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Maximum torque condition and T-s curves for different voltages

Short answer

Maximum torque occurs when R2 = s X2, i.e., sm = R2/X2, giving Tmax = K E2^2/(2X2). Tmax is independent of R2 and proportional to V^2. T–s curves for lower voltages retain the same sm but show reduced Tmax and starting torque as V^2.

Derive the condition for maximum torque of an induction motor. Draw the torque-slip characteristics of induction motor for different supply voltage.

The developed torque of a 3-phase induction motor at slip ss is

T=K s E22R2R22+(sX2)2,K=32πns (ns in rps)T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2}, \qquad K = \frac{3}{2\pi n_s}\ (n_s \text{ in rps})

where E2E_2, X2X_2 are rotor emf and reactance per phase at standstill and R2R_2 is rotor resistance per phase.

Condition for maximum torque

For a given motor and supply, E2E_2, R2R_2 and X2X_2 are constant; only ss varies. Divide numerator and denominator by ss:

T=KE22R2R22s+sX22T = \frac{K E_2^2 R_2}{\dfrac{R_2^2}{s} + s X_2^2}

TT is maximum when the denominator y=R22s+sX22y = \dfrac{R_2^2}{s} + sX_2^2 is minimum:

dyds=−R22s2+X22=0s2X22=R22sm=R2X2\begin{aligned} \frac{dy}{ds} &= -\frac{R_2^2}{s^2} + X_2^2 = 0 \\ s^2 X_2^2 &= R_2^2 \\ s_m &= \frac{R_2}{X_2} \end{aligned}

Since d2yds2=2R22s3>0\dfrac{d^2y}{ds^2} = \dfrac{2R_2^2}{s^3} > 0, yy is minimum and TT is maximum.

Condition: torque is maximum when R2=sX2R_2 = s X_2, i.e. rotor resistance equals rotor reactance under running conditions.

Maximum torque:

Tmax=K(R2/X2)E22R2R22+R22=KE222X2\begin{aligned} T_{max} &= \frac{K (R_2/X_2) E_2^2 R_2}{R_2^2 + R_2^2} = \frac{K E_2^2}{2 X_2} \end{aligned}

Conclusions:

  1. TmaxT_{max} is independent of R2R_2; R2R_2 only changes the slip at which it occurs.
  2. Tmax∝1/X2T_{max} \propto 1/X_2.
  3. Tmax∝E22∝V2T_{max} \propto E_2^2 \propto V^2.
  4. For maximum starting torque, R2=X2R_2 = X_2 (sm=1s_m = 1).

(With stator impedance included: sm=R2′/R12+(X1+X2′)2s_m = R_2'/\sqrt{R_1^2 + (X_1 + X_2')^2}.)

T–s characteristics for different supply voltages

Since E2∝VE_2 \propto V, at any slip T∝V2T \propto V^2, while sm=R2/X2s_m = R_2/X_2 does not depend on VV.

 T            V1 > V2 > V3
 |         .--.        V1
 |       .'    '.
 |      /  .--.  '.    V2
 |     / .'    '. '.
 |    / /  .--.  '. '.  V3
 |   / / .'    '. '. '.
 |  / / /        '.'. '.
 |_/_/_/_____:_____'-'-'--- s
 0           sm             1
          (same for all V)
Effect of reducing VVResult
Maximum torqueFalls as V2V^2
Starting torqueFalls as V2V^2
Slip at max torqueUnchanged
Full-load slipIncreases (speed falls)
Rotor current at same loadIncreases, heating rises

Example: a 10% fall in voltage reduces TmaxT_{max} and TstT_{st} to 0.92=0.810.9^2 = 0.81 of their values, i.e. by 19%. This is why induction motors are sensitive to supply voltage dips, and why reduced-voltage starters (star–delta, auto-transformer) reduce starting torque.

✓ Answer checked by Claude Opus 5.5 (high effort)

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