### Blocked Rotor Test On Induction Motor

This test is also called as Locked Rotor test or short circuit test.In the previous articles we had discussions on construction of induction motor,working of induction motor.We know induction motor is functionally similar to transformer.Like short circuit test on transformer blocked rotor test on induction motor is conducted.Quick Point:Blocked rotor test on induction motor is to find out stator copper loss, rotor copper loss without friction and windage losses.

In this test, the rotor is blocked by a belt-pulley mechanism,so it is not allowed to rotate.So rotor speed will be zero (Nr = 0).So slip (s) =1 and RL' = R2' (1-s)/s is zero.If the induction motor whichever is using for this test is slip ring induction motor then the winding are short circuited at the slip rings.

As RL' = 0, the equivalent circuit is exactly similar to that of a transformer and hence the calculations are similar to that of short circuit test on a transformer.

### Procedure of Blocked Rotor Test of Induction Motor

**1.**First block the rotor of induction motor by pulley-belt mechanism.

**2.**Apply 10 to 15 % of rated voltage to stator of induction motor.(Because if we apply even more than 30% rated voltage rotor will be short circuited.)

**3.**Now, slowly increase the voltage in the stator winding so that current reaches to its rated value. At this point, note down the readings of the voltmeter, wattmeter and ammeter to know the values of voltage, power and current.

**4.**Now the applied voltage Vsc, the input power Wsc and a short circuit current Isc are measured.

As RL' = 0, the equivalent circuit is exactly similar to that of a transformer and hence the calculations are similar to that of short circuit test on a transformer.

Vsc = Short circuit reduced voltage (line value)

Isc = Short circuit current (line value)

Wsc = Short circuit input power

Now Wsc = √3Vsc Isc cosΦsc ----------- Line values

**cosΦsc=Wsc/√3Vsc Isc (**This gives us short circuit power factor of a motor.

**)**

Now the equivalent circuit is as shown in above figure.

**Wsc=3(Isc)²R1e**

where Isc = Per phase value

**R1e = Wsc/3(Isc)²**

This is equivalent resistance referred to stator.

Z1e = Vsc (per phase)/ Isc (per phase) = Equivalent impedance referred to stator.

**X**

**1e**

**= √Z²1e-R²1e**= equivalent reactance reffered to stator.

During this test, the stator carries rated current hence the stator copper loss is also dominant. Similarly the rotor also carries short circuit current to produce dominant rotor copper loss. As the voltage is reduced, the iron loss which is proportional to voltage is negligibly small. The motor is at standstill hence mechanical loss i.e. friction and windage loss is absent. Hence we can write,

**Wsc = Stator copper loss + Rotor copper loss**

But it is necessary to obtain short circuit current when normal voltage is applied to the motor. This is practically not possible. But the reduced voltage test results can be used to find current ISN which is short circuit current if normal voltage is applied.

If VL = Normal rated voltage (line value)

Vsc = Reduced short circuit voltage (line voltage)

then ISN = (VL * ISC)

**/**Vsc
where Isc = Short circuit current at reduced voltage

Thus, ISN = Short circuit current at normal voltage

Now power input is proportional to square of the current.

So WSN = Short circuit input power at normal voltage

This cab be obtained as,

But at normal voltage core loss can not be negligible hence,

**WSN = Wsc (ISN / ISC)²**

**WSN = Core loss + Stator and rotor copper loss**