Induction motor equivalent circuit from no-load and blocked rotor tests

The no-load test gives the shunt (magnetizing) branch; the blocked-rotor test gives the series (leakage) branch. Star connection: phase voltage $= V_L/\sqrt{3}$, phase current $=$ line current.

No-load test (400 V, 10 A, 1400 W)

$$\begin{aligned} V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ \cos\phi_0 &= \frac{P_0}{\sqrt{3}V_L I_0} = \frac{1400}{\sqrt{3}\times 400 \times 10} = 0.2021 \\ \sin\phi_0 &= 0.9794 \\ I_w &= I_0\cos\phi_0 = 10 \times 0.2021 = 2.021\ \text{A} \\ I_\mu &= I_0\sin\phi_0 = 10 \times 0.9794 = 9.794\ \text{A} \\ R_0 &= \frac{V_{ph}}{I_w} = \frac{230.94}{2.021} = 114.29\ \Omega \\ X_0 &= \frac{V_{ph}}{I_\mu} = \frac{230.94}{9.794} = 23.58\ \Omega \end{aligned}$$

($R_0$ represents core plus friction and windage loss; the small stator Cu loss at no load is neglected.)

Blocked-rotor test (200 V, 55 A, 7000 W)

$$\begin{aligned} Z_{eq} &= \frac{200/\sqrt{3}}{55} = \frac{115.47}{55} = 2.0995\ \Omega \\ R_{eq} &= \frac{P_{sc}}{3 I_{sc}^2} = \frac{7000}{3 \times 55^2} = 0.7713\ \Omega \\ X_{eq} &= \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{2.0995^2 - 0.7713^2} = 1.9526\ \Omega \end{aligned}$$

Separating stator and rotor values

$$\begin{aligned} R_1 &= 1.2 \times 0.6 = 0.72\ \Omega \\ R_2' &= R_{eq} - R_1 = 0.7713 - 0.72 = 0.0513\ \Omega \\ X_1 &= X_2' = \frac{X_{eq}}{2} = \frac{1.9526}{2} = 0.976\ \Omega \end{aligned}$$

(Equal split of leakage reactance is the usual assumption.)

Equivalent circuit referred to stator (per phase)

 I1  R1=0.72   jX1=0.976    jX2'=0.976  R2'=0.0513
 o--[====]----[~~~~]---+---+---[~~~~]---[====]--+
                       |   |                    |
 V=230.94 V      R0=114.3  jX0=23.58     R2'(1-s)/s
                       |   |              (load)
 o---------------------+---+--------------------+

Answer: $R_0 = 114.29\ \Omega$, $X_0 = 23.58\ \Omega$, $R_1 = 0.72\ \Omega$, $R_2' = 0.0513\ \Omega$, $X_1 = X_2' = 0.976\ \Omega$ per phase.