--- title: "Draw and explain torque-slip characteristics" subject: "Electrical Machines" chapter: 5 key: "semester_4/electrical_machines:ts-characteristics-regions" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ts-characteristics-regions site: BCT Engineering --- # Draw and explain torque-slip characteristics Electrical Machines · Chapter 5: Three Phase Induction Machines - Asked in: 2068 Magh, 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/ts-characteristics-regions - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5 ## Question Draw and explain torque-slip characteristics of 3-phase induction motor, showing clearly the starting torque, maximum torque and normal operating region. ## Short answer Torque-slip curve starts at starting torque at s=1, rises to maximum at sm=R2/X2, then falls to zero at s=0. Starting torque must exceed load; maximum torque is 2-3 times full-load torque; normal operating region is the linear stable part below sm, where T is proportional to slip and slip is 1-5%. ## Answer The torque-slip characteristic shows how the electromagnetic torque of a three-phase induction motor varies with slip, from $s=1$ (standstill) to $s=0$ (synchronous speed). $$ T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2} $$ ```text T | Tmax | _.-*-._ | Tst _-' `-._ | *-' `-. <- rated | / `* load T_fl +--+--------+-------------+--> s s=1 s_m s=0 (unstable) (stable region) ``` (slip decreases to the right, so speed increases to the right) #### Regions of the curve 1. **Starting torque ($s = 1$):** $$ T_{st} = \frac{3}{\omega_s}\cdot\frac{E_2^2 R_2}{R_2^2 + X_2^2} $$ It is the torque at standstill and must exceed the load torque for the motor to start. 2. **Maximum torque ($s = s_m = R_2/X_2$):** $$ T_{max} = \frac{3E_2^2}{2\omega_s X_2} $$ It is independent of $R_2$ and is also called pull-out or breakdown torque, typically 2 to 3 times full-load torque. 3. **Normal operating region ($0 < s < s_m$):** here $sX_2 \ll R_2$, so $T \approx \frac{3E_2^2}{\omega_s R_2}\,s$, which is a straight line. The torque is proportional to slip. A small drop in speed gives a large rise in torque, so the speed stays nearly constant (shunt-like behaviour). This region is **stable** and the motor runs here with slip of about 1 to 5 %. 4. **Unstable region ($s_m < s < 1$):** the torque falls as slip increases (reactance dominates, $T \propto 1/s$). If the load torque is more than the electromagnetic torque, the motor slows further and stalls. For stability, the load torque line must cut the motor curve in the normal region. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Maximum torque T_max = 3E_2^2/(2 omega_s X_2) depends on: A. R_2 strongly B. Neither X_2 nor E_2 C. X_2 and E_2 but not R_2 D. Rotor speed only Answer: C. R_2 cancels so T_max is independent of rotor resistance. (Key point: Max torque formula) 2. Why is the normal operating region nearly a straight line? A. R_2 is much smaller than sX_2 B. sX_2 is much smaller than R_2, so T is roughly proportional to s C. Slip is large there D. The flux is zero Answer: B. For small s, T ≈ 3E_2^2 s/(omega_s R_2). (Key point: Linear region) 3. In the region s_m < s < 1: A. Torque rises with slip and it is stable B. Torque is constant C. Torque falls as slip rises (about 1/s), so extra load causes stalling D. Torque is negative Answer: C. Reactance dominates, so T falls with slip and the motor stalls. (Key point: Unstable region) 4. For the motor to start, the starting torque must: A. Be zero B. Exceed the load torque C. Equal T_max exactly D. Be below full-load torque Answer: B. T_st must be greater than the load torque. 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