--- title: "Induction motor torque and condition for maximum torque" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-torque-derive-max-condition" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-torque-derive-max-condition site: BCT Engineering --- # Induction motor torque and condition for maximum torque Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2075 Bhadra, 8 marks; 2074 Bhadra, 8 marks (asked 2 times) - Video: [IM16 - Condition for Maximum Torque of Induction Motor](https://www.youtube.com/watch?v=Ir8-xhRMpUE) — Pradeep Yemula - 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-derive-max-condition - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Derive the relationship for torque developed by a 3-phase induction motor. Draw a typical torque-slip characteristic and deduce the condition for maximum torque. ## Short answer Torque T = (K s E2^2 R2)/(R2^2 + (sX2)^2). Maximum torque occurs when rotor resistance equals rotor reactance at that slip, i.e., sm = R2/X2, giving Tmax = K E2^2/(2X2), independent of R2. ## Answer The torque of an induction motor is proportional to the product of the stator flux, the rotor current and the rotor power factor: $$ T \propto \phi\, I_2 \cos\phi_2 $$ #### Derivation of running torque Let at standstill: rotor emf per phase $= E_2$, rotor resistance $= R_2$, rotor reactance $= X_2$. At slip $s$: - rotor frequency $f_2 = s f$ - rotor emf $= sE_2$ - rotor reactance $= sX_2$ Rotor current and power factor: $$ I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} $$ Since $E_2 \propto \phi$, write $\phi = E_2/k_1$: $$ \begin{aligned} T &= k\,\phi\, I_2 \cos\phi_2 \\ &= k\,\frac{E_2}{k_1}\cdot\frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} \end{aligned} $$ The same result follows from power: air-gap power $P_{ag} = 3I_2^2 R_2/s$ and $T = P_{ag}/\omega_s$, giving $K = 3/\omega_s = 3/(2\pi n_s)$ with $n_s$ in rps. #### Torque–slip characteristic ```text T | Tmax | .--. | .' '. | / '. | / '-._ | / '-- Tst | / T ~ s | T ~ 1/s |/__________|___________ s 0 sm 1 N=Ns N=0 ``` - Small $s$: $T \propto s$ (straight line) — stable running region. - Large $s$: $T \propto 1/s$ (hyperbola) — unstable region. - At $s = 1$: starting torque. #### Condition for maximum torque $E_2$, $R_2$, $X_2$ are constant, so $T$ is maximum when $\dfrac{dT}{ds} = 0$. Write $$ T = \frac{K E_2^2 R_2}{\dfrac{R_2^2}{s} + sX_2^2} $$ $T$ is maximum when the denominator $y = R_2^2/s + sX_2^2$ is minimum: $$ \begin{aligned} \frac{dy}{ds} &= -\frac{R_2^2}{s^2} + X_2^2 = 0 \\ \Rightarrow\ R_2 &= s X_2 \\ \Rightarrow\ s_m &= \frac{R_2}{X_2} \end{aligned} $$ ($d^2y/ds^2 = 2R_2^2/s^3 > 0$, so it is a minimum of $y$, i.e. maximum of $T$.) **Condition:** torque is maximum when rotor resistance equals rotor reactance at that slip, $R_2 = s X_2$. Substituting $s_m$: $$ T_{max} = \frac{K (R_2/X_2) E_2^2 R_2}{R_2^2 + R_2^2} = \frac{K E_2^2}{2X_2} $$ Results: 1. $T_{max}$ does not depend on $R_2$; only the slip at which it occurs does. 2. $T_{max} \propto E_2^2 \propto V^2$ and $\propto 1/X_2$. 3. For maximum starting torque, $s_m = 1$, i.e. $R_2 = X_2$ (done with external rotor resistance in slip-ring motors). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the basic proportionality for induction motor torque? A. T proportional to phi x I2 x cos(phi2) B. T proportional to phi x I2 x sin(phi2) C. T proportional to I2 only D. T proportional to E2 / R2 Answer: A. Torque depends on flux, rotor current and rotor power factor. (Key point: Starting relation) 2. At slip s, what are the rotor emf and reactance? A. E2/s and X2/s B. sE2 and sX2 C. E2 and X2 D. sE2 and X2 Answer: B. Rotor frequency is sf, so emf and reactance scale with s. (Key point: Rotor quantities) 3. Maximum torque occurs when: A. R2 = sX2 B. R2 = X2 at any slip C. s = 1 always D. R2 = 0 Answer: A. Minimizing R2^2/s + sX2^2 gives s_m = R2/X2. (Key point: Condition for Tmax) 4. What is Tmax? A. K E2^2 R2/X2 B. K E2^2/(2 X2), independent of R2 C. K E2^2/R2 D. K E2^2 X2/2 Answer: B. Substituting sm gives KE2^2/(2X2); R2 only decides where the peak occurs. 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