Starting, no-load and full-load current of slip ring motor

Given: 4 poles, 50 Hz, star stator and rotor, $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$ per phase at standstill, emf between slip rings at standstill $= 400$ V, stator/rotor turns ratio $= 4$, $N_0 = 1490$ rpm, $N_{fl} = 1300$ rpm.

Rotor emf per phase at standstill:

$$E_2 = \frac{400}{\sqrt{3}} = 230.94\ \text{V}, \qquad N_s = \frac{120 \times 50}{4} = 1500\ \text{rpm}$$

Rotor current at slip $s$:

$$I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}$$

Stator current (load component, neglecting no-load current) $I_1 \approx I_2/4$.

(i) Starting current ($s = 1$)

$$\begin{aligned} I_{2,st} &= \frac{230.94}{\sqrt{0.8^2 + 4^2}} = \frac{230.94}{4.0792} = 56.61\ \text{A} \\ I_{1,st} &\approx \frac{56.61}{4} = 14.15\ \text{A} \end{aligned}$$

(ii) No-load current ($N = 1490$ rpm)

$$\begin{aligned} s_0 &= \frac{1500 - 1490}{1500} = 0.006667 \\ I_{2,0} &= \frac{0.006667 \times 230.94}{\sqrt{0.8^2 + (0.006667 \times 4)^2}} = \frac{1.5396}{0.80044} = 1.923\ \text{A} \\ I_{1,0} &\approx \frac{1.923}{4} = 0.481\ \text{A} \end{aligned}$$

(iii) Full-load current ($N = 1300$ rpm)

$$\begin{aligned} s_{fl} &= \frac{1500 - 1300}{1500} = 0.13333 \\ I_{2,fl} &= \frac{0.13333 \times 230.94}{\sqrt{0.8^2 + (0.13333 \times 4)^2}} = \frac{30.792}{0.96148} = 32.03\ \text{A} \\ I_{1,fl} &\approx \frac{32.03}{4} = 8.01\ \text{A} \end{aligned}$$

Answer (rotor currents): starting 56.61 A, no load 1.92 A, full load 32.03 A (stator side $\approx$ 14.15 A, 0.48 A and 8.01 A, ignoring the magnetizing current).