--- title: "Determining single phase IM equivalent circuit parameters" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:equivalent-circuit-parameter-tests" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/equivalent-circuit-parameter-tests site: BCT Engineering --- # Determining single phase IM equivalent circuit parameters Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2070 Chaitra, 8 marks - Video: [Equivalent Circuit Parameters - Single Phase Induction Motor - Electrical Machines 3](https://www.youtube.com/watch?v=IyXR3IRAS1I) — Ekeeda - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/equivalent-circuit-parameter-tests - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question Discuss the procedure to determine the parameters of equivalent circuit of one phase induction motor. ## Short answer DC test gives stator resistance R1 (corrected for skin effect). Blocked-rotor test with locked rotor and reduced voltage at rated current yields R2', X1, X2' via measured Vsc, Isc, Psc. No-load test at rated voltage gives magnetising reactance Xm and rotational losses. ## Answer The parameters of the single-phase induction motor equivalent circuit ($R_1$, $X_1$, $R_2'$, $X_2'$, $X_m$ and rotational loss) are found from three simple tests on the **main winding** (auxiliary winding left open): a DC resistance test, a **blocked-rotor test** and a **no-load test**. ```text A W o--(A)--+--[W]--+-------+ | | | 1-ph (V) Main rotor variac | winding (free or supply | | blocked) o-------+-------+-------+ ``` #### 1. DC resistance test Pass DC through the main winding and measure $V_{dc}$ and $I_{dc}$: $$ R_1 = \frac{V_{dc}}{I_{dc}} \quad(\text{multiply by about } 1.1\text{–}1.2 \text{ for AC/skin effect}) $$ #### 2. Blocked-rotor (short-circuit) test **Procedure:** hold the rotor stationary, apply a **reduced voltage** through a variac until **rated current** flows. Record $V_{sc}$, $I_{sc}$, $P_{sc}$. At $s = 1$ the forward and backward halves are equal, and since $X_m \gg$ rotor impedance, the magnetising branch is neglected. The circuit becomes $R_1 + jX_1 + R_2' + jX_2'$. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}}, \qquad R_{sc} = \frac{P_{sc}}{I_{sc}^2} \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} \\ R_2' &= R_{sc} - R_1 \\ X_1 &= X_2' = \frac{X_{sc}}{2} \end{aligned} $$ #### 3. No-load test **Procedure:** run the motor at **rated voltage** with no load (start it with the auxiliary winding, then open it). Record $V_0$, $I_0$, $P_0$. At no load $s \approx 0$: - Forward branch: $0.5R_2'/s \to \infty$, so only $j0.5X_m$ remains. - Backward branch: $0.5R_2'/(2-s) \approx R_2'/4$; since $0.5X_m \gg$ this, only $R_2'/4 + j0.5X_2'$ remains. ```text R1 X1 0.5Xm R2'/4 0.5X2' o-/\/-mmm---mmm---+--/\/\---mmm---o ``` $$ \begin{aligned} Z_0 &= \frac{V_0}{I_0}, \qquad \cos\phi_0 = \frac{P_0}{V_0 I_0} \\ X_0 &= Z_0 \sin\phi_0 = X_1 + 0.5X_m + 0.5X_2' \\ X_m &= 2\left(X_0 - X_1 - 0.5X_2'\right) = 2(X_0 - 1.5X_1) \end{aligned} $$ #### Rotational (core + friction + windage) loss $$ P_{rot} = P_0 - I_0^2\left(R_1 + \frac{R_2'}{4}\right) $$ This loss is assumed constant and is subtracted from the mechanical power when finding output. #### Summary | Test | Condition | Gives | |---|---|---| | DC test | DC supply | $R_1$ | | Blocked rotor | Rotor locked, reduced V, rated I | $R_2'$, $X_1$, $X_2'$ | | No load | Rated V, no load | $X_m$, rotational loss | With these values, the full forward/backward equivalent circuit is drawn, and current, power factor, torque and efficiency at any slip can be calculated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which tests are used to find single-phase motor equivalent circuit parameters? A. DC resistance, blocked-rotor and no-load tests on the main winding B. Only open-circuit test C. Swinburne and Hopkinson tests D. Brake test only Answer: A. R_1 comes from the DC test, rotor and leakage values from blocked rotor, X_m and rotational loss from no-load. (Key point: Tests required) 2. From the blocked-rotor test, R_2' is found as: A. R_sc + R_1 B. R_sc - R_1 C. R_sc/2 D. Z_sc - R_1 Answer: B. R_sc = R_1 + R_2', so R_2' = R_sc - R_1. (Key point: Blocked rotor) 3. X_1 and X_2' are taken as: A. X_sc each B. X_sc/2 each C. Zero D. X_m/2 each Answer: B. They are assumed equal, so each is half of X_sc. (Key point: Leakage reactances) 4. At no load, why does the forward branch reduce to j0.5X_m? A. 0.5R2'/s tends to infinity as s tends to 0 B. The rotor is blocked C. R_1 is zero D. X_m is small Answer: A. With s near 0, the forward rotor branch is effectively open, leaving only the magnetising half. 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