. **High-slip region ($s_m < s \le 1$):** $sX_2 \gg R_2$, so $$ T \approx \frac{K E_2^2 R_2}{s X_2^2} \quad\Rightarrow\quad T \propto \frac{1}{s} $$ The curve is a rectangular hyperbola, ending at starting torque $T_{st}$ at $s = 1$ (C). 4. For $s < 0$ the machine generates; for $s > 1$ it brakes (plugging). #### Why the motor operates only in the linear portion 1. **Stability:** in the linear region, if load torque increases, the speed falls slightly, slip rises and the motor torque **rises** to meet the new load; a new steady point is reached. So $dT/ds > 0$ gives stable operation. In the region beyond $s_m$, an increase in load reduces speed, slip increases and the motor torque **falls**; the speed keeps falling until the motor stalls. Operation there is unstable. 2. **Near-constant speed:** in the linear region slip is only 2–5%, so speed changes very little from no load to full load, which most drives need. 3. **Efficiency:** rotor copper loss $= s \times P_{ag}$. At low slip this loss is small; at high slip most air-gap power is wasted as heat in the rotor. 4. **Current and heating:** rotor current rises sharply with slip; at high slip the current is near starting value (5–7 × FL), which would overheat the machine. 5. **Power factor:** rotor power factor $R_2/\sqrt{R_2^2 + (sX_2)^2}$ is high at low slip and poor at high slip. Hence the motor is designed so that full-load torque lies on the linear part, well below $T_{max}$, leaving a margin for temporary overloads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the T-s curve nearly a straight line at low slip? A. sX2 is much smaller than R2, so T is approximately proportional to s B. sX2 is much larger than R2, so T is proportional to 1/s C. R2 is zero at low slip D. E2 becomes proportional to s Answer: A. When sX2 << R2 the denominator is about R2^2, so T = K E2^2 s / R2, i.e. T is proportional to s. (Key point: Low-slip region) 2. At which slip does the maximum (pull-out) torque occur and what separates the stable and unstable regions? A. s = 1, the starting point B. s = X2/R2 C. s = 0, synchronous speed D. s = R2/X2, beyond which the curve falls with slip Answer: D. T is maximum at sm = R2/X2; to the right of it T falls as slip rises (T proportional to 1/s). (Key point: Maximum torque point) 3. Why is operation beyond sm unstable? A. Torque rises as load rises, so speed overshoots B. A load increase lowers speed, raising slip but lowering motor torque, so the motor keeps slowing and stalls C. The supply voltage falls automatically D. The rotor frequency becomes zero Answer: B. With dT/ds < 0 beyond sm, extra load slows the motor, torque drops further, and it stalls. (Key point: Stability) 4. Why is operation at high slip avoided apart from stability? A. Rotor copper loss s x Pag is small at high slip B. Rotor power factor improves at high slip C. Rotor copper loss s x Pag is large and the currents are near starting value, causing overheating and poor efficiency D. Speed becomes exactly constant Answer: C. High slip wastes much air-gap power as rotor heat and draws 5-7 times full-load current. 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