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Torque-slip characteristic and effect of rotor resistance

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.

Explain the torque-slip characteristics of a 3-phase induction motor and explain the effect of rotor circuit resistance on the characteristic.

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=K s E22 R2R22+(sX2)2T = \frac{K\, s\, E_2^2\, R_2}{R_2^2 + (s X_2)^2}

where E2E_2 and X2X_2 are rotor emf and reactance per phase at standstill, R2R_2 is rotor resistance, ss is slip and K=3/(2πns)K = 3/(2\pi n_s) with nsn_s in rps.

Shape of the curve

 T
 |          Tmax
 |         .--.
 |       .'    '.
 |      /        '.
 |     /           '-.
 |    /               '-. Tst
 |   /  stable  |  unstable
 |  /           |
 | /            |
 |/_____________|____________ s
 0             sm           1
(N=Ns)                    (N=0)
  1. Low-slip region (s≪sms \ll s_m, normal running): sX2≪R2sX_2 \ll R_2, so T≈KsE22/R2T \approx K s E_2^2/R_2, i.e. T∝sT \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 sm=R2/X2s_m = R_2/X_2, with Tmax=KE22/(2X2)T_{max} = K E_2^2/(2X_2). It is also called pull-out or breakdown torque.
  3. High-slip region (s>sms > s_m): sX2≫R2sX_2 \gg R_2, so T≈KE22R2/(sX22)T \approx K E_2^2 R_2/(s X_2^2), i.e. T∝1/sT \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=1s = 1 the torque is the starting torque Tst=KE22R2/(R22+X22)T_{st} = K E_2^2 R_2/(R_2^2 + X_2^2).

Effect of rotor circuit resistance

When R2R_2 is increased, the whole curve keeps the same peak height but the peak moves to the right (towards s=1s = 1).

Using Tst/Tmax=2a/(1+a2)T_{st}/T_{max} = 2a/(1 + a^2) with a=R2/X2a = R_2/X_2, for X2=1 ΩX_2 = 1\ \Omega:

R2R_2 (Ω)sms_mTst/TmaxT_{st}/T_{max}
0.10.10.198
0.30.30.550
0.50.50.800
1.01.01.000
  • Maximum torque is unchanged: Tmax=KE22/(2X2)T_{max} = KE_2^2/(2X_2) does not contain R2R_2.
  • Slip at maximum torque increases: sm=R2/X2s_m = R_2/X_2, so higher R2R_2 shifts the peak towards s=1s = 1 (lower speed).
  • Starting torque increases up to R2=X2R_2 = X_2, when sm=1s_m = 1 and Tst=TmaxT_{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 =sPag= 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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