--- title: "Induction motor equivalent circuit from two-wattmeter tests" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-400v-noload-blocked-two-wattmeter" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-400v-noload-blocked-two-wattmeter site: BCT Engineering --- # Induction motor equivalent circuit from two-wattmeter tests Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2071 Bhadra, 8 marks - 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-400v-noload-blocked-two-wattmeter - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question The data obtained from the test of a 3-phase star connected, 400 V, 50 Hz induction motor are as follows: No-Load Test: V1 = 400 V, I0 = 20 A, W1 = 5000 W and W2 = −3200 W Blocked rotor test: VSC = 50 V, ISC = 60 A, W1 = 2300 W and W2 = 750 W Calculate the equivalent circuit parameters referred to stator side. ## Short answer Per‑phase equivalent circuit parameters (referred to stator): R0 = 88.89 Ω, X0 = 11.65 Ω; R1 = R2' = 0.1412 Ω, X1 = X2' = 0.1948 Ω. ## Answer In the two-wattmeter method the total 3-phase power is the algebraic sum of the readings. The no-load test gives the shunt branch, the blocked-rotor test the series branch. Star connection: $V_{ph} = V_L/\sqrt{3}$, $I_{ph} = I_L$. #### No-load test $$ \begin{aligned} P_0 &= W_1 + W_2 = 5000 + (-3200) = 1800\ \text{W} \\ V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ \cos\phi_0 &= \frac{P_0}{\sqrt{3}V_L I_0} = \frac{1800}{\sqrt{3}\times 400 \times 20} = 0.1299, \quad \sin\phi_0 = 0.9915 \\ I_w &= I_0\cos\phi_0 = 20 \times 0.1299 = 2.598\ \text{A} \\ I_\mu &= I_0\sin\phi_0 = 20 \times 0.9915 = 19.83\ \text{A} \\ R_0 &= \frac{V_{ph}}{I_w} = \frac{230.94}{2.598} = 88.89\ \Omega \\ X_0 &= \frac{V_{ph}}{I_\mu} = \frac{230.94}{19.83} = 11.65\ \Omega \end{aligned} $$ (The negative reading $W_2$ confirms pf below 0.5, as expected at no load.) #### Blocked-rotor test $$ \begin{aligned} P_{sc} &= W_1 + W_2 = 2300 + 750 = 3050\ \text{W} \\ Z_{eq} &= \frac{V_{sc}/\sqrt{3}}{I_{sc}} = \frac{28.87}{60} = 0.4811\ \Omega \\ R_{eq} &= \frac{P_{sc}}{3I_{sc}^2} = \frac{3050}{3 \times 60^2} = 0.2824\ \Omega \\ X_{eq} &= \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{0.4811^2 - 0.2824^2} = 0.3895\ \Omega \end{aligned} $$ Stator resistance is not given, so the usual split is assumed: $$ \begin{aligned} R_1 &= R_2' = \frac{R_{eq}}{2} = 0.1412\ \Omega \\ X_1 &= X_2' = \frac{X_{eq}}{2} = 0.1948\ \Omega \end{aligned} $$ Note: the power factors were found from $P/(\sqrt{3}VI)$ using the measured V, I and total power; the ratio formula $\tan\phi = \sqrt{3}(W_1 - W_2)/(W_1 + W_2)$ gives different values ($\cos\phi_0 \approx 0.126$ at no load and $\cos\phi_{sc} \approx 0.75$ in the blocked-rotor test, against 0.130 and 0.587 here) because the given readings are not consistent with the measured V and I. ```text I1 0.1412 j0.1948 j0.1948 0.1412 o--[====]--[~~~~]--+---+--[~~~~]--[====]--+ | | | V=230.94 V R0=88.89 jX0=11.65 0.1412(1-s)/s | | | o------------------+---+------------------+ ``` **Answer (per phase, referred to stator): $R_0 = 88.89\ \Omega$, $X_0 = 11.65\ \Omega$, $R_1 = R_2' = 0.1412\ \Omega$, $X_1 = X_2' = 0.1948\ \Omega$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With W1 = 5000 W and W2 = -3200 W in the no-load test, what is the total input power? A. 1800 W B. 8200 W C. 3200 W D. 5000 W Answer: A. Total power is the algebraic sum of the readings: 5000 + (-3200) = 1800 W. (Key point: Two-wattmeter power) 2. How are R0 and X0 obtained from the no-load test? A. R0 = P0/I0^2 and X0 = V/I0 B. R0 = Vph/Iw and X0 = Vph/Imu, using Iw = I0 cos(phi0) and Imu = I0 sin(phi0) C. R0 = Vph/Imu and X0 = Vph/Iw D. R0 = Vsc/Isc and X0 = Psc/Isc Answer: B. The no-load current splits into a core-loss component Iw and a magnetising component Imu, each with the phase voltage across it. (Key point: Shunt branch from no-load test) 3. How is R_eq found from the blocked rotor test? A. R_eq = Vsc/Isc = 0.4811 ohm B. R_eq = P_sc/Isc = 50.8 ohm C. R_eq = P_sc/(3 Isc^2) = 0.2824 ohm D. R_eq = sqrt(Zeq^2 + Xeq^2) Answer: C. Total copper loss is 3 Isc^2 R_eq, so R_eq = 3050/(3 x 3600) = 0.2824 ohm. (Key point: Series branch from blocked rotor test) 4. Stator values are not given, so how are R1, R2' and X1, X2' obtained? A. Assume R1 = Req and R2' = 0 B. Assume X1 = Xeq and X2' = 0 C. Measure them from the no-load test D. Assume R1 = R2' = Req/2 and X1 = X2' = Xeq/2 Answer: D. The usual assumption splits the blocked-rotor values equally, giving 0.1412 ohm and 0.1948 ohm. 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