--- title: "Induction motor equivalent circuit from no-load and blocked rotor tests" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-noload-blocked-400v-equivalent" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-noload-blocked-400v-equivalent site: BCT Engineering --- # Induction motor equivalent circuit from no-load and blocked rotor tests Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2079 Chaitra, 8 marks - Video: [Electrical Machines | Tutorial - 18 | Three-Phase Induction Motors | No-Load & Blocked-Rotor Tests](https://www.youtube.com/watch?v=k4VC2OQKOIs) — Engineering Devotion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-noload-blocked-400v-equivalent - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question Following data were obtained from no-load and blocked rotor test of a 400 V, star connected induction motor (line values). | Test | Voltage | Current | Power | |---|---|---|---| | No load test | 400 V | 10 A | 1400 W | | Blocked rotor test | 200 V | 55 A | 7000 W | Calculate the equivalent circuit parameters per phase and draw the equivalent circuit referred to stator. Take DC resistance per phase of stator as 0.6 Ω. The effective stator resistance per phase is taken as 1.2 times its d.c value. ## Short answer No‑load test (400 V, 10 A, 1400 W) gives shunt branch R0 = 114.29 Ω, X0 = 23.58 Ω. Blocked‑rotor test (200 V, 55 A, 7000 W) gives Req = 0.7713 Ω, Xeq = 1.9526 Ω. With R1 = 1.2×0.6 = 0.72 Ω: R2' = 0.0513 Ω, X1 = X2' = 0.976 Ω per phase. ## Answer 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) ```text 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---------------------+---+--------------------+ ``` | Parameter | Value (Ω/phase) | |---|---| | $R_1$ | 0.72 | | $X_1$ | 0.976 | | $R_2'$ | 0.0513 | | $X_2'$ | 0.976 | | $R_0$ | 114.29 | | $X_0$ | 23.58 | **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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which parameters does each test give? A. No-load gives series leakage branch; blocked rotor gives shunt branch B. Both tests give the shunt branch C. No-load gives the shunt (R0, X0) branch; blocked rotor gives the series (leakage) branch D. No-load gives only R1; blocked rotor gives only X1 Answer: C. At no load slip is almost zero so the rotor branch is nearly open and the shunt branch dominates; with a blocked rotor the shunt branch is negligible. (Key point: Which test gives which branch) 2. How are Iw and Iμ found from the no-load test? A. Iw = I0 cos(phi0) with cos(phi0) = P0/(sqrt3 VL I0); Iμ = I0 sin(phi0) B. Iw = I0 sin(phi0); Iμ = I0 cos(phi0) C. Iw = P0/VL; Iμ = I0 - Iw D. Iw = I0 / 2; Iμ = I0 / 2 Answer: A. The working (loss) component is I0 cos(phi0) = 2.021 A and the magnetizing component is I0 sin(phi0) = 9.794 A. (Key point: No-load components) 3. In the blocked-rotor test, how is Req per phase calculated? A. Req = V/I using line voltage B. Req = Psc/(sqrt3 Vsc) C. Req = Psc/Isc^2 (single phase) D. Req = Psc/(3 Isc^2) = 0.7713 Ω Answer: D. The wattmeter reading is the total three-phase loss, so per phase Req = Psc/(3 Isc^2). (Key point: Blocked-rotor Req) 4. How are the stator and rotor values separated here? A. R1 = 1.2 x 0.6 = 0.72 Ω, R2' = Req - R1, and X1 = X2' = Xeq/2 B. R1 = 0.6 Ω, R2' = Req - R1, X1 = Xeq C. R2' = R1 = Req/2 and X1 = 0 D. R2' = Req and R1 = 0 Answer: A. The effective stator resistance is 1.2 times DC, rotor resistance is the remainder of Req, and leakage reactance is split equally. 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