--- title: "Induction motor working and torque equation" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-working-torque-equation" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-working-torque-equation site: BCT Engineering --- # Induction motor working and torque equation Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2080 Chaitra, 3+5 marks - Video: [#Electrical Machines-II | L24 | Torque Equation (3-Phase Induction Motor)](https://www.youtube.com/watch?v=lLBNOIkEo74) — Shivanand Shakya - 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-working-torque-equation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question How does three phase induction motor work? Derive the torque equation and show the relationship between speed and torque. ## Short answer The stator's rotating magnetic field induces rotor currents; interaction produces torque T = K s E2^2 R2 / (R2^2 + (sX2)^2). Near synchronous speed T ∝ s (linear); at high slip T ∝ 1/s. The motor runs stably only in the linear low-slip region. ## Answer A 3-phase induction motor works on **electromagnetic induction**: the stator's rotating field induces current in the rotor, and the interaction produces torque. #### Working 1. A balanced 3-phase supply to the 3-phase stator winding produces a **rotating magnetic field** of constant magnitude moving at $N_s = 120f/P$. 2. The field cuts the rotor conductors and induces emf in them. 3. The rotor circuit is closed, so rotor current flows. 4. Current-carrying conductors in the field experience force; by Lenz's law the rotor turns **in the direction of the field** to reduce relative motion. 5. The rotor settles at speed $N < N_s$; slip $s = (N_s - N)/N_s$. If $N = N_s$, no emf, no torque. #### Torque equation At slip $s$: rotor emf $= sE_2$, rotor reactance $= sX_2$. $$ I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} $$ Torque $T \propto \phi I_2\cos\phi_2$ and $E_2 \propto \phi$: $$ \begin{aligned} T &= k\,\phi\,I_2\cos\phi_2 \\ &= k' E_2 \cdot \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} \end{aligned} $$ Using air-gap power, $T = \dfrac{P_{ag}}{\omega_s} = \dfrac{3I_2^2R_2/s}{2\pi N_s/60}$, so $K = \dfrac{3}{2\pi N_s/60}$. Starting torque ($s = 1$): $T_{st} = \dfrac{K E_2^2 R_2}{R_2^2 + X_2^2}$. Maximum torque: $s_m = R_2/X_2$, $T_{max} = \dfrac{KE_2^2}{2X_2}$. #### Relation between speed and torque Put $s = (N_s - N)/N_s$: - **Near synchronous speed** (small $s$): $sX_2 \ll R_2$, so $T \approx \dfrac{K E_2^2}{R_2}\, s = \dfrac{K E_2^2}{R_2}\cdot\dfrac{N_s - N}{N_s}$. Torque rises linearly as speed falls — stable region. - **Low speeds** (large $s$): $T \approx \dfrac{K E_2^2 R_2}{s X_2^2}$, i.e. $T \propto 1/s$. Torque falls as speed falls — unstable region. ```text N Ns|'-._ | '-. normal running | '. Nm|--------: Tmax (pull-out) | .' | .' unstable | .' 0|_.'___________ T Tst ``` So the motor runs at almost constant speed from no load to full load, like a DC shunt motor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is T in terms of slip? A. K s E2^2 R2/(R2^2 + (sX2)^2) B. K E2^2/R2 C. K s E2/R2 D. K R2/(sX2) Answer: A. From T = k phi I2 cos(phi2). (Key point: Torque equation) 2. Near synchronous speed, how does T vary? A. T is proportional to s, linear and stable B. T is proportional to 1/s C. T is constant D. T is proportional to s^2 Answer: A. sX2 << R2. (Key point: Small slip) 3. At low speeds, why is operation unstable? A. T is proportional to 1/s, so torque falls as speed falls B. Torque rises C. Slip is zero D. Rotor stops rotating field Answer: A. Hyperbolic region. (Key point: Large slip) 4. How does speed change from no load to full load? A. Almost constant, like a DC shunt motor B. Falls greatly C. Rises D. Zero Answer: A. The linear region is steep. 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