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Slip ring motor: speed at max torque, resistance

Short answer

Speed at max torque = 1350 rpm (slip 0.1). T_max / T_st = (1+0.1²)/(2×0.1) = 5.05. For T_st = 0.5 T_max, required rotor resistance per phase = 1.072 Ω (add 0.672 Ω external).

The rotor resistance and reactance of a 4-pole, 50 Hz, 3-phase slip ring induction motor are 0.4 and 4 ohm/phase respectively at stand still. Calculate the speed at maximum torque and the ratio (max torque)/(starting torque). What value should the resistance per phase have so that the starting torque is half of maximum torque?

Given: 4-pole, 50 Hz, R2=0.4 ΩR_2 = 0.4\ \Omega, X2=4 ΩX_2 = 4\ \Omega per phase (standstill).

Speed at maximum torque

Ns=120×504=1500 rpm,sm=R2X2=0.44=0.1N_s = \frac{120\times 50}{4} = 1500\ \text{rpm}, \qquad s_m = \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1 Nm=Ns(1−sm)=1500×0.9=1350 rpmN_m = N_s(1-s_m) = 1500\times 0.9 = 1350\ \text{rpm}

Ratio Tmax/TstT_{max}/T_{st}

TstTmax=2sm1+sm2=0.21.01=0.198  ⇒  TmaxTst=1+sm22sm=1.010.2=5.05\frac{T_{st}}{T_{max}} = \frac{2s_m}{1+s_m^2} = \frac{0.2}{1.01} = 0.198 \;\Rightarrow\; \frac{T_{max}}{T_{st}} = \frac{1+s_m^2}{2s_m} = \frac{1.01}{0.2} = 5.05

Resistance for Tst=12TmaxT_{st} = \frac12 T_{max}

Let the new total rotor resistance be R2′R_2', so sm′=R2′/X2s_m' = R_2'/X_2:

2sm′1+sm′2=0.5  ⇒  sm′2−4sm′+1=0\frac{2s_m'}{1+s_m'^2} = 0.5 \;\Rightarrow\; s_m'^2 - 4s_m' + 1 = 0 sm′=2−3=0.2679(the root<1)s_m' = 2 - \sqrt3 = 0.2679 \quad (\text{the root} < 1) R2′=sm′X2=0.2679×4=1.0718 ΩR_2' = s_m' X_2 = 0.2679\times 4 = 1.0718\ \Omega

Additional resistance to be added per phase =1.0718−0.4=0.6718 Ω= 1.0718 - 0.4 = 0.6718\ \Omega.

Answer: speed at TmaxT_{max} = 1350 rpm; Tmax/Tst=5.05T_{max}/T_{st} = 5.05; rotor resistance per phase =1.072 Ω= 1.072\ \Omega (external 0.672 Ω0.672\ \Omega added).

✓ Answer checked by Claude Opus 5.5 (high effort)

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