--- title: "T-S characteristics and condition for maximum torque" subject: "Electrical Machines" chapter: 5 key: "semester_4/electrical_machines:ts-characteristics-max-torque-condition" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ts-characteristics-max-torque-condition site: BCT Engineering --- # T-S characteristics and condition for maximum torque Electrical Machines · Chapter 5: Three Phase Induction Machines - Asked in: 2070 Bhadra, 8 marks; 2071 Magh, 3+3 marks (asked 2 times) - Video: [Maximum Torque Equation Derivation](https://www.youtube.com/watch?v=0bagD_YDL58) — Engineering Fundamentals - 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-max-torque-condition - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5 ## Question Explain the torque-slip characteristics of an induction motor. Show the condition for which the maximum torque develops in the induction motor. ## Short answer Torque-slip curve is linear at low slip, peaks at sm, then decreases. Maximum torque condition is dT/ds=0, giving R2 = sm X2, so sm = R2/X2. Maximum torque equals 3E2^2/(2 omega_s X2) and is independent of rotor resistance. ## Answer #### Torque-slip characteristics The torque-slip curve shows the variation of the torque of a 3-phase induction motor with slip, at constant supply voltage and frequency. Torque of a 3-phase induction motor at slip $s$ (per-phase equivalent): $$ T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2 + (sX_2)^2}\ \text{N·m} $$ where $E_2$ and $X_2$ are the rotor emf and reactance per phase at standstill and $\omega_s$ is the synchronous speed in rad/s. ```text Torque T | Tmax | .--*--. | Tst / | \ Operating | *--' | \ region | | \___ ___ (stable) | | +------------+-----------+----> slip s 1 (standstill) sm 0 (sync) ``` - **At standstill ($s = 1$):** starting torque $T_{st}$, with high rotor current but a poor power factor. - **At small slip:** $sX_2\ll R_2$, so $T\propto s$. The curve is nearly a straight line, and this is the **stable operating region** (full-load slip 2–5%). Load increases give a small fall in speed. - **At $s = s_m$:** the torque reaches its maximum value $T_{max}$. - **At large slip:** $T\propto 1/s$, so torque decreases as $s$ rises. This part is unstable for normal loads. - **At synchronous speed ($s = 0$):** torque is zero. #### Condition for maximum torque Let $K = \dfrac{3E_2^2}{\omega_s}$. Then $$ T = K\,\frac{sR_2}{R_2^2 + s^2X_2^2} $$ For the maximum value, $\dfrac{dT}{ds} = 0$: $$ \frac{dT}{ds} = K\,\frac{(R_2^2 + s^2X_2^2)R_2 - sR_2(2sX_2^2)}{(R_2^2 + s^2X_2^2)^2} = 0 $$ $$ R_2^2 + s^2X_2^2 - 2s^2X_2^2 = 0 \;\Rightarrow\; R_2^2 = s^2X_2^2 $$ $$ s_m = \frac{R_2}{X_2} $$ So the torque is maximum when the **rotor resistance equals the rotor reactance at slip $s_m$**, that is $R_2 = s_mX_2$ (the rotor power factor is then $0.707$ lagging). Substituting $s = s_m$: $$ T_{max} = K\,\frac{s_mR_2}{2R_2^2} = \frac{3E_2^2}{2\omega_sX_2} $$ $T_{max}$ is independent of the rotor resistance, is proportional to $V^2$ and inversely proportional to $X_2$. Also: $\dfrac{T}{T_{max}} = \dfrac{2ss_m}{s^2 + s_m^2}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the condition for maximum torque obtained? A. Setting T = 0 B. Setting dT/ds = 0 C. Setting s = 1 D. Setting R2 = 0 Answer: B. Differentiate T with respect to slip and equate to zero. (Key point: Method) 2. The result of dT/ds = 0 is: A. R2 = 2 s X2 B. R2 = s^2 X2 C. R2 = s_m X2, i.e. s_m = R2/X2 D. R2 = X2 always Answer: C. R2^2 = s^2 X2^2 gives s_m = R2/X2. (Key point: Condition) 3. Tmax equals: A. 3E2^2/(2 ws X2) B. 3E2^2 R2/ws C. 3E2^2/(ws R2) D. E2/X2 Answer: A. Substituting s_m gives Tmax = 3E2^2/(2 ws X2), independent of R2. (Key point: Tmax value) 4. At maximum torque the rotor power factor is: A. 1 B. 0.5 C. 0 D. 0.707 lagging Answer: D. R2 = s_m X2 means equal resistance and reactance, so cos(phi) = 0.707. 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