Slip at max torque and full load, starting current

Given: $T_{st} = 1.25\,T_{fl}$, $T_{max} = 2.5\,T_{fl}$, stator resistance and rotational losses neglected, constant $R_2$.

(i) Slip at maximum torque

$$\frac{T_{st}}{T_{max}} = \frac{1.25}{2.5} = 0.5 = \frac{2s_m}{1+s_m^2} \;\Rightarrow\; s_m^2 - 4s_m + 1 = 0$$ $$s_m = 2-\sqrt3 = 0.2679$$

(ii) Slip at full load

$$\frac{T_{fl}}{T_{max}} = \frac{1}{2.5} = 0.4 = \frac{2 s s_m}{s^2+s_m^2} \;\Rightarrow\; s^2 - 5 s_m s + s_m^2 = 0$$ $$s = s_m\,\frac{5 \pm\sqrt{21}}{2}$$

Taking the smaller root (the stable region, $s < s_m$):

$$s_{fl} = 0.2679\times 0.2087 = 0.0559$$

(iii) Starting current in per unit of full-load current

Rotor current: $I_2 = \dfrac{E_2}{\sqrt{(R_2/s)^2 + X_2^2}}$. With $R_2 = s_mX_2$:

$$\frac{I_{st}}{I_{fl}} = \frac{\sqrt{(R_2/s_{fl})^2 + X_2^2}}{\sqrt{R_2^2 + X_2^2}} = \frac{\sqrt{(s_m/s_{fl})^2+1}}{\sqrt{s_m^2+1}}$$ $$= \frac{\sqrt{(0.2679/0.0559)^2 + 1}}{\sqrt{0.2679^2 + 1}} = \frac{\sqrt{23.96}}{1.0353}= 4.73$$

Answer: (i) $s_m = 0.268$; (ii) $s_{fl} = 0.0559$; (iii) $I_{st} = 4.73$ p.u. of full-load current.