Max torque speed and Tmax/Tst of slip ring motor

Given: 4 poles, 50 Hz, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ per phase at standstill. Stator impedance neglected.

Speed at maximum torque

$$\begin{aligned} N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm} \\ s_m &= \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1 \\ N_m &= (1 - 0.1) \times 1500 = 1350\ \text{rpm} \end{aligned}$$

Ratio of maximum to starting torque

$$\begin{aligned} \frac{T_{max}}{T_{st}} &= \frac{R_2^2 + X_2^2}{2R_2X_2} = \frac{0.16 + 16}{2 \times 0.4 \times 4} = \frac{16.16}{3.2} = 5.05 \end{aligned}$$

Rotor resistance for $T_{st} = \tfrac{1}{2} T_{max}$

Let total rotor resistance per phase be $R$ and $a = R/X_2$:

$$\frac{T_{st}}{T_{max}} = \frac{2RX_2}{R^2 + X_2^2} = \frac{2a}{1 + a^2} = \frac{1}{2}$$ $$\begin{aligned} a^2 - 4a + 1 &= 0 \\ a &= \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} \end{aligned}$$

Take the smaller root (the larger one, $a = 3.73$, also works but wastes much more power and gives lower running speed):

$$\begin{aligned} a &= 2 - \sqrt{3} = 0.2679 \\ R &= 0.2679 \times 4 = 1.072\ \Omega \end{aligned}$$

External resistance to be added per phase $= 1.072 - 0.4 = 0.672\ \Omega$.

Check: $2 \times 1.072 \times 4/(1.072^2 + 16) = 8.574/17.149 = 0.5$.

Answer: Speed at $T_{max}$ = 1350 rpm; $T_{max}/T_{st}$ = 5.05; rotor resistance needed = 1.072 Ω per phase (i.e. 0.672 Ω extra per phase).