--- title: "T-S characteristics, max torque, rotor resistance effect" subject: "Electrical Machines" chapter: 5 key: "semester_4/electrical_machines:ts-max-torque-rotor-resistance-variation" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ts-max-torque-rotor-resistance-variation site: BCT Engineering --- # T-S characteristics, max torque, rotor resistance effect Electrical Machines · Chapter 5: Three Phase Induction Machines - Asked in: 2070 Magh, 8 marks - Video: [Electrical Machines | Lec 6C | Three-Phase Induction Motor | Rotor Quantities: Slip, Torque, etc](https://www.youtube.com/watch?v=s-wCI5jDu3g) — 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/ts-max-torque-rotor-resistance-variation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5 ## Question Explain the torque-slip characteristics of 3 phase induction motor. Show the condition for which the maximum torque develops in the induction motor. Discuss the effect of variation of rotor resistance on this maximum torque. ## Short answer Maximum torque occurs when rotor resistance equals rotor reactance at that slip, sm = R2/X2, giving Tmax = 3E2^2/(2 omega_s X2). Tmax is independent of rotor resistance; increasing R2 shifts the peak to higher slip and increases starting torque up to Tmax when R2 = X2. ## Answer #### Torque-slip characteristics The torque of a three-phase induction motor is $$ T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2} $$ where $E_2$ is the rotor emf per phase at standstill, and $R_2$ and $X_2$ are the rotor resistance and standstill reactance per phase. ```text T | R2 small R2 larger | Tmax ______ Tmax | .-*-. .-*-. | / \ / \ | Tst1 Tst2 (higher) +--+---------+--------------> s s=1 sm1 sm2 0 ``` The curve rises from $T_{st}$ at $s=1$ to a maximum and then falls to zero at $s = 0$. It is nearly linear near $s=0$ (the stable region), and the motor runs at low slip there. #### Condition for maximum torque Differentiate $T$ with respect to $s$ and equate it to zero: $$ \frac{dT}{ds} = 0 \;\Rightarrow\; R_2^2 - s^2X_2^2 = 0 \;\Rightarrow\; s_m = \frac{R_2}{X_2} $$ So the torque is maximum when the rotor resistance equals the rotor reactance at that slip ($R_2 = s_mX_2$). Substituting: $$ T_{max} = \frac{3E_2^2}{2\omega_s X_2} $$ #### Effect of rotor resistance - $T_{max}$ is **independent of $R_2$**; only its position changes. - $s_m \propto R_2$: with higher $R_2$ the maximum torque occurs at a higher slip (lower speed). - Starting torque rises as $R_2$ increases (up to $R_2 = X_2$, where $T_{st} = T_{max}$). - The full-load slip increases, so the speed falls; the rotor copper loss rises and the efficiency falls. This is why slip-ring motors use external rotor resistance for starting and speed control. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Setting dT/ds = 0 gives the condition: A. s_m = X_2/R_2 B. s_m = R_2 X_2 C. s_m = R_2/X_2, i.e. R_2 = s_m X_2 D. s_m = 1/(R_2 X_2) Answer: C. R_2^2 = s^2 X_2^2 gives s_m = R_2/X_2. (Key point: Maximum torque condition) 2. Increasing rotor resistance changes T_max: A. It raises T_max B. It lowers T_max C. It leaves T_max unchanged D. It reverses T_max Answer: C. T_max is independent of R_2. (Key point: Effect on Tmax) 3. Increasing R_2 moves the peak torque: A. To higher slip (lower speed) B. To lower slip C. Not at all D. Beyond synchronous speed Answer: A. s_m is proportional to R_2. (Key point: Effect on position) 4. Starting torque rises with R_2 up to: A. R_2 = 2X_2 B. R_2 = 0 C. R_2 = X_2 where T_st = T_max D. Infinite resistance Answer: C. At R_2 = X_2, s_m = 1 and T_st = T_max. 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