---
title: "No-load and blocked rotor tests of induction motor"
subject: "Electrical Machines I"
chapter: 5
key: "semester_4/electrical_machines_i:im-noload-blocked-rotor-test-explain"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-noload-blocked-rotor-test-explain
site: BCT Engineering
---

# No-load and blocked rotor tests of induction motor

Electrical Machines I · Chapter 5: Three-Phase Induction Machines

- Asked in: 2069 Bhadra, 8 marks
- Video: [No Load Test And Block Rotor Test | Testing Of Induction machine  || Latest Upload 2018](https://www.youtube.com/watch?v=VmHW01M3F1E) — Electrical Engineering Idea's
- 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-rotor-test-explain
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5

## Question

Explain the no-load test and blocked rotor test of a three-phase induction motor. How the data obtained from these tests can be used to calculate the equivalent circuit parameters of the motor?

## Short answer

No-load test runs motor at rated voltage unloaded, measuring Po, Io, Vo to find core-loss resistance Rc and magnetizing reactance Xm after subtracting copper losses. Blocked-rotor test applies reduced voltage to pass rated current with rotor stalled, giving total leakage impedance Zeq=R1+R2' and Xeq=X1+X2'; R1 from DC test. These form the per-phase equivalent circuit.

## Answer

The **no-load test** and the **blocked rotor test** of a three-phase induction motor are the counterparts of the open-circuit and short-circuit tests of a transformer. Together they give the parameters of the per-phase equivalent circuit ($R_1, X_1, R_2', X_2', R_c, X_m$) without loading the motor.

#### Circuit used

```text
 3-ph       +--[A]--[W1]--+
 variable --|             |--- 3-ph
 supply  ---+--[ ]--[W2]--+--- induction
 (autotr.)  |     [V]     |--- motor
            +-------------+
 Two-wattmeter method: P = W1 + W2
```

An ammeter, a voltmeter and two wattmeters (two-wattmeter method) are connected between a variable three-phase supply (autotransformer) and the stator.

#### No-load test

**Procedure:** The motor runs uncoupled from any load at **rated voltage and rated frequency**. Readings: line voltage $V_0$, line current $I_0$ and input power $P_0 = W_1 + W_2$.

**Theory:** At no load the slip is very small ($s \approx 0$), so $R_2'/s$ is very large and the rotor branch is practically open. The current $I_0$ (25–40 % of rated, because of the air gap) is mainly magnetising current. Input power covers:

$$
P_0 = \text{stator Cu loss} + \text{core loss} + \text{friction and windage loss}
$$

**Calculations (per phase, star equivalent):**

$$
\begin{aligned}
V_{ph} &= \frac{V_0}{\sqrt{3}}, \qquad \cos\phi_0 = \frac{P_0}{\sqrt{3}\,V_0 I_0} \\
P_{rot} &= P_0 - 3 I_0^2 R_1 \quad (\text{core + mechanical loss}) \\
I_w &= I_0 \cos\phi_0, \qquad I_\mu = I_0 \sin\phi_0 \\
R_c &= \frac{V_{ph}}{I_w}, \qquad X_m = \frac{V_{ph}}{I_\mu}
\end{aligned}
$$

Also $Z_0 = V_{ph}/I_0 \approx X_1 + X_m$, so a more accurate value is $X_m = X_0 - X_1$ once $X_1$ is known from the blocked rotor test. Friction and windage loss can be separated from core loss by repeating the test at reduced voltages and extrapolating the $P_0$–$V^2$ curve to zero voltage.

#### Blocked rotor test

**Procedure:** The rotor is held stationary (blocked) and a **reduced voltage** (about 10–20 % of rated) is applied, raised until **rated current** flows. Readings: $V_{sc}$, $I_{sc}$, $P_{sc}$. For accurate rotor values, the test is ideally done at about 25 % of rated frequency, since the rotor normally works at slip frequency.

**Theory:** With the rotor at rest, $s = 1$, so $R_2'/s = R_2'$ is small. The magnetising branch has a much higher impedance and, at low voltage, draws very little current; it is neglected. Core loss is negligible at low voltage and there is no mechanical loss, so $P_{sc}$ is almost entirely the full-load copper loss of stator and rotor.

**Calculations (per phase):**

$$
\begin{aligned}
Z_{eq} &= \frac{V_{sc}/\sqrt{3}}{I_{sc}}, \qquad R_{eq} = \frac{P_{sc}}{3 I_{sc}^2} = R_1 + R_2' \\
X_{eq} &= \sqrt{Z_{eq}^2 - R_{eq}^2} = X_1 + X_2' \\
R_2' &= R_{eq} - R_1
\end{aligned}
$$

$R_1$ is measured by a DC test on the stator winding (DC resistance multiplied by about 1.2 for skin effect; for a star winding, $R_1 = R_{dc,\,line}/2$).

**Splitting $X_{eq}$:** Usually $X_1 = X_2' = X_{eq}/2$ (IEEE gives other ratios, e.g. 0.4 : 0.6 for class B and 0.3 : 0.7 for class C designs).

#### Using the results: equivalent circuit

```text
  R1    jX1          jX2'     R2'
 -^^^---mmm----+------mmm------^^^--+
               |    |               |
 V1          Rc[]  []jXm            [] R2'(1-s)/s
               |    |               |    (load)
 --------------+------+-------------+
```

| Parameter | Obtained from |
|---|---|
| $R_1$ | DC resistance test |
| $R_2'$ | Blocked rotor test minus $R_1$ |
| $X_1,\ X_2'$ | Blocked rotor test ($X_{eq}$ split) |
| $R_c$ | No-load test ($I_w$) |
| $X_m$ | No-load test ($I_\mu$) |
| Rotational loss | No-load test |

With these values the equivalent circuit can predict current, power factor, torque, efficiency and slip at any load, and the data can also be used to draw the circle diagram.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the no-load test run and why is the rotor branch ignored?
   A. Rated voltage with no load; s is about 0 so R2'/s is very large and the rotor branch is practically open
   B. Reduced voltage with rotor blocked; s = 1
   C. Rated voltage with rotor blocked
   D. Rated load with the supply frequency halved
   Answer: A. At no load slip is near zero, so the rotor draws almost no current and I0 is mostly magnetising current. (Key point: No-load test conditions)

2. What does the no-load input power P0 cover?
   A. Only rotor copper loss
   B. Stator copper, core and friction and windage losses
   C. Only the full-load copper loss
   D. Only the core loss
   Answer: B. The rotational loss is P0 - 3 I0^2 R1, i.e. core plus mechanical loss. (Key point: Losses in no-load power)

3. In the blocked rotor test, P_sc mainly represents:
   A. Core loss only
   B. Friction and windage loss
   C. Full-load stator and rotor copper loss, since core and mechanical losses are negligible
   D. No-load stator copper loss
   Answer: C. With reduced voltage the flux is small, the rotor is stationary, so almost all power is copper loss. (Key point: Blocked rotor test)

4. Where does R2' come from?
   A. The no-load test I_w
   B. The no-load test I_mu
   C. The synchronous speed
   D. Blocked rotor R_eq minus R1, where R1 comes from the DC test
   Answer: D. R_eq = R1 + R2', so R2' = R_eq - R1, with R1 measured by a DC resistance test. (Key point: Source of parameters)

## Related in Electrical Machines I

- [Induction motor equivalent circuit from no-load and blocked rotor tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-noload-blocked-400v-equivalent) — Electrical Machines I, chapter 5
- [Induction motor equivalent circuit from two-wattmeter tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-400v-noload-blocked-two-wattmeter) — Electrical Machines I, chapter 5

## Related in other subjects

- [Determining single phase IM equivalent circuit parameters](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aequivalent-circuit-parameter-tests) — Electric Machines II, chapter 3