--- title: "Torque-slip characteristic and effect of rotor resistance" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-ts-curve-rotor-resistance-effect" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-ts-curve-rotor-resistance-effect site: BCT Engineering --- # Torque-slip characteristic and effect of rotor resistance Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2077 Chaitra, 8 marks; 2076 Bhadra, 8 marks; 2075 Bhadra, 8 marks; 2073 Magh, 8 marks; 2071 Bhadra, 8 marks; 2070 Magh, 8 marks (asked 6 times) - Video: [effect of rotor resistance variation on torque slip characteristics | torque speed characteristics](https://www.youtube.com/watch?v=-KGIlyWVJAg) — Mind of Engineer - 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-ts-curve-rotor-resistance-effect - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Explain the torque-slip characteristics of a 3-phase induction motor and explain the effect of rotor circuit resistance on the characteristic. ## Short answer The torque–slip characteristic has a stable low-slip region where T ∝ s, a peak at slip sm = R2/X2, and an unstable high-slip region. Increasing rotor resistance shifts the peak to higher slip without changing maximum torque, raising starting torque (up to R2 = X2) but increasing full-load slip and losses. ## Answer The **torque–slip characteristic** of a 3-phase induction motor is the curve of developed torque against slip (or speed) at constant supply voltage and frequency. #### Torque equation $$ T = \frac{K\, s\, E_2^2\, R_2}{R_2^2 + (s X_2)^2} $$ where $E_2$ and $X_2$ are rotor emf and reactance per phase at standstill, $R_2$ is rotor resistance, $s$ is slip and $K = 3/(2\pi n_s)$ with $n_s$ in rps. #### Shape of the curve ```text T | Tmax | .--. | .' '. | / '. | / '-. | / '-. Tst | / stable | unstable | / | | / | |/_____________|____________ s 0 sm 1 (N=Ns) (N=0) ``` 1. **Low-slip region ($s \ll s_m$, normal running):** $sX_2 \ll R_2$, so $T \approx K s E_2^2/R_2$, i.e. $T \propto s$. The curve is nearly a straight line through the origin. This is the stable operating region; speed falls only slightly as load rises. 2. **Maximum torque:** occurs at $s_m = R_2/X_2$, with $T_{max} = K E_2^2/(2X_2)$. It is also called pull-out or breakdown torque. 3. **High-slip region ($s > s_m$):** $sX_2 \gg R_2$, so $T \approx K E_2^2 R_2/(s X_2^2)$, i.e. $T \propto 1/s$. The curve is a rectangular hyperbola. Operation here is unstable: if load increases, speed falls, torque falls further, and the motor stalls. 4. **At $s = 1$** the torque is the starting torque $T_{st} = K E_2^2 R_2/(R_2^2 + X_2^2)$. #### Effect of rotor circuit resistance When $R_2$ is increased, the whole curve keeps the same peak height but the peak moves to the right (towards $s = 1$). Using $T_{st}/T_{max} = 2a/(1 + a^2)$ with $a = R_2/X_2$, for $X_2 = 1\ \Omega$: | $R_2$ (Ω) | $s_m$ | $T_{st}/T_{max}$ | |---|---|---| | 0.1 | 0.1 | 0.198 | | 0.3 | 0.3 | 0.550 | | 0.5 | 0.5 | 0.800 | | 1.0 | 1.0 | 1.000 | - **Maximum torque is unchanged:** $T_{max} = KE_2^2/(2X_2)$ does not contain $R_2$. - **Slip at maximum torque increases:** $s_m = R_2/X_2$, so higher $R_2$ shifts the peak towards $s = 1$ (lower speed). - **Starting torque increases** up to $R_2 = X_2$, when $s_m = 1$ and $T_{st} = T_{max}$. Beyond that, starting torque falls again. - **Starting current decreases** and starting power factor improves. - **Full-load slip increases**, so running efficiency falls (rotor copper loss $= s P_{ag}$) and speed regulation becomes poorer. This is why slip-ring motors use external rotor resistance only during starting (or for speed control) and short it out in normal running. Double-cage and deep-bar motors get the same effect automatically. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the low-slip region of the torque-slip curve, how does torque vary? A. T is proportional to 1/s B. T is proportional to s, approximately linear and stable C. T is constant D. T is proportional to s^2 Answer: B. For small s, sX2 << R2 and T is approximately K s E2^2/R2. (Key point: Low-slip region) 2. Why is operation at s greater than sm unstable? A. Torque rises as speed falls B. Torque falls as speed falls (T proportional to 1/s), so extra load stalls the motor C. Rotor resistance is zero D. Frequency is zero Answer: B. In this region T is proportional to 1/s, so a slowdown reduces torque further. (Key point: Unstable region) 3. What happens to maximum torque when rotor resistance is increased? A. It increases B. It decreases C. It stays the same D. It becomes zero Answer: C. Tmax = K E2^2/(2 X2) does not contain R2. (Key point: Effect on Tmax) 4. What happens to slip at maximum torque (sm = R2/X2) when R2 is increased? A. It decreases towards zero B. It increases, shifting the peak towards s = 1 C. It stays constant D. It becomes negative Answer: B. sm is proportional to R2, so the peak moves right and the starting torque increases up to R2 = X2. 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