--- title: "Induction motor torque equation and torque-slip characteristic" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-torque-equation-ts-characteristic" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-torque-equation-ts-characteristic site: BCT Engineering --- # Induction motor torque equation and torque-slip characteristic Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2069 Poush, 8 marks - Video: [IM 12 Torque Slip Characteristics](https://www.youtube.com/watch?v=qoeOtMZzhkc) — Lectures in Electrical Engineering - 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-equation-ts-characteristic - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Derive the torque equation (TR = K·s·E2²·R2 / (R2² + s²X2²)) for three-phase induction motor. Draw and explain the torque-slip characteristic. ## Short answer The rotor emf sE2, impedance √(R2²+s²X2²) give I2, cosφ2. With T ∝ φ I2 cosφ2, substituting yields T = K s E2² R2/(R2²+s²X2²). The torque-slip characteristic is zero at synchronous speed, rises to maximum at s=R2/X2, then falls as 1/s; stable only below s_m. Rotor resistance shifts starting torque up without changing peak torque. ## Answer The torque developed by a 3-phase induction motor is proportional to the stator flux, the rotor current and the rotor power factor: $T \propto \phi I_2\cos\phi_2$. #### Derivation At standstill let rotor emf $= E_2$, rotor resistance $= R_2$, rotor reactance $= X_2$ (per phase). When running at slip $s$: - rotor emf $E_{2r} = sE_2$ - rotor reactance $X_{2r} = sX_2$ (since $f_2 = sf$) - rotor resistance remains $R_2$ ```text R2 sX2 +-[===]----[~~~]--+ | | (~) sE2 I2 -> | +-----------------+ ``` $$ I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} $$ Since $E_2 \propto \phi$, $\phi = E_2/k_1$: $$ \begin{aligned} T_R &= k\,\phi\, I_2\cos\phi_2 \\ &= k\frac{E_2}{k_1}\cdot\frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + s^2 X_2^2} \end{aligned} $$ From the power method, $T_R = \dfrac{P_{ag}}{\omega_s} = \dfrac{3I_2^2R_2/s}{2\pi N_s/60}$, so $K = \dfrac{3}{2\pi N_s/60}$ (N·m when $E_2$ in volts, $R_2, X_2$ in ohms). #### Torque–slip characteristic ```text T | Tmax | .--. | .' '. | Tfl / '. | ...* '-._ | /| '-- Tst | / | T ~ 1/s | / | (unstable) | / T~s |/(stable) |_____|_______________|__ s 0 sm 1 N=Ns N=0 ``` 1. **At $s = 0$** ($N = N_s$): $T = 0$. 2. **Low slip ($s < s_m$):** $sX_2 \ll R_2$, so $T \approx \dfrac{KE_2^2}{R_2}s$, i.e. $T \propto s$. Nearly straight line. Normal operating region; stable; speed nearly constant (full-load slip 2–5%). 3. **Maximum torque:** $dT/ds = 0$ gives $s_m = R_2/X_2$ and $T_{max} = \dfrac{KE_2^2}{2X_2}$, independent of $R_2$, proportional to $V^2$. Usually 2–3 times full-load torque. 4. **High slip ($s > s_m$):** $sX_2 \gg R_2$, so $T \approx \dfrac{KE_2^2R_2}{sX_2^2}$, i.e. $T \propto 1/s$ (rectangular hyperbola). Unstable: extra load reduces speed and torque further, and the motor stalls. 5. **At $s = 1$:** starting torque $T_{st} = \dfrac{KE_2^2R_2}{R_2^2 + X_2^2}$. Increasing $R_2$ moves $s_m$ towards 1 (higher starting torque) without changing $T_{max}$; reducing supply voltage lowers the whole curve as $V^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the rotor runs at slip s, which quantities become s times their standstill value? A. Rotor emf and rotor reactance B. Rotor resistance and emf C. Rotor resistance and reactance D. Only the rotor resistance Answer: A. E2r = sE2 and X2r = sX2 because f2 = sf, while R2 remains unchanged. (Key point: Rotor quantities at slip s) 2. What is the final torque expression derived? A. T = K E2^2 R2/(R2^2 + X2^2) B. T = K s E2^2 R2/(R2^2 + s^2 X2^2) C. T = K s E2 R2/(R2 + sX2) D. T = K E2^2 X2/(R2^2 + s^2 X2^2) Answer: B. Substituting phi = E2/k1, I2 = sE2/sqrt(R2^2 + s^2X2^2) and cos(phi2) = R2/sqrt(R2^2 + s^2X2^2) gives this expression. (Key point: Torque equation result) 3. What are the slip and value of maximum torque? A. s_m = X2/R2 and Tmax depends on R2 B. s_m = 1 and Tmax = K E2^2/R2 C. s_m = R2/X2 and Tmax = K E2^2/(2 X2), independent of R2 D. s_m = R2/X2 and Tmax proportional to R2 Answer: C. Differentiating gives s_m = R2/X2; Tmax does not depend on rotor resistance but varies with V^2. (Key point: Maximum torque) 4. In which region of the torque-slip curve is operation stable, and why? A. High slip, s > s_m, where T is proportional to 1/s B. At s = 1 only C. Near s_m only where dT/ds is infinite D. Low slip, s < s_m, where T is proportional to s Answer: D. For s < s_m, extra load slows the rotor, raising slip and torque, restoring balance; for s > s_m it gets worse and the motor stalls. 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