--- title: "Torque-slip curve: starting, maximum and operating region" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-torque-slip-starting-max-operating" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-torque-slip-starting-max-operating site: BCT Engineering --- # Torque-slip curve: starting, maximum and operating region Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2072 Asoj, 8 marks - Video: [torque slip characteristics of 3 phase induction motor | torque speed characteristics |motor control](https://www.youtube.com/watch?v=g4jfjVvgFrs) — Mind of Engineer - 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-starting-max-operating - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Draw and explain the torque-slip (speed) characteristics of 3-phase induction motor, showing clearly the starting torque, maximum torque and normal operating region. ## Short answer Torque-slip curve: stable region s < sm (T ∝ s, sX2 ≪ R2), unstable region s > sm (T ∝ 1/s). Start torque Tst = K E2² R2/(R2²+X2²). Maximum torque at sm = R2/X2, Tmax = K E2²/(2X2), independent of R2. Normal operation is on linear part with slip 2–5%. ## Answer The **torque–slip (speed) characteristic** of a 3-phase induction motor is the graph of developed torque against slip (or speed) at constant supply voltage and frequency. It follows from $$ T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} $$ ```text T | B Tmax | .---. | .' '. | Tfl / '. | ...../ A '._ | /| '-. C | / | '-- Tst | / | unstable | | / | <-- region --> | | / | | | / | | |_/______|______________________|__ s 0 sfl sm 1 N=Ns N=0 |<-normal->| operating ``` #### Starting torque (point C, $s = 1$, $N = 0$) $$ T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2} $$ At standstill the rotor reactance $X_2$ is large compared with $R_2$, so rotor power factor is low and starting torque is moderate (about 1.5–2 times full-load torque for cage motors), even though starting current is 5–7 times full-load. Increasing $R_2$ (slip-ring motors) raises $T_{st}$, up to $T_{max}$ when $R_2 = X_2$. #### Maximum torque (point B) Putting $dT/ds = 0$ gives $R_2 = sX_2$: $$ s_m = \frac{R_2}{X_2}, \qquad T_{max} = \frac{K E_2^2}{2X_2} $$ $T_{max}$ is the **pull-out (breakdown) torque**, usually 2–3 times full-load torque. It is independent of $R_2$ and proportional to $V^2$. If load torque exceeds $T_{max}$, the motor stalls. #### Regions of the curve 1. **Normal operating (stable) region, $0 < s < s_m$:** $sX_2 \ll R_2$, so $T \approx (KE_2^2/R_2)s$, i.e. $T \propto s$ — almost a straight line. The full-load point A lies here, at slip 2–5%. If load increases, speed falls a little and torque rises to match: stable, nearly constant speed. 2. **Unstable region, $s_m < s < 1$:** $sX_2 \gg R_2$, so $T \approx KE_2^2R_2/(sX_2^2)$, i.e. $T \propto 1/s$ — a rectangular hyperbola. A rise in load lowers speed, which lowers torque further, so the motor slows to a stop. The motor only passes through this region while accelerating. 3. **Beyond the motoring range:** $s < 0$ ($N > N_s$) gives generating action; $s > 1$ (rotor driven against the field) gives braking (plugging). | Point | Slip | Torque | |---|---|---| | Synchronous speed | 0 | 0 | | Full load | 0.02–0.05 | $T_{fl}$ | | Maximum | $R_2/X_2$ | $KE_2^2/(2X_2)$ | | Starting | 1 | $KE_2^2R_2/(R_2^2 + X_2^2)$ | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the starting torque of a cage motor only moderate despite a starting current of 5-7 times full load? A. Rotor reactance is large compared with R2 at standstill, so rotor power factor is low B. The air-gap flux is zero at start C. Rotor emf is zero at start D. R2 is infinite at start Answer: A. At s = 1 X2 >> R2, giving a low rotor power factor and so only about 1.5-2 times full-load torque. (Key point: Starting torque) 2. Where does maximum torque occur and what is it called? A. At s = 1; starting torque B. At s = R2/X2; pull-out (breakdown) torque, independent of R2 C. At s = 0; no-load torque D. At s = X2/R2; pull-out torque proportional to R2 Answer: B. dT/ds = 0 gives sm = R2/X2, and Tmax = KE2^2/(2X2) is independent of R2. (Key point: Maximum torque) 3. Where is the normal stable operating region and why? A. Between sm and 1, where T falls as 1/s B. Beyond s = 1, in braking C. Below sm (full load at 2-5% slip), where an increase in load raises torque as speed falls D. At s = sm exactly Answer: C. For 0 < s < sm torque rises with slip, so load changes are met with a small speed drop. (Key point: Normal operating region) 4. Why is the region sm < s < 1 unstable for steady running? A. Torque is proportional to s there B. Rotor current is zero C. Voltage collapses D. More load lowers speed, which lowers torque further so the motor stalls; it is only passed through while accelerating Answer: D. T is proportional to 1/s there, so slip rise reduces torque. 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